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 \begin{document}
%\frontmatter
 \title{K\"umeler kuram\i}
 \author{David Pierce}
%\date{\today, \printtime}
\date{\today}
 \publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\uppertitleback{\centering
Bu eser\\
 Creative Commons Attribution--Gayriticari--Share-Alike\\
3.0 Unported Lisans\i\ ile lisansl\i d\i r.\\
Lisans\i n bir kopyas\i n\i\ g\"orebilmek i\c cin,\\
\url{http://creativecommons.org/licenses/by-nc-sa/3.0/deed.tr}\\
adresini ziyaret edin ya da a\c sa\u g\i daki adrese yaz\i n:\\
Creative Commons,
444 Castro Street,
Suite 900,\\
Mountain View,
California, 94041, USA.\\
\mbox{}\\
\cc \ccby David Austin Pierce \ccnc \ccsa}

\lowertitleback{\centering
%\chapter*{\"Ons\"oz}
Bu notlar\i, MAT 340 kodlu Aksiyomatik K\"umeler Kuram\i\ dersi i\c cin
yaz\i yorum.  L\"utfen hatalar\i\ bana bildirin.
}

 \maketitle

\tableofcontents

\listoffigures

%\mainmatter

\chapter{Giri\c s}

\section{Sayma ve ordinaller}

Bir torbada birka\c c tane satran\c c ta\c s\i m\i z var, 
onlar\i\ teker teker \c cekiyoruz, ve ayn\i\ zamanda say\i lar
diyoruz:
\begin{compactenum}[1:]
\item
piyade (\eng{pawn});
\item
kale (\eng{rook});
\item
at (\eng{knight});
\item
f{}il (\eng{bishop});
\item
vezir (\eng{queen});
\item
\c sah (\eng{king}).
\end{compactenum}
Bu \c sekilde ta\c slar\i\ \emph{saym\i\c s olduk.}
Sonu\c c olarak 6 tane ta\c s\i m\i z var deriz.  
Ama ta\c slar\i\ belli bir \emph{s\i rada} \c cektik.  
Ba\c ska bir s\i ra m\"umk\"und\"u.  
Ta\c slar\i\ tekrar \c cantaya koyup \c cekiyoruz:
\begin{compactenum}[1:]
\item
piyade;
\item
at;
\item
vezir;
\item
kale;
\item
f{}il;
\item
\c sah.
\end{compactenum}
Son ta\c s\i\ \c cekince yine 6 numaras\i n\i\ diyoruz.  
Her zaman \"oyle olacak:  
her zaman ta\c slar\i\ say\i nca 6'ya kadar sayaca\u g\i z.  
Ama nas\i l biliyoruz?\label{nasil}

Saymak nedir?  Sayman\i n nesnesi, bir
\textbf{topluluktur}\index{topluluk}
(\eng{collection}).\footnote{\textbf{K\"umeler}\index{k\"ume}
  (\eng{sets}), \"ozel topluluk olacak.}  Bir toplulu\u gu say\i nca
asl\i nda onu 
\textbf{s\i ral\i yoruz}%
\index{s\i ra}\index{s\i ra!---lama}
(\eng{order}). 

$A$ bir topluluk olsun, ve $R$, onun bir 
\textbf{s\i ralamas\i}
(\eng{ordering}) olsun.  
O zaman $A$ toplulu\u gunun \textbf{elemanlar\i}%
\index{eleman} (\eng{elements}) 
veya \textbf{\"o\u geleri}%
\index{\"o\u ge} (\eng{members}) vard\i r; 
ve bu toplulu\u gun t\"um $b$, $c$, ve $d$ elemanlar\i\ i\c cin
\begin{compactenum}[1)]
\item
$b\mathrel Rb$ de\u gil, yani
\begin{equation*}
\lnot\;b\mathrel Rb;
\end{equation*}
\item
$b\mathrel Rc$ ve $c\mathrel Rd$ ise $b\mathrel Rd$, yani
\begin{equation*}
b\mathrel Rc\land c\mathrel Rd\lto b\mathrel Rd;
\end{equation*}
\item
$b$ ve $c$ birbirinden farkl\i ysa ya $b\mathrel Rc$ ya da $c\mathrel Rb$, yani
\begin{equation*}
b=c\lor b\mathrel Rc\lor c\mathrel Rb.
\end{equation*}
\end{compactenum}
B\"oylece $R$,
\begin{compactenum}[1)] 
\item
\textbf{yans\i mas\i z} veya \textbf{d\"on\"u\c ss\"uz}
(\eng{irreflexive}),\footnote{I\c s\i k, bir aynadan yans\i r; ses,
  bir kayal\i ktan yans\i r.  \emph{Y\i kanmak} f{}iili, \emph{kendi
    kendini y\i kamak} \"obe\u ginin anlam\i na gelirse, d\"on\"u\c
  sl\"ud\"ur; \emph{y\i kan\i lma} f{}iilinin anlam\i na gelirse,
  edilgendir \cite{Lewis2,Ozkirimli-T}.} 
\item
\textbf{ge\c ci\c sli} veya \textbf{ge\c ci\c sken} (\eng{transitive}),\footnote{\emph{Kaynatmak} f{}iili ge\c ci\c slidir, \c c\"unk\"u bir nesne ister; \emph{kaynamak} ge\c ci\c ssizdir.} ve 
\item
\textbf{do\u grusal} (\eng{linear}) veya \textbf{tam} (\eng{total})
\end{compactenum}
bir ba\u g\i nt\i d\i r.  
O zaman $(A,R)$ s\i ral\i\ ikilisi, 
bir \textbf{s\i rad\i r.}\index{s\i ra}  
Bu s\i ra, $A$ \textbf{toplulu\u gunun bir s\i ras\i d\i r.} 

\c Simdi $A$, satran\c c ta\c slar\i\ torbam\i z olsun.  O zaman $A$
toplulu\u gunun t\"um s\i ralar\i, birbiriyle
\textbf{izomorftur}\index{izomorf} (\eng{isomorphic}).  
Demek ki $R$ ile $S$, 
$A$ toplulu\u gunun iki s\i ralamas\i ysa, 
o zaman $A$ toplulu\u gundan kendisine giden 
\"oyle bir birebir ve \"orten $f$ g\"ondermesi vard\i r---yani 
$A$ toplulu\u gunun \"oyle bir $f$
\textbf{perm\"utasyonu} (\eng{permutation}) veya \textbf{e\c sle\c
  smesi}\index{e\c sle\c sme} vard\i r---ki $A$ toplulu\u gunun t\"um
$b$ ile $c$ elemanlar\i\ i\c cin 
\begin{equation*}
b\mathrel Rc\liff f(b)\mathrel Sf(c)
\end{equation*}
denkli\u gi do\u grudur.  Ama bunu nas\i l biliyoruz?

\c Simdi $A$, pozitif \emph{tamsay\i lar}\index{say\i!tam---lar\i}
toplulu\u gu olsun.  Yani $A=\N$ 
olsun.  Bu toplulu\u gun al\i\c s\i lm\i \c s ``do\u gal'' $<$ s\i
ralamas\i\ vard\i r.  Ama ba\c ska s\i ralamalar\i\ da vard\i r.
Mesela $\N$ toplulu\u gunun \"oyle bir $R$ \textbf{ba\u g\i nt\i s\i}%
\index{ba\u g\i nt\i} 
veya \textbf{ili\c skisi}%
\index{ili\c ski} 
(\eng{relation}) vard\i r ki toplulu\u gun t\"um $k$ ile $m$
elemanlar\i\ i\c cin 
\begin{equation*}
k\mathrel Rm\liff (1<k\land k<m)\lor(1=m\land m<k)
\end{equation*}
denkli\u gi do\u grudur.  
\"Oyleyse $R$ ba\u g\i nt\i s\i, $\N$ toplulu\u gunu s\i ral\i yor; 
asl\i nda $R$ s\i ralamas\i, $<$ s\i ras\i\ ile hemen hemen ayn\i d\i r, 
ancak $R$ s\i ras\i na g\"ore $1$ eleman\i, 
$\N$ toplulu\u gunun \emph{son} eleman\i d\i r.  
O zaman $(\N,<)$ ile $(\N,R)$, birbirine izomorf de\u gildir:
\begin{equation*}
\begin{array}{c|*5c}
<&1,&2,&3,&\dots;&?\\\hline
R&2,&3,&4,&\dots;&1
\end{array}
\end{equation*}
\c Simdi
\begin{equation*}
k\mathrel Sm\liff(2\divides k+m\land k<m)\lor(2\ndivides k\land 2\divides m)
\end{equation*}
olsun.  O zaman $k\mathrel Sm$ ancak ve ancak
\begin{compactenum}[1)]
\item
hem $k$ hem $m$ ya tek ya \c cift, ve $k<m$, veya
\item
$k$ tek ve $m$ \c cift.
\end{compactenum}
O zaman $S$ ba\u g\i nt\i s\i\ da, $\N$ toplulu\u gunu s\i ral\i yor, ama $(\N,<)$ ile $(\N,S)$ s\i ralar\i, birbirine izomorf de\u gildir:
\begin{equation*}
\begin{array}{c|*8c}
<&1,&2,&3,&\dots;& ?& ?& ?&\dots\\\hline
S&1,&3,&5,&\dots;&2,&4,&6,&\dots
\end{array}
\end{equation*}

$\N$ toplulu\u gu say\i labilir mi?  
Normalde, sayarken, say\i lar diyoruz.  
$R$ s\i ralamas\i na g\"ore $\N$ toplulu\u gunu say\i nca
$1$ i\c cin hangi say\i y\i\ diyebiliriz?  
Yani yukar\i daki ilk tablonun alt sat\i r\i ndaki 
$1$ numaras\i n\i n \"ust\"unde, 
soru i\c saretinin yerine hangi say\i y\i\ koyabiliriz?  
Bu say\i\, $\upomega+1$ olacak.
%\glossary{$\upomega$}
Ondan sonra $\upomega+2$, $\upomega+3$, vesaire
say\i lar\i\ olacak; bunlardan sonra, $\upomega+\upomega$, yani
$\upomega\cdot2$, $\upomega\cdot2+1$, vesaire say\i lar\i\ olacak.
Ama $\N$ toplulu\u gunun sadece $\upomega$ tane eleman\i\ olacak. 

Asl\i nda k\"umeler kuramc\i lar\i\ olarak sayarken, 
$0$'dan ba\c slayaca\u g\i z:
\begin{equation*}
\begin{array}{c|*8c}
 &0,&1,&2,&\dots;&\upomega,&\upomega+1,&\upomega+2,&\dots\\\hline
S&1,&3,&5,&\dots;&       2,&         4,&         6,&\dots
\end{array}
\end{equation*}
Burada $0$, $1$, $2$, $3$, \dots; $\upomega$, $\upomega+1$,
$\upomega+2$, \dots; $\upomega\cdot2$, $\upomega\cdot2+1$, \dots
numaralar\i, \textbf{ordinal say\i lar}\index{ordinal} veya
\textbf{ordinallerdir.}  (Her ordinal, bu s\i rada bulunacak.)  
Ayr\i ca $0$, $1$, $2$, $3$, \dots, $\upomega$ numaralar\i,
\textbf{kardinal} (\eng{cardinal}) \textbf{say\i lar}\index{kardinal}
veya \textbf{kardinaldirler} (ba\c ska kardinaller olacak); ama
$\upomega+1$, bir kardinal de\u gildir. 

Her kardinal, bir ordinal olacak, ama her ordinal, bir kardinal olmayacak.

Her ordinal, bir \textbf{k\"ume}\index{k\"ume} olacak; ama
baz\i\ k\"umeler, ordinal olmayacak. 

Her \textbf{k\"ume,} bir topluluk olacak; ve her k\"umenin her
eleman\i, bir k\"ume olacak.  O zaman $a$ ile $b$ k\"umeyse, ya $a$
k\"umesi, $b$ k\"umesinin eleman\i d\i r, ya da eleman\i\ de\u gildir.
\.Ilk durumda $b$ k\"umesi, $a$ k\"umesini \textbf{i\c
  cerir}\index{i\c cerme} (\eng{contains}), yani $a$ k\"umesi, $b$
k\"umesi taraf\i ndan \textbf{i\c cerilir,} ve 
\begin{equation*}
a\in b
\end{equation*}
%\glossary{$\in$}%
ifadesini yazar\i z;\footnote{\label{epsilon}Buradaki $\in$ i\c
  sareti, Yunan \foreignlanguage{polutonikogreek}{e} (epsilon)
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  \foreignlanguage{polutonikogreek}{>est'i} kelimesinin ilk harf\/idir,
  ve $A$ \foreignlanguage{polutonikogreek}{>est'i} $B$ c\"umlesi,
  ``$A$, $B$'dir'' (\eng{$A$ is $B$}) anlam\i na gelir.  Epsilonun bu kullan\i\c s\i n\i, Peano \cite{Peano} ortaya koymu\c stur.} ikinci
durumda $b$ k\"umesi, $a$ k\"umesini \textbf{i\c cermez,} ve 
\begin{equation*}
a\notin b
\end{equation*}
ifadesini yazar\i z.  Genelde $C$ bir topluluk ise, ya $a\in C$ ya da
$a\notin C$. 

Bize g\"ore \textbf{bo\c s bir topluluk}---elemanlar\i\ olmayan bir
topluluk---vard\i r, ve bu topluluk, bir k\"umedir.  Bu varsay\i m, 
\textbf{Bo\c s K\"ume Aksiyomudur}%
\label{boskume}%
\index{aksiyom!Bo\c s K\"ume A---u}%
\index{bo\c s k\"ume}%
\index{k\"ume!bo\c s ---} 
(\eng{Empty Set Axiom}).  
Bo\c s k\"umenin i\c sareti, 
\begin{equation*}
\emptyset.
\end{equation*}%
%\glossary{$\emptyset$}
Ayr\i ca $a$ ile $b$ k\"umeyse, o zaman \"oyle bir k\"ume vard\i r ki
her eleman\i, ya $a$ k\"umesinin bir eleman\i, ya da $b$ k\"umesinin
kendisidir.  Bu yeni k\"umenin ifadesi, 
\begin{equation*}
a\cup\{b\}.
\end{equation*}
%\glossary{$a\cup\{b\}$}%
Bu toplulu\u gun k\"ume oldu\u gu, 
\textbf{Biti\c stirme Aksiyomudur}%
\label{bitistirme}%
\index{aksiyom!Biti\c stirme A---u} 
(\eng{Adjunction Axiom}).%
\footnote{Bu aksiyom, Tarski ve Givant \cite[p.~223, QIII]{MR920815}
  kayna\u g\i nda bulunur; \.Ingilizce ad\i, Boolos
  \cite[p.~100]{Boolos-again} kayna\u g\i nda bulunur.}  
Burada $a$
bo\c s ise, yeni $a\cup\{b\}$ k\"umesi, 
\begin{equation*}
\{b\}
\end{equation*}
olarak yaz\i l\i r.  O zaman a\c sa\u g\i daki gibi k\"umelerimiz vard\i r:
\begin{align*}
&\emptyset,&
&\{\emptyset\},&
&\{\emptyset\}\cup\bigl\{\{\emptyset\}\bigr\},&
&\Bigl(\{\emptyset\}\cup\bigl\{\{\emptyset\}\bigr\}\Bigr)\cup\Bigl\{\{\emptyset\}\cup\bigl\{\{\emptyset\}\bigr\}\Bigr\}.
%\end{align*}
\intertext{Bu ifadelerin yerine}
%\begin{align*}
&\emptyset,&
&\{\emptyset\},&
&\bigl\{\emptyset,\{\emptyset\}\bigr\},&
&\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\}
\end{align*}
ifadelerini yazabiliriz.  Asl\i nda $0$%
%\glossary{$0$} 
say\i s\i n\i\ $\emptyset$
olarak tan\i mlar\i z, yani 
\begin{equation*}
0=\emptyset.
\end{equation*}
Bu say\i, \textbf{ilk ordinaldir.}  Her $\alpha$ ordinali i\c cin bir
sonraki ordinal olacak, ve bu ordinal, $\alpha\cup\{\alpha\}$ olacak.
Mesela $0$'dan bir sonraki ordinal $\{0\}$ olacak; yani
\begin{equation*}
1=\{0\}
\end{equation*}
%\glossary{$1$}%
olacak.  Ayr\i ca her $\alpha$ ordinal i\c cin
\begin{equation*}
\alpha+1=\alpha\cup\{\alpha\}
\end{equation*}
olacak.  Ama bildi\u gimiz gibi
\begin{align*}
1+1&=2,&
2+1&=3,&
3+1&=4,
\end{align*}
%\glossary{$2$, $3$, $4$, \dots}%
vesaire.  O zaman
\begin{gather*}
2=1\cup\{1\}=\{0,1\},\\
3=2\cup\{2\}=\{0,1,2\},\\
4=3\cup\{3\}=\{0,1,2,3\},
\end{gather*}
vesaire.  B\"oyle tan\i mlanm\i\c s say\i lar, \textbf{von Neumann
  do\u gal say\i lar\i d\i r}%
\index{say\i!von Neumann do\u gal ---lar\i}%
\index{von Neumann do\u gal say\i lar\i}\label{vnn}
(\eng{von Neumann natural numbers} \cite{von-Neumann}).  Bu say\i lar, bir
toplulu\u gu olu\c sturacak, ve bu topluluk, $\upomega$ olacak.  Yani
$\upomega$, \"oyle bir topluluktur ki 
\begin{compactenum}[1)]
\item
$0\in\upomega$,
\item
$\alpha\in\upomega$ ise $\alpha+1\in\upomega$, ve
\item
$\upomega$ toplulu\u gunun ba\c ska eleman\i\ yoktur.
\end{compactenum}
\"Oyleyse $\upomega$ toplulu\u gunun tan\i m\i, 
\textbf{\"ozyineli}%
\index{\"ozyineli tan\i m} 
veya 
\textbf{rek\"ursiftir}%
\index{rek\"ursif tan\i m} 
(\eng{recursive}).

\section{Ordinaller Hesaplar\i}

\textbf{Sonsuzluk Aksiyomuna}%
\index{aksiyom!Sonsuzluk A---u}%
\label{sonsuzluk}%
\footnote{Veya \textbf{Sonsuz K\"ume Aksiyomu} \cite{Nesin-SKK}.} 
(\eng{Axiom of Infinity} \cite{Zermelo-invest})
g\"ore $\upomega$ toplulu\u gu, bir k\"ume olacak.  O zaman $\upomega$
bir ordinal olacak, ve bu ordinalin her $k$ eleman\i\ i\c cin
$\upomega+k$ k\"umesi, bir ordinal olacak. 

Asl\i nda t\"um $\alpha$ ile $\beta$ ordinaller i\c cin
\begin{align*}
&\alpha+\beta\text{ toplam\i n\i,}&
&\alpha\cdot\beta\text{ \c carp\i m\i n\i, ve}&
&\alpha^{\beta}\text{ kuvvetini}
\end{align*}
tan\i mlayaca\u g\i z.  O zaman
\begin{gather*}
1+\upomega=\upomega<\upomega+1,\\
2\cdot\upomega=\upomega<\upomega\cdot2,\\
(\upomega+1)^{\upomega}=\upomega^{\upomega}<\upomega^{\upomega+1}
\end{gather*}
olacak.  Asl\i nda:
\begin{asparaitem}
\item
$1+\upomega$ toplam\i,
\begin{equation*}
(0,0,1,2,3,\dots)
\end{equation*}
s\i ras\i n\i n ordinalidir, ama $\upomega+1$,
\begin{equation*}
(0,1,2,3,\dots,0)
\end{equation*}
s\i ras\i n\i n ordinalidir.
\item
$2\cdot\upomega$ \c carp\i m\i,
\begin{equation*}
(0,1,0,1,0,1,\dots)
\end{equation*}
s\i ras\i n\i n ordinalidir, ama $\upomega\cdot2$,
\begin{equation*}
(0,1,2,3,\dots,0,1,2,3,\dots)
\end{equation*}
s\i ras\i n\i n ordinalidir; ayr\i ca
\begin{gather*}
	2\cdot\upomega=2+2+2+\dotsb,\\
	\upomega\cdot2=\upomega+\upomega=\upomega+1+1+1+\dotsb
\end{gather*}
(\sayfaya{o+o} bak\i n).
\item
$(\upomega+1)^{\upomega}$ kuvveti,
\begin{equation*}
((\upomega+1)^2,(\upomega+1)^3,(\upomega+1)^4,\dots)
\end{equation*}
dizisinin \textbf{limitidir,}\index{limit} ve
\begin{gather*}\allowdisplaybreaks
\begin{aligned}
(\upomega+1)^2
&=(\upomega+1)\cdot(\upomega+1)\\
&=(\upomega+1)\cdot\upomega+(\upomega+1)\cdot1\\
&=(\upomega+1+\upomega+1+\upomega+1+\dotsb)+\upomega+1\\
&=(\upomega+\upomega+\upomega+\dotsb)+\upomega+1\\
&=\upomega^2+\upomega+1,
\end{aligned}\\
\begin{aligned}
(\upomega+1)^3
&=(\upomega+1)^2\cdot(\upomega+1)\\
&=(\upomega^2+\upomega+1)\cdot(\upomega+1)\\
&=(\upomega^2+\upomega+1)\cdot\upomega+\upomega^2+\upomega+1\\
&=(\upomega^2+\upomega+1+\upomega^2+\upomega+1+\upomega^2+\dotsb) +\upomega^2+\upomega+1\\
&=(\upomega^2+\upomega^2+\dotsb)+\upomega^2+\upomega+1\\
&=\upomega^3+\upomega^2+\upomega+1,
\end{aligned}
\end{gather*}
ve genelde
\begin{equation*}
(\upomega+1)^n=\upomega^n+\upomega^{n-1}+\dots+\upomega+1.
\end{equation*}
\end{asparaitem}
Ayr\i ca her pozitif $\alpha$ ordinali i\c cin \"oyle bir $\ell$ do\u
gal say\i s\i, ve $\alpha_0$, \dots, $\alpha_{\ell}$ ordinalleri, ve
$a_0$, \dots, $a_{\ell}$ pozitif do\u gal say\i lar\i\ vard\i r ki 
\begin{align*}
\alpha_0&>\dots>\alpha_{\ell},&
\alpha&=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_{\ell}}\cdot a_{\ell}.
\end{align*}
Burada $\upomega^{\alpha_0}\cdot
a_0+\dots+\upomega^{\alpha_{\ell}}\cdot a_{\ell}$ ifadesi, $\alpha$
ordinalinin \textbf{Cantor normal bi\c cimidir}%
\index{Cantor normal bi\c cimi}%
\label{Cnf}%
\index{normal!Cantor --- bi\c cimi}
(\eng{Cantor normal form}).  
Her pozitif ordinalin tek
bir Cantor normal bi\c cimi vard\i r.  Bundan hesaplama
kurallar\i\ t\"ureyebilir. 

\section{K\"umeler ve S\i n\i flar}

Her topluluk, bir k\"ume de\u gildir.\label{Russell}  \"Orne\u gin
\"oyle bir $R$ toplulu\u gu vard\i r ki her eleman\i\ bir k\"ume, ama
bu k\"ume, kendisinin eleman\i\ de\u gildir.  Yani 
\begin{equation*}
R=\{x\colon x\notin x\}.
\end{equation*}
%\glossary{$x$}%
Burada $x$ de\u gi\c skeni her zaman bir k\"ume olacak.  \c
Simdi $a$ bir k\"ume olsun.  E\u ger $a\in a$ ise, o zaman $a\notin
R$, dolay\i s\i yla $a\neq R$.  E\u ger $a\notin a$ ise, o zaman $a\in
R$, dolay\i s\i yla $a\neq R$.  Her durumda $R$ toplulu\u gu,
$a$ k\"umesi de\u gildir.  Yani $R$, bir k\"ume de\u gildir.  Bu
teoreme 
\textbf{Russell Paradoksu}%
\index{paradoks!Russell P---u}%
\index{teorem!Russell Paradoksu}
denir \cite{Russell-letter}. 

Elemanlar\i\ k\"ume olan baz\i\ topluluklar, 
\textbf{s\i n\i f}%
\index{s\i n\i f}
olacak.  Her k\"ume, bir s\i n\i ft\i r, ancak baz\i\ s\i n\i flar,
k\"ume de\u gildir.  Mesela yukar\i daki gibi $\{x\colon x\notin x\}$
toplulu\u gu, bir s\i n\i ft\i r, ama g\"osterdi\u gimiz gibi k\"ume
de\u gildir.  Tan\i ma g\"ore her s\i n\i f, 
\begin{equation*}
\{x\colon\phi(x)\}
\end{equation*}
%\glossary{$\{x\colon\phi(x)\}$}%
bi\c ciminde yaz\i labilir.  Burada $\phi(x)$, k\"umeler kuram\i n\i n
mant\i\u g\i nda bir \textbf{form\"uld\"ur.}\index{form\"ul}  E\u ger
$a$ bir k\"umeyse, o zaman $\phi(a)$ ifadesi, bir
\textbf{c\"umledir.}\index{c\"umle}  Her c\"umle, ya do\u gru ya
yanl\i\c st\i r.  Bir $\{x\colon\phi(x)\}$ s\i n\i f\i n\i n
elemanlar\i, $\phi(a)$ c\"umesini do\u gru yapan $a$ k\"umeleridir.
Bu s\i n\i f, $\phi(x)$ form\"ul\"u taraf\i ndan  
\textbf{tan\i mlan\i r.}%
\index{tan\i mlama}

Bir $\phi(x)$ form\"ul\"un\"un bir ve tek bir \textbf{serbest de\u
  gi\c skeni}\index{ de\u gi\c sken!serbest} vard\i r, ve bu de\u gi\c
sken, $x$ olur.  Ancak bir form\"ul\"un birden fazla serbest de\u gi\c
skeni olabilir.  \"Orne\u gin 
\begin{equation*}
\Forall z(z\in x\liff z\in y)
\end{equation*}
ifadesi, bir form\"uld\"ur, ve serbest de\u gi\c skenleri, $x$ ile $y$
olur.  Bu form\"ulde $z$, \textbf{ba\u glant\i l\i\ de\u gi\c
  skendir.}\index{de\u gi\c sken!ba\u glant\i l\i}  Form\"ul,
k\"umelerin \textbf{e\c sitlik}\index{e\c sitlik} ba\u g\i nt\i s\i
n\i\ tan\i mlar.  Yani $a$ ile $b$ k\"umeleri birbirine e\c sittir,
ancak ve ancak 
\begin{equation*}
\Forall z(z\in a\liff z\in b),
\end{equation*}
yani elemanlar\i\ ayn\i d\i r.  K\"ume olmayan bir s\i n\i f\i n
oldu\u gunu kan\i tlarken, bu kural\i\ kulland\i k.  Yukar\i daki
$\Forall z(z\in x\liff z\in y)$ form\"ul\"un\"un yerine 
\begin{equation*}\label{=}
x=y
\end{equation*}
%\glossary{$=$}%
ifadesini yazar\i z.  O halde bir $\{x\colon x=x\}$ s\i n\i
f\i\ vard\i r, ve bu s\i n\i f, t\"um k\"umelerin s\i n\i f\i d\i r.
Bu s\i n\i f, 
\textbf{evrensel s\i n\i ft\i r}%
\index{evrensel s\i n\i f}
(\eng{universal class}),
ve i\c sareti,
\begin{equation*}\label{universe}
\universe
\end{equation*}
%\glossary{$\universe$}%
olacak.  Ayr\i ca $a$ bir k\"umeyse, o zaman bir $\{x\colon x\in a\}$
s\i n\i f\i\ vard\i r, ama bu s\i n\i f, $a$ k\"umenin kendisidir,
yani 
\begin{equation*}
a=\{x\colon x\in a\}.
\end{equation*}
\"Oyleyse, dedi\u gimiz gibi, her k\"ume, bir s\i n\i ft\i r.

Sonsuzluk Aksiyomunu kullanmadan $\upomega$ toplulu\u gunun s\i n\i f
oldu\u gu apa\c c\i k de\u gildir, ama s\i n\i f olacakt\i r.  Ondan sonra
Sonsuzluk Aksiyomu, $\Exists xx=\upomega$ bi\c ciminde olabilecektir.

Asl\i nda $\upomega$ s\i n\i f\i\ bir k\"ume oldu\u gundan,
\textbf{Yerle\c stirme Aksiyomuna}%
\index{aksiyom!Yerle\c stirme A---u}%
\index{Yerle\c stirme Aksiyomu}%
\label{yerlestirme}
(\eng{Replacement Axiom})\footnote{Skolem
  \cite{Skolem-some-remarks}, 1922 y\i l\i nda bu aksiyomu tavsiye
  etti; ayn\i\ y\i lda Fraenkel, benzer bir aksiyomu tavsiye etmi\c s.
  Ayr\i ca Cantor'a \cite[p.~114]{Cantor-letter} bak\i n.}
g\"ore $\{y\colon\Exists x(x\in\upomega\land y=\upomega+x)\}$ s\i n\i
f\i, bir k\"ume olacakt\i r.
Bu k\"ume
\begin{equation*}
\{\upomega+x\colon x\in\upomega\}
\end{equation*}
olarak yaz\i labilir. 
\textbf{Bile\c sim Aksiyomuna}%
\label{bilesim}%
\index{aksiyom!Bile\c sim A---u}%
\index{bile\c sim!B--- Aksiyomu}
(\eng{Union Axiom} \cite{Zermelo-invest})
g\"ore bu k\"umenin
\begin{align*}
&\bigcup\{\upomega+x\colon x\in\upomega\}&
&\text{ veya }&
&\bigcup_{x\in\upomega}(\upomega+x)
\end{align*}
bile\c simi de bir k\"umedir; tan\i ma g\"ore bu bile\c sim,
$\upomega+\upomega$%
\label{o+o} 
toplam\i d\i r.

K\"umelerden olu\c sturulmu\c s baz\i\ topluluklar, 
s\i n\i f de\u gildir.  
Bu sonu\c c, 
\textbf{G\"odel'in Eksiklik Teoremi}%
\index{teorem!G\"odel Eksiklik T---i} 
(\eng{G\"odel's Incompleteness Theorem}\linebreak 
\cite{Goedel-incompl}%\nocite{MR1890980}
)
veya 
\textbf{Tarski'nin Do\u grulu\u gun Tan\i mlanamamas\i\ Teoremi}%
\index{teorem!Tarski Do\u grulu\u gun Tan\i mlanamamas\i\ T---i} 
(\eng{Tarski's Theorem on the Indefinability of Truth}
\cite{Tarski-truth}%\nocite{MR736686}
)
gibidir.  
Bu teoremlerin as\i l bi\c cimleri,
$\N$ toplulu\u gu hakk\i ndad\i r, 
ve bu bi\c cimde teoremlerini kan\i tlamak zordur.  
Fakat bu teoremler,  
$\universe$ hakk\i nda yaz\i labilir; 
ve bu bi\c cimde onlar\i\ kan\i tlamak daha kolayd\i r. 

T\"um ordinallerin toplulu\u gu, bir s\i n\i f olacak, 
ve bu s\i n\i f\i n i\c sareti
\begin{equation*}
\on
\end{equation*}
%\glossary{$\on$}%
olacak.  Asl\i nda bu s\i n\i f, bir $a$ k\"umesiyse, o zaman
$a\in\on$ olurdu, yani $a\in a$ olurdu; ama bir ordinal i\c cin bu i\c
cerme imk\^ans\i zd\i r.  Sonu\c c olarak $\on$, bir k\"ume de\u
gildir.  Bu teorem, \textbf{Burali-Forti Paradoksu}%
\index{paradoks!Burali-Forti P---u}%
\index{teorem!Burali-Forti Paradoksu}\label{BFP}
\cite{Burali-Forti}
olarak bilinir.

\section{Kardinaller}

$\on$ s\i n\i f\i n\i n bir s\i ralamas\i\ vard\i r, ve bu s\i ralama,
\emph{i\c cerilmedir,} yani $\in$ ile g\"osterilen s\i ralamad\i r.  
\textbf{Se\c cim Aksiyomuna}%
\label{secim}%
\index{aksiyom!Se\c cim A---u} 
(\eng{Axiom of Choice} \cite{Zermelo-invest}) 
g\"ore, her $a$ k\"umesinden bir $\beta$
ordinaline giden bir \textbf{e\c sleme}\index{e\c sleme}\index{e\c slenik} (yani bir birebir \"orten g\"onderme) vard\i r.  O halde
\begin{equation*}
a\approx\beta
\end{equation*}
%\glossary{$\approx$}%
ifadesini yazal\i m, 
ve $a$ ile $\beta$ k\"umelerine \textbf{e\c slenik} densin 
\cite[s.~82]{Nesin-SKK}.  E\u ger $a$ verilirse, ve
$a\approx\beta$ ko\c sulunu sa\u glayan $\beta$ ordinallerinin en
k\"u\c c\"u\u g\"u $\kappa$ (``kappa'') ise, o zaman $\kappa$, $a$ k\"umesinin 
\textbf{kardinalidir.}%
\index{kardinal}\label{kardinal}  
T\"um kardinallerden olu\c sturulmu\c s topluluk, bir s\i n\i f
olacak, ve bu s\i n\i f\i n i\c sareti 
\begin{equation*}
\cn
\end{equation*}
%\glossary{$\cn$}%
olacak.  En k\"u\c c\"uk \emph{sonsuz} kardinal, $\upomega$ olur.
$\on$ s\i n\i f\i ndan $\cn$ s\i n\i f\i na giden bir
\begin{equation*}
\xi\mapsto\aleph_{\xi}
\end{equation*}
%\glossary{$\aleph_{\xi}$}%
g\"ondermesi vard\i r.  Burada
\begin{align*}
\aleph_0&=\upomega&
&\text{ ve }&
\alpha<\beta\liff\aleph_{\alpha}<\aleph_{\beta},
\end{align*}
ve her sonsuz kardinal, bir $\alpha$ ordinali i\c cin,
$\aleph_{\alpha}$ bi\c cimindedir.  \.Iki kardinalin \emph{kardinal}
toplam\i\ ve \emph{kardinal} \c carp\i m\i\ vard\i r, ama
\begin{equation*}
\aleph_{\alpha}\cardsum\aleph_{\beta}=
\aleph_{\alpha}\cardprod\aleph_{\beta}=\aleph_{\max\{\alpha,\beta\}}
\end{equation*}
Ayr\i ca $1\leq k<\upomega$ ise 
$\aleph_{\alpha}\cardsum k=k\cardsum \aleph_{\alpha}=\aleph_{\alpha}\cardprod
k=k\cardprod\aleph_{\alpha}=\aleph_{\alpha}$.

Genelde siyah harfler, s\i n\i flar\i\ g\"osterecek.  
\c Simdi $\bm A$ ile $\bm B$, s\i n\i f olsun.%
%\glossary{$\bm A$, $\bm B$, \dots}
E\u ger $\bm A$ s\i n\i f\i n\i n
her eleman\i, $\bm B$ s\i n\i f\i n\i n eleman\i ysa, o zaman $\bm A$ s\i
n\i f\i na $\bm B$ 
\textbf{s\i n\i f\i n\i n alts\i n\i f\i}%
\index{alts\i n\i f}
(\eng{subclass of the class $B$}) denir, ve
\begin{equation*}
\bm A\included\bm B
\end{equation*}
%\glossary{$\included$}%
ifadesi yaz\i l\i r.  Bu durumda $\bm B$ s\i n\i f\i, $\bm A$ s\i n\i
f\i n\i\ 
\textbf{kapsar}%
\index{kapsama}  
(\eng{includes}).
\.I\c cerilme ($\in$) ve kapsanma ($\included$) ili\c skileri,
birbirinden tamamen farkl\i d\i r.

\textbf{Ay\i rma Aksiyomuna}%
\label{ayirma}%
\index{aksiyom!Ay\i rma A---u} 
(\eng{Separation Axiom} \cite{Zermelo-invest}) 
g\"ore, her \emph{k\"umenin} her alts\i n\i
f\i, bir k\"umedir.  \c Simdi, e\u ger $\phi(x)$ bir form\"ul ise, ve
$a$ bir k\"umeyse, o zaman \"oyle bir s\i n\i f vard\i r ki her
eleman\i, hem $a$ k\"umesinin eleman\i d\i r, hem de $\phi(x)$
form\"ul\"un\"u sa\u glar.  Bu s\i n\i f, 
\begin{equation*}
\{x\in a\colon\phi(x)\}
\end{equation*}
%\glossary{$\{x\in a\colon\phi(x)\}$}%
olarak yaz\i l\i r.  Ay\i rma Aksiyomuna g\"ore, bu s\i n\i f, bir
k\"umedir.  O zaman bu k\"ume, 
\textbf{$a$ k\"umesinin bir altk\"umesidir}%
\index{altk\"ume}
(\eng{a subset of the set $a$}).

Bir $a$ k\"umesinin t\"um altk\"umeleri, bir s\i n\i f olu\c sturur.  Bu s\i n\i f, $a$ k\"umesinin \textbf{kuvvet s\i n\i f\i d\i r} (\eng{power class}), ve
\begin{equation*}
\pow a
\end{equation*}
%\glossary{$\pow a$}%
olarak yaz\i l\i r.  
\textbf{Kuvvet K\"umesi Aksiyomuna}%
\label{kuvvetkumesi}%
\index{aksiyom!Kuvvet K\"umesi A---u} 
(\eng{Power Set Axiom} \cite{Zermelo-invest}) 
g\"ore, bu s\i n\i f, her zaman bir
k\"umedir.  
\textbf{Cantor'un Teoremine}%
\index{teorem!Cantor'un T---i}%
\index{Cantor'un Teoremi}%
\footnote{Levy'ye \cite{MR1924429} g\"ore Cantor, bu teoremi 1892 y\i
  l\i nda yay\i mlad\i.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
g\"ore,
her k\"umenin kuvvet k\"umesi, k\"umeden kesinlikle daha
b\"uy\"ukt\"ur, yani kardinali daha b\"uy\"ukt\"ur.  Bu teorem, 
\begin{equation*}
a\prec\pow a
\end{equation*}
%\glossary{$\prec$}
ifadesiyle s\"oylenir.

E\u ger $a$ ile $b$, iki k\"umeyse, 
o zaman $a$ k\"umesinden $b$ k\"umesine giden 
g\"ondermeler toplulu\u gu, bir k\"umedir, ve bu k\"ume
\begin{equation*}
{}^ab
\end{equation*}
%\glossary{${}^ab$}%
olarak yaz\i labilir.  O zaman
${}^a2\approx\pow a$.
E\u ger $\kappa$ ile $\lambda$, iki kardinal ise, tan\i ma g\"ore
\begin{equation*}
\kappa^{\lambda}
\end{equation*}
kuvveti, ${}^{\lambda}\kappa$ k\"umesinin kardinalidir.  
E\u ger $2\leq\kappa\leq\lambda$ ise, o zaman
\begin{equation*}
2^{\lambda}\leq\kappa^{\lambda}\leq(2^{\kappa})^{\lambda}=2^{\kappa\cdot\lambda}=2^{\lambda};
\end{equation*}
\"ozel olarak $\kappa^{\lambda}=2^{\lambda}$.

\c Simdi $\Z$,
%\glossary{$\Z$}
\textbf{tamsay\i lar}\index{say\i!tam---lar\i} toplulu\u gu olsun.  
O zaman
$\Z\approx\upomega$,
\c c\"unk\"u tamsay\i lar, sonsuz bir
\begin{equation*}
0,1,-1,2,-2,3,-3,4,\dots
\end{equation*}
listesinde yaz\i labilir.  Ayr\i ca her tamsay\i, $\upomega$ k\"umesinin
elemanlar\i\ gibi, bir k\"ume olarak d\"u\c s\"un\"ulebilir.  Bunu
g\"ostermek i\c cin, e\u ger $a$ ile $b$, herhangi iki k\"umeyse, o zaman
\begin{equation*}
  (a,b)
\end{equation*}
%\glossary{$(a,b)$}%
 \textbf{s\i ral\i\ ikilisi}%
\index{ikili}%
\index{s\i ra!---l\i\ ikili}
(\eng{ordered pair}),
$\big\{\{a\},\{a,b\}\bigr\}$
k\"umesi olarak tan\i mlan\i r.\footnote{\label{Kuratowski}Bu,
  Kuratowski'nin tan\i m\i d\i r \cite{Kuratowski}.  Daha \"once, Wiener 
  \cite{Wiener} daha karma\c s\i k bir tan\i m verdi.}  O zaman
$n\in\upomega$ ve $n>0$ ise, o zaman $-n$ tamsay\i s\i, $(0,n)$ olarak
tan\i mlanabilir.

Ba\c ska y\"ontemle $\Z$ toplulu\u gunun her $r$ eleman\i n\i,
\begin{equation*}
\{(x,y)\in\upomega\times\upomega\colon x=y+r\}
\end{equation*}
olarak tan\i
mlanabiliriz.  Bu tan\i ma g\"ore $\Z$ toplulu\u gunun her eleman\i,
bir 
\textbf{denklik s\i n\i f\i d\i r.}%
\index{s\i n\i f!denklik ---\i}%
\index{denk!---lik s\i n\i f\i, ba\u g\i nt\i s\i}
Asl\i nda $\upomega\times\upomega$ \c carp\i m\i nda \"oyle bir $E$
\textbf{denklik ba\u g\i nt\i s\i}%
\index{ba\u g\i nt\i!denklik ---s\i} 
vard\i r ki
\begin{equation*}
  (a,b)\mathrel E(c,d)\liff a+d=b+c,
\end{equation*}
ve $\Z$ toplulu\u gu, $(\upomega\times\upomega)/E$ b\"ol\"um\"u olarak
tan\i mlanabilir.

  \"Oyleyse
$\Z$ toplulu\u gu, bir s\i n\i ft\i r.  O zaman 
Yerle\c stirme Aksiyomuna
g\"ore $\Z$, bir k\"ume
olmal\i, \c c\"unk\"u $\Z\approx\upomega$. 

Benzer \c sekilde $\Q$
%\glossary{$\Q$}
\textbf{kesirli say\i lar}\index{say\i!kesirli
  ---lar} toplulu\u gu, \"oyle bir
$(\Z\times\Z)/F$ b\"ol\"um\"ud\"ur ki 
\begin{equation*}
  (a,b)\mathrel F(c,d)\liff ad=bc.
\end{equation*}
Asl\i nda $\Q\approx\upomega$,
\c c\"unk\"u kesirli say\i lar, 
\Sekilde{fig:SB}ki ``Stern--Brocot a\u gac\i'' olarak, 
ve ondan sonra bir liste olarak, yaz\i labilir. 
\begin{figure}[ht]
\centering
\newcommand{\SB}[1]{\TR{\psframebox[linestyle=none]{\ensuremath{#1}}}}
\pstree[treemode=D,treesep=0mm%,showbbox=true,treenodesize=10pt
]{\SB{0}}
{\pstree{\SB{-1}}
       {\pstree{\SB{-2}}
               {\pstree{\SB{-3}}
                       {\SB{-4}
                        \SB{-\frac52}}
                \pstree{\SB{-\frac32}}
                       {\SB{-\frac53}
                        \SB{-\frac43}}}
        \pstree{\SB{-\frac12}}
               {\pstree{\SB{-\frac23}}
                       {\SB{-\frac34}
                        \SB{-\frac35}}
                \pstree{\SB{-\frac13}}
                       {\SB{-\frac25}
                        \SB{-\frac14}}}}
\pstree{\SB{1}}
       {\pstree{\SB{\frac12}}
               {\pstree{\SB{\frac13}}
                       {\SB{\frac14}
                        \SB{\frac25}}
                \pstree{\SB{\frac23}}
                       {\SB{\frac35}
                        \SB{\frac34}}}
        \pstree{\SB{2}}
               {\pstree{\SB{\frac32}}
                       {\SB{\frac43}
                        \SB{\frac53}}
                \pstree{\SB{3}}
                       {\SB{\frac52}
                        \SB{4}}}}}
\caption{Stern--Brocot A\u gac\i}\label{fig:SB}
\end{figure}

\c Simdi $\R$,
%\glossary{$\R$}
\textbf{ger\c cel say\i lar}\index{say\i!gercel ---lar}
toplulu\u gu olsun.  Her 
kesirli say\i, ger\c cel say\i\ olarak d\"u\c s\"un\"ulebilir.  Ayr\i ca
her iki farkl\i\ ger\c cel say\i n\i n aras\i nda bir kesirli
say\i\ vard\i r.  O zaman $\R$ toplulu\u gundan $\pow{\Q}$ kuvvet
k\"umesine giden \"oyle bir $h$ g\"ondermesi vard\i r ki her $a$ ger\c
cel say\i s\i\ i\c cin 
\begin{equation*}
h(a)=\{x\in\Q\colon x<a\},
\end{equation*}
ve bu g\"onderme, birebirdir.  \"Oyleyse $a$ say\i s\i, $h(a)$
k\"umesi olarak d\"u\c s\"un\"ulebilir, ve $\R$, bir k\"umedir.  Ayr\i
ca 
\begin{align*}
\R\preccurlyeq\pow{\Q}&\approx\pow{\upomega}&
&\text{ ve }&
\pow{\upomega}&\preccurlyeq\R.
\end{align*}
\"Orne\u gin
\begin{equation*}
\pow{\upomega}\approx{}^{\upomega}2
\end{equation*}
\c c\"unk\"u ${}^{\upomega}2$ k\"umesinden $\pow{\upomega}$ k\"umesine giden bir
\begin{equation*}
f\mapsto\{x\in\upomega\colon f(x)=1\}
\end{equation*}
e\c slemesi vard\i r, ve ayr\i ca ${}^{\upomega}2$ k\"umesinden $\R$ k\"umesine giden bir birebir
\begin{equation*}
f\mapsto\sum_{k=0}^{\infty}\frac{2\cdot f(k)}{3^{k+1}}
\end{equation*}
g\"ondermesi vard\i r.  Sonu\c c olarak, 
\textbf{Schr\"oder--Bernstein Teoremine}%
\index{teorem!Schr\"oder--Bernstein T---i}\label{SBT}
g\"ore
\begin{equation*}
\R\approx\pow{\upomega},
\end{equation*}
\c c\"unk\"u bu teoreme g\"ore t\"um $a$ ile $b$ k\"umeleri i\c cin
\begin{equation*}
a\preccurlyeq b\preccurlyeq a\lto a\approx b.
\end{equation*}

\c Simdi Cantor'un Teoreminden
$\upomega\prec\R$.
\"Ozel olarak \"oyle bir $\alpha$ olacak ki $\alpha>0$ ve
$\R\approx\aleph_{\alpha}$.
Ama $\alpha$ ordinalinin $1$ olup olmad\i\u g\i n\i\ bilmiyoruz.
\textbf{Kontin\"u Hipotezine}%
\index{Kontin\"u Hipotezi}
(\eng{Continuum Hypothesis})
g\"ore $\alpha=1$, yani $\upomega\preccurlyeq
a\prec\pow{\upomega}$ ise $a\approx\upomega$.
\textbf{Genelle\c stirilmi\c s Kontin\"u Hipotezine}
(\eng{Generalized Continuum Hypothesis})
g\"ore her sonsuz $b$ k\"umesi i\c cin $b\preccurlyeq c\prec\pow{b}$
ise $b\approx c$.

Se\c cim Aksiyomu hari\c c k\"umeler kuram\i n\i n kullanaca\u g\i m\i
z aksiyomlar\i, 
\emph{Zermelo--Fraenkel Aksiyomlar\i d\i r.}%
\index{aksiyom!Zermelo--Fraenkel A---lar\i}%
\index{aksiyom!Se\c cim A---u}
Asl\i nda Zermelo'nun verdi\u gi aksiyomlar \cite{Zermelo-invest}, a\c
sa\u g\i dad\i r.
\begin{compactenum}[I.]
  \item
\emph{Uzama}%
\index{aksiyom!Uzama A---u}
(\sayfada{uzama}).
\item
\textbf{Temel K\"umeler}%
\index{aksiyom!Temel K\"umeler A---u}
(\eng{Elementary Sets}):
 $\emptyset$, $\{a\}$, ve $\{a,b\}$
topluluklar\i, k\"umedir.
\item
Ay\i rma%
\index{aksiyom!Ay\i rma A---u}
 (\sayfada{ayirma}).
\item
Kuvvet K\"umesi (\sayfada{kuvvetkumesi}).
\item
Bile\c sim (\sayfada{bilesim}).
\item
Se\c cim%
\index{aksiyom!Se\c cim A---u}
  (\sayfada{secim}).
\item
Sonsuzluk%
\index{aksiyom!Sonsuzluk A---u}
 (\sayfada{sonsuzluk}).
\end{compactenum}
(\sayfada{bitistirme}ki Biti\c stirme Aksiyomumuz,%
\index{aksiyom!Biti\c stirme A---u}
Zermelo'nun II.\ ve V.\ aksiyomlar\i\ taraf\i ndan gerektirilir.  Ters
olarak Biti\c stirme ve Bo\c s K\"ume Aksiyomlar\i m\i z, Zermelo'nun
II.\ aksiyomunu gerektirir.)  Sonra iki aksiyom daha verildi:
\begin{compactenum}[I.]
\setcounter{enumi}7
  \item
Yerle\c stirme%
\index{aksiyom!Yerle\c stirme A---u}
 (\sayfada{yerlestirme}).
\item
\textbf{Temellendirme}%
\label{temellendirme}%
\index{aksiyom!Temellendirme A---u}
(\eng{Foundation} \cite{Skolem-some-remarks}):  Her bo\c s olmayan $a$ k\"umesinin \"oyle bir $b$
eleman\i\ vard\i r ki $a\cap b=\emptyset$ 
(\sayfaya{kesisim} bak\i n).
\end{compactenum}
I--V ile VII--IX \nolu\ aksiyomlar, 
\textbf{Zermelo--Fraenkel Aksiyomlar\i d\i r.} 

Birka\c c tane k\i saltmalar kullan\i l\i r:
\begin{align*}
\ac&=\text{Se\c cim Aksiyomu,}\\
\zf&=\text{Zermelo--Fraenkel Aksiyomlar\i,}\\
\zfc&=\text{Zermelo--Fraenkel Aksiyomlar\i yla Se\c cim Aksiyomu,}\\
\ch&=\text{Kontin\"u Hipotezi,}\\
\gch&=\text{Genelle\c stirilmi\c s Kontin\"u Hipotezi.}
\end{align*}
%\glossary{\ac, \zf, \zfc, \ch, \gch}%
O zaman
\begin{equation*}
\zfc=\zf+\ac.
\end{equation*}
G\"odel'in kan\i tlad\i\u g\i\ teoreme g\"ore $\zf$ \textbf{tutarl\i
  ysa}\index{tutarl\i} (yani ondan bir \c celi\c ski \c c\i kmazsa), o
zaman $\zfc$ aksiyomlar\i\ da tutarl\i d\i r, ve ayr\i ca $\zfc$
aksiyomlar\i yla $\gch$ tutarl\i d\i r
\cite{Goedel-pnas-1,Goedel-pnas-2}.  
Sierpi\'nski \cite{MR0020121},
\begin{equation*}
  \zf+\gch\lto\ac
\end{equation*}
gerektirmesinin g\"osterdi.\footnote{Sierpi\'nski'ye g\"ore 1926 y\i
  l\i nda Lindenbaum ve Tarski, bu gerektirmesini ilan ettiler, ama
  kan\i t\i n\i\ vermediler.}
Cohen'in \cite{MR0232676}
kan\i tlad\i\u g\i\ teoreme g\"ore $\zf$ tutarl\i ysa, o zaman
$\zf+\lnot\ac$ aksiyomlar\i\ da tutarl\i d\i r, ve ayr\i ca
$\zfc+\lnot\ch$ tutarl\i d\i r.  Sierpi\'nski'nin teoremi, a\c sa\u
g\i daki \Teorem{thm:Sierpinski} olacakt\i r;
G\"odel'in ve Cohen'in teoremlerini 
kan\i tlamayaca\u g\i z.

\chapter{Mant\i k}\label{mantik}

\section{Form\"uller}

Form\"ullerde kullanaca\u g\i m\i z simgelerin birka\c c tane t\"ur\"u vard\i r:
\begin{compactenum}[1)]
\item
\textbf{de\u gi\c skenler}\index{de\u gi\c sken} (\eng{variables}):
$z$, $y$, $x$, \dots; $x_0$, $x_1$, $x_2$, \dots;%
\glossary{$x$, $y$, $z$, \dots}
\item
\textbf{sabitler}\index{sabit} (\eng{constants}): $a$, $b$, $c$,
\dots; $a_0$, $a_1$, $a_2$, \dots;%
\glossary{$a$, $b$, $c$, \dots}\footnote{Bilinen de\u gerler i\c
  cin Latin alfabesinin ba\c slang\i c\i ndan harflerin kullan\i l\i\c
  s\i, ve bilinmeyen de\u gerler i\c cin Latin alfabesinin sonundan
  harflerin kullan\i l\i\c s\i, Descartes'te \cite{Descartes-Geometry}
  g\"or\"un\"ur.} 
\item
\textbf{ikili ba\u glay\i c\i lar}\index{ba\u glay\i c\i} (\eng{binary connectives}): $\land$, $\lor$, $\lto$, $\liff$;\footnote{Bazen $\lto$ ile $\liff$ oklar\i n\i n yerine $\to$ ile $\leftrightarrow$ i\c saretleri yaz\i l\i r.  Bunlar\i\ kalemle yazmak daha kolayd\i r.  Ama bu notlarda, $\bm F\colon\bm A\to\bm B$ ifadesi, $\bm F$ g\"ondermesinin $\bm A$ s\i n\i f\i ndan $\bm B$ s\i n\i f\i na gitti\u ginin anlam\i na gelecek.  
A\c sa\u g\i daki \sayfaya{to} bak\i n.}
\item
bir \textbf{birli ba\u glay\i c\i} (\eng{singulary connective}): $\lnot$;
\item
\textbf{niceleyiciler}\index{niceleyici} (\eng{quantifiers}): $\exists$, $\forall$;
\item
\textbf{ayra\c clar}\index{ayra\c c} (\eng{parentheses, brackets}): $($, $)$;
\item
bir \textbf{y\"uklem}\index{y\"uklem} (\eng{predicate}): $\in$ (epsilon).\footnote{Yukar\i daki \sayfada{epsilon}ki dipnota bak\i n.}
\end{compactenum}
Bir \textbf{terim}%
\index{terim}\label{terim}
(\eng{term}), ya de\u gi\c sken ya da sabittir.  E\u ger $t$ ile $u$, iki terim ise, o zaman
\begin{equation*}
t\in u
\end{equation*}
ifadesi, bir \textbf{b\"ol\"unemeyen form\"uld\"ur}%
\index{form\"ul} (\eng{atomic formula}).  
Genelde \textbf{form\"ullerin} tan\i m\i, \"ozyinelidir:
\begin{compactenum}
\item
B\"ol\"unemeyen bir form\"ul, bir form\"uld\"ur.
\item
E\u ger $\phi$, bir form\"ul ise, o zaman
\begin{equation*}
\lnot\phi
\end{equation*}
ifadesi de bir form\"uld\"ur.
\item
E\u ger $\phi$ ile $\psi$, iki form\"ul ise, o zaman
\begin{align*}
&(\phi\land\psi),&
&(\phi\lor\psi),&
&(\phi\lto\psi),&
&(\phi\liff\psi)
\end{align*}
 ifadeleri de, form\"uld\"ur.
\item
E\u ger $\phi$ bir form\"ul ise, ve $x$ bir de\u gi\c sken ise, o zaman
\begin{align*}
&\Exists x\phi,&\Forall x\phi
\end{align*}
ifadeleri de form\"uld\"ur.
\end{compactenum}
Form\"ullerin her t\"ur\"un\"un ad\i\ vard\i r:
\begin{compactenum}
\item
$\lnot\phi$ form\"ul\"u, bir 
\textbf{de\u gillemedir}%
\index{de\u gilleme}
(\eng{negation}).
\item
$(\phi\land\psi)$ form\"ul\"u, bir 
\textbf{birle\c sme}%
\index{birle\c sme}
veya \textbf{t\"umel evetlemedir}
\index{t\"umel evetleme}%
\index{evetleme}
(\eng{conjunction}).
\item
$(\phi\lor\psi)$ form\"ul\"u, bir 
\textbf{ayr\i lma}%
\index{ayr\i lma}
veya \textbf{tikel evetlemedir} 
(\eng{disjunction}). 
\item
$(\phi\lto\psi)$ form\"ul\"u, bir 
\textbf{gerektirme}%
%\textbf{kar\i\c st\i rmad\i r}
\index{gerektirme}
(\eng{implication}). 
\item
$(\phi\liff\psi)$ form\"ul\"u, bir 
\textbf{denkliktir}%
\index{denklik}
(\eng{equivalence}). 
\item
$\Exists x\phi$ form\"ul\"u, bir 
\textbf{\"orneklemedir}%
\index{\"ornekleme}
(\eng{instantiation}). 
\item
$\Forall x\phi$ form\"ul\"u, bir 
\textbf{genelle\c stirmedir}%
\index{genelle\c stirme}
(\eng{generalization}). 
\end{compactenum}
Bu t\"urlerin adlar\i, \c cok \"onemli de\u gildir.  Fakat a\c sa\u
g\i daki teorem \c cok \"onemlidir. 

\begin{theorem}
Her form\"ul\"un tek bir \c sekilde tek bir t\"ur\"u vard\i r.
\end{theorem}

Mesela ayn\i\ form\"ul, hem gerektirme, hem \"ornekleme olamaz:
$\Exists x(\phi\lto\psi)$ form\"ul\"u, gerektirme de\u gil,
\"orneklemedir; $(\Exists x\phi\lto\psi)$ form\"ul\"u, \"ornekleme de\u gil, gerektirmedir.

Ayr\i ca $(\phi\land(\psi\land\theta))$ form\"ul\"u, tek bir \c
sekilde birle\c smedir.  Asl\i nda sadece $\phi$ ile
$(\psi\land\theta)$ form\"ullerinin birle\c smesidir.  E\u ger $A$
harf\/i, $\phi\land(\psi$ ifadesini g\"osterirse ve $B$ harf\/i,
$\theta)$ ifadesini g\"osterirse, o zaman $(A\land B)$ ifadesi,
$(\phi\land(\psi\land\theta))$ form\"ul\"un\"u g\"osterir; ama tan\i
ma g\"ore bu form\"ul, $A$ ile $B$ ifadelerinin birle\c smesi de\u
gildir, \c c\"unk\"u $A$ ile $B$ ifadeleri (yani $A$ ile $B$ taraf\i
ndan g\"osterilen ifadeler), form\"ul de\u gildir. 

Teoremi kan\i tlamayaca\u g\i z.  
Fakat teoremi kullanarak a\c sa\u g\i daki \"ozyineli tan\i m\i\ yapabiliriz.  
Bir de\u g\i\c skenin bir form\"ulde birka\c c tane 
\textbf{ge\c ci\c si}%
\index{ge\c cis}
(\eng{occurrence})
olabilir.  Mesela $\Forall x(x\in y\liff x\in z)$ form\"ul\"unde $x$ de\u gi\c skeninin \"u\c c tane ge\c ci\c si vard\i r (ve $y$ ile $z$ de\u gi\c skenlerinin birer ge\c ci\c si vard\i r).
\begin{compactenum}
\item
B\"ol\"unemeyen bir form\"ulde bir de\u gi\c skenin her ge\c ci\c si,
\textbf{serbest} bir ge\c ci\c stir.
\item
Bir de\u gi\c skenin $\phi$ form\"ul\"undeki her serbest ge\c ci\c si,
$\lnot\phi$, $(\phi*\psi)$, ve $(\psi*\phi)$ form\"ullerinde de
serbesttir.  (Burada $*$ i\c sareti, herhangi bir ikili ba\u glay\i c\i
d\i r.) 
\item
E\u ger $x$ ile $y$, iki \emph{farkl\i} de\u gi\c sken ise, o zaman $x$ de\u
gi\c skeninin $\phi$ form\"ul\"unde her serbest ge\c ci\c si, $\Exists
y\phi$ ile $\Forall y\phi$ form\"ullerinde de serbesttir. 
\item
$\Exists x\phi$ ile $\Forall x\phi$ form\"ullerinde $x$ de\u gi\c
  skeninin hi\c c serbest ge\c ci\c si yoktur.
\end{compactenum}
Bir form\"ulde bir de\u gi\c skenin serbest ge\c ci\c si varsa, bu
de\u gi\c sken, form\"ul\"un bir \textbf{serbest de\u gi\c skenidir.}
Serbest de\u gi\c skeni olmayan bir form\"ul, bir
\textbf{c\"umledir.}\index{c\"umle}  C\"umleler i\c cin $\sigma$, $\tau$,
ve $\rho$ gibi Yunan harflerini kullanaca\u g\i z. 

\section{Do\u gruluk ve Yanl\i\c sl\i k}

Bir $\phi$ form\"ul\"un\"un tek serbest de\u gi\c skeni $x$ ise, o
zaman form\"ul 
\begin{equation*}
\phi(x)
\end{equation*}
olarak yaz\i labilir.  O halde $a$ bir sabit ise, ve $x$ de\u gi\c
skeninin $\phi$ form\"ul\"undeki her \emph{serbest} ge\c ci\c sinin yerine
$a$ konulursa, \c c\i kan c\"umle 
\begin{equation*}
\phi(a)
\end{equation*}
olarak yaz\i labilir.  \c Simdi 
\textbf{do\u grulu\u gu}%
\index{do\u gruluk}\label{truth}
(\eng{truth}) ve
\textbf{yanl\i\c sl\i\u g\i}%
\index{yanl\i\c sl\i k}
(\eng{falsehood})
tan\i mlayabiliriz: 
\begin{compactenum}
\item
E\u ger $b$ k\"umesi, $a$ k\"umesini i\c cerirse, o zaman $a\in b$
c\"umlesi do\u grudur; i\c cermezse, yanl\i\c st\i r. 
\item
E\u ger $\sigma$ c\"umlesi do\u gruysa, o zaman $\lnot\sigma$ de\u
gillemesi yanl\i\c st\i r; $\sigma$ yanl\i\c s ise, $\lnot\sigma$ do\u
grudur. 
\item
E\u ger hem $\sigma$ hem $\tau$ do\u gruysa, o zaman
$(\sigma\land\tau)$ birle\c smesi de do\u grudur; $\sigma$ ile $\tau$
c\"umlelerinin biri yanl\i\c s ise, birle\c smesi de yanl\i\c st\i r. 
\item
E\u ger bir $a$ k\"umesi i\c cin $\phi(a)$ c\"umlesi do\u gruysa, o
zaman $\Exists x\phi(x)$ \"orneklemesi de do\u grudur; hi\c c \"oyle
bir $a$ yoksa, \"ornekleme yanl\i\c st\i r. 
\item
$(\sigma\lor\tau)$ c\"umlesi, $\lnot(\lnot\sigma\land\lnot\tau)$
  c\"umlesinin anlam\i na gelir, yani bu iki c\"umle ayn\i\ zamanda ya
  do\u grudur, ya da yanl\i\c st\i r. 
\item
$(\sigma\lto\tau)$ c\"umlesi, $(\lnot\sigma\lor\tau)$ c\"umlesinin anlam\i na gelir.
\item
$(\sigma\liff\tau)$ c\"umlesi, $\bigl((\sigma\lto\tau)\land(\tau\lto\sigma)\bigr)$ c\"umlesinin anlam\i na gelir.
\item
$\Forall x\phi(x)$ c\"umlesi, $\lnot\Exists x\lnot\phi(x)$ c\"umlesinin anlam\i na gelir.
\end{compactenum}
\"Ozel olarak form\"ullerde $\lor$, $\lto$, $\liff$, ve $\forall$
simgeleri gerekmez; sadece kolayl\i k i\c cin kullanaca\u g\i z.  Ama
$(\sigma\lto\tau)$ c\"umlesi do\u grudur ancak ve ancak $\tau$ do\u
gru veya $\sigma$ yanl\i\c st\i r; ve $(\sigma\liff\tau)$ c\"umlesi
do\u grudur ancak ve ancak hem $\sigma$ hem $\tau$ ya do\u gru ya
yanl\i\c st\i r.  Ayr\i ca $\Forall x\phi(x)$ do\u grudur ancak ve
ancak her $a$ k\"umesi i\c cin $\phi(a)$ do\u grudur. 

Birka\c c tane k\i saltma daha kullan\i r\i z:
\begin{compactenum}
\item
$\lnot\; t\in u$ form\"ul\"un\"un yerine $t\notin u$ ifadesini yazar\i z;
\item
Bir $(\phi*\psi)$ form\"ul\"un\"un en d\i\c staki ayra\c clar\i
n\i\ yazmay\i z.
\item
$\lto$ ile $\liff$ ba\u glay\i c\i lar\i na g\"ore $\land$ ile
$\lor$ ba\u glay\i c\i lar\i na \"onceli\u gi veririz:  Mesela
$\phi\land\psi\lto\chi$ ifadesi, $(\phi\land\psi)\lto\chi$
form\"ul\"un\"un anlam\i na gelir.   
\item
$\phi\lto\psi\lto\chi$ ifadesi, $\phi\lto(\psi\lto\chi)$
form\"ul\"un\"un anlam\i na gelir. 
\end{compactenum}
Bir $\phi$ form\"ul\"un\"un serbest de\u gi\c skenleri $x$ ile $y$
ise, o zaman form\"ul 
\begin{equation*}
\phi(x,y)
\end{equation*}
olarak yaz\i labilir.  O halde $a$ ile $b$, iki sabit ise, ve $x$ de\u
gi\c skeninin $\phi$ form\"ul\"undeki her serbest ge\c ci\c sinin
yerine $a$ konulursa, ve benzer \c sekilde $y$ de\u gi\c skeninin her
serbest ge\c ci\c sinin yerine $b$ konulursa, \c c\i kan c\"umle
\begin{equation*}
\phi(a,b)
\end{equation*}
olarak yaz\i labilir.  

Genelde $\phi$ form\"ul\"un\"un serbest de\u gi\c skenleri, bir $\vec
x$ listesini olu\c sturursa, o zaman form\"ul 
\begin{equation*}
\phi(\vec x)
\end{equation*}
olarak yaz\i labilir; ayr\i ca
\begin{align*}
\Forall{\vec x}&\phi(\vec x),&
\Exists{\vec x}&\phi(\vec x)
\end{align*}
c\"umleleri yaz\i labilir.  E\u ger $\vec a$, uzunlu\u gun $\vec x$
listesinin uzunlu\u gu olan bir sabit listesiyse, o zaman 
\begin{equation*}
\phi(\vec a)
\end{equation*}
c\"umlesi de \c c\i kar.
E\u ger $\phi(\vec x)$ ile $\psi(\vec x)$, iki form\"ul ise, ve \emph{sadece
do\u grulu\u gun tan\i m\i n\i\ kullanarak}
\begin{equation*}
\Forall{\vec x}\bigl(\phi(\vec x)\liff\psi(\vec x)\bigr)
\end{equation*}
c\"umlesinin do\u grulu\u gu kan\i tlanabilirse, o zaman $\phi$ ile
$\psi$ birbirine 
\textbf{(mant\i\u ga g\"ore) denktir}\index{denk} (\eng{logically
  equivalent}):
k\i saca
\begin{equation*}
  \phi\denk\psi.
\end{equation*}
\"Oyleyse $\phi$ ile $\psi$ birbirine denktir, ancak
ve ancak her $\vec a$ sabit listesi i\c cin, \emph{do\u grulu\u gun tan\i
m\i na g\"ore}
\begin{equation*}
\phi(\vec a)\liff\psi(\vec a)
\end{equation*}
c\"umlesi do\u grudur.  \"Orne\u gin, yukar\i daki tan\i mlara g\"ore
\begin{gather*}
	\phi\lor\psi\denk\lnot(\lnot\phi\land\lnot\psi),\\
	\phi\lto\psi\denk\lnot\phi\lor\psi,\\
	\phi\liff\psi\denk(\phi\lto\psi)\land(\psi\lto\phi),\\
	\Forall x\phi\denk\lnot\Exists x\lnot\phi.
\end{gather*}
Ama $\Exists y\Forall x\bigl(\phi(x)\lto x\in y\bigr)$ ile
$\Exists y\Forall x\bigl(\phi(x)\liff x\in y\bigr)$, denk de\u gildir.

\begin{theorem}\label{thm:denklik}
\mbox{}
\begin{compactenum}
\item
Her form\"ul, kendisine denktir.
\item
E\u ger $\phi$ ile $\psi$ denk ise, o zaman $\psi$ ile $\phi$ denktir.
\item
E\u ger $\phi$ ile $\psi$ denk ise, ve $\psi$ ile $\chi$ denk ise, o zaman $\phi$ ile $\chi$ denktir.
\end{compactenum}
\begin{comment}
  

Yani
\begin{gather*}
	\phi\denk\phi,\\
	\phi\denk\psi\lto\psi\denk\phi,\\
	\phi\denk\psi\land\psi\denk\chi\lto\phi\denk\chi.
\end{gather*}



\end{comment}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
$\sigma\liff\sigma$ her zaman do\u grudur.
\item
$\sigma\liff\tau$ do\u gru olsun.  O zaman hem $\sigma$ hem $\tau$ ya do\u gru ya yanl\i\c st\i r.  \"Oyleyse hem $\tau$ hem $\sigma$ ya do\u gru ya yanl\i\c st\i r; yani $\tau\liff\sigma$ do\u grudur.
\item
$\sigma\liff\tau$ ve $\tau\liff\rho$ do\u gru olsun.  E\u ger $\sigma$ do\u gruysa, o zaman $\tau$ do\u gru olmal\i, ve sonu\c c olarak $\rho$ do\u gru olmal\i, dolay\i s\i yla $\sigma\liff\rho$ do\u grudur.  Benzer \c sekilde $\sigma$ yanl\i\c s ise $\sigma\liff\rho$ tekrar do\u grudur.\qedhere
\end{asparaenum}
\end{proof}

\begin{theorem}\mbox{}\label{thm:lto}
\begin{compactenum}
\item
$\phi\lto\psi\lto\chi$ ile $\phi\land\psi\lto\chi$ denktir.
\item
E\u ger $x$ de\u gi\c skeni, $\phi$ form\"ul\"unde serbest de\u gilse, 
o zaman
\begin{equation*}
\Forall x(\phi\lto\psi)\denk\phi\lto\Forall x\psi.
\end{equation*}
\end{compactenum}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
$\sigma\lto\tau\lto\rho$ do\u gru olsun.  E\u ger $\sigma\land\tau$
  c\"umlesi de do\u gruysa, o zaman hem $\sigma$ hem $\tau$ do\u
  grudur, ve sonu\c c olarak $\tau\lto\rho$ do\u grudur, ve $\rho$
  do\u grudur.  Yani $\sigma\land\tau\lto\rho$ do\u grudur. 

Tersi i\c cin $\sigma\land\tau\lto\rho$ do\u gru olsun.  O zaman
$\sigma\land\tau$ yanl\i\c s veya $\rho$ do\u grudur.  Yani $\sigma$
yanl\i\c s, veya $\tau$ yanl\i\c s, veya $\rho$ do\u grudur.  E\u ger
$\sigma$ do\u gruysa, o zaman $\tau$ yanl\i\c s, veya $\rho$ do\u
grudur, yani $\tau\lto\rho$ do\u grudur.  Sonu\c c olarak
$\sigma\lto\tau\lto\rho$ do\u grudur. 

\item
$\Forall x(\sigma\lto\phi(x))$ do\u gru olsun.  
O zaman her $a$ i\c cin $\sigma\lto\phi(a)$ do\u grudur.  
Sonu\c c olarak $\sigma$ do\u gruysa, 
o zaman her $a$ i\c cin $\phi(a)$ do\u gru\-dur.  
Yani $\sigma\lto\Forall x\phi(x)$ do\u grudur.

Benzer \c sekilde $\sigma\lto\Forall x\phi(x)$ do\u gruysa 
$\Forall x(\sigma\lto\phi(x))$ do\u grudur.\qedhere
\end{asparaenum}
\end{proof}

\section{E\c sitlik}

Yukar\i daki \sayfada{=} dedi\u gimiz gibi\label{==}
\begin{equation*}
  t=u
\end{equation*}%
\glossary{$=$}
ifadesi, $\Forall x(x\in t\liff
x\in u)$ form\"ul\"un\"un k\i saltmas\i\ olarak kullan\i labilir.
O zaman $=$ i\c sareti, yeni bir y\"uklemdir, 
ve tan\i m\i na g\"ore 
\begin{equation*}
t=u\denk\Forall x(x\in t\liff x\in u).
\end{equation*}
Burada $x$, herhangi bir de\u gi\c sken olabilir, ama $t$ ile $u$
terimlerinden farkl\i\ olmal\i d\i r.  \"Orne\u gin $x=y$ ifadesi,
$\Forall z(z\in x\liff z\in y)$ form\"ul\"un\"un k\i saltmas\i d\i r,
ama $\Forall x(x\in x\liff x\in y)$ form\"ul\"un\"un k\i
saltmas\i\ de\u gildir.

O zaman
\begin{equation}\label{eqn:=}
\Forall x\Forall y(x=y\liff\Forall z(z\in x\liff z\in y))
\end{equation}
c\"umlesi do\u grudur.  Yani t\"um $a$ ile $b$ k\"umeleri i\c cin
\begin{equation*}
a=b\liff\Forall x(x\in a\liff x\in b)
\end{equation*}
c\"umlesi do\u grudur.  Bu c\"umle, $\liff$ simgesinin tan\i m\i na
g\"ore, iki c\"umlenin birle\c smesine denktir, ve bu c\"umleler,
\begin{align*}
a=b&\lto\Forall x(x\in a\liff x\in b),&
\Forall x(x\in a\liff x\in b)&\lto a=b.
\end{align*}
O zaman t\"um $a$ ile $b$ k\"umeleri i\c cin, hem
\begin{equation*}
\Forall x(x\in a\liff x\in b)\lto a=b
\end{equation*}
do\u grudur, hem de, \Teoreme{thm:lto} g\"ore, her $c$ k\"umesi i\c cin,
\begin{equation*}
a=b\land c\in a\lto c\in b
\end{equation*}
do\u grudur.

Bizim i\c cin, \eqref{eqn:=} c\"umlesinin do\u grulu\u gu, bir tan\i
md\i r.  Yani, simgesi $\in$ olan \textbf{i\c cerilme}\index{i\c
  cerilme} ba\u g\i nt\i s\i, temel bir ba\u g\i nt\i d\i r, ama
\textbf{e\c sitlik}\index{e\c sitlik} ba\u g\i nt\i s\i, yukar\i daki
\eqref{eqn:=} c\"umlesini sa\u glayan bir $=$ ba\u g\i nt\i s\i d\i r. 

\begin{theorem}\label{thm:=-equiv}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{align*}
&a=a,&
a=b&\lto b=a,&
a=b\land b=c&\lto a=c
\end{align*}
c\"umleleri do\u grudur.
\end{theorem}

\ktk 

Teoreme g\"ore e\c sitlik ba\u g\i nt\i s\i, 
\textbf{d\"on\"u\c sl\"u} (\eng{reflexive}), 
\textbf{simetrik} (\eng{symmetric}), ve
\textbf{ge\c ci\c sli} (\eng{transitive}) bir ba\u g\i nt\i d\i r;
k\i saca bir 
\textbf{denklik ba\u g\i nt\i s\i d\i r}%
\index{denklik}\label{denklik}
(\eng{equivalence relation}). 

Teoremin dolay\i s\i yla 
$a=b\land b=c$ c\"umlesinin k\i saltmas\i\ olarak $a=b=c$ ifadesi yaz\i l\i r; 
yani
\begin{equation*}
a=b=c\denk a=b\land b=c.
\end{equation*}

\.Ilk resmi aksiyomumuz \c su:

\begin{axiom}[E\c sitlik]\index{aksiyom!E\c sitlik A---u}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{equation*}
a=b\land a\in c\lto b\in c
\end{equation*}
c\"umlesi do\u grudur.  Yani
\begin{equation*}
  \Forall x\Forall y\Forall z(x=y\land x\in z\lto y\in z)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

Bu aksiyomun ba\c ska bi\c cimleri vard\i r, mesela:
\begin{compactenum}
\item
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin $a=b\lto a\in c\lto b\in c$.
\item
T\"um $a$ ile $b$ k\"umeleri i\c cin $\Forall x(a=b\lto a\in x\lto b\in x)$.
\item
T\"um $a$ ile $b$ k\"umeleri i\c cin $\Forall x(a=b\land a\in x\lto b\in x)$.
\item
T\"um $a$ ile $b$ k\"umeleri i\c cin $a=b\lto\Forall x(a\in x\lto b\in x)$.
\item
$\Forall x\Forall y(x=y\lto\Forall z(x\in z\lto y\in z))$.
\item
$\Forall x\Forall y\Forall z(x=y\lto x\in z\lto y\in z)$.
\end{compactenum}

\begin{xca}
$a=b\land\Forall x(a\in x\lto b\in x)$ c\"umlesi, E\c sitlik Aksiyomundan kan\i tlanabilir mi?
\end{xca}

\begin{theorem}
Her $\phi(x)$ tek serbest de\u gi\c skenli form\"ul\"u i\c cin
\begin{equation}\label{eqn:a=b}
a=b\land\phi(a)\lto\phi(b)
\end{equation}
c\"umlesi do\u grudur.
\end{theorem}

\begin{proof}
Form\"ullerin \"ozyineli tan\i m\i\ nedeni ile, 
\emph{t\"umevar\i m} kullanabiliriz.
\begin{asparaenum}
\item
\.Ilk olarak $\phi$ b\"ol\"unemesin.  Yani $\phi(x)$, ya $c\in x$ veya $x\in c$ bi\c ciminde olsun.  O zaman \eqref{eqn:a=b} c\"umlesi, ya e\c sitli\u gin tan\i m\i ndan, ya da E\c sitlik Aksiyomundan, do\u grudur.
\item
E\u ger $\phi$, ya $\psi$ ya da $\chi$ ise, \eqref{eqn:a=b} do\u gru
olsun.  \c Simdi $a=b\land(\psi(a)\land\chi(a))$ do\u gru olsun.  O
zaman hem $a=b\land\psi(a)$ hem $a=b\land\chi(a)$ do\u gru olmal\i.
Sonu\c c olarak varsay\i m\i m\i zdan hem $\psi(b)$ hem $\chi(b)$ do\u
gru olmal\i, yani $\psi(b)\land\chi(b)$ do\u gru olmal\i.  \"Oyleyse
$\phi$, $\psi\land\chi$ ise \eqref{eqn:a=b} do\u grudur. 
\item
Son olarak, t\"um $c$ i\c cin $\phi(x)$, $\psi(x,c)$ ise, \eqref{eqn:a=b} do\u gru olsun.  \c Simdi $a=b\land\Exists y\phi(a,y)$ do\u gru olsun.  O zaman bir $c$ i\c cin $a=b\land\phi(a,c)$ do\u gru olmal\i, dolay\i s\i yla $\phi(b,c)$ do\u gru olmal\i.  Sonu\c c olarak $\Exists y\phi(b,y)$ do\u grudur.  \"Oyleyse $\phi(x)$, $\Exists y\phi(x,y)$ ise \eqref{eqn:a=b} do\u grudur.\qedhere
\end{asparaenum}
\end{proof}

Kitaplar\i n \c co\u gunda hem $\in$ hem $=$, temel ba\u g\i nt\i d\i
r, ve yukar\i daki \sayfada{eqn:=}ki \eqref{eqn:=} c\"umlesi, tan\i m de\u gil, 
\textbf{Uzama Aksiyomudur}%
\label{uzama}%
\footnote{Veya \textbf{K\"ume E\c sitli\u gi Aksiyomu}
\cite{Nesin-SKK}.}%
\index{aksiyom!Uzama A---u} 
(\eng{Axiom of Extensionality} \cite{Zermelo-invest}). 
Bu kitaplarda her $\phi(x)$ tek serbest de\u gi\c skenli form\"ul\"u i\c cin \eqref{eqn:a=b} c\"umlesi, bir
\textbf{mant\i ksal aksiyomd\i r.}%
\index{aksiyom!mant\i ksal ---}%
\index{mant\i ksal aksiyom} 

\section{S\i n\i flar}

Bir $\phi(x)$ form\"ul\"u ve bir $a$ k\"umesi i\c cin $\phi(a)$ c\"umlesi do\u gruysa $a$ k\"umesi, $\phi(x)$ form\"ul\"un\"u 
\textbf{sa\u glar}%
\index{sa\u glamak}
(\eng{satisfies}).
  O zaman $\phi$ form\"ul\"un\"u sa\u glayan k\"umeler toplulu\u gu vard\i r.  Bu topluluk
\begin{equation*}
\{x\colon\phi(x)\}
\end{equation*}
olarak yaz\i l\i r, ve ona 
\textbf{$\phi$ taraf\i ndan tan\i mlanm\i\c s s\i n\i f}%
\index{s\i n\i f}%
\index{tan\i mlama}
(\eng{class defined by $\phi$}) denir.

Yukar\i daki \sayfada{terim}ki tan\i ma g\"ore bir de\u gi\c sken veya sabit, bir \emph{terimdir.}  Daha kesinlikle bir
\textbf{k\"ume terimidir}%
\index{k\"ume!--- terimi}\index{terim!k\"ume ---i}
(\eng{set term}).  \c Simdi, e\u ger $x$ de\u gi\c skeni, $\phi$ form\"ul\"un\"un serbest bir de\u gi\c skeniyse, $\phi$ form\"ul\"un\"u
\begin{equation*}
\phi(\dots x\dots)
\end{equation*}
olarak yazar\i z.  O zaman
\begin{equation*}
\{x\colon\phi(\dots x\dots)\}
\end{equation*}
ifadesi, bir
\textbf{s\i n\i f terimi}%
\index{s\i n\i f!--- terimi}\index{terim!s\i n\i f ---i}
(\eng{class term})
olacak.
S\i n\i f terimlerini form\"ullerde kullanabiliriz, ama \c simdilik, sadece $\in$ i\c saretinin sa\u g\i nda.  Bir $x$ de\u gi\c skeninin bir $\phi(\dots y\dots)$ form\"ul\"undeki serbest ge\c ci\c si, bir
\begin{equation*}
t\in\{y\colon\phi(\dots y\dots)\}
\end{equation*}
form\"ul\"unde (h\^al\^a) serbesttir.
E\u ger $x$ de\u gi\c skeninin $\phi(\dots x\dots)$ form\"ul\"un\"undeki her serbest ge\c ci\c sinin yerine $a$ sabitini koyarsak $\phi(\dots a\dots)$ form\"ul\"u \c c\i kar.  \c Simdi tan\i ma g\"ore
\begin{equation*}
a\in\{x\colon\phi(\dots x\dots)\}\denk\phi(\dots a\dots).
\end{equation*}

Bir sabit veya bir $\{x\colon\phi(x)\}$ s\i n\i f terimi,
\textbf{kapal\i}%
\index{kapal\i}\index{terim!kapal\i\ ---}
(\eng{closed})
bir terimdir.  Kapal\i\ bir terim, bir k\"umenin veya bir s\i n\i f\i
n ad\i d\i r.  $\bm A$, $\bm B$, $\bm C$%
\glossary{$\bm A$, $\bm B$, $\bm C$, \dots}
gibi b\"uy\"uk siyah harfleri kapal\i\ s\i n\i f terimleri olarak
kullanaca\u g\i z.  O zaman \sayfada{eqn:=}ki tan\i
ma g\"ore
\begin{gather*}
\bm A=\bm B\denk\Forall x(x\in\bm A\liff x\in\bm B),\\
a=\bm B\denk\Forall x(x\in a\liff x\in\bm B),\\
\bm B=a\denk a=\bm B.
\end{gather*}
Sonu\c c olarak
\begin{equation*}
a=\{x\colon x\in a\}.
\end{equation*}
Yani her k\"ume, bir s\i n\i fa e\c sitt\i r.  Ama tersi yanl\i\c st\i r;
bildi\u gimiz gibi baz\i\ s\i n\i flar hi\c cbir k\"umeye e\c sit
de\u gildir: 

\begin{theorem}[Russell Paradoksu \cite{Russell-letter}]%
\index{paradoks!Russell P---u}%
\index{teorem!Russell Paradoksu}
$\{x\colon x\notin x\}$ s\i n\i f\i, hi\c cbir k\"umeye e\c sit de\u gildir.
\end{theorem}

\begin{proof}
Bu teoremi zaten \sayfada{Russell} kan\i tlad\i k.  \c
Simdi bir kan\i t daha verece\u giz.
$x\notin x$ form\"ul\"u taraf\i ndan tan\i mlanm\i\c s s\i n\i f, $\bm
  A$ olsun.  O zaman her $b$ k\"umesi i\c cin
  \begin{equation*}
    b\in\bm A\liff b\notin b
  \end{equation*}
do\u grudur.  O zaman $\Forall x(x\in\bm A\liff x\in b)$ c\"umlesi yanl\i\c st\i r.  E\c sitli\u gin tan\i m\i na g\"ore $b\neq\bm A$.
\end{proof}

\c Simdi s\i n\i f terimlerini $\in$ i\c saretinin solunda
kullanabiliriz, ama \c c\i kan c\"umle do\u gru olaca\u g\i\ i\c cin
s\i n\i f terimi bir k\"umeyi adland\i rmal\i: 
\begin{equation*}
\bm A\in\bm B\denk\Exists x(x=\bm A\land x\in\bm B).
\end{equation*}

E\u ger $\Forall x(x\in\bm A\lto x\in\bm B)$ do\u gruysa, o
zaman $\bm A$, $\bm B$ s\i n\i f\i n\i n \textbf{alts\i n\i f\i d\i
  r}\index{alts\i n\i f}
(\eng{subclass}), ve $\bm A\included\bm B$ ifadesini yazar\i z.  Yani
\begin{equation*}
\bm A\included\bm B\denk\Forall x(x\in\bm A\lto x\in\bm B).
\end{equation*}

\begin{theorem}\label{thm:=}\mbox{}
\begin{compactenum}
\item
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{equation*}
\bm A=\bm B\denk\bm A\included\bm B\land\bm B\included\bm A.
\end{equation*}
\item
T\"um $\bm A$, $\bm B$, ve $\bm C$ s\i n\i flar\i\ i\c cin
\begin{equation*}
\bm A\included\bm B\land\bm B\included\bm C\lto\bm A\included\bm C
\end{equation*}
c\"umlesi (mant\i\u ga g\"ore) do\u grudur.
\end{compactenum}
\end{theorem}

\ktk

\section{\.I\c slemler}

S\i n\i flarla birka\c c tane ikili i\c slem vard\i r.  \"Once
\begin{gather*}\label{kesisim}
  \bm A\cap\bm B=\{x\colon x\in\bm A\land x\in\bm B\},\\
  \bm A\cup\bm B=\{x\colon x\in\bm A\lor x\in\bm B\}.
\end{gather*}%
\glossary{$\bm A\cap\bm B$}%
\glossary{$\bm A\cup\bm B$}
Bunlar s\i ras\i yla $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
\textbf{kesi\c simi}%
\index{kesi\c sim}
(\eng{intersection}) ve
\textbf{bile\c simi}%
\index{bile\c sim}
(\eng{union}).

\begin{theorem}
  T\"um $\bm A$, $\bm B$, ve $\bm C$ s\i n\i flar\i\ i\c cin
  \begin{gather*}
\bm A\cap\bm B=\bm B\cap\bm A,\\      
\bm A\cup\bm B=\bm B\cup\bm A,\\
      \bm A\cap(\bm B\cap\bm C)=(\bm A\cap\bm B)\cap\bm C,\\
      \bm A\cup(\bm B\cup\bm C)=(\bm A\cup\bm B)\cup\bm C,\\
    \bm A\cap(\bm B\cup\bm C)=(\bm A\cap\bm B)\cup(\bm A\cap\bm C),\\
    \bm A\cup(\bm B\cap\bm C)=(\bm A\cup\bm B)\cap(\bm A\cup\bm C).
  \end{gather*}
\end{theorem}

\begin{proof}
  $x\in\bm A\land x\in\bm B\denk x\in\bm B\land x\in\bm A$, vesaire.
\end{proof}

Ondan sonra
\begin{equation*}
  \bm A\setminus\bm B=\{x\colon x\in\bm A\land x\notin\bm B\};
\end{equation*}
bu s\i n\i f, $\bm A$ s\i n\i f\i n\i n $\bm B$ s\i n\i f\i ndan
\textbf{fark\i d\i r}%
\index{fark}
(\eng{difference}).  O zaman
\begin{equation*}
    \bm A\symdiff\bm B=(\bm A\setminus\bm A)\cup(\bm B\setminus\bm A);  
\end{equation*}
bu s\i n\i f, $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
\textbf{simetrik fark\i d\i r} (\eng{symmetric difference}).

\begin{comment}
  




\begin{theorem}
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{align*}
\bm A\symdiff\bm B
&=(\bm A\setminus\bm B)\cup(\bm B\setminus\bm A)\\
&=(\bm A\cup\bm B)\setminus(\bm A\cap\bm B).
\end{align*}
\end{theorem}

\ktk


\end{comment}

\Teorem{thm:=} sayesinde bir $\bm A\included\bm
B\land\bm B\included\bm C$ c\"umlesinin yerine
\begin{equation*}
\bm A\included\bm B\included\bm C
\end{equation*}
ifadesini yazabiliriz.  \"Orne\u gin sonraki teoremi yazabiliriz.

\begin{theorem}
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{align*}
\bm A\cap\bm B&\included\bm A\included\bm A\cup\bm B,&
\bm A\cap\bm B&\included\bm B\included\bm A\cup\bm B.
\end{align*}
\end{theorem}

\ktk

S\i n\i flarda bir \emph{birli} i\c slem vard\i r:
\begin{equation*}
\bm A\comp=\{x\colon x\notin\bm A\};
\end{equation*}
bu s\i n\i f, $\bm A$ s\i n\i f\i n\i n \textbf{t\"umleyenidir}%
\index{t\"umleyen}
(\eng{complement}).

\begin{theorem}[De Morgan Kurallar\i\protect{\footnote{Asl\i nda bu kurallar\i,
    Augustus De Morgan'\i n (1806--71) eserlerinde bulamad\i m, ama
    Venedikli Paulus'un ($\sim$1369--1429) eserlerinde
    \cite[31.35]{MR0120142} buldum.}}]%
\index{teorem!De Morgan Kurallar\i}%
\index{De Morgan Kurallar\i}
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{align*}
	(\bm A\cap\bm B)\comp&=\bm A\comp\cup\bm B\comp,&
	(\bm A\cup\bm B)\comp&=\bm A\comp\cap\bm B\comp.
\end{align*}
\end{theorem}



\ktk

\.I\c cerilme ba\u g\i nt\i s\i n\i\ kullanarak birka\c c tane birli
i\c slemi daha tan\i mlayabiliriz:
\begin{gather*}
  \bigcap\bm A=\{x\colon\Forall y(y\in\bm A\lto x\in y)\},\\
\bigcup\bm A=\{x\colon\Exists y(x\in y\land y\in\bm A)\},\\
\begin{aligned}
\pow{\bm A}
&=\{x\colon\Forall y(y\in x\lto y\in\bm A)\}\\
&=\{x\colon x\included\bm A\};
\end{aligned}
\end{gather*}%
\glossary{$\bigcap\bm A$}%
\glossary{$\bigcup\bm A$}%
\glossary{$\pow{\bm A}$}
bunlar s\i ras\i yla $\bm A$ s\i n\i f\i n\i n \textbf{kesi\c simi}
(\eng{intersection}), \textbf{bile\c simi} (\eng{union}), ve
\textbf{kuvvet s\i n\i f\i d\i r} (\eng{power class}).

\begin{theorem}\label{thm:cap-cup}
E\u ger $a\in\bm B$ ise
\begin{equation*}
\bigcap\bm B\included a\included\bigcup\bm B.
\end{equation*}
\end{theorem}

\ktk

Son olarak \sayfada{universe}ki gibi
\begin{equation*}
\universe=\{x\colon x=x\},
\end{equation*}%
\glossary{$\universe$}
ve
\begin{gather*}
  \emptyset=\{x\colon x\neq x\},\\
\{a\}=\{x\colon x=a\},\\
\{a,b\}=\{x\colon x=a\lor x=b\},\\
\{a,b,c\}=\{x\colon x=a\lor x=b\lor x=c\},\\
\parbox{7cm}{\dotfill}
\end{gather*}
Buradaki $\emptyset$ s\i n\i f\i, \textbf{bo\c s s\i n\i ft\i r.}%
\index{s\i n\i f!bo\c s ---}%
\index{bo\c s s\i n\i f}
%Ancak birka\c c aksiyom daha verilmeden bunlar sadece s\i n\i ft\i rlar.

\begin{theorem}\label{thm:cap-empty}
  \begin{align*}
    \bigcap\emptyset&=\universe,&
\bigcup\emptyset&=\emptyset.
  \end{align*}
\end{theorem}

\ktk[\footnote{Baz\i\ kitaplarda $\bm A$ bo\c s ise $\bigcap\bm A$ kesi\c simi tan\i
mlanmaz.  \"Orne\u gin \cite[s.~51 \&\ 285]{Nesin-SKK} kayna\u g\i na
bak\i n.}]

Bu altb\"ol\"um\"un
\begin{align*}
  &\begin{gathered}
\bm A\cap\bm B,\\
\bm A\cup\bm B,\\
\bm A\symdiff\bm B,\\
\bm A\setminus\bm B,
   \end{gathered}&
&\begin{gathered}
    \bm A\comp,\\
\bigcap\bm A,\\
\bigcup\bm A,\\
\pow{\bm A},
  \end{gathered}&
&\universe,&
&\begin{gathered}
 \emptyset,\\
\{a\},\\
\{a,b\},\\
\{a,b,c\}   
  \end{gathered}
\end{align*}
ifadeleri, 
\emph{s\i n\i f} terimidir.%
\index{terim!s\i n\i f ---i}
Her $\bm A$ veya $\bm B$
teriminin yerine ba\c ska bir terimi koyabiliriz.  Zaten bu \c sekilde $(\bm
A\setminus\bm B)\cup(\bm B\setminus\bm A)$ gibi ifadeleri yazd\i k.
Fakat \c simdilik k\"u\c c\"uk harfler hari\c c, k\"ume terimlerimiz
yoktur.  Bu durum hemen de\u gi\c secek.

%\begin{comment}
  


\chapter{Do\u gal Say\i lar}

\section{Do\u gal say\i lar k\"umesi}

Do\u grulu\u gun \sayfada{truth}ki tan\i m\i na
g\"ore $\Exists xx=a$
c\"umlesi do\u gru mudur?  Yani $\Exists x\Forall y(y\in x\liff y\in a)$
c\"umlesi do\u gru mudur?  E\u ger bir $b$ k\"umesi i\c cin $b=a$
c\"umlesi, yani $\Forall y(y\in b\liff y\in a)$ c\"umlesi, do\u
gruysa, o zaman $\Exists xx=a$ c\"umlesi de do\u grudur.  Asl\i nda
\Teoreme{thm:=-equiv} g\"ore $a=a$ c\"umlesi do\u gru,
de\u gil mi?  O halde $\Exists xx=a$ c\"umlesi do\u gru olmal\i. 

Ama bu iddia pek do\u gru de\u gildir.  Bir $a$ k\"umesi varsa, o zaman
$\Exists xx=a$ c\"umlesi do\u grudur.  Bir k\"ume varsa, bu k\"umeye
$a$ denilebilir, ve sonu\c c olarak $\Exists xx=a$ c\"umlesi do\u
gru oluyor.  Bu ana kadar hi\c c kesin bir k\"umemiz olmad\i.  Ama k\"umeler olmal\i, ve birini zaten biliyoruz: 

\begin{axiom}[Bo\c s K\"ume]\index{aksiyom!Bo\c s K\"ume A---u}
  $\emptyset$ bo\c s s\i n\i f, bir k\"umedir:
  \begin{equation*}
    \Exists x\Forall y(y\notin x)
  \end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

Bu aksiyom sayesinde $\emptyset$ i\c sareti, bir 
k\"ume terimidir.%
\index{terim!k\"ume ---i}
Bu y\"uzden
$\{\emptyset\}$ ve $\{\emptyset,a\}$ gibi s\i n\i f terimlerini
yazabiliriz.  Bu terimler de, k\"ume terimi olacak.  Bo\c s k\"ume
gibi bilinen k\"umelerden yeni k\"umeler olu\c sturulabilir:

\begin{axiom}[Biti\c stirme]\index{aksiyom!Biti\c stirme A---u}
T\"um $a$ ile $b$ k\"umeleri i\c cin $a\cup\{b\}$ s\i n\i f\i, bir
k\"umedir:
\begin{equation*}
 \Forall x\Forall y\Exists z\Forall w(w\in z\liff w\in x\lor w=y)
\end{equation*}
c\"umlesi do\u grudur.  
\end{axiom}

\begin{theorem}[Temel K\"umeler]
  T\"um $a$ ile $b$ k\"umeleri i\c cin $\{a\}$ ile $\{a,b\}$ s\i n\i flar\i, 
k\"umedir:
\begin{gather*}
  \Forall x\Exists y\Forall z(z\in y\liff z=x),\\
\Forall x\Forall y\Exists z\Forall w(w\in z\liff w=x\lor w=y)
\end{gather*}
c\"umleleri do\u grudur.
\end{theorem}

\begin{proof}
Bo\c s K\"ume ile Biti\c stirme Aksiyomlar\i na g\"ore
  $\{a\}$ s\i n\i f\i, $\emptyset\cup\{a\}$ k\"umesine e\c sittir, ve
$\{a,b\}$ s\i n\i f\i, $\{a\}\cup\{b\}$ k\"umesine e\c sittir. 
\end{proof}

\"Ozel olarak her $a$ k\"umesi i\c cin $a\cup\{a\}$ bir k\"umedir.  Bu
son k\"ume, $a'$ olsun.  Yani her $a$ k\"umesi i\c cin
\begin{equation*}
a'=a\cup\{a\}
\end{equation*}%
\glossary{$a'$}
olsun.  $a'$ k\"umesi, $a$ k\"umesinin
\textbf{ard\i l\i d\i r}%
\index{ard\i l}
(\eng{successor}).
S\i k s\i k ard\i llar\i\ alarak\label{0123}
\begin{align*}
&\emptyset,&
&\emptyset',&
&\emptyset'',&
&\emptyset''',&
&\dots
\\
%\end{align*}
\intertext{
k\"ume dizisini olu\c
sturabiliriz.  Bu dizi,
}
%\begin{align*}
  &\emptyset,&
  &\{\emptyset\},&
  &\bigl\{\emptyset,\{\emptyset\}\bigr\},&
  &\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\},&
&\dots
\\
%\end{align*}
\intertext{Yukar\i daki \sayfada{vnn}ki gibi bu k\"umeler,
}
%\begin{align*}
&0,&
&1,&
&2,&
&3,&
&\dots
\end{align*}%
\glossary{$0$}
do\u gal say\i lar\i\ olacak.
Elemanlar\i\ \emph{t\"um} do\u gal say\i lar olan bir s\i n\i f var
m\i d\i r? 

Do\u gal say\i lar\i n \emph{toplulu\u gunun} iki \"ozelli\u gi vard\i
r:
\begin{compactenum}
  \item
$0$, bu topluluktad\i r.
\item
E\u ger $a$, bu topluluktaysa, $a\cup\{a\}$ k\"umesi de, bu
topluluktad\i r.
\end{compactenum}
Bu \"ozellikleri olan \emph{k\"umeler,} bir s\i n\i f olu\c sturur.  Yani
\begin{equation*}
\bm{\Omega}=\{x\colon 0\in x\land\Forall y(y\in x\lto y\cup\{y\}\in x)\}
\end{equation*}%
\glossary{$\bm{\Omega}$}
e\c sitli\u gini sa\u glayan bir $\bm{\Omega}$ s\i n\i f\i\ vard\i r.

\begin{theorem}\label{thm:Omega}
\mbox{}
  \begin{compactenum}
    \item
$0\in\bigcap\bm{\Omega}$.
\item
E\u ger $a\in\bigcap\bm{\Omega}$ ise, o zaman
$a\cup\{a\}\in\bigcap\bm{\Omega}$ .
\item
E\u ger $a\included\bigcap\bm{\Omega}$ ise, ve $a$,
\begin{align*}
  0&\in a,&
\Forall x(x\in a&\lto x\cup\{x\}\in a)
\end{align*}
\"ozelliklerini sa\u glarsa, o zaman $a=\bigcap\bm{\Omega}$.
  \end{compactenum}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
E\u ger $a\in\bm{\Omega}$ ise, o zaman $0\in a$.  Sonu\c c olarak $0\in\bigcap\bm{\Omega}$.
\item
$a\in\bigcap\bm{\Omega}$ olsun.  O zaman $\bm{\Omega}$ s\i n\i f\i n\i n her $b$ eleman\i\ i\c cin $a\in b$.  Ayr\i ca $b\in\bm{\Omega}$ y\"uz\"unden $\Forall y(y\in b\lto y\cup\{y\}\in b)$ c\"umlesi do\u grudur.  O zaman $a\cup\{a\}\in b$ olmal\i.  Sonu\c c olarak $a\cup\{a\}\in\bigcap\bm{\Omega}$.
\item
$0\in a$ ve
$\Forall x(x\in a\lto x\cup\{x\}\in a)$ do\u gru olsun.  O zaman $a\in\bm{\Omega}$.  Bu y\"uzden \Teoreme{thm:cap-cup} g\"ore $\bigcap\bm{\Omega}\included a$ olmal\i.  E\u ger ayr\i ca
$a\included\bigcap\bm{\Omega}$ ise, o zaman \Teoreme{thm:=} g\"ore $a=\bigcap\bm{\Omega}$.\qedhere
\end{asparaenum}
\end{proof}

Bu teoreme ra\u gmen e\u ger
\begin{align}\label{eqn:induction}
\bm A&\included\bigcap\bm{\Omega},&
0&\in\bm A,&
\Forall x(x\in\bm A&\lto x\cup\{x\}\in\bm A)
\end{align}
ise $\bm A=\bigcap\bm{\Omega}$ c\"umlesini sonu\c cland\i ram\i yoruz.
Neden?  \c C\"unk\"u \Teoreme{thm:cap-empty} g\"ore 
\begin{equation*}
\bigcap0=\universe
\end{equation*}
(yani $\bigcap\emptyset=\universe$), ve $\bm{\Omega}$ s\i n\i f\i n\i
n bo\c s olmad\i\u g\i n\i\ \c simdilik bilmiyoruz.  Bu durumu hemen
de\u gi\c stirebiliriz: 

\begin{axiom}[Sonsuzluk]%
\index{aksiyom!Sonsuzluk A---u}
$\bm{\Omega}\neq0$, yani
\begin{equation*}
\Exists x\bigl(0\in x\land\Forall y(y\in x\lto y\cup\{y\}\in x)\bigr)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

H\^al\^a yukar\i daki \eqref{eqn:induction} sat\i r\i ndaki varsay\i lardan $\bm A=\bigcap\bm{\Omega}$ c\"umlesini sonu\c cland\i ram\i yoruz.  Neden?  Bir tane aksiyomu daha kullanarak bunu sonu\c cland\i rabiliriz:

\begin{axiom}[Ay\i rma]%
\index{aksiyom!Ay\i rma A---u}
Bir k\"umenin her alts\i n\i f\i, bir k\"umedir, yani her $\phi(x)$ form\"ul\"u i\c cin
\begin{equation*}
\Forall x\Exists y\Forall z\bigl(z\in y\liff z\in x\land\phi(z)\bigr)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

\c Simdi her $a$ k\"umesi ve $\phi(x)$ form\"ul\"u i\c cin $\{x\colon x\in a\land\phi(x)\}$ s\i n\i f\i, bir k\"umedir, ve bu k\"ume
\begin{equation*}
\{x\in a\colon\phi(x)\}
\end{equation*}
olarak yaz\i l\i r.

\begin{theorem}
Bir s\i n\i f bo\c s de\u gilse, kesi\c simi bir k\"umedir.
\end{theorem}

\begin{proof}
$a\in\bm B$ olsun.  \Teoreme{thm:cap-cup} g\"ore $\bigcap\bm B\included a$.  Ay\i rma Aksiyomuna g\"ore $\bigcap\bm B$ kesi\c simi, bir k\"ume olmal\i.
\end{proof}

\"Ozel olarak
\begin{equation*}
\upomega=\bigcap\bm{\Omega}
\end{equation*}%
\glossary{$\upomega$}
e\c sitli\u gini sa\u glayan bir $\upomega$ k\"umesi vard\i r.  Bu k\"umenin elemanlar\i, \textbf{von Neumann do\u gal say\i lar\i d\i r.}%
\index{say\i!von Neumann do\u gal ---lar\i}%
\index{von Neumann do\u gal say\i lar\i}
$\upomega$ i\c sareti, yeni bir k\"ume terimidir.  Bundan sonra $\bm{\Omega}$ s\i n\i f terimini kullanmayaca\u g\i z.

\c Simdi \Teoremi{thm:Omega} a\c sa\u g\i daki bi\c cimde yazabiliriz:
  \begin{compactenum}\label{induction}
    \item
$0\in\upomega$.
\item
E\u ger $a\in\upomega$ ise, o zaman $a'\in\upomega$.
\item
E\u ger $a\included\upomega$ ise, ve $a$,
\begin{align*}
  0&\in a,&
\Forall x(x\in a&\lto x'\in a)
\end{align*}
\"ozelliklerini sa\u glarsa, o zaman $a=\upomega$.
  \end{compactenum}
Ayr\i ca her k\"umeninki gibi $\upomega$ k\"umesinin de her alts\i n\i f\i, bir k\"umedir.  Sonu\c c olarak $\upomega$ k\"umesinin baz\i\ \"ozelliklerini 
\textbf{t\"umevar\i m}%
\index{t\"umevar\i m}
(\eng{induction})
y\"ontemiyle kan\i tlayabilece\u giz.

Asl\i nda bazen $\upomega$ k\"umesinin iki \"ozelli\u gininin
daha kullan\i lmas\i\ gerekecek.  $\Forall xx'\neq0$ apa\c c\i kt\i r.
Ama $k$ ile $m$, do\u gal say\i lar ise, ve $k'=m'$ ise, $k=m$ e\c
sitli\u gini elde etmek, biraz daha zor olacak. 

M\"umk\"unse $k'=m'$ ama $k\neq m$ olsun.  O zaman $k\in m$ ve $m\in k$
olmal\i.  Bundan $k\in k$ c\"umlesini sonu\c cland\i rmak istiyoruz.

E\u ger bir $\bm A$ s\i n\i f\i,
\begin{equation*}
  \Forall x\Forall y(x\in\bm A\land y\in x\lto y\in\bm A)
\end{equation*}
c\"umlesini sa\u glarsa, o zaman $\bm A$ s\i n\i f\i na 
\textbf{ge\c ci\c sli}%
\index{ge\c ci\c sli}
(\eng{transitive}) denir.  \"Oyleyse her ge\c ci\c sli s\i n\i f\i n
her eleman\i, s\i n\i f\i n bir altk\"umesidir de.

\begin{theorem}\label{thm:n-trans}
$\upomega$ k\"umesinin her eleman\i, ge\c ci\c slidir.
\end{theorem}

\begin{proof}
$a$, $\upomega$ k\"umesinin ge\c ci\c sli elemanlar\i\ k\"umesi olsun.  Yani
\begin{align*}
a
&=\{x\in\upomega\colon\Forall y\Forall z(y\in x\land z\in y\lto z\in x)\}\\
&=\{x\in\upomega\colon\Forall y(y\in x\lto y\included x)\}
\end{align*}
olsun.  O zaman $0\in a$.  \emph{T\"umevar\i m hipotezi} olarak $b\in
a$ olsun.  $b'\in a$ c\"umlesinin do\u grulu\u gunu g\"osterece\u giz.
$c\in b'$ olsun.  Ya $c\in b$ ya da $c=b$.  E\u ger $c\in b$ ise, o
zaman hipotezimize g\"ore $c\included b$.  Her durumda $b\included
b'$.  \"Oyleyse $c\included b'$.  Ama $c$, $b'$ k\"umesinin herhangi
bir eleman\i d\i r.  Sonu\c c olarak $b'\in a$.  T\"umevar\i mdan
(yani \Teoremin{thm:Omega} \sayfada{induction}ki bi\c ciminden) $a=\upomega$. 
\end{proof}

\begin{theorem}\label{thm:w-trans}
$\upomega$ k\"umesi, ge\c ci\c slidir.
\end{theorem}

\ktk

\begin{xca}
$\bigl\{0,1,\{1\}\bigr\}$ k\"umesinin ge\c ci\c sli oldu\u gunu kan\i tlay\i n.
\end{xca}

\begin{theorem}\label{thm:irr}
$\upomega$ k\"umesinin hi\c cbir eleman\i, kendisini i\c cermez.
\end{theorem}

\begin{proof}
Tekrar t\"umevar\i m\i\ kullanaca\u g\i z.  \c C\"unk\"u bo\c s
k\"umenin hi\c cbir eleman\i\ yok, $0\notin0$.  \c Simdi
$a\in\upomega$ ve $a\notin a$ olsun.  E\u ger $a'\in a'$ ise, ya
$a'\in a$ ya da $a'=a$.  Her durumda, ge\c cen teoreme g\"ore,
$a'\included a$, dolay\i s\i yla $a\in a$ (\c c\"unk\"u $a\in a'$).
Bu sonu\c c, varsay\i m\i m\i zla \c celi\c sir.  O zaman $a'\notin
a'$ olmal\i.  T\"umevar\i mdan kan\i t\i m\i z bitti. 
\end{proof}

\begin{theorem}
$\upomega$ k\"umesinin t\"um $k$ ile $m$ elemanlar\i\ i\c cin $k'=m'$
  ise $k=m$. 
\end{theorem}

\begin{proof}
M\"umk\"unse $k'=m'$ ama $k\neq m$ olsun.  
Dedi\u gimiz gibi $k\in m$ ve $m\in k$ olmal\i.  
\Teorem{thm:n-trans} ve \numaraya{thm:irr} g\"ore 
$k\in k$ ve $k\notin k$, bir \c celi\c skidir. 
\end{proof}

\c Simdi, \Teoremde{thm:Omega}kiler dahil, $\upomega$
k\"umesinin be\c s tane \"ozelli\u gi vard\i r: 
%\begin{minipage}{\textwidth}
\begin{compactenum}
\item
$0\in\upomega$.
\item
$\Forall x(x\in\upomega\lto x'\in\upomega)$.
\item
$\Forall x\bigl(x\included\upomega\land 0\in x\land\Forall y(y\in x\lto y'\in x)\lto x=\upomega\bigr)$.
\item
$\Forall x(x\in\upomega\lto x'\neq0)$.
\item
$\Forall x\Forall y(x\in\upomega\land y\in\upomega\land x'=y'\lto x=y)$.
\end{compactenum}
%\end{minipage}
Bu \"ozelliklerin \"onemi, 1887 y\i l\i nda Dedekind \cite[II, \P71]{MR0159773}
taraf\i ndan, ve 1889 y\i l\i nda Peano \cite{Peano} taraf\i ndan,
fark edilmi\c stir.  S\i k s\i k 
\textbf{Peano Aksiyomlar\i,}%
\index{aksiyom!Peano A---lar\i, Dedekind--Peano A---lar\i}\label{PA}
bu \"ozelliklere denir, ama
\textbf{Dedekind--Peano Aksiyomlar\i}
de kullan\i labilir.  Asl\i nda bizim i\c cin aksiyomlar de\u gil,
teoremdirler. 

Peano Aksiyomlar\i ndan do\u gal say\i lar\i n t\"um \"ozellikleri
elde edilebilir.  Mesela \emph{iyi\-s\i ralama} \"ozelli\u gi elde
edilebilir.  Asl\i nda $\upomega$, i\c cerilme ($\in$) ba\u g\i nt\i
s\i\ taraf\i ndan iyi\-s\i ralan\i r.  Ama bir ba\u g\i nt\i\ nedir? 

\section{Ba\u g\i nt\i lar}

Herhangi $a$ ile $b$ k\"umeleri i\c cin $\bigl\{\{a\},\{a,b\}\bigr\}$
k\"umesi $(a,b)$ \textbf{s\i ral\i\ ikilisi}%
\index{ikili}%
\index{s\i ra!---l\i\ ikili}
(\eng{ordered pair})
olarak yaz\i l\i r.  Yani%%%%%
\footnote{\sayfada{Kuratowski}ki notta 
dedi\u gimiz gibi bu tan\i m,
  Kuratowski'nin \cite{Kuratowski} 1921 y\i l\i nda verdi\u gi tan\i
  md\i r.} 
\begin{equation*}
(a,b)=\bigl\{\{a\},\{a,b\}\bigr\}.
\end{equation*}%
\glossary{$(a,b)$}

\begin{theorem}
  T\"um $a$, $b$, $c$, ve $d$ k\"umeleri i\c cin
  \begin{equation*}
    (a,b)=(c,d)\liff a=c\land b=d
  \end{equation*}
c\"umlesi do\u grudur.
\end{theorem}

\ktk

\begin{xca}
A\c sa\u g\i daki denkli\u gini%%%%%
\footnote{Heijenoort'a
  \cite[s.~224]{MR1890980} g\"ore bu denklikte, Hausdorff'un 1914 y\i
  l\i nda verdi\u gi s\i ral\i\ ikili tan\i m bulunmu\c stur.} 
kan\i tlay\i n:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{equation*}
 \bigl\{\{a,1\},\{b,2\}\bigr\}
=\bigl\{\{c,1\},\{d,2\}\bigr\}\liff
a=c\land b=d. 
\end{equation*}
\end{xca}

\begin{xca}
A\c sa\u g\i daki denkli\u gini%%%%%
\footnote{Bu denklikte,
  Wiener'in \cite{Wiener} 1914 y\i l\i nda verdi\u gi s\i ral\i\ ikili
  tan\i m bulunmu\c stur.} 
kan\i tlay\i n:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{equation*}
\Bigl\{\bigl\{\{a\},0\bigr\},\bigl\{\{b\}\bigr\}\Bigr\}
=\Bigl\{\bigl\{\{c\},0\bigr\},\bigl\{\{d\}\bigr\}\Bigr\}
\liff a=c\land b=d.
\end{equation*}
\end{xca}

\c Simdi her ikili $\phi(x,y)$ form\"ul\"u i\c cin
\begin{equation*}
  \bigl\{z\colon\Exists x\Exists y\bigl(z=(x,y)\land\phi(x,y)\bigr)\bigr\}
\end{equation*}
s\i n\i f\i,
\begin{equation*}
  \{(x,y)\colon\phi(x,y)\}
\end{equation*}
olarak yaz\i labilir.  \"Oyle bir s\i n\i f, bir 
\textbf{ikili ba\u g\i nt\i d\i r}%
\index{ba\u g\i nt\i}
(\eng{binary relation}).
\"Orne\u gin:
\begin{asparaenum}
\item
\.I\c cerilme ba\u g\i nt\i s\i, $\{(x,y)\colon x\in y\}$ s\i n\i f\i d\i r.
\item
E\c sitlik ba\u g\i nt\i s\i, $\{(x,y)\colon x=y\}$ s\i n\i f\i d\i r.
\end{asparaenum}

Ayn\i\ \c sekilde, e\u ger $\bm R$, bir ikili ba\u g\i nt\i ysa, o
zaman $(x,y)\in\bm R$ form\"ul\"un\"un k\i saltmas\i\ olarak
$x\mathrel{\bm R}y$ ifadesini yazar\i z, yani 
\begin{equation*}
  x\mathrel{\bm R}y\denk(x,y)\in\bm R.
\end{equation*}

$\bm A$ ile $\bm B$, iki s\i n\i f ise, o zaman tan\i ma g\"ore
\begin{equation*}
  \bm A\times\bm B=\{(x,y)\colon x\in\bm A\land y\in\bm B\};
\end{equation*}%
\glossary{$\bm A\times\bm B$}
bu ba\u g\i nt\i, $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
\textbf{\c carp\i m\i d\i r}%
\index{\c carp\i m}
(\eng{product}).
E\u ger $\bm R\included\bm A\times\bm B$, 
o zaman $\bm R$, 
\textbf{$\bm A$ s\i n\i f\i ndan $\bm B$ s\i n\i f\i na giden} 
bir ba\u g\i nt\i d\i r (\eng{relation from $\bm A$ to $\bm B$}).

S\i n\i flar aras\i ndaki bir ba\u g\i nt\i n\i n kendisi,
bir s\i n\i ft\i r.  S\i ral\i\ ikililerin tan\i m\i, s\i n\i flarla
ba\u g\i nt\i lar\i\ birle\c stirir.  Benzer \c sekilde Newton'un A\u
g\i rl\i k Kanunu, Ay'\i n Yer'in etraf\i nda d\"on\"u\c s\"u ile
nesnelerin Yer'e d\"u\c s\"u\c s\"un\"u birle\c stirir. 

E\u ger $\bm F$,
\begin{equation}\label{eqn:F}
\Forall x\Forall y\Forall z(x\mathrel{\bm F}y\land x\mathrel{\bm F}z\lto y=z)
\end{equation}
c\"umlesini sa\u glayan bir ikili ba\u g\i nt\i ysa, o zaman
\begin{compactenum}[(1)]
\item
 $\bm F$ ba\u g\i nt\i s\i na \textbf{g\"onderme}%
\index{g\"onderme}
veya \textbf{fonksiyon}%
\index{fonksiyon}
denir; 
\item
$\{x\colon\Exists yx\mathrel{\bm F}y\}$ s\i n\i f\i na $\bm F$ g\"ondermesinin 
\textbf{tan\i m s\i n\i f\i}%
\index{tan\i m s\i n\i f\i}
(\eng{domain})
denir;
\item
$\{y\colon\Exists xx\mathrel{\bm F}y\}$ s\i n\i f\i na $\bm F$ g\"ondermesinin 
\textbf{de\u ger s\i n\i f\i}%
\index{de\u ger s\i n\i f\i}
(\eng{range})
denir.\footnote{Bu notlarda bir g\"onderme, sadece \eqref{eqn:F} c\"umlesini sa\u glayan bir $\bm F$ ikili ba\u g\i nt\i s\i d\i r.  Fakat baz\i\ kaynaklarda (\"orne\u gin \cite[s.~70]{Nesin-SKK} kayna\u g\i nda) bir g\"onderme veya fonksiyon,
\begin{inparaenum}[(1)]
\item
\eqref{eqn:F} c\"umlesini sa\u glayan bir $\bm F$ ikili ba\u g\i nt\i s\i,
\item
$\{y\colon\Exists xx\mathrel{\bm F}y\}$ s\i n\i f\i na e\c sit bir $\bm A$ s\i n\i f\i, ve
\item
$\{y\colon\Exists xx\mathrel{\bm F}y\}$ s\i n\i f\i n\i\ \emph{kapsayan} bir $\bm B$ s\i n\i f\i\
\end{inparaenum}
taraf\i ndan olu\c sturulmu\c s bir \"u\c cl\"ud\"ur.  O halde (a\c
sa\u g\i daki \sayfada{to}ki gibi) $\bm F\colon\bm
A\to\bm B$ ifadesi yaz\i l\i r.  Ayr\i ca, $\bm B$ s\i n\i f\i na
\emph{g\"ondermenin de\u ger s\i n\i f\i} (veya \emph{var\i\c s s\i
  n\i f\i}) denilebilir.  \.Ingilizcede \eng{codomain} kullan\i l\i r.
Ama buradaki $\bm B$ s\i n\i f\i, sadece $\bm F$ s\i n\i
f\i\ taraf\i ndan belirtilmez, ve buna hi\c cbir ad vermiyoruz.} 
\end{compactenum}
Bu durumda $x\mathrel{\bm F}y$ form\"ul\"un\"un yerine
\begin{equation*}
  y=\bm F(x)
\end{equation*}
ifadesini yazar\i z, \c c\"unk\"u $a\mathrel{\bm F}b$ do\u gruysa, o
zaman $b$ k\"umesi, $a$ k\"umesi taraf\i ndan belirtilir.  Buradaki $\bm F(x)$ ifadesi, yeni bir k\"ume terimidir.  O zaman $\bm F$, 
\begin{equation*}
x\mapsto\bm F(x)
\end{equation*}
\glossary{$x\mapsto\bm F(x)$}%
olarak yaz\i labilir; yani
\begin{equation*}
(x\mapsto\bm F(x))=\{(x,y)\colon y=\bm F(x)\}.
\end{equation*}
\"Orne\u gin:
\begin{asparaenum}
\item
Her $a$ k\"umesi i\c cin $x\mapsto a$ g\"ondermesi,
\textbf{sabit g\"ondermedir}%
\index{sabit!--- g\"onderme}\index{g\"onderme!sabit ---}
(\eng{constant function}). 
Mesela $x\mapsto0$, $x\mapsto1$, \dots, $x\mapsto\upomega$, \dots\
sabit g\"ondermeleri vard\i r.
\item
$x\mapsto x$, \textbf{\"ozde\c slik g\"ondermesidir}%
\index{\"ozde\c slik g\"ondermesi}\index{g\"onderme!\"ozde\c slik ---si}
(\eng{identity function}).
\item
$x\mapsto x'$,
\textbf{ard\i l g\"ondermesidir}
(\eng{successor function}) veya 
\textbf{ard\i llamad\i r}%
\index{g\"onderme!ard\i l ---si}%
\index{ard\i l}
(\eng{succession}).
\end{asparaenum}
E\u ger $\bm F$ g\"ondermesinin tan\i m s\i n\i f\i\ $\bm A$ ise, ve
de\u ger s\i n\i f\i n\i, bir $\bm B$ s\i n\i f\i\ taraf\i ndan
kapsan\i rsa, o zaman
\begin{equation*}\label{to}
  \bm F\colon\bm A\to\bm B
\end{equation*}
ifadesini yazar\i z.  Yani bu ifade,
\begin{multline*}
  \Forall x\Forall y(x\mathrel{\bm F}y\lto x\in\bm A\land
  y\in\bm B)\\
\land\Forall x\bigl(x\in\bm A\lto\Exists y(x\mathrel{\bm F}y)\bigr)\\
\land\Forall x\Forall y\Forall z(x\mathrel{\bm F}y\land x\mathrel{\bm
  F}z\lto y=z)
\end{multline*}
c\"umlesinin k\i saltmas\i d\i r.


\section{S\i ralamalar}

\textbf{S\i ralama}%
\index{s\i ralama}
(\eng{ordering}),
\begin{align*}
\Forall x\lnot\; x&\mathrel{\bm R}x,&
\Forall x\Forall y\Forall z(x\mathrel{\bm R}y\land y\mathrel{\bm R}z&\lto x\mathrel{\bm R}z)
\end{align*}
c\"umlelerini sa\u glayan bir $\bm R$ ikili ba\u g\i nt\i s\i d\i r.  
\"Orne\u gin \sayfada{SBT} bahsedilen 
ve \sayfada{thm:SB} kan\i tlanan 
Schr\"oder--Bernstein Teoremine g\"ore $\prec$ ba\u g\i nt\i s\i, 
bir s\i ralama olacakt\i r.  Ayr\i ca
\begin{equation*}
\bm A\pincluded\bm B\denk\bm A\included\bm B\land\bm A\neq\bm B
\end{equation*}
olsun; o zaman $\pincluded$ ba\u g\i nt\i s\i\ da, bir s\i ralamad\i r. 

Belki bir $\bm R$ ba\u g\i nt\i s\i, bir s\i ralama de\u gildir, 
ama bir $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}
\bm R\cap(\bm A\times\bm A)
\end{equation*}
kesi\c simi, bir s\i ralama olabilir.  
O zaman $\bm A$, $\bm R$ taraf{}\i ndan s\i ralan\i r.  
\"Orne\u gin $\in$, s\i ralama de\u gil; 
ama \Teorem{thm:n-trans} ve \numaraya{thm:irr} g\"ore 
$\in$ ba\u g\i nt\i s\i\ $\upomega$ k\"umesini s\i ralar.

E\u ger $\bm A$ s\i n\i f\i, $\bm R$ taraf{}\i ndan s\i ralan\i rsa, ve \"ustelik
\begin{equation*}
\Forall x\Forall y(x\in\bm A\land y\in\bm A\land x\neq y\lto x\mathrel{\bm R}y\lor y\mathrel{\bm R}x)
\end{equation*}
do\u gruysa, o zaman $\bm R$, $\bm A$ s\i n\i f\i n\i n bir
\textbf{do\u grusal}%
\index{do\u grusal}
(\eng{linear})
s\i ralamas\i\-d\i r.

\begin{theorem}
$\in$ ba\u g\i nt\i s\i, her do\u gal say\i n\i n do\u grusal s\i ralamas\i d\i r.
\end{theorem}

\ktk

E\u ger
\begin{compactenum}[1)]
  \item
$\bm R$, $\bm A$ s\i n\i f\i n\i n do\u grusal s\i ralamas\i ysa,
\item
$\bm A$ s\i n\i f\i n\i n her bo\c s olmayan 
$b$ altk\"umesinin $\bm R$ s\i ralamas\i na g\"ore
\textbf{en k\"u\c c\"uk eleman\i}%
\index{en k\"u\c c\"uk eleman}
(\eng{least element})
veya \textbf{minimumu}%
\index{minimum}
(\eng{minimum})%%%%%
\footnote{Latincede
``k\"u\c c\"uk'' \emph{parvus, -a, -um};
``daha k\"u\c c\"uk'' \emph{minor, -us};
``en k\"u\c c\"uk'' \emph{minimus, -a, -um}
\cite[Lectio 12, s.~114]{LatinDili}.}
%%%%%%%%%%%%%%%%%%%%
varsa, yani
\begin{equation*}
\Forall x\Bigl(x\included\bm A\land x\neq0\lto
\Exists y\bigl(y\in x\land\Forall z(z\in x\setminus\{y\}
\lto y\mathrel{\bm R}z)\bigr)\Bigr)
\end{equation*}
do\u gruysa, ve
\item\label{item:ln}
$\bm A$ s\i n\i f\i n\i n her $b$ eleman\i\ i\c cin 
$\{x\colon x\in\bm A\land x\mathrel{\bm R}b\}$ s\i n\i f\i\ bir k\"ume ise,
\end{compactenum}
o zaman $\bm A$, $\bm R$ taraf{}\i ndan
\textbf{iyi\-s\i ralan\i r}%
\index{iyi\-s\i ralama}\index{s\i ra!iyi---lama}
(\eng{well-ordered}).
$\bm A$ zaten bir k\"ume ise, ko\c sul \ref{item:ln} 
do\u grudan do\u gruya sa\u glan\i r;
ama bu ko\c sul,
\Teoremde{thm:on-wo} 
%ve Al\i\c st\i rma \numarada{ex:rec} 
kullan\i lacakt\i r.

\begin{theorem}\label{thm:n-wo}
$\in$ ba\u g\i nt\i s\i, her do\u gal say\i n\i n iyi\-s\i ralamas\i d\i r.
\end{theorem}

\ktk

\begin{theorem}\label{thm:in-pinc}
$\upomega$ k\"umesinde $\in$ ile $\pincluded$, ayn\i\ ba\u g\i nt\i d\i r, yani
\begin{equation*}
\Forall x\Forall y\bigl(x\in\upomega\land y\in\upomega\lto(x\in y\liff x\pincluded y)\bigr)
\end{equation*}
do\u grudur.
\end{theorem}

\begin{proof}
$k$ ile $m$, do\u gal say\i lar olsun.  
\Teorem{thm:n-trans} ve \numaraya{thm:irr} g\"ore 
$k\in m$ ise $k\pincluded m$.  

\c Simdi $k\pincluded m$ olsun.  \"Onceki teoreme g\"ore $m\setminus
k$ fark\i n\i n en k\"u\c c\"uk $\ell$ eleman\i\ vard\i r.  O zaman
$\ell\in m$, dolay\i s\i yla $\ell\included m$.  Ayr\i ca $a\in\ell$
ise $a\in k$ olmal\i\ (\c c\"unk\"u $a\in m$, ama i\c cerilmeye g\"ore
$\ell$, $m\setminus k$ fark\i n\i n en k\"u\c c\"uk eleman\i d\i r).
\"Oyleyse $\ell\included k$.  Ama $b\in k$ ise $b\in m$, dolay\i
s\i yla $\ell\in b$ veya $\ell=b$ veya $b\in\ell$.  Ancak
$\ell\notin b$ ve $\ell\neq b$ (\c c\"unk\"u $b\included k$ ve
$\ell\notin k$).  \"Oyleyse $b\in\ell$.  Sonu\c c olarak
$k\included\ell$.  Fakat $\ell\included k$.  O zaman $k=\ell$, dolay\i
s\i yla $k\in m$. 
\end{proof}

\begin{theorem}\label{thm:w-wo}
$\upomega$, i\c cerilme taraf{}\i ndan iyi\-s\i ralan\i r.
\end{theorem}

\begin{proof}
$\upomega$ k\"umesinde $m\notin k$ ve $m\neq k$ olsun.  Yani (\"onceki teoremi kullanarak) $m\not\included k$ olsun.  O zaman $m\setminus k$ fark\i n\i n en k\"u\c c\"uk $\ell$ eleman\i\ vard\i r.  Ge\c cen kan\i ttaki gibi $\ell\included k$, yani $\ell\in k$ veya $\ell=k$.  Fakat $\ell\notin k$.  Sonu\c c olarak $\ell=k$, dolay\i s\i yla $k\in m$.  \"Oyleyse i\c cerilme, $\upomega$ k\"umesinin bir do\u grusal s\i ralamas\i d\i r.

Ayr\i ca $a\included\upomega$ ve $n\in a$ ise, 
ya $n$ k\"umesi $a$ k\"umesinin en k\"u\c c\"uk eleman\i d\i r, 
ya da $n\cap a$ kesi\c simi bo\c s de\u gildir.  
Son durumda bu kesi\c simin en k\"u\c c\"uk eleman\i\ vard\i r, 
ve bu eleman, $a$ k\"umesinin en k\"u\c c\"uk eleman\i d\i r.
\end{proof}

\section{Ordinaller}

\"Onceki iki teoremin kan\i tlar\i, do\u gal say\i lar\i n sadece ge\c ci\c slilik ve iyi\-s\i ralama \"ozelliklerini kullanmaktad\i r.  Bir 
\textbf{ordinal,}%
\index{ordinal}
\begin{compactenum}[1)]
\item
ge\c ci\c sli ve
\item
$\in$ taraf\i ndan iyi\-s\i ralanm\i\c s
\end{compactenum}
bir k\"umedir.  Ordinaller,
\begin{equation*}
\on
\end{equation*}%
\glossary{$\on$}
s\i n\i f\i n\i\ olu\c sturur.  
O zaman \Teorem{thm:n-trans} ve \numaraya{thm:n-wo} g\"ore
\begin{equation*}
\upomega\included\on.
\end{equation*}
\"Ustelik \Teorem{thm:w-trans} ve \numaraya{thm:w-wo} g\"ore
\begin{equation*}
\upomega\in\on.
\end{equation*}
Dolay\i s\i yla $\upomega'\included\on$.

\begin{theorem}
Her ordinalin ard\i l\i, bir ordinaldir.
\end{theorem}

\ktk

\begin{theorem}\label{thm:on-in-pinc}
$\on$ s\i n\i f\i nda $\in$ ve $\pincluded$, ayn\i\ ba\u g\i nt\i d\i r.
\end{theorem}

\begin{xca}
\Teoremin{thm:in-pinc} kan\i t\i n\i\ kullanarak bu teoremi kan\i tlay\i n.
\end{xca}

\begin{theorem}\label{thm:on-trans}
$\on$ ge\c ci\c slidir.
\end{theorem}

\begin{proof}
$\alpha$ bir ordinal olsun, ve $\beta\in\alpha$ olsun.  O zaman
  $\beta\included\alpha$.  Bu durumda $\beta$, $\in$ taraf\i ndan
  iyi\-s\i ralan\i r.  \c Simdi $\gamma\in\beta$ olsun.  O zaman
  $\gamma\in\alpha$, dolay\i s\i yla $\gamma\included\alpha$.  O
  zaman $\delta\in\gamma$ ise $\delta\in\alpha$.  $\alpha$, $\in$
  taraf\i ndan iyi\-s\i raland\i\u g\i ndan, $\delta\in\beta$, \c
  c\"unk\"u $\beta$, $\gamma$, ve $\delta$, hepsi $\alpha$
  k\"umesindedir, ve $\delta\in\gamma$, ve $\gamma\in\beta$.  K\i saca
  $\delta\in\gamma\lto\delta\in\beta$, yani $\gamma\included\beta$.
  Ama $\gamma$, $\beta$ k\"umesinin herhangi bir eleman\i d\i r.
  \"Oyleyse $\beta$, ge\c ci\c slidir.  Sonu\c c olarak $\beta$, bir
  ordinaldir.  Ama $\beta$, $\alpha$ ordinalinin herhangi bir eleman\i
  d\i r.  O zaman $\alpha\included\on$.  Ve $\alpha$, herhangi bir
  ordinaldir.  \"Oyleyse $\on$ ge\c ci\c slidir. 
\end{proof}

\begin{theorem}\label{thm:on-wo}
$\on$ s\i n\i f\i, $\in$ taraf\i ndan iyi\-s\i ralan\i r.
\end{theorem}

\ktk[ (\Teoreme{thm:w-wo} bak\i n.)]

\begin{theorem}[Burali-Forti Paradoksu \cite{Burali-Forti}]%
\index{teorem!Burali-Forti Paradoksu}%
\index{paradoks!Burali-Forti P---u}
$\on$ s\i n\i f\i, k\"ume de\u gildir.
\end{theorem}

\begin{proof}
\sayfada{BFP} dedi\u gimiz gibi 
$\on$ bir k\"ume olsayd\i,
\Teorem{thm:on-trans} ve \numaraya{thm:on-wo} g\"ore
$\on\in\on$, ki bu sa\c cmad\i r 
(\c c\"unk\"u $\on$ s\i n\i f\i nda $\in$ d\"on\"u\c ss\"uzd\"ur).
\end{proof}

$\on$ s\i n\i f\i n\i n bir $a$ alt\"umesinin en k\"u\c c\"uk eleman\i
\begin{equation*}
  \min a
\end{equation*}
olarak yaz\i labilir.
\c Su andan itibaren
\begin{align*}
  &\alpha,&&\beta,&&\gamma,&&\delta,&&\theta,&&\iota
\end{align*}
%$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$, ve $\iota$%
\glossary{$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$, $\iota$}
k\"u\c c\"uk Yunan 
harfleri, her zaman \emph{ordinal} sabiti olacakt\i r.  Yani
\begin{equation*}
\alpha\in\on,
\end{equation*}
vesaire.  Ayr\i ca
\Teorem{thm:on-in-pinc} sayesinde $\alpha\in\beta$ veya
$\alpha\pincluded\beta$ form\"ul\"un\"un yerine 
\begin{equation*}
\alpha<\beta
\end{equation*}
ifadesini yazabiliriz, 
ve $\alpha\in\beta\lor\alpha=\beta$ 
veya $\alpha\included\beta$ form\"ul\"un\"un yerine
\begin{equation*}
  \alpha\leq\beta
\end{equation*}
ifadesini yazabiliriz.
Ayr\i ca%
\label{xi}
\begin{align*}
  &\xi,&&\eta,&&\zeta
\end{align*}%
\glossary{$\xi$}\glossary{$\eta$}\glossary{$\zeta$}%
Yunan harfleri, ordinal de\u gi\c skeni olacakt\i r.
\"Orne\u gin
\begin{equation*}
\{\xi\colon\phi(\xi)\}=\{x\colon x\in\on\land\phi(x)\}.
\end{equation*}

\begin{theorem}\label{thm:'}
$\alpha<\beta\lto\alpha'\leq\beta$, dolay\i s\i yla
  \begin{equation*}
 \alpha'=\min\{\xi\colon\alpha<\xi\},   
  \end{equation*}
yani her ordinal i\c cin
daha b\"uy\"uk ordinaller s\i n\i f\i n\i n en k\"u\c c\"uk
eleman\i, ordinalin ard\i l\i d\i r.
\end{theorem}

\ktk

Son teoreme g\"ore,
e\u ger $\alpha$ bo\c s veya ard\i l de\u gilse, ve $\beta\in\alpha$ ise, 
o zaman $\beta'<\alpha$ olmal\i d\i r.  Bu durumda, $\alpha$ ordinaline 
\textbf{limit}%
\index{limit}%
\index{ordinal!limit}
denir.

\begin{theorem}\label{thm:o-lim}
$\upomega$, bir limittir.
Asl\i nda en k\"u\c c\"uk limittir.
\end{theorem}

\ktk

\begin{theorem}\label{thm:on-a}
  $\min(\on\setminus\alpha)=\alpha$.
\end{theorem}

\ktk

\begin{theorem}
$\upomega$, 
hem limit olmayan hem limit i\c cermeyen ordinaller s\i n\i f\i d\i r.  
Yani
\begin{equation}\label{eqn:omega}
\upomega=
\{\xi\colon(\xi=0\lor\Exists yy'=\xi)
\land\Forall z(z\in \xi\lto z=0\lor\Exists yy'=z)\}.
\end{equation}
\end{theorem}

\begin{proof}
Verilen s\i n\i f, $\bm A$ olsun.
T\"umevar\i mla 
(veya \Teorem{thm:o-lim} ile)
her do\u gal say\i, ne limittir ne limit i\c cerir.
B\"oylece $\upomega\included\bm A$.
\"Ote yandan, $\alpha\in\bm A\setminus\upomega$ ise,
o zaman son teoreme g\"ore $\upomega\leq\alpha$,
dolay\i s\i yla $\alpha$ ya bir limittir ya da bir limit i\c cerir,
ve sonu\c c olarak $\alpha\notin\bm A$.
K\i saca $\bm A\included\upomega$,
ve sonu\c c olarak $\bm A=\upomega$.
\end{proof}

Bu teoremin kan\i t\i, Sonsuzluk Aksiyomunu kullanmaz, dolay\i s\i yla
\eqref{eqn:omega} e\c sitli\u gi, $\upomega$ s\i n\i f\i n\i n tan\i
m\i\ olarak kullan\i labilir.  
O halde \sayfada{PA}ki Peano Aksiyomlar\i\ yeniden kan\i tlanmal\i d\i r.

\section{\"Ozyineleme}

$\upomega$ k\"umesinde toplama, bir \textbf{ikili i\c slem}%
\index{i\c slem}
olacak, yani $\upomega\times\upomega$ \c carp\i m\i ndan $\upomega$ k\"umesine giden bir g\"onderme.  Bu i\c slem,
\begin{equation*}
(x,y)\mapsto x+y
\end{equation*}
olarak yaz\i l\i r.  O zaman her $k$ do\u gal say\i s\i\ i\c cin bir
$x\mapsto k+x$ \textbf{birli i\c slemi} olacakt\i r.  Bu i\c slemin
\"ozelliklerinden ikisi, 
\begin{align}\label{eqn:+}
k+0&=k,&\Forall x\bigl(x\in\upomega\lto k+x'&=(k+x)'\bigr)
\end{align}
olacakt\i r.  Asl\i nda $\upomega$ k\"umesindeki birli i\c slemlerden en \c cok birinin bu \"ozellikleri vard\i r.  \c C\"unk\"u $f\colon\upomega\to\upomega$, $f(0)=k$, ve $\Forall x\bigl(x\in\upomega\lto f(x')=f(x)'\bigr)$ olsun.  O zaman $f(0)=k+0$, ve $f(m)=k+m$ ise $f(m')=f(m)'=(k+m)'=k+m'$.  T\"umevar\i mla her $n$ do\u gal say\i s\i\ i\c cin $f(n)=k+n$.

Neden $\upomega$ k\"umesindeki birli i\c slemlerden \emph{en az} birinin \eqref{eqn:+} sat\i r\i ndaki \"ozellikleri vard\i r?  $k=0$ durumunda her $n$ i\c cin $k+n=n$ olsun.  O zaman $k+0=0$, ve $k+m'=m'=(k+m)'$.  \"Ustelik $k=\ell$ durumunda \eqref{eqn:+} sat\i r\i ndaki gibi $x\mapsto k+x$ i\c slemi varsa $\ell'+n=(\ell+n)'$ olsun.  O zaman $\ell'+0=(\ell+0)'=\ell'$, ve $\ell'+m'=(\ell+m')'=(\ell+m)''=(\ell'+m)'$.  Yani $k=\ell'$ durumunda \eqref{eqn:+} sat\i r\i ndaki gibi $x\mapsto k+x$ i\c slemi vard\i r.

T\"umevar\i mla $\upomega$ k\"umesindeki her $k$ i\c cin \eqref{eqn:+}
sat\i r\i ndaki gibi $x\mapsto k+x$ i\c sleminin oldu\u gu sonucuna
varabilir miyiz?  T\"umevar\i mla bir \emph{k\"umenin} $\upomega$
k\"umesine e\c sit oldu\u gu kan\i tlanabilir.  \c Simdiki durumda
hangi k\"ume, $\upomega$ k\"umesine e\c sit olmal\i d\i r?
M\"umk\"unse $a$, $\upomega$ k\"umesinin \"oyle $k$
elemanlar\i\ taraf\i ndan olu\c sturulsun ki \eqref{eqn:+} sat\i r\i
ndaki \"ozelliklerini sa\u glayan bir i\c slem olsun.  O halde
g\"osterdi\u gimiz gibi $a=\upomega$ olmal\i d\i r.  Ama \"oyle bir
$a$ k\"umesi var m\i d\i r?  Hangi form\"ul, bu k\"umeyi tan\i
mlayabilir? 

Sayfa \pageref{yerlestirme} ve \sayfada{Yer}ki Yerle\c stirme Aksiyomuna g\"ore 
bir k\"umede birli bir i\c slemin kendisi, bir k\"umedir.  
O halde istedi\u gimiz $a$ k\"umesi tan\i mlanabilir, 
dolay\i s\i yla $\upomega$ k\"umesindeki toplaman\i n kendisi tan\i mlanabilir.
Asl\i nda \eqref{eqn:+} sat\i r\i ndaki \"ozellikleri, toplaman\i n 
\textbf{\"ozyineli tan\i m\i n\i}%
\index{\"ozyineli tan\i m}
(\eng{recursive definition})
sa\u glar.

Benzer \c sekilde her $k$ do\u gal say\i s\i\ i\c cin
\begin{align}\label{eqn:.}
k\cdot0&=0,&\Forall x(x\in\upomega\lto k\cdot x'&=k\cdot x+k)
\end{align}
\"ozellikleri olan $x\mapsto k\cdot x$ i\c slemi vard\i r.  (Burada tabii ki $k\cdot x+k=(k\cdot x)+k$.)  \c C\"unk\"u $0\cdot n=0$ ise $0\cdot0=0$ ve $0\cdot m'=0=0+0=0\cdot m+0$.  Ayr\i ca istedi\u gimiz gibi $x\mapsto\ell\cdot x$ varsa $\ell'\cdot n=\ell\cdot n+n$ olsun.  O halde
\begin{align*}
\ell'\cdot m'
&=\ell\cdot m'+m'\\
&=(\ell\cdot m+\ell)+m'\\
&=\ell\cdot m+(\ell+m')\\
&=\ell\cdot m+(\ell+m)'\\
&=\ell\cdot m+(m+\ell)'\\
&=\ell\cdot m+(m+\ell')\\
&=(\ell\cdot m+m)+\ell'\\
&=\ell'\cdot m+\ell'.
\end{align*}
Ama burada toplaman\i n birle\c sme ve de\u gi\c sme \"ozelliklerini kulland\i k; bunlar kan\i tlanmal\i d\i r.

Buraya kadar gelmek i\c cin t\"umevar\i m yeter.  Yani \sayfada{PA}ki ilk \"u\c c Peano Aksiyomu yeter.  Say\i lar teorisinde, her pozitif $n$ mod\"ul\"use g\"ore tamsay\i lar, bu aksiyomlar\i\ sa\u glar.  Yani e\u ger bir $a$ k\"umesinin elemanlar\i\, tamsay\i\ ise, ve
\begin{equation*}
x\equiv 0\pmod n
\end{equation*}
denkli\u ginin $a$ k\"umesinden \c c\"oz\"um\"u varsa, ve ayr\i ca her $\ell$ tamsay\i s\i\ i\c cin
\begin{equation*}
x\equiv \ell\lto y\equiv \ell'\pmod n
\end{equation*}
kara\c st\i rmas\i n\i n $a$ k\"umesinden \c c\"oz\"um\"u varsa, o zaman her $k$ tamsay\i s\i\ i\c cin
\begin{equation*}
x\equiv k\pmod n
\end{equation*}
denkli\u ginin $a$ k\"umesinden \c c\"oz\"um\"u vard\i r.  \"Orne\u gin $p$, bir asal say\i\ olsun.  O zaman
\begin{equation*}
0^p\equiv0\pmod p,
\end{equation*}
ve
\begin{equation*}
a^p\equiv a\lto(a+1)^p\equiv a+1\pmod p,
\end{equation*}
\c c\"unk\"u
\begin{align*}
(a+1)^p&\equiv a^p+pa^{p-1}+\binom p2a^{p-2}+\dots+\binom p{p-2}a^2+pa+1\\
       &\equiv a^p+1\pmod p.
\end{align*}
Sonu\c c olarak Fermat'n\i n Teoremi%
\index{teorem!Fermat'n\i n T---i}
do\u grudur, yani her $p$ asal say\i s\i\ i\c cin, her $a$ tamsay\i
s\i\ i\c cin\footnote{Gauss'a \cite[\P50]{Gauss} g\"ore verdi\u gimiz
  kan\i t, Euler'indir.} 
\begin{equation*}
a^p\equiv a\pmod p.
\end{equation*}
Ayn\i\ sebeple t\"um $a$, $b$, $c$, ve $d$ tamsay\i lar\i\ i\c cin, her pozitif $n$ say\i s\i\ i\c cin
\begin{equation*}
a\equiv b\land c\equiv d\lto a+c\equiv b+d\land a\cdot c\equiv b\cdot d\pmod n.
\end{equation*}

Sadece t\"umevar\i m\i\ kullanarak 
$(x,y)\mapsto x^y$ ikili \"ustel i\c slemi tan\i mlanabilir mi?  
\"Ozyineli tan\i m varsa $\upomega$ k\"umesindeki her $k$ i\c cin
\begin{align}\label{eqn:exp}
k^0&=1,&\Forall x(x\in\upomega\lto k^{x'}=k^x\cdot k).
\end{align}
\"Ozel olarak $0^0=1$, ama $n>0$ ise $0^n=0$.  \"Oyleyse
$0\equiv n$ ama $0^0\not\equiv 0^n\pmod n$.  \"Ustel i\c slem i\c cin
t\"umevar\i m yetmez.\footnote{\cite{Pierce-IR} makalesine bak\i n.}

\begin{theorem}[\"Ozyineleme \protect{[\eng{Recursion}]}]%
\index{teorem!\"Ozyineleme T---i}\label{thm:rec}
$\bm A$, bir s\i n\i f olsun, 
ve $b\in\bm A$ ile $\bm F\colon\bm A\to\bm A$ olsun.  
O zaman $\upomega$ k\"umesinden $\bm A$ s\i n\i f\i na giden
\begin{align*}
\bm G(0)&=b,&
\Forall x\Bigl(x\in\upomega\lto\bm G(x')=\bm F\bigl(\bm G(x)\bigr)\Bigr)
\end{align*}
\"ozellikleri olan bir ve tek bir $\bm G$ g\"ondermesi vard\i r.
\end{theorem}

\begin{proof}%[Kan\i t\i n fikri.]
T\"umevar\i mla en \c cok bir $\bm G$ g\"ondermesi vard\i r.  
En az biri varsa, ba\u g\i nt\i\ olarak, 
$\upomega\times\bm A$ \c carp\i m\i n\i n alts\i n\i f\i d\i r, 
ve her $(\ell,d)$ eleman\i\ i\c cin,
\begin{compactitem}
\item
ya $(\ell,d)=(0,b)$,
\item
ya da bir $(k,c)$ eleman\i\ i\c cin, $k'=\ell$ ve $\bm F(c)=d$.
\end{compactitem}
Bu \"ozelli\u gi olan \emph{k\"umeler} vard\i r, mesela
\begin{gather*}
\{(0,b)\},\\
\bigl\{(0,b),\bigl(1,\bm F(b)\bigr)\bigr\},\\
\Bigl\{(0,b),\bigl(1,\bm F(b)\bigr),
\Bigl(2,\bm F\bigl(\bm F(b)\bigr)\Bigr)\Bigr\},\\
\biggl\{(0,b),\bigl(1,\bm F(b)\bigr),
\Bigl(2,\bm F\bigl(\bm F(b)\bigr)\Bigr),
\biggl(3,\bm F\Bigl(\bm F\bigl(\bm F(b)\bigr)\Bigr)\biggr)\biggr\}.
\end{gather*}
\"Ozelli\u gi olan k\"umelerin olu\c sturdu\u gu s\i n\i f, $\bm C$
olsun.  O zaman $\bigcup\bm C$, istedi\u gimiz $\bm G$ g\"ondermesi
olacakt\i r.  Bunu g\"ostermek i\c cin, t\"um Peano
Aksiyomlar\i\ kullan\i lmal\i d\i r.

Hemen $\{(0,b)\}\in\bm C$, dolay\i s\i yla $(0,b)\in\bigcup\bm C$.  \c Simdi $(k,c)\in\bigcup\bm C$ varsay\i ls\i n.  O zaman $\bm C$ s\i n\i f\i n\i n bir $a$ eleman\i\ i\c cin $(k,c)\in a$.  O halde $a\cup\{(k',\bm F(c))\}\in\bm C$.  B\"oylece $(k',\bm F(c))\in\bigcup\bm C$.  T\"umevar\i mla her $k$ do\u gal say\i s\i\ i\c cin $\bm A$ s\i n\i f\i n\i n $(k,c)\in\bigcup\bm C$ i\c cerilmesini sa\u glayan $c$ eleman\i\ vard\i r, yani
\begin{equation*}
\{x\colon\Exists y(x,y)\in\bigcup\bm C\}=\upomega.
\end{equation*}
\c Simdi
\begin{equation*}
\bigl\{x\colon\Forall y\Forall z\bigl((x,y)\in\bigcup\bm C\land(x,z)\in\bigcup\bm C\lto y=z)\bigr)\bigr\}=\upomega
\end{equation*}
e\c sitli\u gini kan\i tlayaca\u g\i z.  Soldaki k\"ume, $a_0$ olsun.  E\u ger $(0,e)\in\bigcup\bm C$ ise, o zaman $\bm C$ s\i n\i f\i n\i n bir $a$ eleman\i\ i\c cin $(0,e)\in a$, dolay\i s\i yla $e=b$ olmal\i d\i r, \c c\"unk\"u $0$, ard\i l de\u gildir.  \"Oyleyse $0\in a_0$.
\c Simdi $k\in a_0$ olsun.  G\"osterdi\u gimiz gibi $\bm A$ s\i n\i f\i n\i n bir $c$ eleman\i\ i\c cin $(k,c)\in\bigcup\bm C$ ve $(k',\bm F(c))\in\bigcup\bm C$.  $(k',d)\in\bigcup\bm C$ varsay\i ls\i n.  O zaman $\bm C$ s\i n\i f\i n\i n bir $a$ eleman\i\ i\c cin $(k',d)\in a$.  O halde $a$ k\"umesinin bir $(j,e)$ eleman\i\ i\c cin $j'=k'$ ve $\bm F(e)=d$.  Bu durumda $j=k$ olmal\i d\i r.  B\"oylece $(k,e)\in\bigcup\bm C$, dolay\i s\i yla $e=c$ ve $d=\bm F(c)$, \c c\"unk\"u $k\in a_0$ varsay\i l\i r.  \"Oyleyse $k'\in a_0$.  T\"umevar\i mla $a_0=\upomega$.

Sonu\c c olarak $\bigcup\bm C$, $\upomega$ k\"umesinden $\bm A$ s\i n\i f\i na giden bir $\bm G$ g\"ondermesidir.  $\bm C$ s\i n\i f\i n\i n tan\i m\i ndan $\bm G$ g\"ondermesinin istedi\u gimiz \"ozellikleri vard\i r.
\end{proof}

\c Simdi, do\u gal say\i larda, \eqref{eqn:+}, \eqref{eqn:.}, ve
\eqref{eqn:exp} sat\i rlar\i 
ndaki b\"ut\"un tan\i mlar ge\c cerlidir.  

\section{E\c sleniklik}

$\bm R$ bir ba\u g\i nt\i\ ise,
bu ba\u g\i nt\i n\i n \textbf{ters ba\u g\i nt\i s\i}%
\index{ba\u g\i nt\i!ters ---} 
veya
\textbf{tersi}% 
\index{ters}
(\eng{converse}),
\begin{equation*}
  \{(y,x)\colon x\mathrel{\bm R}y\}
\end{equation*}
ba\u g\i nt\i s\i d\i r.  
Bu ters ba\u g\i nt\i, $\conv{\bm R}$ olarak
yaz\i l\i r; yani
\begin{equation*}
  x\mathrel{\conv{\bm R}}y\denk y\mathrel{\bm R}x.
\end{equation*}%
\glossary{$\conv{\bm R}$}
$\bm S$ bir ba\u g\i nt\i\ daha ise, o zaman tan\i m olarak
\begin{equation*}
\bm R/\bm S=\{(x,z)\colon\Exists y(x\mathrel{\bm R}y\land y\mathrel{\bm S}z)\}.
\end{equation*}%
\glossary{$\bm R/\bm S$}
Bu yeni ba\u g\i nt\i, $\bm R$ ile $\bm S$ ba\u g\i nt\i lar\i n\i n 
\textbf{bile\c skesidir}%
\index{bile\c ske}
(\eng{composite}).
$\bm S/\bm R$ bile\c skesi, $\bm R/\bm S$ bile\c skesinden farkl\i\ olabilir.

\begin{theorem}
E\u ger $\bm F\colon\bm A\to\bm B$ ve $\bm G\colon\bm B\to\bm C$ ise, o zaman
\begin{equation*}
\bm F/\bm G\colon\bm A\to\bm C,  
\end{equation*}
ve ayr\i ca
$\Forall x\bigl(x\in\bm A\lto(\bm F/\bm G)(x)=\bm G(\bm F(x))\bigr)$.
\end{theorem}

\ktk

Teoremdeki durumda $\bm F/\bm G$ g\"ondermesi,
\begin{equation*}
\bm G\circ\bm F
\end{equation*}
\glossary{$\bm B\circ\bm F$}%
olarak yaz\i l\i r.

Tekrar $\bm F\colon\bm A\to\bm B$ olsun.  E\u ger $\bm F$ ba\u g\i
nt\i s\i n\i n $\conv{\bm F}$ ters ba\u g\i nt\i s\i, $\bm B$ s\i n\i f\i
ndan $\bm A$ s\i n\i f\i na giden bir g\"ondermeyse, o zaman bu
g\"onderme, $\bm F$ g\"ondermesinin 
\textbf{ters g\"ondermesi}%
\index{g\"onderme!ters ---}
veya \textbf{tersidir}%
\index{ters}
(\eng{inverse}), ve
\begin{equation*}
\inv{\bm F}
\end{equation*}%
\glossary{$\inv{\bm F}$}
olarak yaz\i l\i r.  Bu durumda $\bm F$, 
$\bm B$ s\i n\i f\i yla
bir \textbf{e\c slemedir}%
\index{e\c sleme}
(\eng{bijection}),
ve $\inv{\bm F}$, $\bm A$ s\i n\i f\i yla bir e\c slemedir.

\begin{theorem}
  $\bm F\colon\bm A\to\bm B$ olsun.
O zaman $\bm F$, 
$\bm B$ s\i n\i f\i yla bir e\c sleme\-dir
ancak ve ancak
$\bm B$ s\i n\i f\i ndan $\bm A$ s\i n\i f\i na giden 
bir $\bm G$ g\"ondermesi i\c cin 
\begin{align*}
  \bm G\circ\bm F&=\{(x,x)\colon x\in\bm A\},&
  \bm F\circ\bm G&=\{(x,x)\colon x\in\bm B\}.
\end{align*}
Bu durumda $\bm G=\inv{\bm F}$.
\end{theorem}

\ktk

E\u ger $\bm F$,
$\bm A$ s\i n\i f\i ndan
$\bm B$ s\i n\i f\i na giden bir e\c sleme ise,
o zaman
\begin{equation*}
  \bm F\colon\bm A\xrightarrow{\approx}\bm B
\end{equation*}
ifadesini yazar\i z.%%%%%
\footnote{$\bm F\colon\bm A\twoheadrightarrowtail\bm B$ ifadesi de 
m\"umk\"und\"ur.}  
%%%%%%%%%%%%%%%
Bu durumda, $\inv{\bm F}\colon\bm B\xrightarrow{\approx}\bm A$.
\"Oyle bir $\bm F$ oldu\u gundan,
$\bm A$ ile $\bm B$ s\i n\i flar\i\
birbirine \textbf{e\c sleniktir}%
\index{e\c slenik}
(\eng{equipollent}).


\begin{theorem}\label{thm:sinif}
Bir s\i n\i f, bir k\"umeye e\c slenikse, s\i n\i f da bir k\"umedir.
\end{theorem}

\ktk

\Teorem{thm:sinif} ve \ref{thm:gonderme} sayesinde
\emph{k\"umelerin} e\c slenikli\u gi, ikili bir ba\u g\i nt\i d\i r.  Bu ba\u
g\i nt\i n\i n i\c sareti
\begin{equation*}
\approx
\end{equation*}
\glossary{$a\approx b$}%
olsun.  
O zaman $=$ gibi $\approx$, yeni bir y\"uklemdir
(\sayfaya{==} bak\i n).
Ayr\i ca
\begin{multline*}
a\approx b\denk\Exists w
\biggl(\Forall x\Exists y\bigl(x\in a\lto y\in b\land(x,y)\in w\bigr)\\
\land\Forall x\Exists y\bigl(x\in b\lto y\in a\land(y,x)\in w\bigr)\\
\land\Forall x\Forall y\Forall z
\Bigl(\bigl((x,y)\in w\land(x,z)\in w\lto y=z\bigr)\\
\land\bigl((x,z)\in w\land(y,z)\in w\lto x=y\bigr)\Bigr)\biggr).
\end{multline*}

E\c sitlik gibi e\c sleniklik, bir denklik ba\u g\i nt\i s\i d\i r
(\sayfaya{denklik} bak\i n):

\begin{theorem}\label{thm:approx-eq}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{align*}
&a\approx a,&
a\approx b&\lto b\approx a,&
a\approx b\land b\approx c&\lto a\approx c
\end{align*}
c\"umleleri do\u grudur.
\end{theorem}

\ktk

\begin{theorem}
  $\bm R$ bir denklik ba\u g\i nt\i s\i\ ise, o zaman
  \begin{equation*}
    \{x\colon x\mathrel{\bm R}a\}\cap\{x\colon x\mathrel{\bm R}b\}\neq\emptyset
\to \{x\colon x\mathrel{\bm R}a\}=\{x\colon x\mathrel{\bm R} b\}.
  \end{equation*}
\end{theorem}

\ktk

$\bm A\cap\bm B=\emptyset$ ise $\bm A$ ve $\bm B$, 
birbirinden \textbf{ayr\i kt\i r}%
\index{ayr\i k} 
(\eng{disjoint}).
Teoremdeki $\{x\colon x\mathrel{\bm R}a\}$ k\"umesi,
$\bm R$ ba\u g\i nt\i s\i na g\"ore $a$ k\"umesinin
\textbf{denklik s\i n\i f\i\-d\i r}%
\index{denklik s\i n\i f\i}
(\eng{equivalance class}).
Teoreme g\"ore birbirinden farkl\i\ olan denklik s\i n\i flar\i,
birbirinden ayr\i kt\i r.

\section{Kardinaller}


E\c slenikli\u ge g\"ore bir $a$ k\"umesinin
$\{x\colon x\approx a\}$ denklik s\i n\i f\i na
$a$ k\"ume\-sinin \textbf{b\"uy\"ukl\"u\u g\"u}%
\index{b\"uy\"ukl\"uk}
(\eng{size)} densin.
E\u ger $a$ bo\c s de\u gilse 
$\{x\colon x\approx a\}$ s\i n\i f\i,
k\"ume de\u gildir. 
Bu $a$ k\"umesinin denklik s\i n\i f\i\
bir ordinal i\c cerse,
i\c cerdi\u gi en k\"u\c c\"uk ordinal vard\i r;
bu ordinale $a$ k\"umesinin \textbf{kardinali}%
\index{kardinal}
(\eng{cardinal})
denir, ve $a$ k\"umesinin kardinali $\card(a)$ olarak yaz\i lacak.
B\"oylece
\begin{equation*}
  \card(a)=\min\{\xi\colon\xi\approx a\}.
\end{equation*}
\glossary{$\card(a)$}%
(Sayfa \sayfanumarada{xi}n $\xi$ harf\/inin 
ordinal de\u gi\c skeni oldu\u gunu hat\i rlay\i n.)
\c Sim\-dilik her k\"umenin 
bir kardinalinin olup olmad\i\u g\i n\i\ bilmiyoruz.
Ama her ordinalin kardinali vard\i r.
Kardinaller, bir
\begin{equation*}
\cn
\end{equation*}%
\glossary{$\cn$}%
s\i n\i f\i n\i\ olu\c sturur.  O zaman
\begin{equation*}
  \cn\included\on.
\end{equation*}

%\section{Sonlu k\"umeler}

Bir do\u gal say\i yla e\c slenik bir k\"ume,
\textbf{sonludur}%
\index{sonlu, sonsuz}
(\eng{finite}); sonlu olmayan bir s\i n\i f,
\textbf{sonsuzdur}
(\eng{infinite}).
O zaman her sonlu kardinal, bir do\u gal say\i d\i r.
Bu sonucun tersini kan\i tlayaca\u g\i z.

Birka\c c tane von Neumann do\u gal say\i s\i n\i n tan\i m\i n\i,
sayfa \pageref{vnn} ve \sayfada{0123}n hat\i rlayal\i m:
\begin{align*}
  0&=\emptyset,&
1&=\{0\},&
2&=\{0,1\},&
3&=\{0,1,2\},&
4&=\{0,1,2,3\}.
\end{align*}
Bir $a$ k\"umesinin
\begin{compactenum}[1)]
  \item
hi\c cbir eleman\i\ yoksa, o zaman $a\approx0$; asl\i
    nda $a=0$;
  \item
tek bir eleman\i\ varsa, o zaman $a\approx1$;
  \item
iki (ve sadece iki) eleman\i\ varsa, o zaman $a\approx2$;
  \item
\"u\c c (ve sadece \"u\c c) eleman\i\ varsa, o zaman
    $a\approx3$. 
\end{compactenum}
Ayr\i ca
\begin{align*}
  0&\not\approx1,&
  0&\not\approx2,&
  0&\not\approx3,&
  1&\not\approx2,&
  1&\not\approx3,&
  2&\not\approx3.
\end{align*}
Ancak herhangi iki e\c slenik do\u gal say\i\ e\c sit olmal\i\ m\i?  
Bu soruyu, \sayfada{nasil} sormu\c stuk.

\begin{theorem}
\.Iki do\u gal say\i\ birbiriyle e\c slenikse, birbirine e\c sittir:
\begin{equation*}
\Forall x\Forall y(x\in\upomega\land y\in\upomega\lto x\approx y\lto x=y).
\end{equation*}
\end{theorem}

\begin{proof}
T\"umevar\i mla her $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
\Forall x(x\in\upomega\land x\approx n\lto x=n)
\end{equation*}
c\"umlesini kan\i tlayaca\u g\i z.  $n=0$ ise do\u grudur.  
$n=m$ ise do\u gru olsun, 
ve bir $\ell$ do\u gal say\i s\i\ i\c cin $m'\approx\ell$ olsun.  
O zaman $\ell$ bo\c s de\u gil,
dolay\i s\i yla bir bir ard\i l olmal\i.  
$\ell=k'$ olsun.  
$m'$ say\i s\i ndan $k'$ say\i s\i na giden bir $f$ e\c slemesi vard\i r.  
E\u ger $f(m)=k$, o zaman $f\setminus\{(m,k)\}$, 
$m$ say\i s\i ndan $k$ say\i s\i na giden bir e\c slemedir.  
E\u ger $f(m)\neq k$, o zaman
\begin{equation*}
\bigl\{(x,y)\colon x\in m\setminus\{\inv f(k)\}\land y=f(x)\bigr\}
\cup\bigl\{(\inv f(k),f(m))\bigr\}
\end{equation*}
ba\u g\i nt\i s\i, $m$ say\i s\i ndan $k$ say\i s\i na giden bir e\c slemedir.  
\"Oyleyse her durumda $m\approx k$.  
Hipotezimize g\"ore $m=k$ olmal\i, dolay\i s\i yla $m'=\ell$.  Kan\i t bitti.
\end{proof}

Sonu\c c olarak her do\u gal say\i, sonlu bir kardinaldir, yani
\begin{equation*}
  \upomega\included\cn.
\end{equation*}
Sonraki b\"ol\"umde $\on$ s\i n\i f\i n\i\ inceleyece\u giz;
\"ob\"ur b\"ol\"umde, $\cn$.

\chapter{Ordinaller}

Do\u gal say\i larda, bir g\"ondermenin \"ozyineli tan\i m\i n\i n iki
tane par\c cas\i\ vard\i r, biri $0$ i\c cin, biri ard\i llar i\c
cin.  Ordinallerde \"u\c c\"unc\"u bir par\c ca gerekir, limitler
i\c cin.

\section{Supremumlar}

E\u ger $a\included\beta'$ ise, o zaman $a\included\on$ ve
\begin{equation*}
  \Forall{\xi}(\xi\in a\lto\xi\leq\beta),
\end{equation*}
dolay\i s\i yla $\beta$, $a$ k\"umesinin
\textbf{\"ust s\i n\i r\i d\i r}%
\index{\"ust s\i n\i r}
(\eng{upper bound}).
$\on$ iyi\-s\i ralanm\i\c s oldu\u gundan
$a$ k\"umesinin
\textbf{en k\"u\c c\"uk \"ust s\i n\i r\i}
(\eng{least upper bound})
olmal\i d\i r.
Analizdeki gibi
en k\"u\c c\"uk \"ust s\i n\i ra
\textbf{supremum}%
\index{supremum}
(\eng{supremum})
denir, ve $a$ k\"umesinin supremumu i\c cin
\begin{equation*}
  \sup(a)
\end{equation*}
yaz\i l\i r.
E\u ger $\sup(a)\in a$ ise,
o zaman $a$ k\"umesin
\textbf{en b\"uy\"uk eleman\i}
(\eng{greatest element})
veya
\textbf{maksimumu}%
\index{maksimum}
(\eng{maximum})
vard\i r,
ve tabii ki bu eleman, supremumdur.
Ama $\sup(a)\notin a$ ise,
$a$ k\"umesinin en b\"uy\"uk eleman\i\ yoktur.%%%%%
\footnote{Latincede
``y\"uksek'' \emph{superus, -a, -um};
``daha y\"uksek'' \emph{superior, -ius};
``en y\"uksek'' \emph{supremus, -a, -um};
``b\"uy\"uk'' \emph{magnus, -a, -um};
``daha b\"uy\"uk'' \emph{maior, -ius};
``en b\"uy\"uk'' \emph{maximus, -a, -um.}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

T\"um $\alpha$ ordinalleri i\c cin
\begin{align}\label{eqn:++}
  \alpha+0&=\alpha,&
\alpha+\beta'&=(\alpha+\beta)'
\end{align}
olacak.  Ama $\beta$ limitse, $\alpha+\beta$ nedir?
Mesela $\alpha+\upomega$ nedir?  
Asl\i nda \Teoreme{thm:rec} g\"ore 
$\upomega$ k\"umesinden $\on$ s\i n\i f\i na
giden, \eqref{eqn:++} sat\i r\i ndaki \"ozellikleri olan
$x\mapsto\alpha+x$ g\"ondermesi vard\i r.
Her $n$ do\u gal
say\i s\i\ i\c cin, 
\begin{equation*}
  \alpha+n<\alpha+\upomega
\end{equation*}
e\c sitsizli\u gini isteriz.  Yani $\alpha+\upomega$,
$\{y\colon\Exists x(x\in\upomega\land y=\alpha+x)\}$ s\i n\i f\i n\i n
\"ust s\i n\i r\i\
olmal\i d\i r 
(s\i n\i f\i n \"ust s\i n\i r\i\ varsa).  
Bu s\i n\i f, $\{\alpha+x\colon x\in\upomega\}$ olarak yaz\i labilir.  

Genelde $\bm F\colon\bm A\to\bm B$ ve $\bm C\included\bm A$ ise
$\{y\colon\Exists x(x\in\bm C\land\bm F(x)=y)\}$
s\i n\i f\i,
\begin{equation*}
\{\bm F(x)\colon x\in\bm C\}\quad\text{ veya }\quad\bm F[\bm C]
\end{equation*}
olarak yaz\i labilir.  
Bu s\i n\i f, $\bm C$ s\i n\i f\i n\i n $\bm F$ alt\i nda 
\textbf{g\"or\"unt\"us\"ud\"ur.}%
\index{g\"or\"unt\"u}
Bu durumda, $\bm F$ g\"ondermesi $\bm C$ s\i n\i f\i nda
\textbf{tan\i mlan\i r,}% 
\index{tan\i mlama}
\c c\"unk\"u $\bm C$, $\bm F$ g\"ondermesinin tan\i m s\i n\i
f\i\ taraf\i ndan kapsan\i r.  E\u ger $\bm F$, $\bm C$ s\i n\i f\i
nda tan\i mlanmazsa, $\bm F[\bm C]$ ifadesini yazmayaca\u g\i z. 

$\upomega$ k\"umesinin 
$\{\alpha+x\colon x\in\upomega\}$ g\"or\"unt\"us\"un\"un 
\"ust s\i n\i r\i\ varsa, en k\"u\c c\"uk \"ust s\i n\i r\i, 
yani supremumu,
vard\i r (\c c\"unk\"u $\on$, iyi\-s\i ralan\i r).  \c Simdi
\eqref{eqn:++} sat\i r\i ndaki \"ozelliklere g\"ore  
\begin{equation*}
  \alpha\pincluded\alpha+1\pincluded\alpha+2\pincluded\dotsb
\end{equation*}
E\u ger $\bigcup\{\alpha+x\colon x\in\upomega\}$ bir ordinalse,
bu ordinal 
$\{\alpha+x\colon x\in\upomega\}$ k\"umesinin \"ust s\i n\i r\i d\i r,
asl\i nda supremumudur.
Bunu g\"osterelim.

\begin{theorem}
$\bm A\included\on$ olsun.
  \begin{enumerate}
  \item 
$\bigcup\bm A$ bir k\"umeyse, $\bigcup\bm A\in\on$.
\item
$\bigcup\bm A$ bir k\"ume de\u gilse, $\bigcup\bm A=\on$.
  \end{enumerate}
\end{theorem}

\begin{proof}
$b\in\bigcup\bm A$ ise, $\bm A$ s\i n\i f\i n\i n 
bir $\alpha$ eleman\i\ i\c cin $b\in\alpha$, 
dolay\i s\i yla $b\included\alpha$ ve onun i\c cin $b\included\bigcup\bm A$.
\"Oyleyse $\bigcup\bm A$ ge\c ci\c slidir.  
Ayr\i ca, $\on$ s\i n\i f\i\ da ge\c ci\c sli oldu\u gundan, 
$\bigcup\bm A\included\on$,
dolay\i s\i yla $\bigcup\bm A$, $\in$ taraf\i ndan iyi\-s\i ralan\i r.  
\"Oyleyse $\bigcup\bm A$ ya bir ordinaldir, 
ya da k\"ume olmayan bir s\i n\i ft\i r.  

\.Ikinci durumda $\bigcup\bm A$ bile\c siminin 
$\on$ oldu\u gunu g\"osterece\u giz.  
E\u ger $\bigcup\bm A\pincluded\on$ ise 
$\beta\in\on\setminus\bigcup\bm A$ olsun. 
E\u ger $\gamma\in\bigcup\bm A$ ise,
$\gamma\included\bigcup\bm A$, dolay\i s\i yla $\beta\notin\gamma$.
Ayr\i ca $\gamma\neq\beta$, ve sonu\c c olarak $\gamma\in\beta$.  
B\"oylece $\bigcup\bm A\included\beta$.
Bu durumda $\bigcup\bm A$ bir k\"umedir,
dolay\i s\i yla ordinaldir. 
\end{proof}

\c Simdi $\bm A$ bir k\"umeyse, $\bigcup\bm A$ bile\c siminin bir
k\"ume oldu\u gunu isteriz:   

\begin{axiom}[Bile\c sim]%
\index{aksiyom!Bile\c sim A---u}
  Her k\"umenin bile\c simi, bir k\"umedir:
  \begin{equation*}
    \Forall x\Exists y\Forall z\bigl(z\in y\liff\Exists w(w\in x\land
    z\in w)\bigr). 
  \end{equation*}
\end{axiom}

\begin{theorem}
  Ordinallerin olu\c sturdu\u gu her k\"umenin bile\c simi, k\"umenin
  supremumudur, yani $a\included\on$ ise
  \begin{equation*}
    \bigcup a=\sup(a).
  \end{equation*}
\end{theorem}

\begin{proof}
  $a\included\on$ olsun.  Son teorem ve Bile\c sim Aksiyomuna g\"ore $\bigcup a$, bir $\alpha$
  ordinalidir.  O zaman $\alpha$, $a$ k\"umesinin bir \"ust s\i n\i r\i
  d\i r.  E\u ger $\beta<\alpha$ ise, o zaman $\beta\in\alpha$,
  dolay\i s\i yla $a$ k\"umesinin bir $\gamma$ eleman\i\ i\c cin
  $\beta\in\gamma$, yani $\beta<\gamma$.  Sonu\c c olarak $\beta$, $a$
  k\"umesinin \"ust s\i n\i r\i\ de\u gildir.  \"Oyleyse $\alpha=\sup(a)$.
\end{proof}

\c Simdi $\{\alpha+x\colon x\in\upomega\}$ gibi g\"or\"unt\"uler, k\"ume
olsun:

\begin{axiom}[Yerle\c stirme]%
\index{aksiyom!Yerle\c stirme A---u}\label{Yer}
Her g\"ondermenin tan\i m s\i n\i f\i n\i n altk\"umesinin g\"onderme alt\i nda g\"or\"unt\"us\"u, bir k\"umedir.  Yani her ikili $\phi(x,y)$
form\"ul\"u i\c cin
\begin{multline*}
\Forall w\biggl(\Forall x\Forall y\Forall z\bigl(\phi(x,y)\land\phi(x,z)\land x\in w\lto y=z\bigr)\\
\lto\Exists z\Forall y\Bigl(y\in z\liff\Exists x\bigl(x\in w\land\phi(x,y)\bigr)\Bigr)\biggr).
\end{multline*}
\end{axiom}

\c Simdi $\beta$ ordinalinde $x\mapsto\alpha+x$ g\"ondermesi tan\i
mlan\i rsa $\{\alpha+x\colon x\in\beta\}$ g\"or\"unt\"us\"u, bir
k\"umedir.  $\beta$ limitse 
\begin{equation}\label{eqn:+lim}
  \alpha+\beta=\sup\{\alpha+\xi\colon\xi<\beta\}
\end{equation}
olsun.  Bu ko\c sul, \eqref{eqn:++} sat\i r\i ndaki ko\c sullarla, $\on$ s\i n\i f\i nda $x\mapsto\alpha+x$ i\c
slemesini tan\i mlayacakt\i r.

\section{T\"umevar\i m ve \"ozyineleme}


\begin{theorem}[T\"umevar\i m]%
\index{teorem!T\"umevar\i m T---i}
$\bm A\included\on$ olsun.  E\u ger
\begin{compactenum}[1)]
\item
$0\in\bm A$,
\item
her $\alpha$ i\c cin $\alpha\in\bm A\lto\alpha'\in\bm A$,
\item
her $\alpha$ limiti i\c cin $\alpha\included\bm A\lto\alpha\in\bm A$
\end{compactenum}
ise, o zaman $\bm A=\on$.
\end{theorem}

\begin{proof}
Hipotez alt\i nda $\on\setminus\bm A$ fark\i n\i n en k\"u\c c\"uk eleman\i\ olamaz.
\end{proof}

E\u ger $\bm F\colon\bm A\to\bm B$ ve $\bm C\included\bm A$ ise
\begin{equation*}
\bm F\cap(\bm C\times\bm B)=\bm F\restriction\bm C
\end{equation*}
olsun.  Bu $\bm F\restriction\bm C$ g\"ondermesi, 
$\bm F$ g\"ondermesinin $\bm C$ s\i n\i f\i na \textbf{s\i n\i rlamas\i d\i r}%
\index{s\i n\i rlama}
(\eng{restriction}).

\begin{theorem}[Ordinaller \"Ozyinelemesi I]%
\index{teorem!Ordinaller \"Ozyinelemesi T---i}\label{thm:rec-ord}
$\bm A$ bir s\i n\i f olsun, ve
\begin{align*}
b&\in\bm A,&\bm F&\colon\bm A\to\bm A,&\bm G&\colon\pow{\bm A}\to\bm A
\end{align*}
olsun.  O zaman
$\on$ s\i n\i f\i ndan $\bm A$ s\i n\i f\i 
na giden
\begin{gather*}
  \bm H(0)=b,\\
\bm H(\alpha')=\bm F(\bm H(\alpha)),\\
\alpha\text{ limit ise }\bm H(\alpha)
=\bm G(\bm H[\alpha])
%=\bm G\bigl(\{\bm H(\xi)\colon\xi<\alpha\}\bigr) 
\end{gather*}
\"ozellikleri olan bir ve tek bir $\bm H$ g\"ondermesi vard\i r.
\end{theorem}

\.Iki kan\i t verece\u giz.

\begin{proof}[Kan\i t 1.]
T\"umevar\i mla en \c cok bir $\bm H$ g\"ondermesi vard\i r.  
\c C\"unk\"u $\bm H_1$ g\"ondermesinin ve $\bm H$ g\"ondermesinin \"ozellikleri 
ayn\i\ olsun.  O zaman
\begin{compactenum}[1)]
\item
$\bm H_1(0)=b=\bm H(0)$;
\item
$\bm H_1(\alpha)=\bm H(\alpha)$ ise
\begin{equation*}
\bm H_1(\alpha')=\bm F(\bm H_1(\alpha))=\bm F(\bm H(\alpha))=\bm H(\alpha');
\end{equation*}
\item
$\alpha$ limit ise ve $\bm H_1\restriction\alpha=\bm H\restriction\alpha$ ise
\begin{equation*}
\bm H_1(\alpha)=\bm G(\bm H_1[\alpha])=\bm G(\bm H[\alpha])=\bm H(\alpha).
\end{equation*}
\end{compactenum}
\Teoremin{thm:rec} kan\i t\i ndaki gibi 
bir $\bm C$ s\i n\i f\i\ i\c cin $\bm H=\bigcup\bm C$ olacakt\i r.  Bu s\i n\i f\i n tan\i m\i na g\"ore her $a$ eleman\i\ i\c cin $a\subseteq\on\times\bm A$, ve $a$ k\"umesinin her $(\alpha,d)$ eleman\i\ i\c cin,
\begin{compactitem}
\item
ya $(\alpha,d)=(0,b)$,
\item
ya da bir $(\beta,c)$ eleman\i\ i\c cin, $\beta'=\alpha$ ve $\bm F(c)=d$,
\item
ya da $\alpha$ limittir, 
ve $a\cap(\alpha\times\bm A)$ kesi\c simi, 
tan\i m k\"umesi $\alpha$ olan bir $f$ g\"ondermesidir, 
ve $\bm G(f[\alpha])=d$.
\end{compactitem}
E\u ger $\bigcup\bm C$ bile\c simi, tan\i m s\i n\i f\i\ $\on$ olan bir g\"onderme de\u gilse, bir \emph{en k\"u\c c\"uk} $\alpha$ i\c cin
\begin{equation*}
\left\{x\colon x\in\bm A\land(\alpha,x)\in\bigcup\bm C\right\}
\end{equation*}
s\i n\i f\i n\i n ya hi\c c eleman\i\ yoktur ya da en az iki
eleman\i\ vard\i r.  O zaman $\alpha\neq0$.  E\u ger
$\alpha=\beta'$ ise, o zaman bir $c$ i\c cin $(\beta,c)\in\bigcup\bm
C$, dolay\i s\i yla $(\alpha,\bm F(c))\in\bigcup\bm C$.  Bu
durumda e\u ger $(\alpha,d)\in\bigcup\bm C$ ise bir $e$ i\c cin $d=\bm
F(e)$ ve $(\beta,e)\in\bigcup\bm C$, dolay\i s\i yla $c=e$ ve $d=\bm
F(c)$ (\c c\"unk\"u $\alpha$ en k\"u\c c\"ukt\"ur).  \"Oyleyse
$\alpha$ ard\i l olamaz.  Benzer \c sekilde $\alpha$ limit olamaz.
Sonu\c c olarak $\bigcup\bm C$ bile\c simi, tan\i m s\i n\i f\i\ $\on$
olan bir g\"onderme olmal\i d\i r.  Bu g\"ondermenin, tan\i m\i ndan
dolay\i\ istedi\u gimiz \"ozellikleri vard\i r.
\end{proof}

\begin{proof}[Kan\i t 2.]
  Her $\beta$ i\c cin
$\beta'$ k\"umesinden $\bm A$ s\i n\i f\i na giden
\begin{gather*}
  h_{\beta}(0)=b,\\
\alpha<\beta\lto h_{\beta}(\alpha')=\bm F(h_{\beta}(\alpha)),\\
\alpha<\beta\land\alpha\text{ limit}\lto h_{\beta}(\alpha)
=\bm G(h_{\beta}[\alpha])
\end{gather*}
\"ozellikleri olan 
bir ve tek bir $h_{\beta}$ g\"ondermesinin oldu\u gunu g\"osterece\u giz.
$\bm B$, istedi\u gimiz \"ozellikleri olan 
$\beta$ ordinalleri olu\c sturdu\u gu s\i n\i f olsun.
E\u ger $\beta\in\bm B$, $\gamma\in\bm B$, ve $\beta\leq\gamma$ ise,
o zaman $g_{\beta}$ ve $g_{\gamma}\restriction\beta'$ 
g\"ondermelerinin ayn\i\ \"ozelli\u gi vard\i r,
dolay\i s\i yla
\begin{equation*}
  g_{\beta}\included g_{\gamma}.
\end{equation*}
\c Simdi t\"umevar\i m kullanaca\u g\i z.
\begin{asparaenum}
  \item
$h_0=\{(0,b)\}$ olabilir ve olmal\i d\i r, dolay\i s\i yla $0\in\bm B$.
\item
$\gamma\in\bm B$ ise
$h_{\gamma'}
=h_{\gamma}\cup\bigl\{\bigl(\gamma',\bm F(h_{\gamma}(\gamma))\bigr)\bigr\}$
olabilir ve olmal\i d\i r, dolay\i s\i yla $\gamma'\in\bm B$.
\item
$\gamma$ limit ve $\gamma\included\bm B$ ise
$\theta\leq\delta<\gamma$ oldu\u gu zaman
$h_{\theta}\included h_{\delta}$ olmal\i d\i r,
dolay\i s\i yla $h_{\gamma}=\bigcup\{h_{\xi}\colon\xi<\gamma\}$ 
olabilir ve olmal\i d\i r, ve sonu\c c olarak $\gamma\in\bm B$.
\end{asparaenum}
Benzer \c sekilde $\bm H=\bigcup\{h_{\xi}\colon\xi\in\on\}$ 
olabilir ve olmal\i d\i r.
\end{proof}

Ordinaller \"Ozyinelemesi Teoreminin farkl\i\ bir bi\c cimi,
\Teorem{thm:rec-ord-again} olacakt\i r.
Ama ordinaller toplamas\i, \c carpmas\i, ve kuvvet almas\i\ i\c cin,
son teorem yeter.

\section{Toplama}

\Teoreme{thm:rec-ord} g\"ore her $\alpha$ i\c cin 
\sayfada{eqn:++}ki \eqref{eqn:++} ve
\sayfada{eqn:+lim}ki \eqref{eqn:+lim} sat\i rlar\i ndaki ko\c sullar 
$\on$ s\i n\i f\i nda
$x\mapsto\alpha+x$ i\c slemini tan\i mlar.  
Yani tan\i ma g\"ore
\begin{gather*}
  \alpha+0=\alpha,\\
\alpha+\beta'=(\alpha+\beta)',\\
\beta\text{ limit}\lto\alpha+\beta=\sup\{\alpha+\xi\colon\xi<\beta\}.
\end{gather*}
\glossary{$\alpha+\beta$}%
(Baz\i\ kitaplarda farkl\i\ ama denk bir tan\i m verilir
\cite[s.~145]{Nesin-AKK}; 
\sayfaya{eqn:+ord} bak\i n.)
\"Ozel durum olarak
\begin{equation*}
  \alpha+1=\alpha'.
\end{equation*}
Ayr\i ca $1+\upomega=\sup\{1+x\colon x\in\upomega\}=\upomega<\upomega+1$,
ve genelde
\begin{equation*}
0<n<\upomega\quad\text{ ise }\quad
n+\upomega=\upomega<\upomega+n.
\end{equation*}
B\"oylece $\on$ s\i n\i f\i nda toplama de\u gi\c smeli de\u gildir.
\"Ozel olarak $\on$ s\i n\i f\i ndaki 
\begin{align*}
\xi&\mapsto\alpha+\xi,&\xi&\mapsto\xi+\alpha
\end{align*}
i\c slemleri birbirinden farkl\i d\i r.
Birincisi, \Teoreme{thm:<} g\"ore 
\emph{kesin artan} olacakt\i r;
ikincisi, \Teoreme{thm:leq} g\"ore 
sadece \emph{artan} olacakt\i r.
\Sekil{fig:y=omega+x} ve \numaraya{fig:y=x+omega} bak\i n.
\begin{figure}
  \centering
\psset{unit=3cm}
  \begin{pspicture}(-0.5,-0.5)(2.5,3.5)
\psdots(0,1)(0.25,1.25)(0.4375,1.4375)(0.578125,1.578125)
(0.684,1.684)(0.763,1.763)(0.822,1.822)
\psdots(1,2)(1.25,2.25)(1.4375,2.4375)(1.578125,2.578125)
(1.684,2.684)(1.763,2.763)(1.822,2.822)
    \psdots(2,3)
%%%%%%%%%%%%%%%%%%%%%%
\uput[d](0,0){$0$}
\uput[d](0.25,0){$1$}
\uput[d](0.4375,0){$2$}
\uput[d](0.578125,0){$3$}
\uput[d](1,0){$\upomega$}
\uput[d](1.25,0){$\upomega+1$}
%\uput[d](1.4375,0){$\upomega+2$}
%\uput[d](1.578125,0){$\upomega+3$}
\uput[d](2,0){$\upomega+\upomega$}
\uput[l](0,0){$0$}
\uput[l](0,0.25){$1$}
\uput[l](0,0.4375){$2$}
\uput[l](0,0.578125){$3$}
\uput[l](0,1){$\upomega$}
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\uput[l](0,2){$\upomega+\upomega$}
\uput[l](0,2.25){$\upomega+\upomega+1$}
\uput[l](0,2.4375){$\upomega+\upomega+2$}
\uput[l](0,2.578125){$\upomega+\upomega+3$}
\uput[l](0,3){$\upomega+\upomega+\upomega$}
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\uput[r](2.2,0){$\xi$}
\uput[u](0,3.2){$\eta$}
%%%%%%%%%%%%%%%%%%%%%%%%%
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  \end{pspicture}
  \caption{$\eta=\upomega+\xi$ denkleminin grafi\u gi}\label{fig:y=omega+x}
  
\end{figure}
\begin{figure}
  \centering
\psset{unit=3cm}
  \begin{pspicture}(-0.5,-0.5)(2.5,3.5)
\psdots(0,1)(0.25,1)(0.4375,1)(0.578125,1)
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(1.684,2)(1.763,2)(1.822,2)
    \psdots(2,3)
%%%%%%%%%%%%%%%%%%%%%%%%%
\uput[d](0,0){$0$}
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\uput[d](1,0){$\upomega$}
\uput[d](1.25,0){$\upomega+1$}
%\uput[d](1.4375,0){$\upomega+2$}
%\uput[d](1.578125,0){$\upomega+3$}
\uput[d](2,0){$\upomega+\upomega$}
\uput[l](0,0){$0$}
\uput[l](0,0.25){$1$}
\uput[l](0,0.4375){$2$}
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\uput[l](0,1){$\upomega$}
\uput[l](0,1.25){$\upomega+1$}
\uput[l](0,1.4375){$\upomega+2$}
\uput[l](0,1.578125){$\upomega+3$}
\uput[l](0,2){$\upomega+\upomega$}
\uput[l](0,2.25){$\upomega+\upomega+1$}
\uput[l](0,2.4375){$\upomega+\upomega+2$}
\uput[l](0,2.578125){$\upomega+\upomega+3$}
\uput[l](0,3){$\upomega+\upomega+\upomega$}
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%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%
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\psline(0,2)(2.1,2)
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  \end{pspicture}
  \caption{$\eta=\xi+\upomega$ denkleminin grafi\u gi}\label{fig:y=x+omega}
  
\end{figure}
Teoremleri kan\i tlayarak 
a\c sa\u g\i daki teoremdeki kurallar\i\ kullanaca\u g\i z.

\begin{theorem}\label{thm:sup}
$a\included\on$, $\beta\in a$, ve $b\included\on$ olsun.
  \begin{enumerate}
  \item\label{item:sup1} 
$\beta\leq\sup(a)$.
\item\label{item:sup2}
$b\included a$ ise $\sup(b)\leq\sup(a)$.
\item\label{item:sup3}
$\Forall{\xi}\bigl(\xi\in b\lto\Exists{\eta}(\eta\in a\land\xi\leq\eta)\bigr)$ 
ise 
$\sup(b)\leq\sup(a)$. 
  \end{enumerate}
\end{theorem}

\ktk

\begin{theorem}\label{thm:<}
$\xi\mapsto\alpha+\xi$ g\"ondermesi, \textbf{kesin artand\i r}%
\index{artan g\"onderme}
(\eng{strictly increasing}),
yani 
\begin{equation*}
\beta<\gamma\lto\alpha+\beta<\alpha+\gamma.
\end{equation*}
\end{theorem}

\begin{proof}
  $\gamma$ \"uzerinden t\"umevar\i m kullanaca\u g\i z.  
  \begin{asparaenum}
    \item
    $\gamma=0$ ise, iddia do\u grudur, \c c\"unk\"u hi\c cbir zaman
  $\beta<0$ de\u gildir.  
\item
$\gamma=\delta$ durumda iddian\i n do\u gru oldu\u gu varsay\i ls\i n.  E\u ger $\beta<\delta'$ ise, o zaman $\beta\leq\delta$, dolay\i s\i yla
  \begin{equation*}
    \alpha+\beta\leq\alpha+\delta<(\alpha+\delta)'=\alpha+\delta'.
  \end{equation*}
\item
$\delta$ limit
ve $\beta<\delta$ ise, o zaman $\beta<\beta'<\delta$, dolay\i s\i yla
(\Teoremin{thm:sup} \ref{item:sup1} \c s\i kk\i n\i\ kullanarak)
\begin{equation*}
\alpha+\beta<\alpha+\beta'\leq\sup_{\xi<\delta}(\alpha+\xi)=\alpha+\delta.\qedhere
\end{equation*}
  \end{asparaenum}
\end{proof}

Son teoremin kan\i t\i n\i n 3.\ ad\i m\i nda 
$\gamma<\delta$ durumu i\c c\i n iddian\i n do\u gru oldu\u gu 
varsay\i m gerekmedi.


\begin{theorem}\label{thm:leq}
$\xi\mapsto\xi+\alpha$ g\"ondermesi, \textbf{artand\i r,}
yani
\begin{equation*}
  \beta\leq\gamma\lto\beta+\alpha\leq\gamma+\alpha.
\end{equation*}
\end{theorem}

\begin{proof}
  \c Simdi $\alpha$ \"uzerinden t\"umevar\i m kullanaca\u g\i z.
  $\beta\leq\gamma$ olsun.
  \begin{asparaenum}
    \item
$\beta+0=\beta\leq\gamma=\gamma+0$.
\item
$\beta+\alpha=\gamma+\alpha$ ise tabii ki
  \begin{equation*}
    \beta+\alpha'=(\beta+\alpha)'=(\gamma+\alpha)'=\gamma+\alpha'.
  \end{equation*}
$\beta+\alpha<\gamma+\alpha$ ise, 
\Teoreme{thm:'} g\"ore
\begin{equation*}
  \beta+\alpha'
  =(\beta+\alpha)'\leq\gamma+\alpha<(\gamma+\alpha)'=\gamma+\alpha'.  
\end{equation*}
\item
$\delta$ limit olsun, ve $\alpha<\delta$ ise,
  $\beta+\alpha\leq\gamma+\alpha$ varsay\i ls\i n.  
O zaman
(\Teoremin{thm:sup} \ref{item:sup3} \c s\i kk\i n\i\ kullanarak)
  \begin{equation*}
    \beta+\delta=\sup_{\xi<\gamma}(\beta+\xi)
\leq\sup_{\xi<\gamma}(\gamma+\xi)=\gamma+\delta.\qedhere
  \end{equation*}
  \end{asparaenum}
\end{proof}

G\"ord\"u\u g\"um\"uz gibi ayn\i\ zamanda $\beta<\gamma$ ama
$\beta+\alpha=\gamma+\alpha$ olabilir, 
mesela \Sekilde{fig:y=x+omega}ki gibi
\begin{equation*}
k<\ell<\upomega\quad\text{ ise }\quad k+\upomega=\ell+\upomega.
\end{equation*}
Ayr\i ca $\xi\mapsto\xi+\alpha$ g\"ondermesi i\c cin
analizdeki gibi ``ara de\u ger teoremi'' yanl\i\c st\i r.
\"Orne\u gin $0+\upomega<\upomega+1<\upomega+\upomega$,
ama
\begin{equation*}
  \xi+\upomega=\upomega+1
\end{equation*}
denkleminin hi\c c \c c\"oz\"um\"u yoktur.
Yine de $\xi\mapsto\alpha+\xi$ g\"ondermesi i\c cin
ara de\u ger teoremi do\u gru olacakt\i r.

\begin{theorem}\label{thm:0+}
Her $\alpha$ i\c cin $0+\alpha=\alpha$.
\end{theorem}

\ktk

Toplaman\i n ba\c ska bir anlay\i\c s\i\ i\c cin,
\Teoremde{thm:ord+}\
(\sayfada{thm:ord+}),
sonraki teorem kullan\i lacakt\i r.

\begin{theorem}[\c C\i karma]\label{thm:subtraction}
  $\alpha\leq\beta$ ise
  \begin{equation}\label{eqn:sub}
    \alpha+\xi=\beta
  \end{equation}
denkleminin bir ve tek bir \c c\"oz\"um\"u vard\i r.  
\end{theorem}

\begin{proof}
Bir ordinal bulmak i\c cin iki y\"ontemimiz vard\i r.
Bu ordinal, bir ordinaller k\"umesinin \emph{minimumu}
veya \emph{supremumu} olabilir.
Burada \eqref{eqn:sub} denkleminin \c c\"oz\"um\"u bir minimum olacakt\i r;
ba\c ska y\"ontemle bir supremumdur
(\sayfaya{subtraction} bak\i n).
  \Teoreme{thm:<} g\"ore denklemin en \c cok bir \c
  c\"oz\"um\"u vard\i r.  
\Teorem{thm:leq} ve \numaraya{thm:0+} g\"ore
  \begin{equation*}
    \alpha+\beta\geq0+\beta=\beta,
  \end{equation*}
dolay\i s\i yla $\{\xi\colon\beta\leq\alpha+\xi\}$ bo\c s de\u
gildir (\c c\"unk\"u $\beta$ eleman\i n\i\ i\c cerir).  
Bu k\"umenin en k\"u\c c\"uk eleman\i, $\delta$ olsun.  
Yani
\begin{equation*}
  \delta=\min\{\xi\colon\beta\leq\alpha+\xi\}
\end{equation*}
olsun.
O zaman
\begin{equation*}
  \beta\leq\alpha+\delta.
\end{equation*}
Biz $\alpha+\delta\leq\beta$ e\c sitsizli\u gini g\"osterece\u giz.
\"U\c c durum vard\i r.
\begin{asparaenum}
  \item
$\delta=0$ ise
$\alpha+\delta=\alpha+0=\alpha\leq\beta$.
\item
$\delta=\gamma'$ ise $\alpha+\gamma<\beta$, dolay\i s\i yla
(\sayfada{thm:'}ki
\Teoremi{thm:'} kullanarak)
\begin{equation*}
    \alpha+\delta=(\alpha+\gamma)'\leq\beta.
\end{equation*}
\item
$\delta$ limit olsun.  E\u ger $\gamma<\delta$ ise
  $\alpha+\gamma<\beta$ olmal\i d\i r, dolay\i s\i yla
  \begin{equation*}
    \alpha+\delta=\sup_{\xi<\delta}(\alpha+\xi)\leq\beta.\qedhere
  \end{equation*}
\end{asparaenum}
\end{proof}

$\alpha\leq\beta$ durumunda $\alpha+\xi=\beta$ denkleminin \c c\"oz\"um\"u i\c cin
\begin{equation*}
\beta-\alpha
\end{equation*}
ifadesi yaz\i labilir.  \"Orne\u gin $\alpha'-1=\alpha$.
\c Simdi ordinaller toplamas\i n\i n 
\emph{birle\c smeli} oldu\u gunu g\"osterece\u giz.

\begin{theorem}\label{thm:lim}
  $\beta$ limitse $\alpha+\beta$ toplam\i\ da limittir.
\end{theorem}

\ktk

\begin{theorem}\label{thm:+assoc}
Ordinaller toplamas\i\ 
\textbf{birle\c smelidir,}
yani
t\"um $\alpha$, $\beta$, ve $\gamma$ i\c cin
  \begin{equation*}
    \alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma.
  \end{equation*}
\end{theorem}

\begin{proof}
  $\gamma$ \"uzerinden t\"umevar\i m kullanaca\u g\i z.
  \begin{asparaenum}
    \item
$\alpha+(\beta+0)=\alpha+\beta=(\alpha+\beta)+0$.
\item
$\alpha+(\beta+\delta)=(\alpha+\beta)+\delta$ ise
  \begin{align*}
    \alpha+(\beta+\delta')=\alpha+(\beta+\delta)'
&=(\alpha+(\beta+\delta))'\\
&=((\alpha+\beta)+\delta)'
=(\alpha+\beta)+\delta'.
  \end{align*}
\item
$\delta$ limit olsun, ve $\gamma<\delta$ ise
$\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$ olsun.  
O zaman
\begin{align*}
  (\alpha+\beta)+\delta
&=\sup_{\xi<\delta}\bigl((\alpha+\beta)+\xi\bigr)\\
&=\sup_{\xi<\delta}\bigl(\alpha+(\beta+\xi)\bigr)\\
&\leq\alpha+\sup_{\xi<\delta}(\beta+\xi).
\end{align*}
$\delta$ limit oldu\u gundan
\begin{equation*}
\sup_{\xi<\delta}(\beta+\xi)=\beta+\delta.
\end{equation*}
Bundan dolay\i
\begin{equation*}
(\alpha+\beta)+\delta\leq\alpha+(\beta+\delta);
\end{equation*}
ayr\i ca $\theta<\beta+\delta$ ise, bir $\gamma$ i\c cin,
$\gamma<\delta$ ve $\theta<\beta+\gamma$,
yani
\begin{equation*}
  \Forall{\eta}\bigl(\eta<\beta+\delta
\lto\Exists{\xi}(\xi<\delta\land\eta<\beta+\xi)\bigr).
\end{equation*}
\Teoreme{thm:lim} g\"ore,
$\beta+\delta$ bir limittir,
dolay\i s\i yla
\begin{equation*}
\alpha+(\beta+\delta)
=\sup_{\eta<\beta+\delta}(\alpha+\eta).
\end{equation*}
O halde
(\Teoremin{thm:sup} \ref{item:sup3} \c s\i kk\i n\i\ kullanarak)
\begin{equation*}
  \sup_{\eta<\beta+\delta}(\alpha+\eta)
\leq\sup_{\xi<\delta}\bigl(\alpha+(\beta+\xi)\bigr)
=(\alpha+\beta)+\delta.
\end{equation*}
Sonu\c c olarak $\alpha+(\beta+\delta)=(\alpha+\beta)+\delta$.\qedhere 
  \end{asparaenum}
\end{proof}

\section{Normal i\c slemleri}

Asl\i nda son teoremin kan\i t\i, genel bir y\"ontem kurur.
$\bm F$,
$\on$ \"uzerinde birli bir i\c slem olsun,
yani $\bm F\colon\on\to\on$ olsun.  Analizdeki gibi,
bir $\alpha$ i\c cin,
e\u ger
\begin{equation*}
  \beta<\bm F(\alpha)<\gamma
\end{equation*}
e\c sitsizli\u gini sa\u glayan her $\beta$ ve $\gamma$ i\c cin,
\begin{align*}
  \delta&<\alpha<\theta,&
  \Forall{\xi}(\delta<\xi<\theta&\lto\beta<\bm F(\xi)<\gamma)
\end{align*}
ko\c sullar\i n\i\ sa\u glayan $\delta$ ve $\theta$ varsa,
o zaman $\bm F$ g\"ondermesi $\alpha$ noktas\i nda \textbf{s\"ureklidir}%
\index{s\"urekli g\"onderme}
(\eng{continuous}).
E\u ger $\alpha=0$ veya $\bm F(\alpha)=0$ ise,
tan\i m bir az de\u gi\c siktir.

\begin{theorem}
$\bm F\colon\on\to\on$ olsun.
  \begin{enumerate}
  \item 
Limit olmayan her ordinalde $\bm F$ s\"ureklidir.
\item
$\bm F$ artan olsun ve $\alpha$ bir limit olsun.  
O zaman $\alpha$ noktas\i nda $\bm F$ s\"ureklidir
ancak ve ancak
\begin{equation*}
\bm F(\alpha)
%=\sup_{\xi<\alpha}\bm F(\xi)
=\sup(\bm F[\alpha]).
\end{equation*}
  \end{enumerate}
\end{theorem}

\ktk

$\on$ \"uzerinde
kesin artan ve s\"urekli birli bir i\c sleme
\textbf{normal}%
\index{normal!--- i\c slem}
(\eng{normal}) diyece\u giz.

\begin{theorem}
Her $\alpha$ i\c cin
\begin{equation*}
 \xi\mapsto\alpha+\xi 
\end{equation*}
i\c slemi normaldir.
\"Ozel olarak $\xi\mapsto\xi$ normaldir.
\end{theorem}

\begin{proof}
Toplaman\i n tan\i m\i\ ve  \Teorem{thm:<} ve \ref{thm:0+}.
\end{proof}

\c Simdilik $\xi\mapsto\alpha+\xi$ i\c slemleri, tek \"orne\u gimizdir:

\begin{xca}
  $\alpha>0$ olsun.  
A\c sa\u g\i daki i\c slemlerin normal olmad\i\u g\i n\i\ g\"osterin:
\begin{align*}
  \xi&\mapsto\xi+\alpha,&
\xi&\mapsto\xi',&
\xi&\mapsto\xi+\xi.
\end{align*}
\end{xca}

\Teoremin{thm:+assoc} kan\i t\i n\i n 3.\ ad\i m\i nda
\begin{equation*}
  \alpha+\sup_{\xi<\delta}(\beta+\xi)
=\sup_{\xi<\delta}(\alpha+(\beta+\xi))
\end{equation*}
e\c sitli\u gini g\"osterdik,
ve bunun i\c cin sadece
\begin{enumerate}[1)]
\item
$\xi\mapsto\alpha+\xi$ i\c sleminin normal oldu\u gunu,
\item\label{item:sup-limit}
$\delta$ limit iken
$\sup\{\beta+\xi\colon\xi<\delta\}$ ordinalinin limit oldu\u gunu
\end{enumerate}
kulland\i k.
Asl\i nda \ref{item:sup-limit}.\ ko\c sulun yerinde
a\c sa\u g\i daki teorem kullan\i labilir.

\begin{theorem}
  E\u ger $0\pincluded a\pincluded\on$ ve $\sup(a)\notin a$ ise, 
o zaman $\sup(a)$, bir limittir.
\end{theorem}

\ktk

\begin{theorem}\label{thm:sup-x}
  $\bm F$ normal bir i\c slem olsun.
Ordinallerin olu\c sturdu\u gu,
bo\c s olmayan t\"um $a$ k\"umeleri i\c cin
  \begin{equation*}
\bm F\bigl(\sup(a)\bigr)=\sup\bigl(\bm F[a]\bigr).
  \end{equation*}
\end{theorem}

\ktk[  \.Iki durum vard\i r.
$\sup(a)\in a$ ise kan\i t kolayd\i r.
$\sup(a)\notin a$ ise limit olmal\i d\i r,
ve bu durumda kan\i t,
\Teoremin{thm:+assoc} kan\i t\i n\i n 3.\ ad\i m\i ndaki gibidir.]

Teoremde $a$ bo\c s olmamal\i d\i r,
\c c\"unk\"u
$\sup(\bm F[0])=\sup(0)=0$, 
ama $\bm F(0)>0$ olabilir. 
\"Orne\u gin $\upomega+0=\upomega>0$.
Teoremin bir sonucu,
\Teorem{thm:max} olacakt\i r.
Kan\i t\i\ i\c cin
sonraki teoremi de kullanaca\u g\i z.

\begin{theorem}\label{thm:inc}
E\u ger $\bm F$, $\on$ s\i n\i f\i nda kesin artan bir
i\c slemse, t\"um $\alpha$ i\c cin 
\begin{equation*}
\alpha\leq\bm F(\alpha).
\end{equation*}
\end{theorem}

\begin{proof}
$\alpha>\bm F(\alpha)$ ise, $\bm F$ kesin artan oldu\u gundan $\bm
  F(\alpha)>\bm F(\bm F(\alpha))$, dolay\i s\i yla $\{\xi\colon\xi>\bm
  F(\xi)\}$ s\i n\i f\i n\i n en k\"u\c c\"uk eleman\i\ yoktur.  $\on$
  iyi\-s\i ralanm\i\c s oldu\u gundan $\{\xi\colon\xi>\bm F(\xi)\}$ s\i
  n\i f\i\ bo\c s olmal\i d\i r. 
\end{proof}

Bu teorem sayesinde $\gamma\leq\alpha+\gamma$,
dolay\i s\i yla $\alpha+\xi=\beta$ denkleminin \c c\"oz\"um\"u
$\beta$ ordinalinden b\"uy\"uk olamaz.
Di\u ger taraftan bunu zaten biliyoruz \c c\"unk\"u
\begin{align*}
  \gamma
&=0+\gamma&&\text{[\Teorem{thm:0+}]}\\
&\leq\alpha+\gamma.&&\text{[\Teorem{thm:leq}]}
\end{align*}

\begin{theorem}\label{thm:max}
  $\bm F$ normal ve $\bm F(0)\leq\alpha$ olsun.
O zaman $\{\xi\colon\bm F(\xi)\leq\alpha\}$ s\i n\i f\i\
bir k\"umedir,
ve bu k\"umenin en b\"uy\"uk eleman\i\ vard\i r.
\end{theorem}

\begin{proof}
  Son teoremi kullanarak
  \begin{equation*}
    \{\xi\colon\bm F(\xi)\leq\alpha\}
\included\alpha'
  \end{equation*}
kapsanmas\i n\i\ biliyoruz,
dolay\i s\i yla $\{\xi\colon\bm F(\xi)\leq\alpha\}$ s\i n\i f\i\ bir k\"umedir.
\Teoremi{thm:sup-x} kullanarak
\begin{equation*}
  \bm F(\sup\{\xi\colon\bm F(\xi)\leq\alpha\})
=\sup\{\bm F(\xi)\colon\bm F(\xi)\leq\beta\}\leq\beta,
\end{equation*}
dolay\i s\i yla 
$\sup\{\xi\colon\bm F(\xi)\leq\alpha\}\in\{\xi\colon\bm F(\xi)\leq\alpha\}$,
yani bu k\"umenin en b\"uy\"uk eleman\i\ vard\i r.
\end{proof}

\c Simdi \Teoremin{thm:subtraction}%
\label{subtraction} 
ba\c ska bir kan\i t\i\ vard\i r.
E\u ger $\alpha\leq\beta$ ise,
o zaman $\{\xi\colon\alpha+\xi\leq\beta\}$ s\i n\i f\i,
en b\"uy\"uk eleman\i\ olan bir k\"ume olmal\i d\i r.
En b\"uy\"uk eleman\i\ $\delta$ olsun.  E\u ger $\alpha+\delta<\beta$ ise,
o zaman
\begin{equation*}
  \alpha+\delta'=(\alpha+\delta)'\leq\beta,
\end{equation*}
ki bu imk\^ans\i z\i r.
Sonu\c c olarak $\alpha+\delta=\beta$.
Benzer \c sekilde
\Teoremi{thm:division} kan\i tlayaca\u g\i z.

\section{\c Carpma}

Her $\alpha$ i\c cin $\on$ s\i n\i f\i nda $x\mapsto\alpha\cdot x$ i\c slemi, 
tan\i m\i na g\"ore,
\begin{gather*}
	\alpha\cdot0=0,\\
	\alpha\cdot\beta'=\alpha\cdot\beta+\alpha,\\
	\gamma\text{ limit}\lto\alpha\cdot\gamma=\sup\{\alpha\cdot\xi\colon\xi<\gamma\}
\end{gather*}%
\glossary{$\alpha\cdot\beta$}
ko\c sullar\i\ sa\u glar.  \"Ozel olarak
\begin{equation*}
\alpha\cdot1=\alpha.
\end{equation*}
O zaman $\upomega\cdot2 
=\upomega\cdot1+\upomega
=\upomega+\upomega
=\sup\{\upomega+x\colon x\in\upomega\}$, ama
\begin{equation*}
2\cdot\upomega=\sup_{x\in\upomega}(2\cdot x)=\upomega,  
\end{equation*}
dolay\i s\i yla $2\cdot\upomega<\upomega\cdot2$.
\"Oyleyse \c carpma de\u gi\c smeli de\u gildir.

\begin{theorem}
$0\cdot\alpha=0$ ve $1\cdot\alpha=\alpha$.
\end{theorem}

\ktk

$\alpha\geq1$ ise, 
$\xi\mapsto\alpha\cdot\xi$ i\c sleminin normal oldu\u gunu 
kan\i tlayaca\u g\i z.  
\.I\c slemin kesin artan oldu\u gunu g\"ostermek yeter.  
\Teoremin{thm:<} kan\i t\i,
genel bir y\"ontem kurur:

\begin{theorem}
E\u ger $\bm F\colon\on\to\on$, ve t\"um $\alpha$ i\c cin
\begin{equation*}
\bm F(\alpha)<\bm F(\alpha'),
\end{equation*}
ve limit olan t\"um $\beta$ i\c cin
\begin{equation*}
\bm F(\beta)=\sup_{\xi<\beta}\bm F(\xi)
\end{equation*}
ise, o zaman $\bm F$ normaldir.
\end{theorem}

\ktk

\begin{theorem}
$\alpha\geq1$ ise $\xi\mapsto\alpha\cdot\xi$ i\c slemi, normaldir.
\end{theorem}

\ktk

\begin{theorem}
Ordinaller \c carpmas\i, toplama \"uzerine \textbf{soldan da\u g\i l\i r,} 
yani
\begin{equation*}
 \alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma. 
\end{equation*}
\end{theorem}

\begin{proof}
  \begin{asparaenum}
    \item
$\gamma=0$ durumunda kan\i t kolayd\i r.  
\item
$\gamma=\delta$ durumunda iddia do\u gruysa
\begin{align*}
\alpha\cdot(\beta+\delta')
&=\alpha\cdot(\beta+\delta)'&&\text{[$+$ tan\i m\i]}\\
&=\alpha\cdot(\beta+\delta)+\alpha&&\text{[${}\cdot{}$ tan\i m\i]}\\
&=(\alpha\cdot\beta+\alpha\cdot\delta)+\alpha&&\text{[hipotez]}\\
&=\alpha\cdot\beta+(\alpha\cdot\delta+\alpha)&&\text{[\Teorem{thm:+assoc}]}\\
&=\alpha\cdot\beta+\alpha\cdot\delta';&&\text{[$+$ tan\i m\i]}
\end{align*}
b\"oylece $\gamma=\delta'$ durumunda iddia do\u grudur.  
\item
Son olarak $\delta$ limit, ve $\gamma<\delta$ durumunda iddia do\u gru olsun.  
$\gamma=\delta$ durumunu kan\i tlayaca\u g\i z.  
$\alpha=0$ ise iddia kolayd\i r.  $\alpha\geq1$ olsun.  
O zaman $\xi\mapsto\alpha\cdot\xi$ ve $\xi\mapsto\alpha\cdot\beta+\xi$ 
i\c slemleri normal oldu\u gundan
\begin{align*}
\alpha\cdot(\beta+\delta)
&=\alpha\cdot\sup_{\xi<\delta}(\beta+\xi)&&\text{[${}\cdot{}$ tan\i m\i]}\\
&=\sup_{\xi<\delta}(\alpha\cdot(\beta+\xi))&&\text{[\Teorem{thm:sup-x}: $\xi\mapsto\alpha+\xi$ normal]}\\
&=\sup_{\xi<\delta}(\alpha\cdot\beta+\alpha\cdot\xi)&&\text{[hipotez]}\\
&=\alpha\cdot\beta+\sup_{\xi<\delta}(\alpha\cdot\xi)&&\text{[\Teorem{thm:sup-x}: $\xi\mapsto\alpha\cdot\beta+\xi$ normal]}\\
&=\alpha\cdot\beta+\alpha\cdot\delta.&&\text{[${}\cdot{}$ tan\i m\i]}\qedhere
\end{align*}
  \end{asparaenum}
\end{proof}

G\"ord\"u\u g\"um\"uz gibi $2\cdot\upomega<\upomega+\upomega$,
dolay\i s\i yla
\begin{equation*}
(1+1)\cdot\upomega<1\cdot\upomega+1\cdot\upomega.  
\end{equation*}
\"Oyleyse \c carpma sa\u gdan da\u g\i lmaz.

\begin{theorem}
Ordinaller \c carpmas\i\ birle\c smelidir, yani
\begin{equation*}
  \alpha\cdot(\beta\cdot\gamma)=(\alpha\cdot\beta)\cdot\gamma.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}
Her $\xi\mapsto\xi\cdot\alpha$ i\c slemi artand\i r, yani
\begin{equation*}
 \beta\leq\gamma\lto\beta\cdot\alpha\leq\gamma\cdot\alpha. 
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}[B\"olme]\label{thm:division}
$1\leq\alpha$ ise $(\xi,\eta)$ i\c cin
  \begin{align}\label{eqn:.sys}
\alpha\cdot\xi+\eta&=\beta&
&\land&
\eta&<\alpha
  \end{align}
sisteminin bir ve tek bir \c c\"oz\"um\"u vard\i r.
\end{theorem}

\begin{proof}
\Teoreme{thm:max} g\"ore 
$\{\xi\colon\alpha\cdot\xi\leq\beta\}$ k\"umesinin 
en b\"uy\"uk $\gamma$ eleman\i\ vard\i r.
\Teoreme{thm:subtraction} g\"ore 
$\alpha\cdot\gamma+\eta=\beta$ denkleminin 
$\delta$ \c c\"oz\"um\"u vard\i r.  
E\u ger $\delta\geq\alpha$ ise 
\begin{equation*}
  \alpha\cdot\gamma'=\alpha\cdot\gamma+\alpha\leq\alpha\cdot\gamma+\delta=\beta,
\end{equation*}
dolay\i s\i yla $\gamma'\leq\gamma$,
ki bu imk\^ans\i zd\i r.  
\"Oyleyse $\delta<\alpha$, 
ve b\"oylece $(\gamma,\delta)$, 
\eqref{eqn:.sys} sisteminin istedi\u gimiz \c c\"oz\"umd\"ur.  
Benzer \c sekilde ba\c ska \c c\"oz\"um yoktur.
Asl\i nda e\u ger $(\gamma_1,\delta_1)$, ba\c ska bir \c c\"oz\"um ise,
o zaman 
$\gamma=\gamma_1\land\delta<\delta_1$ 
veya 
$\gamma<\gamma_1$ 
varsay\i labilir.
Birinci durumda
\begin{equation*}
  \alpha\cdot\gamma+\delta
<\alpha\cdot\gamma+\alpha
=\alpha\cdot\gamma'
\leq\alpha\cdot\gamma_1
\leq\alpha\cdot\gamma_1+\delta_1;
\end{equation*}
ve ikinci durumda
$\alpha\cdot\gamma+\delta=\alpha\cdot\gamma_1+\delta
<\alpha\cdot\gamma_1+\delta_1$;
her sonu\c c imk\^ans\i zd\i r.
\end{proof}

\section{Kuvvet alma}

Her $\alpha$ i\c cin, $\alpha>0$ ise, $\on$ s\i n\i f\i nda $x\mapsto\alpha^x$ i\c slemi, tan\i m\i na g\"ore,
\begin{gather*}
	\alpha^0=1,\\
	\alpha^{\beta'}=\alpha^\beta\cdot\alpha,\\
	\gamma\text{ limit}\lto\alpha^\gamma=\sup\{\alpha^{\xi}\colon\xi<\gamma\}
\end{gather*}%
\glossary{$\alpha^{\beta}$}
ko\c sullar\i n\i\ sa\u glar.  \"Ozel olarak
\begin{equation*}
\alpha^1=\alpha.
\end{equation*}
Ayr\i ca, tan\i ma g\"ore,
\begin{align*}
0^0&=1,&\beta>0&\lto 0^{\beta}=0.
\end{align*}
\"Oyleyse $\gamma$ limit ise $0^{\gamma}$ kuvveti, $\sup\{0^{\xi}\colon\xi<\gamma\}$ de\u gildir, ama
\begin{equation*}
0^{\gamma}=\sup\{0^{\xi}\colon 0<\xi<\gamma\}.
\end{equation*}

\begin{theorem}\mbox{}\label{thm:powers}
  \begin{enumerate}
  \item 
$1^{\alpha}=1$.
\item
$\alpha\geq1$ ise $\xi\mapsto\xi^{\alpha}$ artand\i r.
\item
$\alpha\geq2$ ise $\xi\mapsto\alpha^{\xi}$ i\c slemi, normaldir.
\item
$\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma}$.
\item
$\alpha^{\beta\cdot\gamma}=(\alpha^{\beta})^{\gamma}$.
  \end{enumerate}
\end{theorem}

\ktk

\subsection{Ordinal tabanlar\i}

\.Ilkokuldan bildi\u gimiz gibi, 
e\u ger $2\leq t<\upomega$ ve $1\leq a<\upomega$ ise, 
o zaman $t$ say\i s\i ndan k\"u\c c\"uk olan
baz\i\ $a_0$, $a_1$, \dots, $a_n$ do\u gal say\i lar\i\ i\c cin
\begin{equation}\label{eqn:base-t}
a=t^n\cdot a_0+t^{n-1}\cdot a_1+\dots+t^0\cdot a_n.
\end{equation}
Bu durumda, $a_0\neq0$ ise, o zaman $a$ say\i s\i,
\begin{align*}
&a_0a_1\dots a_n&
&\text{ veya }&
&(a_0a_1\dots a_n)_t
\end{align*}
olarak yaz\i labilir; bu ifade, $a$ say\i s\i n\i n 
$t$ \textbf{taban\i nda yaz\i l\i m\i d\i r}%
\index{taban}\index{yaz\i l\i m}
\cite[17.\ b\"ol.]{Nesin-AKK}
(\eng{base-$t$ numeral}).  $0$ olan $a_i$ 
\textbf{rakamlar\i}%
\index{rakam}
(\eng{digits}) 
\c c\i kart\i l\i rsa, 
bir $m$ do\u gal say\i s\i\ i\c cin, 
\begin{gather*}
a=t^{b_0}\cdot c_0+t^{b_1}\cdot c_1+\dots+t^{b_m}\cdot c_m,\\
  b_0>b_1>\dots>b_m,\\
\{c_0,c_1,\dots,c_m\}\included\{1,\dots,t-1\}
\end{gather*}
ko\c sullar\i n\i\ sa\u glayan $b_i$ ve $c_i$ do\u gal say\i lar\i\ vard\i r.  
B\"oylece $t$ taban\i\ ve $a$ say\i s\i,
\begin{equation*}
\{(b_0,c_0),(b_1,c_1),\dots,(b_m,c_m)\}
\end{equation*}
g\"ondermesini belirtir.  
G\"osterece\u gimiz gibi 
$1$'den b\"uy\"uk olan her ordinal taban\i\ i\c cin,
$0$ olmayan her ordinal, 
b\"oyle bir g\"onderme belirtir.

Tabii ki \eqref{eqn:base-t} sat\i r\i ndaki gibi ifadelerin
\"ozyinelemeli tan\i m\i\ vard\i r:
\begin{enumerate}
\item
  $n=0$ ise $\alpha_0+\dots+\alpha_{n-1}=0$.
\item
$\alpha_0+\dots+\alpha_n=(\alpha_0+\dots+\alpha_{n-1})+\alpha_n$.
\end{enumerate}

\c Simdi $\xi$ ve $\eta$ gibi $\zeta$ harf\/inin 
ordinal de\u gi\c sken oldu\u gunu hat\i rlay\i n. 

\begin{theorem}\label{thm:exp-sol}
$2\leq\alpha$ ve $1\leq\beta$ ise
$(\xi,\eta,\zeta)$ i\c cin
  \begin{align*}
\alpha^{\xi}\cdot\eta+\zeta&=\beta&
&\land&
0<\eta&<\alpha&
&\land&
\zeta&<\alpha^{\xi}
  \end{align*}
sisteminin bir ve tek bir $(\gamma,\delta,\theta)$ \c c\"oz\"um\"u vard\i r, 
ve ayr\i ca $\gamma\leq\beta$.
\end{theorem}

\ktk[ (\Teoreme{thm:division} bak\i n.)]

Sonu\c c olarak $\alpha>1$ ise, 
$0$ olmayan her $\beta$ i\c cin
\begin{gather*}
\beta=\alpha^{\gamma_0}\cdot\delta_0+\alpha^{\gamma_1}\cdot\delta_1+\dotsb,\\
\{\delta_0,\delta_1,\dots\}\included\alpha\setminus\{0\},
%0<\delta_0<\alpha,\qquad
%0<\delta_1<\alpha,\qquad
%\dots,
\\
  \gamma_0>\gamma_1>\dotsb
\end{gather*}
ko\c sullar\i n\i\ sa\u glayan $\gamma_i$ ve $\delta_i$ ordinalleri vard\i r.  
Ayr\i ca, $\on$ iyi\-s\i ralan\-m\i\c s oldu\u gundan, 
kesin azalan $(\gamma_0,\gamma_1,\dots)$ dizisi sona ermelidir.  
Yani bir $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
\beta=\alpha^{\gamma_0}\cdot\delta_0
+\alpha^{\gamma_1}\cdot\delta_1+\dots+\alpha^{\gamma_n}\cdot\delta_n.
\end{equation*}
Buradaki 
$\{(\gamma_0,\delta_0),(\gamma_1,\delta_1),\dots,(\gamma_n,\delta_n)\}$ 
k\"umesi, bir g\"ondermedir, ve \Teoremde{thm:base}
\begin{equation*}
\beta_{\alpha}
\end{equation*}
\glossary{$\beta_{\alpha}$}%
ad\i\ ona verilecek.%%%%%
\footnote{Bu $\beta_{\alpha}$ ifadesi, benimdir; 
ba\c ska kitaplarda g\"ormedim.}  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Genel olarak bu k\"umeyi tan\i mlamak
i\c cin (yani $\xi\mapsto\xi_{\alpha}$ g\"ondermesini tan\i mlamak i\c cin), 
\"ozyineleme y\"ontemini \Teoremde{thm:rec-ord}n
farkl\i\ bir \c sekilde kullanaca\u g\i z. 

\begin{theorem}\label{thm:gonderme}
Tan\i m s\i n\i f\i\ bir k\"ume olan her g\"onderme bir k\"umedir.
\end{theorem}

\ktk

Herhangi $\bm A$ s\i n\i f\i\ ve $b$ k\"umesi i\c cin
\begin{equation*}
{}^b\bm A,
\end{equation*}%
\glossary{${}^b\bm A$}
$b$ k\"umesinden $\bm A$ s\i n\i f\i na giden 
g\"ondermelerin s\i n\i f\i\ olsun.  

\begin{theorem}[Ordinaller \"Ozyinelemesi II]\label{thm:rec-ord-again}%
\index{teorem!Ordinaller \"Ozyinelemesi T---i}
$\bm A$ bir s\i n\i f olsun, ve
\begin{equation*}
\bm F\colon\{x\colon\Exists{\eta}x\in{}^{\eta}\bm A\}\to\bm A
\end{equation*}
olsun.%%%%%
\footnote{$\{x\colon\Exists{\eta}x\in{}^{\eta}\bm A\}$ s\i n\i f\i, 
$\bigcup_{\eta}{}^{\eta}\bm A$ olarak yaz\i labilir; 
ama ${}^{\beta}\bm A$ s\i n\i flar\i\, k\"ume olmayabilir.}   
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
O zaman $\on$ s\i n\i f\i ndan 
$\bm A$ s\i n\i f\i na giden ve her $\alpha$ ordinali i\c cin
\begin{equation*}
\bm G(\alpha)=\bm F(\bm G\restriction\alpha)
\end{equation*}
ko\c sulunu sa\u glayan bir ve tek bir $\bm G$ g\"ondermesi vard\i r.
\end{theorem}

\begin{proof}
  Her $\beta$ i\c cin $\beta'$ k\"umesinden $\bm A$ s\i n\i fa giden
  \begin{equation*}
    \alpha<\beta\lto g_{\beta}(\alpha)=\bm F(g_{\beta}\restriction\alpha)
  \end{equation*}
ko\c sulunu sa\u glayan bir ve tek bir $g_{\beta}$ g\"ondermesi 
oldu\u gunu g\"osterece\u giz.
$\bm B$, istedi\u gimiz \"ozelli\u gi olan 
$\beta$ ordinallerinin olu\c sturdu\u gu s\i n\i f olsun.
\Teoremin{thm:rec-ord} ikinci kan\i t\i ndaki gibi
$\gamma\included\bm B$ ise
$\theta\leq\delta<\gamma$ oldu\u gu zaman
\begin{equation*}
 g_{\theta}\included g_{\delta}. 
\end{equation*}
Bu durumda $h_{\gamma}=\bigcup_{\xi<\gamma}g_{\xi}$ olsun.
O zaman $h_{\gamma}\colon\gamma\to\bm A$,
ve $\delta<\gamma$ ise
$h_{\gamma}(\delta)=g_{\delta}(\delta)$,
ve
\begin{equation*}
g_{\gamma}
=h_{\gamma}\cup\{(\gamma,\bm F(h_{\gamma}))\}
\end{equation*}
olabilir ve olmal\i d\i r,
dolay\i s\i yla $\gamma\in\bm B$.
B\"oylece $\on\setminus\bm B$ fark\i n\i n en k\"u\c c\"uk eleman\i\ yoktur,
dolay\i s\i yla $\bm B=\on$.
Sonu\c c olarak $\bm G=\bigcup\{g_{\xi}\colon\xi\in\on\}$
olabilir ve olmal\i d\i r.
\end{proof}

\begin{theorem}\label{thm:base}
$\alpha\geq2$ olsun.
$0_{\alpha}=\emptyset$ ve $0$ olmayan her $\beta$ i\c cin
\begin{align}\label{eqn:base}
\alpha^{\gamma}\cdot\delta+\theta&=\beta,&
0<\delta&<\alpha,&
\theta&<\alpha^{\gamma}
\end{align}
ko\c sullar\i\ sa\u gland\i\u g\i\ zaman
\begin{equation*}
\beta_{\alpha}=\{(\gamma,\delta)\}\cup\theta_{\alpha}
\end{equation*}
ko\c sulunu sa\u glayan
bir ve tek bir $\xi\mapsto\xi_{\alpha}$ g\"ondermesi vard\i r.
Her $\beta$ i\c cin,
\begin{itemize}
\item
$\beta_{\alpha}$ k\"umesi,
de\u ger k\"umesi $\on$ s\i n\i f\i n\i n bir altk\"umesi olan 
bir g\"ondermedir;
\item
bir $n$ do\u gal say\i s\i\ i\c cin,
$\beta_{\alpha}$ g\"ondermesinin tan\i m k\"umesi,
tan\i m k\"umesi $n$ olan, 
kesin azalan bir $f$ g\"ondermesinin de\u ger k\"umesidir;
\item
$i<n$ ko\c sulu sa\u gland\i\u g\i\ zaman 
$f(i)=\gamma_i$ ve $\beta_{\alpha}\bigl(f(i)\bigr)=\delta_i$ ise
  \begin{equation*}
    \beta
=\alpha^{\gamma_0}\cdot\delta_0+\dots+\alpha^{\gamma_{n-1}}\cdot\delta_{n-1}.
  \end{equation*}
\end{itemize}
\end{theorem}

\begin{proof}
\Teoreme{thm:exp-sol} g\"ore, 
\eqref{eqn:base} ko\c sullar\i n\i\ sa\u glayan 
bir ve tek bir $(\gamma,\delta,\theta)$ \"u\c cl\"us\"u vard\i r.  
Ayr\i ca $\theta<\beta$.  
O zaman \"oyle bir $\bm F$ g\"ondermesi vard\i r ki 
her $\beta$ i\c cin, e\u ger $g$, 
tan\i m k\"umesi $\beta$ olan 
(ve de\u ger k\"umesi herhangi bir k\"ume olan) bir g\"onderme, 
ve $(\gamma,\delta,\theta)$ \"u\c cl\"us\"u, 
\eqref{eqn:base} sat\i r\i ndaki gibiyse, o zaman
\begin{equation*}
\bm F(g)=\{(\gamma,\delta)\}\cup g(\theta).
\end{equation*}
\"Oyleyse istedi\u gimiz $\beta\mapsto\beta_{\alpha}$ g\"ondermesi, 
son teoreme g\"ore
\begin{equation*}
\bm G(\beta)=\bm F(\bm G\restriction\beta)
\end{equation*}
ko\c sulunu sa\u glayan $\bm G$ g\"ondermesidir.
\end{proof}

\section%[$\upomega$ taban\i]
{$\upomega$ taban\i\ (Cantor normal bi\c cimi)}

\Teoreme{thm:base} g\"ore, 
her $\alpha$ i\c cin, bir $n$ do\u gal say\i s\i\ i\c cin,
tan\i m k\"umesi $n$ olan,
%de\u ger k\"umesi 
%$\on\times(\upomega\setminus\{0\})$ \c carp\i m\i n\i n bir altk\"umesi olan, 
%ve
\begin{align*}
  \alpha_0&>\dots>\alpha_{n-1},&
\{a_0,\dots,a_{n-1}\}&\included\upomega\setminus\{0\}
\end{align*}
ko\c sullar\i n\i\ sa\u glayan bir $x\mapsto(\alpha_x,a_x)$ g\"ondermesi i\c cin
\begin{equation}\label{eqn:Cnf}
  \alpha=\upomega^{\alpha_0}\cdot a_0+\upomega^{\alpha_1}\cdot a_1+\dots
+\upomega^{\alpha_n}\cdot a_n.
\end{equation}
Sayfa \sayfanumarada{Cnf} dedi\u gimiz gibi bu toplam, $\beta$ ordinalinin
\textbf{Cantor normal bi\c cimidir}%
\index{Cantor!--- normal bi\c cimi}%
\index{normal!Cantor --- bi\c cimi}
(\eng{Cantor normal form}).  
Bu b\"ol\"umde,
Cantor normal bi\c cimleriyle hesaplama kurallar\i n\i\ kuraca\u g\i z.
\.Ilk olarak \eqref{eqn:Cnf} sat\i r\i ndaki $\alpha_0$ \"uss\"une
$\alpha$ ordinalinin \textbf{derecesi}%
\index{derece}
(\eng{degree}) denecektir, ve bu derece
\begin{equation*}
\deg(\alpha)
\end{equation*}%
\glossary{$\deg(\alpha)$}
olarak yaz\i lacakt\i r.

\subsection{Toplama}

Cantor normal bi\c ciminde
\begin{gather*}
  \alpha=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_m}\cdot a_m,\\
\beta=\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n
\end{gather*}
olsun.
O zaman ordinaller toplamas\i\ birle\c smeli oldu\u gundan
\begin{equation}\label{eqn:a+b}
  \alpha+\beta=
\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_m}\cdot a_m
+\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n.
\end{equation}
Ama $\alpha+\beta$ toplam\i n\i n Cantor normal bi\c cimi nedir?
\begin{compactitem}
  \item
$\alpha_m>\beta_0$ ise
\eqref{eqn:a+b} sat\i r\i ndaki ifade 
zaten $\alpha+\beta$ toplam\i n\i n Cantor normal bi\c cimidir.
\item
$\alpha_m=\beta_0$ ise
ordinaller \c carpmas\i n\i n soldan da\u g\i lmas\i n\i\ kullanarak
\begin{multline*}
  \alpha+\beta=
\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_{m-1}}\cdot a_{m-1}\\
{}+\upomega^{\alpha_m}\cdot(a_m+b_0)
+\upomega^{\beta_1}\cdot b_1+\dots+\upomega^{\beta_n}\cdot b_n,
\end{multline*}
ve bu ifade $\alpha+\beta$ toplam\i n\i n Cantor normal bi\c cimidir.
\item
$\alpha_m<\beta_0$ ise
\begin{equation*}
  \alpha+\beta=
\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_{m-1}}\cdot a_{m-1}
+\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n
\end{equation*}
e\c sitli\u gini g\"osterece\u giz.
\end{compactitem}

\begin{theorem}
$\alpha>0$ ise
\begin{equation*}
1+\upomega^{\alpha}=\upomega^{\alpha}.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}\label{thm:deg-a<deg-b}
$\deg(\alpha)<\deg(\beta)$ ise
\begin{equation*}
\alpha+\beta=\beta.
\end{equation*}
\end{theorem}

\begin{proof}
$\alpha<\beta$ ise $\upomega^{\alpha}+\upomega^{\beta}=\upomega^{\beta}$ e\c sitli\u gini kan\i tlayaca\u g\i z.  Bu durumda bir $\gamma$ i\c cin, $\beta=\alpha+\gamma$ ve $\gamma>0$, dolay\i s\i yla
\begin{equation*}
\upomega^{\alpha}+\upomega^{\beta}
=\upomega^{\alpha}+\upomega^{\alpha+\gamma}
=\upomega^{\alpha}\cdot(1+\upomega^{\gamma})
=\upomega^{\alpha}\cdot\upomega^{\gamma}
=\upomega^{\alpha+\gamma}
=\upomega^{\beta}.\qedhere
\end{equation*}
\end{proof}

\"Orne\u gin
\begin{multline*}
(\upomega^{\upomega^2}+\upomega^{\upomega\cdot4+7}\cdot5+\upomega^{\upomega\cdot2}\cdot 7+2)
+(\upomega^{\upomega\cdot4+7}\cdot 8+\upomega\cdot 3+16)\\
=\upomega^{\upomega^2}+\upomega^{\upomega\cdot4+7}\cdot13+\upomega\cdot3+16.
\end{multline*}

\subsection{\c Carpma}

Tekrar
Cantor normal bi\c ciminde
\begin{gather*}
  \alpha=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_m}\cdot a_m,\\
\beta=\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n
\end{gather*}
olsun.
O zaman
\begin{equation*}
  \alpha\cdot\beta
=\alpha\cdot\upomega^{\beta_0}\cdot b_0+\dots+\alpha\cdot\upomega^{\beta_n}\cdot b_n.
\end{equation*}
\c Simdi $\alpha\cdot\upomega^{\beta_k}\cdot b_k$ \c carp\i mlar\i n\i n 
Cantor normal bi\c cimini bulaca\u g\i z.
\.Iki durum vard\i r.
\begin{compactenum}
  \item
$\beta_k=0$ olsun.
O zaman $\alpha\cdot\upomega^{\beta_k}\cdot b_k
=\alpha\cdot b_k$.
\c Simdi $\alpha$, 
\begin{equation*}
  \upomega^{\alpha_0}\cdot a_0
+(\upomega^{\alpha_1}\cdot a_1+\dots +\upomega^{\alpha_m}\cdot a_m)
\end{equation*}
olarak d\"u\c s\"un\"ulebilir.
Burada $\upomega^{\alpha_0}\cdot a_0$ \c carp\i m\i, 
$\alpha$ ordinalinin \textbf{``ba\c s\i d\i r''}%
\index{ba\c s}
(\eng{head}) ve
$\upomega^{\alpha_1}\cdot a_1+\dots +\upomega^{\alpha_m}\cdot a_m$ toplam\i,
$\alpha$ ordinalinin \textbf{``kuyru\u gudur''}%
\index{kuyruk}
(\eng{tail}).
Bu kuyruk, $\gamma$ olsun.
O zaman \Teoreme{thm:deg-a<deg-b} g\"ore 
\begin{equation*}
  \gamma+\upomega^{\alpha_0}=\upomega^{\alpha_0}, 
\end{equation*}
dolay\i s\i yla
\begin{align*}
  \alpha\cdot b_k
&=(\upomega^{\alpha_0}\cdot a_0+\gamma)\cdot b_k\\
&=(\upomega^{\alpha_0}\cdot a_0+\gamma)+\dots+(\upomega^{\alpha_0}\cdot a_0+\gamma)\\
&=\upomega^{\alpha_0}\cdot a_0
+(\gamma+\upomega^{\alpha_0}\cdot a_0)+\cdots
+(\gamma+\upomega^{\alpha_0}\cdot a_0)+\gamma\\
&=\upomega^{\alpha_0}\cdot a_0
+\upomega^{\alpha_0}\cdot a_0+\cdots+\upomega^{\alpha_0}\cdot a_0+\gamma\\
&=\upomega^{\alpha_0}\cdot a_0\cdot b_k+\gamma\\
&=\upomega^{\alpha_0}\cdot a_0\cdot b_k
+\upomega^{\alpha_1}\cdot a_1+\dots +\upomega^{\alpha_m}\cdot a_m.
\end{align*}
\item
$\beta_k\geq1$ olsun.  O zaman bir $\delta$ i\c cin $\beta_k=1+\delta$,
dolay\i s\i yla
\begin{equation*}
  \alpha\cdot\upomega^{\beta_k}\cdot b_k
=\alpha\cdot\upomega\cdot\upomega^{\delta}\cdot b_k.
\end{equation*}
Ayr\i ca
\begin{align*}
  \alpha\cdot\upomega
&=\sup_{x<\upomega}(\alpha\cdot x)\\
&=\sup_{x<\upomega}(\upomega^{\alpha_0}\cdot a_0\cdot x+\gamma)\\
&\leq\sup_{x<\upomega}\bigl(\upomega^{\alpha_0}\cdot a_0\cdot(x+1)\bigr)\\
&=\upomega^{\alpha_0}\cdot\sup_{x<\upomega}\bigl(a_0\cdot(x+1)\bigr)\\
&=\upomega^{\alpha_0}\cdot\upomega\\
&=\upomega^{\alpha_0+1},
\end{align*}
ve benzer \c sekilde
\begin{equation*}
  \upomega^{\alpha_0+1}
=\sup_{x<\upomega}(\upomega^{\alpha_0}\cdot x)
\leq\sup_{x<\upomega}(\alpha\cdot x)
=\alpha\cdot\upomega,
\end{equation*}
dolay\i s\i yla $\alpha\cdot\upomega=\upomega^{\alpha_0+1}$ ve
\begin{equation*}
  \alpha\cdot\upomega^{\beta_k}\cdot b_k=\upomega^{\alpha_0+\beta_k}\cdot b_k.
\end{equation*}
\end{compactenum}
Buldu\u gumuz kurallar\i,
resmi teoremler olarak yaz\i p kan\i tlayal\i m.

\begin{theorem}
$\alpha>\deg(\beta)$ ve $1\leq k<\upomega$ ve $1\leq n<\upomega$ ise
\begin{equation*}
(\upomega^{\alpha}\cdot k+\beta)\cdot n=\upomega^{\alpha}\cdot k\cdot n+\beta.
\end{equation*}
\end{theorem}

\begin{proof}
$n=1$ durumunda iddia do\u grudur.  $n=m$ durumunda do\u gru ise
\begin{align*}
(\upomega^{\alpha}\cdot k+\beta)\cdot(m+1)
&=(\upomega^{\alpha}\cdot k+\beta)\cdot m+\upomega^{\alpha}\cdot k+\beta\\
&=\upomega^{\alpha}\cdot k\cdot m+\beta+\upomega^{\alpha}\cdot k+\beta\\
&=\upomega^{\alpha}\cdot k\cdot m+\upomega^{\alpha}\cdot k+\beta\\
&=\upomega^{\alpha}\cdot k\cdot(m+1)+\beta,
\end{align*}
dolay\i s\i yla $n=m+1$ durumunda da do\u grudur.  T\"umevar\i mdan $1\leq n<\upomega$ ise iddia do\u grudur.
\end{proof}

\"Orne\u gin
\begin{equation*}
(\upomega^{\upomega}\cdot2+\upomega+5)\cdot7
=\upomega^{\upomega}\cdot14+\upomega+5.
\end{equation*}

\begin{theorem}
$1\leq\alpha$ ve $0<\beta$ ise
\begin{equation*}
\alpha\cdot\upomega^{\beta}=\upomega^{\deg(\alpha)+\beta}.
\end{equation*}
\end{theorem}

\begin{proof}
\"Once $\beta=1$ durumunda iddiay\i\ kan\i tlayaca\u g\i z.  $1\leq \alpha<\upomega$ ise $\deg(\alpha)=0$, dolay\i s\i yla
\begin{equation*}
\alpha\cdot\upomega^1
=\alpha\cdot\upomega
=\sup_{x\in\upomega}\alpha\cdot x=\upomega=\upomega^{\deg(\alpha)+1}.
\end{equation*}
\c Simdi $\upomega\leq\alpha$ olsun.  O zaman
\begin{align*}
\alpha&=\upomega^{\gamma}\cdot k+\delta,&
\alpha&>\deg(\gamma),&
1&\leq k<\upomega
\end{align*}
ko\c sullar\i n\i\ sa\u glayan $\gamma$, $k$, ve $\delta$ vard\i r.
O halde $1\leq n<\upomega$ ise
\begin{gather*}
	\upomega^{\gamma}\cdot k\leq\alpha\leq\alpha\cdot n
	=\upomega^{\gamma}\cdot k\cdot n+\delta
	<\upomega^{\gamma}\cdot(k\cdot n+1)<\upomega^{\gamma+1},\\
	\upomega^{\gamma+1}
	=\sup_{1\leq\xi<\upomega}(\upomega^{\gamma}\cdot\xi)
	\leq\sup_{1\leq\xi<\upomega}(\alpha\cdot\xi)
	\leq\upomega^{\gamma+1},\\
	\alpha\cdot\upomega=\upomega^{\gamma+1}=\upomega^{\deg(\alpha)+1}.
\end{gather*}
Genelde $\alpha\geq1$ ve $\beta\geq1$ ise, bir $\theta$ i\c cin $\beta=1+\theta$, dolay\i s\i yla
\begin{equation*}
\alpha\cdot\upomega^{\beta}=\alpha\cdot\upomega\cdot\upomega^{\theta}
=\upomega^{\deg(\alpha)+1+\theta}
=\upomega^{\deg(\alpha)+\beta}.\qedhere
\end{equation*}
\end{proof}

\"Orne\u gin
\begin{gather*}
5\cdot(\upomega^2\cdot 3+\upomega\cdot16+7)=\upomega^2\cdot 3+\upomega\cdot16+35,\\
(\upomega^{\upomega}\cdot2+\upomega+5)
\cdot(\upomega^2\cdot 3+\upomega\cdot16)
=\upomega^{\upomega+2}\cdot 3+\upomega^{\upomega+1}\cdot16,
\end{gather*}
ve
\begin{multline*}
(\upomega^{\upomega}\cdot2+\upomega+5)
\cdot(\upomega^2\cdot 3+\upomega\cdot16+7)\\
=\upomega^{\upomega+2}\cdot 3+\upomega^{\upomega+1}\cdot16+\upomega^{\upomega}\cdot14+\upomega+5.
\end{multline*}

\subsection{Kuvvet alma}

Tekrar
Cantor normal bi\c ciminde
\begin{gather*}
  \alpha=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_m}\cdot a_m,\\
\beta=\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n
\end{gather*}
olsun.
O zaman
\begin{equation*}
  \alpha^{\beta}
=\alpha^{\upomega^{\beta_0}\cdot b_0}\cdots\alpha^{\upomega^{\beta_n}\cdot b_n}.
\end{equation*}
\c Simdi $\alpha^{\upomega^{\beta_k}\cdot b_k}$ kuvvetlerini bulaca\u g\i z.
\.Iki durum vard\i r.
\begin{compactenum}
  \item
$\beta_k=0$ ise $\alpha^{\upomega^{\beta_k}\cdot b_k}=\alpha^{b_k}=\alpha\cdots\alpha$.
Ayr\i ca $m>0$ 
ve $\gamma=\upomega^{\alpha_1}\cdot a_1+\dots+\upomega^{\alpha_m}\cdot a_m$ ise
\begin{gather*}
  \begin{aligned}
\alpha^2
&=\alpha\cdot\alpha\\
&=\alpha\cdot(\upomega^{\alpha_0}\cdot a_0+\gamma)\\
&=\alpha\cdot\upomega^{\alpha_0}\cdot a_0+\alpha\cdot\gamma\\
&=(\upomega^{\alpha_0}\cdot a_0+\gamma)\cdot\upomega^{\alpha_0}\cdot a_0
+\alpha\cdot\gamma\\
&=\upomega^{\alpha_0+\alpha_0}\cdot a_0+\alpha\cdot\gamma\\
&=\upomega^{\alpha_0\cdot2}\cdot a_0+\alpha\cdot\gamma,
  \end{aligned}\\
  \begin{aligned}
\alpha^3
&=\alpha\cdot\alpha^2\\
&=\alpha\cdot(\upomega^{\alpha_0\cdot2}\cdot a_0+\alpha\cdot\gamma)\\
&=\alpha\cdot\upomega^{\alpha_0\cdot2}\cdot a_0+\alpha\cdot\alpha\cdot\gamma\\
&=(\upomega^{\alpha_0}\cdot a_0+\gamma)\cdot\upomega^{\alpha_0\cdot2}\cdot a_0
+\alpha\cdot\alpha\cdot\gamma\\
&=\upomega^{\alpha_0+\alpha_0\cdot2}\cdot a_0
+\alpha\cdot\alpha\cdot\gamma\\
&=\upomega^{\alpha_0\cdot3}\cdot a_0+\alpha^2\cdot\gamma,
  \end{aligned}
\end{gather*}
ve genelde
\begin{equation*}
  \alpha^{n+1}=\upomega^{\alpha_0\cdot(n+1)}\cdot a_0+\alpha^n\cdot\gamma.
\end{equation*}
\item
$\beta_k\geq1$ olsun.  O zaman bir $\delta$ i\c cin $\beta_k=1+\delta$,
dolay\i s\i yla
\begin{equation*}
  \alpha^{\upomega^{\beta_k}\cdot b_k}
=\left(\alpha^{\upomega}\right)^{\upomega^{\delta}\cdot b_k}.
\end{equation*}
\c Simdi iki durum daha vard\i r.
\begin{compactenum}
  \item
 $\alpha$ sonlu (yani $\alpha_0=0$) olsun.
O zaman
\begin{equation*}
  \alpha^{\upomega}
=a_0{}^{\upomega}
=\sup_{x<\upomega}(a_0{}^x)=\upomega,
\end{equation*}
dolay\i s\i yla
\begin{equation*}
  \alpha^{\upomega^{\beta_k}\cdot b_k}=\upomega^{\upomega^{\delta}\cdot b_k}.
\end{equation*}
\item
$\alpha\geq\upomega$ (yani $\alpha_0>0$) olsun.
O zaman
\begin{align*}
  \alpha^{\upomega}
&=\lim_{x<\upomega}(\alpha^x)\\
&=\lim_{x<\upomega}(\upomega^{\alpha_0\cdot x}\cdot a_0+\alpha^{n-1}\cdot\gamma)\\
&\leq\lim_{x<\upomega}\left(\upomega^{\alpha_0\cdot(x+1)}\right)\\
&=\upomega^{\sup_{x<\upomega}\bigl(\alpha_0\cdot(x+1)\bigr)}\\
&=\upomega^{\alpha_0\cdot\upomega},
\end{align*}
ve
\begin{equation*}
  \upomega^{\alpha_0\cdot\upomega}\leq\alpha^{\upomega},
\end{equation*}
dolay\i s\i yla $\alpha^{\upomega}=\upomega^{\alpha_0\cdot\upomega}$ ve
\begin{equation*}
  \alpha^{\upomega^{\beta_k}\cdot b_k}=\upomega^{\alpha_0\cdot\upomega^{\beta_k}\cdot b_k}.
\end{equation*}
\end{compactenum}
\end{compactenum}

\.Iki teorem \c c\i kar.

\begin{theorem}
$1\leq k<\upomega$ ve $n<\upomega\leq\alpha$ ise
\begin{align*}
k^{\upomega^{n+1}}&=\upomega^{\upomega^n},&
k^{\upomega^{\alpha}}&=\upomega^{\upomega^{\alpha}}.
\end{align*}
\end{theorem}

\ktk[ \emph{\.Ipucu:}  $n+1=1+n$ ve $\alpha=1+\alpha$.]

\"Orne\u gin
\begin{equation*}
2^{\upomega^{\upomega}\cdot3+\upomega^5\cdot4+\upomega\cdot7+5}
=\upomega^{\upomega^{\upomega}\cdot3+\upomega^4\cdot4+7}\cdot32.
\end{equation*}

\begin{theorem}
$\upomega\leq\alpha$ ve $\beta$ limit ve $n<\upomega$ ise
\begin{equation*}
\alpha^{\beta+n}=\upomega^{\deg(\alpha)\cdot\beta}\cdot\alpha^n.
\end{equation*}
\end{theorem}

\ktk[ \emph{\.Ipucu:}  \"Once $\alpha^{\upomega}=\upomega^{\deg(\alpha)\cdot\upomega}$ denkli\u gini kan\i tlay\i n.]

\"Orne\u gin
\begin{multline*}
(\upomega^{\upomega+1}+\upomega^2+1)^{\upomega^2+\upomega\cdot3+2}\\
\begin{aligned}
&=\upomega^{(\upomega+1)\cdot(\upomega^2+\upomega\cdot3)} \cdot(\upomega^{\upomega+1}+\upomega^2+1)^2\\
&=\upomega^{\upomega^3+\upomega^2\cdot3}\cdot(\upomega^{\upomega+1}+\upomega^2+1)^2\\
&=\upomega^{\upomega^3+\upomega^2\cdot3} \cdot(\upomega^{\upomega+1+\upomega+1}+\upomega^{\upomega+1+2}+\upomega^{\upomega+1}+\upomega^2+1)\\
&=\upomega^{\upomega^3+\upomega^2\cdot3} 
\cdot(\upomega^{\upomega\cdot2+1}
+\upomega^{\upomega+3}
+\upomega^{\upomega+1}
+\upomega^2+1)\\
&=\upomega^{\upomega^3+\upomega^2\cdot3+\upomega\cdot2+1}
+\upomega^{\upomega^3+\upomega^2\cdot3+\upomega+3}
\end{aligned}\\
+\upomega^{\upomega^3+\upomega^2\cdot3+\upomega+1}
+\upomega^{\upomega^3+\upomega^2\cdot3+2}
+\upomega^{\upomega^3+\upomega^2\cdot3}.
\end{multline*}


\chapter{Kardinaller}

\section{\.Iyis\i ralanm\i\c s k\"umeler}

\Teoremin{thm:sinif} daha genel bi\c cimi vard\i r:

\begin{theorem}\label{thm:class-embed}
Bir s\i n\i f, bir k\"umeye g\"om\"ulebilirse, bu s\i n\i f da bir k\"umedir.
\end{theorem}

\ktk

\begin{theorem}
T\"um $a$ ile $b$ k\"umeleri i\c cin $a\times b$ \c carp\i m\i, bir k\"umedir.
\end{theorem}

\begin{proof}
Yerle\c stirme Aksiyomuna g\"ore her $c$ i\c cin $a\times\{c\}$ bir k\"umedir, 
dolay\i s\i yla 
bir $\bigl\{a\times\{x\}\colon x\in b\bigr\}$ s\i n\i f\i\ vard\i r, 
ve bu s\i n\i f da bir k\"umedir.  Ayr\i ca
\begin{equation*}
a\times b=\bigcup\bigl\{a\times\{x\}\colon x\in b\bigr\},
\end{equation*}
ve Bile\c sim Aksiyomuna g\"ore bu bile\c sim, bir k\"umedir.
\end{proof}

E\u ger bir $a$ k\"umesi, 
bir $\bm S$ ba\u g\i nt\i s\i\ taraf\i ndan iyi\-s\i ran\i rsa,
o zaman $\bm S\cap(a\times a)$ ba\u g\i nt\i s\i\ da taraf\i ndan 
iyi\-s\i ralan\i r, 
ve asl\i nda $a$ k\"umesinin t\"um $b$ ve $c$ elemanlar\i\ i\c cin
\begin{equation*}
b\mathrel{\bm S}c\liff b\mathrel{\bigl(\bm S\cap(a\times a)\bigr)}c.  
\end{equation*}
Son teoreme g\"ore $\bm S\cap(a\times a)$ bir k\"ume oldu\u gundan
$\bigl(a,\bm S\cap(a\times a)\bigr)$ s\i ral\i\ ikilisi
olu\c sturulabilir,
ve bu s\i ral\i\ ikilisine
\textbf{iyi\-s\i ralanm\i\c s k\"ume}%
\index{iyi\-s\i ralanm\i\c s k\"ume}
(\eng{well ordered set})
denebilir.
O halde $\bigl(a,\bm S\cap(a\times a)\bigr)$,
\begin{equation*}
(a,\bm S)
\end{equation*}
olarak yaz\i labilir.  
Mesela her $\alpha$ ordinali i\c cin
$(\alpha,\in)$, iyi\-s\i ralanm\i\c s bir k\"umedir.
Asl\i nda $\alpha$, $(\alpha,\in)$ olarak d\"u\c s\"un\"ur.
Genelde $(a,\bm S)$ bir iyi\-s\i ralanm\i\c s k\"ume 
ve $\emptyset\pincluded c\included a$ ise,
o zaman $\bm S$ ba\u g\i nt\i sina g\"ore
$c$ k\"umesinin en k\"u\c c\"uk eleman\i
\begin{equation*}
  \min_{\bm S}(c)
\end{equation*}
olarak yaz\i labilir.

\c Simdi $(a,\bm S)$ ve $(b,\bm T$), iyi\-s\i ralanm\i\c s k\"ume olsun,
ve $h\colon a\to b$ olsun.
E\u ger $a$ k\"umesinin t\"um $c$ ve $d$ elemanlar\i\ i\c cin
\begin{equation*}
  c\mathrel{\bm S}d\lto h(c)\mathrel{\bm T}h(d)
\end{equation*}
ise,
o zaman (\Teoremde{thm:<}ki gibi) $h$,
\textbf{kesin artand\i r}%
\index{artan g\"onderme}
(\eng{strictly increasing}).
Bu durumda $h$, $b$ k\"umesini \"orterse,
o zaman $h$,
$(a,\bm S)$ iyi\-s\i ralanm\i\c s k\"umesinden
$(b,\bm T)$ iyi\-s\i ralanm\i\c s k\"umesine giden
bir \textbf{e\c s\-yap\i\ e\c slemesi}%
\index{e\c syap\i} \cite{Nesin-AKK}
veya
\textbf{izomorfizmdir}%
\index{izomorfizm}
(\eng{isomorphism}),
ve
\begin{equation*}
  h\colon(a,\bm S)\xrightarrow{\cong}(b,\bm T)
\end{equation*}
ifadesini yazabiliriz.
$(a,\bm S)$ iyi\-s\i ralanm\i\c s k\"umesinden
$(b,\bm T)$ iyi\-s\i ra\-lanm\i\c s k\"umesine giden
bir e\c syap\i\ e\c slemesi varsa,
\begin{equation*}
  (a,\bm S)\cong(b,\bm T)
\end{equation*}
ifadesini yazabiliriz.

\begin{theorem}
$\cong$ ba\u g\i nt\i s\i, bir denklik s\i n\i f\i d\i r, ve
\begin{equation*}
 h\colon(a,\bm S)\xrightarrow{\cong}(b,\bm T)\lto
\inv h\colon(b,\bm T)\xrightarrow{\cong}(a,\bm S).
\end{equation*}
\end{theorem}

\ktk

E\u ger $(a,\bm S)\cong(b,\bm T)$ ise,
bu iki iyi\-s\i ralanm\i\c s k\"ume
birbirine \textbf{e\c syap\i\-sald\i r}
(\eng{isomorphic}).

\begin{theorem}
$(a,\bm S)$ bir iyis\i ralanm\i\c s k\"ume,
$b$ bir k\"ume, $\bm T$ bir ba\u g\i nt\i,
ve $f\colon a\to b$ olsun.
E\u ger $a$ k\"umesinin t\"um $c$ ve $d$ elemanlar\i\ i\c cin
  \begin{equation*}
    c\mathrel{\bm S}d\liff f(c)\mathrel{\bm T}f(d)
  \end{equation*}
ise, o zaman $(f[a],\bm T)$ bir iyis\i ralanm\i\c s k\"umedir
ve
\begin{equation*}
 f\colon(a,\bm S)\xrightarrow{\cong}(f[a],\bm T). 
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}[\.Iyis\i ralanm\i\c s \"Ozyineleme]\label{thm:s}
Her $(a,<)$ iyis\i ralanm\i\c s\linebreak k\"umesi i\c cin,
bir ve tek bir $\alpha$ i\c cin,
\begin{equation*}
  (a,<)\cong(\alpha,\in).
\end{equation*}
Ayr\i ca bir ve tek bir $f$ i\c cin
\begin{equation}\label{eqn:f}
  f\colon(a,<)\xrightarrow{\cong}(\alpha,\in).
\end{equation}
Asl\i nda $a$ k\"umesinin t\"um $b$ eleman\i\ i\c cin
\begin{equation}\label{eqn:ba}
f(b)=\{f(x)\colon x\in a\land x<b\}.
\end{equation}
\end{theorem}

\begin{proof}
E\u ger \eqref{eqn:f} c\"umlesi do\u gru, $\{b,c\}\included a$, ve $c<b$ ise,
o zaman $f(c)\in f(b)$,
dolay\i s\i yla,
$f$ \"orten de oldu\u gundan, 
\eqref{eqn:ba} c\"umlesi de do\u grudur.

\c Simdi
\ref{thm:rec-ord-again} numaral\i\
\"Ozyineleme Teoremindeki gibi
$a$ k\"umesinin her $b$ eleman\i\ i\c cin
tan\i m k\"umesi $\{x\in a\colon x\leq b\}$ olan
\begin{equation*}
  c\in x\land c\leq b\lto f_b(c)=\{f_b(x)\colon x\in a\land x<c\}
\end{equation*}
ko\c sulunu sa\u glayan 
bir ve tek bir $f_b$ g\"ondermesinin oldu\u gunu g\"osterece\u giz.
O zaman $a_0$,
istedi\u gimiz \"ozelli\u gi olan
$a$ k\"umesinin $b$ elemanlar\i n\i n olu\c sturdu\u gu k\"ume olsun.
E\u ger $\{x\in a\colon x<b\}\included a_0$ ise
$\{c,d\}\included a$ ve $d<c<b$ oldu\u gu zaman
  \begin{equation*}
    f_d\included f_c.
  \end{equation*}
Bu durumda $x\mapsto f_x(x)$,
tan\i m k\"umesi $\{x\in a\colon x<b\}$ olan bir $g_b$ g\"ondermedir, ve
\begin{equation*}
  f_b=g_b\cup\{(b,\{f_x(x)\colon x\in a\land x<b\})\}
\end{equation*}
olabilir ve olmal\i d\i r,
dolay\i s\i yla $b\in a_0$.
B\"oylece $a\setminus a_0$ fark\i n\i n en k\"u\c c\"uk eleman\i\ yoktur,
dolay\i s\i yla $a_0=a$.
Sonu\c c olarak $f$, tan\i m k\"umesi $a$ olan
$x\mapsto f_x(x)$ g\"ondermesi
olabilir ve olmal\i d\i r.
Ayr\i ca $a$ k\"umesinin t\"um $b$ ve $c$ elemanlar\i\ i\c cin
\begin{equation*}
c<b\liff f(c)\in f(b),
\end{equation*}
dolay\i s\i yla,
son teorem sayesinde, $(f[a],\in)$ bir iyis\i ralanm\i\c s k\"umedir ve
\begin{equation*}
  f\colon(a,<)\xrightarrow{\cong}(f[a],\in).
\end{equation*}
Ayr\i ca $b\in a$ ise $f(b)\included f[a]$,
ve sonu\c c olarak $f[a]\in\on$.
\end{proof}

$(a,\bm S)$, bir iyi\-s\i ralanm\i\c s k\"ume ise, 
ve $(a,\bm S)\cong(\alpha,\in)$ ise,
o zaman $\alpha$, $(a,\bm S)$ ikilisinin \textbf{ordinalidir}%
\index{ordinal}
(\eng{$\alpha$ is the ordinal of $(a,\bm S)$}), ve
\begin{equation*}
\alpha=\ord(a,\bm S)
\end{equation*}%
\glossary{$\ord(a,\bm S)$}
ifadesini yazabiliriz.  O halde
\begin{equation*}
a\approx\ord(a,\bm S).
\end{equation*}
\"Ozel olarak $a$ k\"umesinin kardinali vard\i r
(\sayfaya{kardinal} bak\i n).

\section{Say\i labilme}


$\bm F\colon\bm A\to\bm B$ ama e\c sleme de\u gilse,
daha zay\i f ko\c sullar sa\u glayabilir.
E\u ger
\begin{equation*}
\bm F[\bm A]=\bm B  
\end{equation*}
ise, o zaman $\bm F$, $\bm B$ s\i n\i f\i
n\i\ \textbf{\"orten}% 
\index{\"orten}
bir g\"ondermedir
(\eng{$\bm F$ is onto $\bm B$}), ve
\begin{equation*}
\bm F\colon\bm A\twoheadrightarrow\bm B
\end{equation*}%
\glossary{$\bm F\colon\bm A\twoheadrightarrow\bm B$}
ifadesini yazabiliriz.
E\u ger
\begin{equation*}
\Forall x\Forall y(x\in\bm A\land y\in\bm A\lto\bm F(x)=\bm F(y)\lto x=y)
\end{equation*}
ise, o zaman $\bm F$, \textbf{birebir}%
\index{g\"onderme!birebir ---}
(\eng{one-to-one}) veya
\textbf{injektif}
\index{g\"onderme!injektif ---}
(\eng{injective})
bir g\"ondermedir; ayr\i ca $\bm F$, bir \textbf{g\"ommedir}%
\index{g\"omme}
(\eng{embedding}).  Bu durumda
\begin{equation*}
  \bm F\colon\bm A\xrightarrow{\preccurlyeq}\bm B
\end{equation*}
\glossary{$\bm F\colon\bm A\xrightarrow{\preccurlyeq}\bm B$}%
ifadesini yazabiliriz.%%%%%
\footnote{$\bm F\colon\bm A\rightarrowtail\bm B$ ifadesi de m\"umk\"und\"ur.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{theorem}
Bir g\"onderme bir e\c slemedir ancak ve ancak birebir ve \"ortendir:
\begin{equation*}
  \bm F\colon\bm A\xrightarrow{\approx}\bm B
\liff
\bm F\colon\bm A\twoheadrightarrow\bm B
\land\bm F\colon\bm A\xrightarrow{\preccurlyeq}\bm B.
\end{equation*}
\end{theorem}

\ktk

Her $a$ \emph{k\"umesi} i\c cin,
%\Teoreme{thm:gonderme} g\"ore,
$a$ k\"umesinden $\bm B$ s\i n\i f\i na giden g\"ommeler,
${}^a\bm B$ s\i n\i f\i n\i n bir alt s\i n\i f\i n\i\ olu\c sturur;
bu alt s\i n\i f bo\c s de\u gilse,
\begin{equation*}
a\preccurlyeq\bm B
\end{equation*}%
\glossary{$a\preccurlyeq\bm B$}
ifadesini yazar\i z.

\begin{theorem}\label{thm:preccurlyeq}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{align*}
&a\preccurlyeq a,&
a\preccurlyeq b\land b\preccurlyeq c&\lto a\preccurlyeq c,&
a\approx b&\lto a\preccurlyeq b
\end{align*}
c\"umleleri do\u grudur.
\end{theorem}

\ktk

\begin{theorem}
$\upomega$, bir kardinaldir, yani
\begin{equation*}
\upomega\in\cn.
\end{equation*}
\end{theorem}

\begin{proof}
T\"umevar\i mla her $m$ do\u gal say\i s\i\ i\c cin 
$f\colon m+1\xrightarrow{\preccurlyeq}\upomega$ ise 
$f[m+1]$ k\"umesinin en b\"uy\"uk $n$ eleman\i\ vard\i r, 
dolay\i s\i yla $n+1\in\upomega\setminus f[m]$; 
\"ozel olarak $f$, $\upomega$ k\"umesini \"orten de\u gildir.
\end{proof}

Teoremin sonucu olarak $\upomega$ k\"umesiyle e\c slenik her k\"ume, sonsuzdur.
\"Oyle bir k\"ume,
\textbf{say\i labilir sonsuzluktad\i r}
(\eng{countably infinite}).
Sonlu veya say\i labilir sonsuzluktaki bir k\"ume,
\textbf{say\i labilir}%
\index{say\i labilir, say\i lamaz}
(\eng{countable}).
Di\u ger k\"umeler ve s\i n\i flar,
\textbf{say\i lamaz sonsuzluktad\i r} 
(\eng{uncountably infinite})
veya \textbf{say\i lamaz}
(\eng{uncountable}).  
O zaman k\"ume olmayan her s\i n\i f, say\i lamaz.

\begin{theorem}\label{thm:cntbl}
Bo\c s olmayan bir $a$ k\"umesi i\c cin a\c sa\u g\i daki ko\c sullar, 
birbirine denktir.
\begin{compactenum}
\item\label{item:countable}
$a$ say\i labilir.
\item\label{item:covered}
$\upomega$ k\"umesinden $a$ k\"umesini \"orten bir g\"onderme vard\i r.
\item\label{item:embeds}
$a\preccurlyeq\upomega$.
  \end{compactenum}
\end{theorem}

\begin{proof}
  \begin{asparadesc}
    \item[$\eqref{item:countable}\lto\eqref{item:covered}.$]
$a$ say\i labilir olsun.
\.Iki durum vard\i r.
      \begin{itemize}
      \item 
$f\colon a\xrightarrow{\approx}\upomega$ ise
$\inv{f}\colon\upomega\twoheadrightarrow a$.
\item
$n\in\upomega$ ve $f\colon a\xrightarrow{\approx}n$ ise,
$a$ k\"umesin her $b$ eleman\i\ i\c cin
\begin{equation*}
\inv{f}\cup\{(x,b)\colon n\leq x<\upomega\}\colon\upomega\twoheadrightarrow a.
\end{equation*}
      \end{itemize}
\item[$\eqref{item:covered}\lto\eqref{item:embeds}.$]
$h\colon\upomega\twoheadrightarrow a$ ise
\begin{equation*}
x\mapsto\min\{y\in\upomega\colon h(y)=x\}\colon a
\xrightarrow{\preccurlyeq}\upomega.
\end{equation*}
\item[$\eqref{item:embeds}\lto\eqref{item:countable}.$]
$f\colon a\xrightarrow{\preccurlyeq}\upomega$ olsun.  
O zaman $f\colon a\xrightarrow{\approx}f[a]$,
ve $(f[a],\in)$ bir iyi\-s\i ralanm\i\c s k\"umedir.
\c Simdi
\begin{equation*}
\ord(f[a],\in)=\alpha  
\end{equation*}
olsun.
O zaman
$a\approx f[a]\approx\alpha$.
Biz $\alpha\leq\upomega$ ko\c sulunu kan\i tlayaca\u g\i z.
Yani $\alpha\notin\upomega$ ise
$\alpha=\upomega$ ko\c sulunu kan\i tlayaca\u g\i z.
Bunun i\c cin
\begin{equation*}
  g\colon(\alpha,\in)\xrightarrow{\cong}(f[a],\in)
\end{equation*}
olsun.
E\u ger $\alpha\notin\upomega$ ise,
o zaman
$0\leq g(0)$ ve
\begin{equation*}
  k\leq g(k)\lto k+1\leq g(k)+1\leq g(k+1),
\end{equation*}
dolay\i s\i yla her $m$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
  m\leq g(m).
\end{equation*}
O zaman $g(\upomega)$ tan\i mlan\i rsa $g(\upomega)\geq m$,
dolay\i s\i yla $g(\upomega)\geq\upomega$, ki bu imk\^ans\i zd\i r.
Sonu\c c olarak $\alpha\leq\upomega$.\qedhere
  \end{asparadesc}
\end{proof}

E\u ger $a\preccurlyeq\bm B$ ama $a\not\approx\bm B$ ise
\begin{equation*}
  a\prec\bm B
\end{equation*}
\glossary{$a\prec\bm B$}%
ifadesini yazar\i z.
Sonu\c c olarak bir $a$ k\"umesi sonludur ancak ve ancak $a\prec\upomega$.

\c Su anda say\i lamaz sonsuzlukta hi\c cbir \emph{k\"ume} bilmiyoruz.

\subsection{Toplama}

Toplama veya ikili bile\c sim i\c slemiyle 
say\i lamaz sonsuzluktaki k\"ume\-ler\i n olu\c sturulamad\i\u g\i n\i\ 
g\"osterece\u giz. 

Tan\i m olarak, t\"um $a$ ile $b$ k\"umeleri i\c cin
\begin{equation*}
  a\sqcup b=(a\times\{0\})\cup(b\times\{1\}).
\end{equation*}
\glossary{$a\sqcup b$}%
Bu bile\c sime, $a$ ile $b$ k\"umelerinin \textbf{ayr\i k bile\c simi}%
\index{ayr\i k bile\c sim}
(\eng{disjoint union}) denir,
\c c\"unk\"u $a\approx a\times\{0\}$, ve $b\approx b\times\{1\}$, 
ve $a\times\{0\}$ ve $b\times\{1\}$ \c carp\i mlar\i,
birbirinden ayr\i kt\i r.
\"Ozel olarak
\begin{equation*}
  a\sqcup a=a\times2.
\end{equation*}
Sonraki teorem i\c cin,
\Sekle{fig:+} bak\i n.
\begin{figure}
  \centering
  \begin{pspicture}(0,-0.5)(5,2.5)
\psline[linestyle=dotted](0,0)(0,2)
\psline[linestyle=dotted](2,0)(2,2)
\psline[linestyle=dotted](3,0)(3,2)
\psline[linestyle=dotted](3,0)(5,0)(5,2)
\psline{*-o}(0,2)(5,2)
\psline{*-o}(0,1)(2,1)
\psline{*-o}(0,0)(3,0)
\psdots(3,2)
\uput[l](0,0){$0$}
\uput[l](0,1){$1$}
\uput[d](0,0){$0$}
\uput[d](2,0){$\beta$}
\uput[d](3,0){$\alpha$}
\uput[d](5,0){$\alpha+\beta$}
\rput(1.25,0.5){$\alpha\sqcup\beta$}
\uput[u](2.5,2){$\alpha+\beta$}
  \end{pspicture}
  \caption{$\alpha+\beta\approx\alpha\sqcup\beta$}\label{fig:+}
  
\end{figure}


\begin{theorem}\label{thm:ord+}
T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin
\begin{equation*}
\alpha+\beta\approx\alpha\sqcup\beta.
%(\alpha\times\{0\})\cup(\beta\times\{1\}).
\end{equation*}
\end{theorem}

\begin{proof}
$\bm F$, tan\i m s\i n\i f\i\ $\universe\times2$ olan  
$(\xi,y)\mapsto\alpha\cdot y+\xi$ g\"ondermesi olsun.
O zaman $\bm F(\xi,0)=\xi$, dolay\i s\i yla
\begin{equation*}
  \bm F\restriction(\alpha\times\{0\})\colon
\alpha\times\{0\}\xrightarrow{\approx}\alpha.
\end{equation*}
Ayr\i ca $\bm F(\xi,1)=\alpha+\xi$, dolay\i s\i yla,
\Teoremi{thm:subtraction} kullanarak,
\begin{equation*}
  \bm F\restriction(\beta\times\{1\})\colon
\beta\times\{1\}\xrightarrow{\approx}
\{\zeta\colon\alpha\leq\zeta<\alpha\cdot\beta\}.
\end{equation*}
Sonu\c c olarak
$\bm F\restriction(\alpha\sqcup\beta)\colon
\alpha\sqcup\beta
\xrightarrow{\approx}\alpha+\beta$.
\end{proof}

Kan\i tta $\bm F\restriction(\alpha\sqcup\beta)$
g\"ondermesinin tersi,
\begin{equation*}
\Bigl\{\bigl(\xi,(\xi,0)\bigr)\colon \xi<\alpha\Bigr\}
\cup\Bigl\{\bigl(\xi,(\xi-\alpha,1)\bigr)\colon
\alpha\leq \xi<\alpha+\beta\Bigr\}.
\end{equation*}
Bu g\"onderme $g$ olsun,
ve 
$s=\Bigl\{\bigl(g(\xi),g(\eta)\bigr)\colon\xi<\eta<\alpha+\beta\Bigr\}$ olsun.
O zaman $(\alpha\sqcup\beta,s)$, bir iyi\-s\i ralanm\i\c s k\"umedir, ve
\begin{equation}\label{eqn:+ord}
  \ord(\alpha\sqcup\beta,s)=\alpha+\beta.
\end{equation}
Ayr\i ca
$(\xi,y)\mathrel s(\xi_1,y_1)\liff\Bigl(y<y_1\lor(y=y_1\land\xi<\xi_1)\Bigr)$.
\"Ozel olarak $g$ e\c slenmesini kullanmadan $s$ tan\i mlanabilir.
Onun i\c cin baz\i\ kitaplarda $\alpha+\beta$ toplam\i,
\eqref{eqn:+ord} e\c sitli\u giyle tan\i mlan\i r.

\begin{theorem}\label{thm:approx-sqcup}
$a\approx b$ ve $c\approx d$ ise
$a\sqcup c\approx b\sqcup d$.
\end{theorem}

\ktk

\begin{theorem}\label{thm:o+o=o}
$\upomega\sqcup\upomega\approx\upomega$.
\end{theorem}

\begin{proof}
$(x,y)\mapsto 2x+y\colon
\upomega\sqcup\upomega\xrightarrow{\approx}\upomega$.
(\Sekle{fig:o+o=o} bak\i n.)
\begin{figure}
  \begin{equation*}
    \begin{array}{*{14}{c}}
1&3&5&7&9&11&13&15&17&19&21&23&25&\dots\\
0&2&4&6&8&10&12&14&16&18&20&22&24&\dots
    \end{array}
  \end{equation*}
  \caption{$\upomega\sqcup\upomega\approx\upomega$}\label{fig:o+o=o}
  
\end{figure}
\end{proof}

Sonu\c c olarak
\begin{equation*}
 \upomega+\upomega\approx\upomega, 
\end{equation*}
ve daha genelde $\alpha$ ve $\beta$ say\i labilirse 
$\alpha+\beta\preccurlyeq\upomega$.

\begin{theorem}
$a$ ile $b$ say\i labilirse $a\cup b$ bile\c simi de say\i labilir.
\end{theorem}

\ktk

\begin{theorem}
  $0$ olmayan her $n$ do\u gal say\i s\i\ i\c cin 
$\upomega\cdot n\approx\upomega$.
\end{theorem}

\begin{proof}
T\"umevar\i m y\"ontemini kullanaca\u g\i z.
\begin{asparaenum}
  \item
$\upomega\cdot1=\upomega$.
\item
$\upomega\cdot m\approx\upomega$ ise
  \begin{align*}
    \upomega\cdot(m+1)
&=\upomega\cdot m+\upomega&&\text{[tan\i m]}\\
&\approx\upomega\cdot m\sqcup\upomega&&\text{[\Teorem{thm:ord+}]}\\
&\approx\upomega\sqcup\upomega&&\text{[hipotez ve \Teorem{thm:approx-sqcup}]}\\
&\approx\upomega.&&\text{[\Teorem{thm:o+o=o}]}\qedhere
  \end{align*}
\end{asparaenum}
\end{proof}

Benzer (ama daha karma\c s\i k) \c sekilde 
$\upomega\times n\approx\upomega$ e\c slenikli\u gi kan\i tlanabilir;
ama bu sonu\c c, sonraki teoremden de \c c\i kar.

\begin{theorem}
T\"um $\alpha$ ordinali ve $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
  \alpha\times n\approx\alpha\cdot n.
\end{equation*}
\end{theorem}

\begin{proof}
T\"umevar\i m y\"ontemini kullanaca\u g\i z.
\begin{asparaenum}
  \item
$\alpha\times0=\emptyset=0=\alpha\cdot0$.
\item
$\alpha\times m\approx\alpha\cdot m$ ise
  \begin{align*}
    \alpha\times(m+1)
&=(\alpha\times m)\cup(\alpha\times\{m\})\\
&\approx\bigl((\alpha\times m)\times\{0\})\cup(\alpha\times\{1\})\\
&=(\alpha\times m)\sqcup\alpha\\
&\approx\alpha\cdot m+\alpha\\
&=\alpha\cdot(m+1).\qedhere
  \end{align*}
\end{asparaenum}
\end{proof}

\subsection{\c Carpma}

\c Carpmayla say\i lamaz sonsuzluktaki k\"umeler olu\c sturulamaz.
Sonraki teorem i\c cin, \Sekle{fig:.} bak\i n.
\begin{figure}
  \centering
\psset{yunit=2cm}
  \begin{pspicture}(0,-0.25)(8,1.75)
    \psline[linestyle=dotted](0,0)(0,1)(2,1)(2,0)
\psline{*-o}(0,1.5)(8,1.5)
\psline{*-o}(0,0)(2,0)
\psline{*-o}(0,0.25)(2,0.25)
\psline{*-o}(0,0.43)(2,0.43)
\psline{*-o}(0,0.58)(2,0.58)
\psline{-*}(0,0)(0,1)
\psdots(2,1.5)(3.5,1.5)(4.63,1.5)(5.47,1.5)
\uput[l](0,0){$0$}
\uput[l](0,0.25){$1$}
\uput[l](0,0.43){$2$}
\uput[l](0,0.58){$3$}
\uput[l](0,1){$\beta$}
\uput[d](0,0){$0$}
\uput[d](2,0){$\alpha$}
\uput[u](0,1.5){$0$}
\uput[u](2,1.5){$\alpha$}
\uput[u](3.5,1.5){$\alpha\cdot2$}
\uput[u](4.63,1.5){$\alpha\cdot3$}
\uput[u](5.47,1.5){$\alpha\cdot4$}
\uput[u](8,1.5){$\alpha\cdot\beta$}
\uput{10pt}[r](2,0.5){$\alpha\times\beta$}
\uput{10pt}[d](4,1.5){$\alpha\cdot\beta$}
  \end{pspicture}
  \caption{$\alpha\cdot\beta\approx\alpha\times\beta$}\label{fig:.}
  
\end{figure}


\begin{theorem}\label{thm:ord.}
T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin
\begin{equation*}
\alpha\cdot\beta\approx\alpha\times\beta.
\end{equation*}
\end{theorem}

\begin{proof}
B\"olme Teoremine (yani \Teoreme{thm:division}) g\"ore
\begin{equation*}
  (\xi,\eta)\mapsto\alpha\cdot\eta+\xi\colon
\alpha\times\universe\xrightarrow{\approx}\universe.
\end{equation*}
Ayr\i ca $\gamma\in\alpha$ ise
$\alpha\cdot\delta+\gamma<\alpha\cdot\beta\liff\delta<\beta$.
Sonu\c c olan
\begin{equation*}
  (\xi,\eta)\mapsto\alpha\cdot\eta+\xi\colon
\alpha\times\beta\xrightarrow{\approx}\alpha\cdot\beta.\qedhere
\end{equation*}
\end{proof}

\begin{xca}
  Tan\i m s\i n\i f\i\ $\universe\times\universe$ olarak 
$(\xi,\eta)\mapsto\alpha\cdot\eta+\xi$ g\"ondermesinin
birebir olmad\i\u g\i n\i\ g\"osterin.
\end{xca}

\begin{theorem}\label{thm:oxo=o}
$\upomega\times\upomega\approx\upomega$.
\end{theorem}

\begin{proof}
$f= \bigl\{\bigl((\xi,\eta),\zeta\bigr)
\in(\upomega\times\upomega)\times\upomega\colon$
 \begin{equation*}
    (\xi>\eta\lto\zeta=\xi^2+\eta)
\land(\xi\leq \eta\lto\zeta=\eta^2+\xi+\eta)\bigr\}   
 \end{equation*}
olsun.
(\Sekle{fig:ooo} bak\i n.)
\begin{figure}
  \begin{equation*}
\newlength{\ooo}
\settowidth{\ooo}{35}
    \begin{matrix}
 0& 2& 6&12&20&30&\makebox[\ooo]{\vdots}\\
 1& 3& 7&13&21&31&\vdots\\
 4& 5& 8&14&22&32&\vdots\\
 9&10&11&15&23&33&\vdots\\
16&17&18&19&24&34&\vdots\\
25&26&27&28&29&35&\vdots\\
     \hdotsfor 6 &\vdots
    \end{matrix}
  \end{equation*}
  \caption{$\upomega\cdot\upomega\approx\upomega$}\label{fig:ooo}
  
\end{figure}
O zaman
$f\colon
\upomega\times\upomega\xrightarrow{\approx}\upomega$.
\end{proof}

Sonu\c c olarak $\upomega\cdot\upomega\approx\upomega$.

\begin{theorem}\label{thm:approx-times}
  $a\approx b$ ve $c\approx d$ ise $a\times c\approx b\times d$.
\end{theorem}

\begin{proof}
  $f\colon a\xrightarrow{\approx}b$
ve $g\colon c\xrightarrow{\approx}d$ ise
\begin{equation*}
  (x,y)\mapsto(f(x),g(y))\colon 
a\times c\xrightarrow{\approx}b\times d.\qedhere
\end{equation*}
\end{proof}

Kan\i ttaki e\c sleme, 
\begin{equation*}
 \biggl\{\Bigl(\bigl(x,f(x)\bigr),\bigl(y,g(y)\bigr)\Bigr)\colon 
x\in a\land y\in b\biggr\} 
\end{equation*}
k\"umesidir.
G\"onderme olarak $f\times g$ bi\c ciminde yaz\i labilir.
\"Oyle bir ifade, \Teoremin{thm:o^o=o} kan\i t\i nda kullan\i lacakt\i r.

\begin{theorem}\label{thm:cxc=c}
$a$ ile $b$ say\i labilirse $a\times b$ \c carp\i m\i\ da say\i labilir.
\end{theorem}

\ktk

Sonu\c c olarak $\alpha$ ve $\beta$ say\i labilirse 
$\alpha\cdot\beta\preccurlyeq\upomega$.


\begin{theorem}
  $0$ olmayan her $n$ do\u gal say\i s\i\ i\c cin
$\upomega^n$ ordinal kuvveti say\i labilir sonsuzluktad\i r, yani
  \begin{equation*}
    \upomega^n\approx\upomega.
  \end{equation*}  
\end{theorem}

\begin{proof}
  T\"umevar\i m y\"ontemini kullanaca\u g\i z.
  \begin{asparaenum}
    \item
Tan\i mdan
$\upomega^1=\upomega$.
\item
$\upomega^m\approx\upomega$ ise
  \begin{align*}
    \upomega^{m+1}
&=\upomega^m\cdot\upomega&&\text{[tan\i m]}\\
&\approx\upomega^m\times\upomega&&\text{[\Teorem{thm:ord.}]}\\
&\approx\upomega\times\upomega&&\text{[hipotez ve \Teorem{thm:approx-times}]}\\
&\approx\upomega.&&\text{[\Teorem{thm:oxo=o}]}\qedhere
  \end{align*}
  \end{asparaenum}
\end{proof}

Son alt b\"ol\"umdeki gibi,
benzer \c sekilde $\upomega\approx{}^n\upomega$ e\c slenli\u gi 
kan\i tlanabilir.
\"Ozel olarak ${}^n\upomega$, bir k\"umedir.
Bu sonu\c clar, sonraki teoremden de \c c\i kar.

\begin{theorem}\label{thm:na=an}
  T\"um $\alpha$ ordinali ve $n$ do\u gal say\i s\i\ i\c cin
  \begin{equation*}
    \alpha^n\approx{}^n\alpha.
  \end{equation*}
\end{theorem}

\begin{proof}
${}^00=\{\emptyset\}=1=0^0$ ve
  ${}^{n+1}0=\emptyset=0=0^{n+1}$.
\c Simdi $\alpha>0$ olsun.
\begin{asparaenum}
  \item
 ${}^0\alpha=\{\emptyset\}=1=\alpha^0$.
\item
$f_m\colon{}^m\alpha\xrightarrow{\approx}\alpha^m$ olsun.
E\u ger $g\in{}^{m+1}\alpha$ ise
\begin{equation*}
  f_{m+1}(g)=\alpha\cdot f_m(g\restriction m)+g(m)
\end{equation*}
olsun.  
O zaman $f_{m+1}\colon{}^{m+1}\alpha\xrightarrow{\approx}\alpha^{m+1}$.\qedhere
\end{asparaenum}
\end{proof}

Kan\i tta buldu\u gumuz
${}^n\alpha$ ile $\alpha^n$ aras\i nda e\c sleme,
\begin{equation*}
  (x_0,\dots,x_{n-1})\mapsto\alpha^{n-1}\cdot x_0+\alpha^{n-2}\cdot x_1+\dots
+\alpha\cdot x_{n-2}+x_{n-1}.
\end{equation*}

\Teoremin{thm:oxo=o} yard\i m\i yla
sonraki teorem kendisini kan\i tlar.

\begin{theorem}\label{thm:cucs=c}
E\u ger $a$ say\i labilir, ve her eleman\i\ da say\i labilir ise,
ve ayr\i ca
$f\colon\upomega\twoheadrightarrow a$,
ve her $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
  g_n\colon\upomega\twoheadrightarrow f(n)
\end{equation*}
ise, o zaman
$\bigcup a=\{g_x(y)\colon(x,y)\in\upomega\times\upomega\}$,
dolay\i s\i yla
\begin{equation*}
  \bigcup a
\end{equation*}
bile\c simi de say\i labilir.
\end{theorem}

\"Ornekler sonraki alt b\"ol\"umdedir.
K\i saca, elemanlar\i\ say\i labilir olan 
say\i labilir bir $a$ k\"umesinin bili\c simi de say\i labilir;
ama $a$ k\"umesinin her eleman\i, 
tan\i m k\"umesi $\upomega$ olan
bilinen bir g\"onderme taraf\i ndan \"ort\"ulmelidir.
Yani $\bm A$, t\"um say\i labilir k\"umeler s\i n\i f\i\ olsun;
o zaman
\begin{equation*}
  \Forall x\left(x\in\bm A\land
\Forall y(y\in x\lto y\in\bm A)\lto\bigcup x\in\bm A\right)
\end{equation*}
c\"umlesini kan\i tlam\i\c s \emph{olmad\i k.}
Bu c\"umlenin do\u grulu\u gu i\c cin
Se\c cim Aksiyomu gerekir.

\subsection{Kuvvet alma}

\emph{Ordinal} kuvvetleri alarak 
say\i lamaz sonsuzluktaki k\"umeler olu\c sturulamaz.

\begin{theorem}\label{thm:o^o=o}
  $\upomega^{\upomega}\approx\upomega$.
\end{theorem}

\begin{proof}
  $\upomega^{\upomega}=\bigcup_{k<\upomega}\upomega^k$.
\Teoremin{thm:oxo=o} kan\i t\i ndaki gibi
\begin{equation*}
  f\colon\upomega\times\upomega\xrightarrow{\approx}\upomega
\end{equation*}
olsun.
Her $k$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
  g_k=\bigl((\xi,\eta)\mapsto\upomega^k\cdot\eta+\xi\bigr)
\restriction(\upomega^k\times\upomega)
\end{equation*}
olsun; o zaman
\Teoremin{thm:ord.} kan\i t\i ndaki gibi
\begin{equation*}
  g_k\colon\upomega^k\times\upomega\xrightarrow{\approx}\upomega^{k+1}.
\end{equation*}
Bu durumda $h\colon\upomega\xrightarrow{\approx}\upomega^k$ ise
\begin{equation*}
  g_k\circ\bigl(h\times(x\mapsto x)\bigr)\circ\inv f\colon
\upomega\xrightarrow{\approx}\upomega^{k+1}.
\end{equation*}
(\Sekle{fig:comp} bak\i n.)
\begin{figure}
  \begin{equation*}
\xymatrix@C=1.8cm{\upomega\ar[r]^{\inv f}&\upomega\times\upomega\ar[r]^{h\times(x\mapsto x)}&\upomega^k\times\upomega\ar[r]^{g_k}&\upomega^{k+1}}
  \end{equation*}
  \caption{$\upomega^{k+1}\approx\upomega$} \label{fig:comp}
\end{figure}
\"Ozyineleme ile \"oyle $m\mapsto f_m$ g\"ondermesi vard\i r ki
\begin{align*}
  f_0&=(x\mapsto0)\restriction\upomega,&
f_1&=(x\mapsto x)\restriction\upomega,&
  f_{k+1}&\colon\upomega\xrightarrow{\approx}\upomega^{k+1}.
\end{align*}
O zaman \Teorem{thm:cucs=c} sayesinde 
$\bigcup_{k<\upomega}\upomega^k$ bile\c simi say\i labilir.
\end{proof}

$\upomega^{\upomega}$ gibi ${}^{\upomega}\upomega$ say\i labilir mi?\label{oo?}
Daha genelde $\alpha^{\upomega}\approx{}^{\upomega}\alpha$?
\Teoremde{thm:na=an} buldu\u gumuz 
${}^n\alpha$ ile $\alpha^n$ aras\i nda e\c sleni\u gi kullanarak
${}^{\upomega}\alpha$ s\i n\i f\i ndan $\alpha^{\upomega}$ k\"umesine giden
bir e\c sleme tan\i mlayamay\i z.
\Teoreme{thm:oo-neq-oo} bak\i n.

T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin, 
tan\i m k\"umeleri $\beta$ ordinalinin sonlu bir altk\"umesi olan 
ve de\u ger k\"umeleri $\alpha\setminus\{0\}$ fark\i n\i n bir altk\"umesi olan 
g\"ondermelerin s\i n\i f\i, 
$\exp(\alpha,\beta)$
\glossary{$\exp(\alpha,\beta)$}%
olsun.%%%%%
\footnote{$\exp(\alpha,\beta)$ ifadesi, 
ve a\c sa\u g\i daki \Teorem{thm:exp}, 
Levy'nin \cite[IV.2.10]{MR1924429} kitab\i ndan al\i nm\i\c st\i r.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Yani
\begin{equation*}
  \exp(\alpha,\beta)
=\Bigl\{f\colon\Exists x\bigl(x\included\beta\land 
x\prec\upomega\land f\colon x\to\alpha\setminus\{0\}\bigr)\Bigr\}
\end{equation*}
olsun.

\begin{theorem}\label{thm:exp}
T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin
\begin{equation*}
\alpha^{\beta}\approx\exp(\alpha,\beta).
\end{equation*}
\end{theorem}

\begin{proof}
$\exp(\alpha,\beta)$ s\i n\i f\i n\i n her eleman\i,
  \begin{equation*}
    \gamma_0>\gamma_1>\dots>\gamma_{n-1}
  \end{equation*}
ko\c sulunu sa\u glayan 
$\{(\gamma_0,\delta_0),\dots,(\gamma_{n-1},\delta_{n-1})\}$ 
bi\c ciminde yaz\i labilir.
Bu durumda
\begin{equation*}
  \bm F\bigl(\{(\gamma_0,\delta_0),\dots,(\gamma_{n-1},\delta_{n-1})\}\bigr)
=\alpha^{\gamma_0}\cdot\delta_1+\dots
+\alpha^{\gamma_{n-1}}\cdot\delta_{n-1}
\end{equation*}
olsun.  O halde
\begin{equation*}
  \bm F\colon\exp(\alpha,\beta)\xrightarrow{\approx}\alpha^{\beta};
\end{equation*}
asl\i nda \Teoreme{thm:base} g\"ore
$\inv{\bm F}$, $\xi\mapsto\xi_{\alpha}$ g\"ondermesidir.
\end{proof}

\begin{theorem}
$\alpha\approx\beta$ ve $\gamma\approx\delta$ ise 
$\exp(\alpha,\gamma)\approx\exp(\beta,\delta)$.
\end{theorem}

\ktk

Sonu\c c olarak $\alpha$ ve $\beta$ say\i labilirse 
$\alpha^{\beta}\preccurlyeq\upomega$.

\section{B\"uy\"ukl\"uklerin s\i ralanmas\i}

\Teoremin{thm:preccurlyeq} \"ozel durumu vard\i r:
\begin{equation*}
a\prec b\land b\prec c\lto a\preccurlyeq c.  
\end{equation*}
Ama $a\prec b\land b\prec c$ ise $a\prec c$ sonucuna varabilir miyiz?
Yani $\prec$ ba\u g\i nt\i s\i\ bir s\i ralama m\i d\i r?

Sayfa \pageref{secim} ve \sayfanumarada{ax:choice}ki Se\c cim Aksiyomuna g\"ore
her k\"ume iyi\-s\i ralana\-bilir,
dolay\i s\i yla
her b\"uy\"ukl\"uk, bir ve tek bir kardinal i\c cerir, 
ve sonu\c c olarak b\"uy\"ukl\"ukler, 
i\c cerilen kardinallere g\"ore iyi\-s\i ralanabilir.  
\"Ozel olarak $\prec$ ba\u g\i nt\i s\i\ bir s\i ralamad\i r.
Asl\i nda,
Se\c cim Aksiyomunun yerinde
\Teorem{thm:<<<} kullanarak,
b\"uy\"ukl\"uklerin s\i ralanabildi\u gini
g\"osterece\u giz.

\begin{theorem}[Schr\"oder--Bernstein]\label{thm:SB}
T\"um $a$ ve $b$ k\"umeleri i\c cin
\begin{equation*}
a\preccurlyeq b\land b\preccurlyeq a\lto a\approx b.
\end{equation*}
\end{theorem}

\begin{proof}
  $f\colon a\xrightarrow{\preccurlyeq}b$ 
ve $g\colon b\xrightarrow{\preccurlyeq}a$ olsun.  
\"Ozyinelemeyle
\begin{align*}
  (a_0,b_0)&=(a,b),&
(a_{n+1},b_{n+1})&=(g[b_n],f[a_n])
\end{align*}
olsun.  O zaman
\begin{align*}
  a_0&\includes a_1\includes\cdots,&
b_0&\includes b_1\includes\cdots
\end{align*}
Ayr\i ca
\begin{gather*}
  f\restriction(a_{2n}\setminus a_{2n+1})\colon 
a_{2n}\setminus a_{2n+1}\xrightarrow{\approx}b_{2n+1}\setminus b_{2n+2},\\
  \inv g\restriction(a_{2n+1}\setminus a_{2n+2})\colon 
a_{2n+1}\setminus a_{2n+2}\xrightarrow{\approx}b_{2n}\setminus b_{2n+1}.
\end{gather*}
O zaman
\begin{equation*}
  f\restriction\left(\bigcup_{n\in\upomega}(a_{2n}\setminus a_{2n+1})
\cup\bigcap_{n\in\upomega}a_n\right)
\cup\inv g\restriction\bigcup_{n\in\upomega}(a_{2n+1}\setminus a_{2n+2}),
\end{equation*}
$a$ k\"umesinden $b$ k\"umesine giden bir e\c slemedir.
\end{proof}

Schr\"oder--Bernstein Teoremi i\c cin Zermelo,
Sonsuzluk Aksiyomunu kullanmayan bir kan\i t\i\ verdi.
Sayfa \sayfanumaraya{Z-SB} bak\i n.

\begin{theorem}\label{thm:<<<}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{equation*}
a\prec b\land b\prec c\lto a\prec c.
\end{equation*}
\end{theorem}

\ktk

\c Simdi b\"uy\"ukl\"ukler, 
$\prec$ ba\u g\i nt\i s\i na g\"ore s\i ralanabilir;
ama Se\c cim Aksiyomunu kullanmadan
bu s\i ralaman\i n do\u grusal olup olmad\i\u g\i n\i\ bil\-miyoruz.


\section{Say\i lamaz sonsuzluk}

Baz\i\ \textbf{sonsuz b\"uy\"ukl\"ukler}
(yani eleman\i\ sonsuz olan b\"uy\"ukl\"ukler),
ba\c ska sonsuz b\"uy\"ukl\"uklerden daha b\"uy\"ukt\"ur.

\begin{theorem}[Cantor]%
\index{Cantor!---'un Teoremi}%
\index{teorem!Cantor'un T---i}%
  Her $a$ k\"umesi i\c cin
  \begin{equation*}
    a\prec\pow a.
  \end{equation*}
\end{theorem}

\begin{proof}
 $\bigl\{\bigl(x,\{x\}\bigr)\colon
  x\in a\bigr\}$ g\"ondermesinin sayesinde $a\preccurlyeq\pow
  a$.  \c Simdi $f\colon a\xrightarrow{\preccurlyeq}\pow a$ ve
  \begin{equation*}
    b=\{x\in a\colon x\notin f(x)\}
  \end{equation*}
olsun.  O zaman $a$ k\"umesinin her $c$ eleman\i\ i\c cin
\begin{equation*}
  c\in b\liff c\notin f(c).
\end{equation*}
\"Oyleyse $b\neq f(c)$.  Dolay\i s\i yla $b\notin f[a]$, ve $f$, e\c
sleme de\u gildir.  
Sonu\c c olarak $a\not\approx\pow a$.
\end{proof}

\c Simdi \sayfada{oo?} sordu\u gumuz soruyu cevaplayaca\u g\i z.

\begin{theorem}\label{thm:pow-a=a2}
Her $a$ k\"umesi i\c cin
\begin{equation*}
x\mapsto\{(y,0)\colon y\in a\setminus x\}\cup\{(y,1)\colon y\in x\}
\colon\pow a\xrightarrow{\approx}{}^a2.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}\label{thm:oo-neq-oo}
  $\upomega^{\upomega}\prec{}^{\upomega}\upomega$.
\end{theorem}

\begin{proof}
\Teorem{thm:o^o=o} ve Cantor'un Teoremi ile
  \begin{equation*}
\upomega^{\upomega}
\approx\upomega\prec\pow{\upomega}.
  \end{equation*}
\Teorem{thm:pow-a=a2} ile, 
${}^{\upomega}2\included{}^{\upomega}\upomega$ oldu\u gundan,
bir $\bm F$ i\c cin
\begin{equation*}
  \bm F\colon\pow{\upomega}
\xrightarrow{\preccurlyeq}{}^{\upomega}\upomega.\qedhere
\end{equation*}
Sonu\c c olarak $\upomega^{\upomega}\prec{}^{\upomega}\upomega$.
\end{proof}

Cantor'un Teoremi, k\"ume olmayan s\i n\i flar i\c cin yanl\i\c st\i r.  
Mesela
\begin{equation*}
\universe=\pow{\universe}.
\end{equation*}

\begin{xca}
Cantor'un Teoremini kan\i tlarken
$a$ k\"umesinin k\"ume ol\-du\u gunu nas\i l kulland\i k?
\end{xca}

\Teorem{thm:na=an} ve \ref{thm:pow-a=a2} sayesinde
$a$ sonlu ise $\pow a$ kuvvet s\i n\i f\i\ da,
sonlu bir k\"umedir.
O zaman bir $b$ k\"umesi i\c cin $\pow b$ sonsuz ise,
$b$ k\"umesi de sonsuzdur,
dolay\i s\i yla,
Cantor'un Teoremi sayesinde, 
$\pow b$ kuvvet s\i n\i f\i\ say\i lamaz.

\begin{axiom}[Kuvvet K\"umesi]\label{ax:pow}%
\index{aksiyom!Kuvvet K\"umesi A---u}
Her k\"umenin kuvvet s\i n\i f\i\ bir\linebreak k\"ume\-dir, yani
\begin{equation*}
\Forall x\Exists y\Forall z\bigl(z\in y\liff\Forall w(w\in z\lto w\in x)\bigr)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

\"Ozel olarak say\i lamaz k\"umeler vard\i r,
mesela $\pow{\upomega}$, $\pow{\pow{\upomega}}$, ve saire.
B\"oylece en b\"uy\"uk k\"ume yoktur.
Ama (Se\c cim Aksiyomunu kullanmadan)
bu kuvvet k\"umelerinin kardinallerinin olup olmad\i\u g\i n\i\
bilmiyoruz.
A\c sa\u g\i daki Hartogs'un Teoremi ile,
daha k\"u\c c\"uk olmayan kardinaller olacakt\i r.

Sonraki teoremler i\c cin
\begin{align*}
\pow[0]a&=a,&
\pow[n+1]a&=\pow{\pow[n]a}
\end{align*}
\"ozyineli tan\i m\i n\i\ yapar\i z.

\begin{theorem}
T\"um $a$ ve $b$ i\c cin $a\times b\included\pow[2]{a\cup b}$.  
\end{theorem}

\ktk

Sonraki teorem, \Teorem{thm:Sierpinski} i\c cin kullan\i lacakt\i r.

\begin{theorem}[Hartogs]\label{thm:Hartogs}
Her $a$ k\"umesi i\c cin, 
bir $\beta$ ordinali i\c cin,
\begin{align*}
  \beta&\not\preccurlyeq a,&
\beta&\preccurlyeq\pow[4]a.
\end{align*}
\end{theorem}

\begin{proof}
$\bm B=\{\xi\colon\xi\preccurlyeq a\}$ olsun.
O zaman $\bm B$ ge\c ci\c slidir,
dolay\i s\i ya $\bm B$ bir k\"ume ise, bir $\beta$ ordinalidir.
Bu durumda $\beta\notin\beta$ oldu\u gundan
\begin{equation*}
\beta\not\preccurlyeq a.
\end{equation*}
Neden $\bm B$ bir k\"ume olmal\i d\i r?
$\pow a\times\pow{a\times a}$ \c carp\i m\i n\i n
\begin{equation*}
(c,s)\in b\liff(\text{$s$ ba\u g\i nt\i s\i, $c$ k\"umesini iyi\-s\i ralar})
\end{equation*}
ko\c sulunu sa\u glayan $b$ altk\"umesi vard\i r.
O zaman
\begin{equation*}
 \bm B=\{\ord(c,s)\colon(c,s)\in b\}=\beta.
\end{equation*}
Ayr\i ca
\begin{multline*}
  \alpha\mapsto
\Bigl\{s\cup\bigl\{(x,x)\colon x\in c\bigr\}\colon\\
(c,s)\in b\land s\included c\times c\land\alpha=\ord(c,s)\Bigr\}\colon\\
\beta\xrightarrow{\preccurlyeq}\pow[4]a.\qedhere
\end{multline*}
\end{proof}

\c Su andan itibaren
\begin{align*}
  &\kappa,&&\lambda,&&\mu,&&\nu
\end{align*}%
\glossary{$\kappa$, $\lambda$, $\mu$, $\nu$}%
k\"u\c c\"uk Yunan harfleri, her zaman kardinalleri g\"osterecektir.

\c Simdi her $\kappa$ kardinali i\c cin, 
bu kardinale g\"om\"ulemeyen \emph{en k\"u\c c\"uk} ordinal vard\i r.  
Bu ordinal, bir kardinal olmal\i d\i r.  
Bu kardinale, $\kappa$ kardinalinin \textbf{ard\i l\i}%
\index{ard\i l}
(veya \emph{kardinal ard\i l\i})
denir, ve bu ard\i l
\begin{equation*}
\kappa^+
\end{equation*}%
\glossary{$\kappa^+$}
ifadesiyle g\"osterilir.  
B\"oylece $\kappa<\kappa^+$, 
ve her $\lambda$ kardinali i\c cin 
ya $\lambda\leq\kappa$ ya da $\kappa^+\leq\lambda$.

$\kappa$ sonsuz ise $\kappa^+$ ordinal olarak ard\i l de\u gildir:

\begin{theorem}
  Her sonsuz kardinal, ordinal olarak bir limittir.
\end{theorem}

\ktk

\begin{theorem}\label{thm:card-lim}
Elemanlar\i\ kardinal olan bir k\"umenin supremumu, bir kardinaldir.
\end{theorem}

\ktk

\c Simdi
\begin{gather*}
	\aleph_0=\upomega,\\
	\aleph_{\beta+1}=(\aleph_{\beta})^+,\\
	\gamma\text{ limit}\lto\aleph_{\gamma}=\sup\{\aleph_{\xi}\colon\xi<\gamma\}
\end{gather*}
\glossary{$\aleph_{\alpha}$}%
ko\c sullar\i, 
$\on$ s\i n\i f\i nda bir $\xi\mapsto\aleph_{\xi}$ i\c slemini tan\i mlar.  
(Burada $\aleph$, \.Ibrani \emph{alef} harf\/idir.)

\begin{theorem}
$\xi\mapsto\aleph_{\xi}$ g\"ondermesi, $\cn\setminus\upomega$
  s\i n\i f\i n\i\ \"orten, normal bir i\c slemdir.
\end{theorem}

\ktk

\section{Toplama ve \c carpma}

Tan\i ma g\"ore
\begin{align*}
\kappa\cardsum\lambda&=\card(\kappa+\lambda),&
\kappa\cardprod\lambda&=\card(\kappa\cdot\lambda).
\end{align*}%
\glossary{$\kappa\cardsum\lambda$}%
\glossary{$\kappa\cardprod\lambda$}
O zaman \Teorem{thm:ord+} ve \ref{thm:ord.} sayesinde
\begin{align*}
\kappa\cardsum\lambda&=\card(\kappa\sqcup\lambda),&
%\bigl((\kappa\times\{0\})\cup(\lambda\times\{1\})\bigr),&
\kappa\cardprod\lambda&=\card(\kappa\times\lambda).
\end{align*}

Sonraki teoremler sayesinde kardinal hesapmalar\i\ kolayd\i r.

\begin{theorem}
  T\"um $\kappa$ ile $\lambda$ kardinalleri i\c cin
  \begin{align*}
    \kappa\cardsum\lambda&=\lambda\cardsum\kappa,&
    \kappa\cardprod\lambda&=\lambda\cardprod\kappa.
  \end{align*}
\end{theorem}

\ktk

\begin{theorem}\label{thm:2kl}
$2\leq\kappa\leq\lambda$ ise
\begin{equation*}
\lambda
\leq\kappa\cardsum\lambda
\leq\kappa\cardprod\lambda
\leq\lambda\cardprod\lambda.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}
  $\bm F$, ordinallerde normal bir i\c slem, ve
  \begin{equation}\label{eqn:F-approx}
    \alpha\approx\beta\lto\bm F(\alpha)\approx\bm F(\beta) 
  \end{equation}
ise, o zaman t\"um sonsuz $\kappa$ kardinali i\c cin
  \begin{equation*}
    \bm F(\kappa)=\kappa.
  \end{equation*}
\end{theorem}

\begin{proof}
$\bm F$ kesin artan oldu\u gundan, 
\Teorem{thm:inc} sayesinde $\kappa\leq\bm F(\kappa)$.  
M\"umk\"unse $\kappa<\bm F(\kappa)$ olsun.  
\Teorem{thm:card-lim} sayesinde $\kappa$ bir ordinal limiti oldu\u gundan
ve $\bm F$ normal oldu\u gundan 
bir $\alpha$ i\c cin
\begin{equation*}
\alpha<\kappa<\bm F(\alpha).
\end{equation*}
O halde $\lambda=\card(\alpha)$ ise, 
\eqref{eqn:F-approx} sat\i r\i ndaki ko\c sula g\"ore
  \begin{equation*}
    \lambda<\kappa\leq\bm F(\lambda).
  \end{equation*}
B\"oylece $\{\xi\colon\xi\in\cn\land\xi<\bm F(\xi)\}$ s\i n\i f\i n\i n 
en k\"u\c c\"uk eleman\i\ yoktur, 
dolay\i s\i yla s\i n\i f bo\c stur.
\end{proof}

$\on$ s\i n\i f\i n\i n bir $a$ altk\"umesinin en b\"uy\"uk eleman\i\ varsa
\begin{equation*}
  \max a
\end{equation*}
olarak yaz\i labilir.

\begin{theorem}\label{thm:l=ll}
$\lambda$ sonsuz ise $\lambda=\lambda\cardprod\lambda$.
\end{theorem}

\begin{proof}
$\on\times\on$ \c carp\i m\i,
\begin{equation*}
\max\{\alpha,\beta\}<\gamma\liff(\alpha,\beta)<(\gamma,0)
\end{equation*}
ko\c sulunu sa\u glayan
bir $<$ ba\u g\i nt\i s\i\ taraf\i ndan iyi\-s\i ralan\i r, ve bu durumda
\begin{equation*}
  \alpha\times\alpha=
  \bigl\{(\xi,\eta)\colon(\xi,\eta)<(\alpha,0)\bigr\}.
\end{equation*}
\"Orne\u gin
\begin{equation}\label{eqn:cnxcn}
\left.
\begin{gathered}
  (\alpha,\beta)<(\gamma,\delta)\liff
  \begin{gathered}[t]
\max\{\alpha,\beta\}<\max\{\gamma,\delta\}\\
{}\lor\beta<\delta<\gamma=\alpha\\
{}\lor(\beta<\alpha=\delta\land\gamma\leq\delta)\\
{}\lor\alpha<\gamma\leq\delta=\beta
  \end{gathered}
\end{gathered}\qquad
\right\}
\end{equation}
olabilir.
O zaman 
\sayfada{thm:s}ki \Teoreme{thm:s} g\"ore,
$\on\times\on$ \c carp\i m\-\i ndan $\on$ s\i n\i f\i na giden
\begin{equation*}
  \bm G(\alpha,\beta)=\{\bm G(\xi,\eta)\colon(\xi,\eta)<(\alpha,\beta)\}
\end{equation*}
ko\c sulunu sa\u glayan bir $\bm G$ g\"ondermesi vard\i r,
ve bu durumda
\begin{equation*}
  \bm G(\alpha,\beta)
=\ord\Bigl(\bigl\{(\xi,\eta)\colon(\xi,\eta)<(\alpha,\beta)\bigr\},<\Bigr).
\end{equation*}
Mesela \eqref{eqn:cnxcn} sat\i r\i ndaki ko\c sullar\i\ alt\i nda
$\bm G$ g\"ondermesinin de\u gerleri,
\Sekilde{fig:cnxcn}ki gibidir.
(\Sayfada{fig:ooo}ki \c Sekil \ref{fig:ooo} ile kar\c s\i la\c st\i r\i n.)
\begin{figure}
  \begin{equation*}
\settowidth{\ooo}{35}
    \begin{array}{c|*6c}
      & 0& 1&      2&               3&              \beta&\gamma\\\hline
     0& 0& 2&      6&\makebox[\ooo]{}&                   &\delta+\gamma\\
     1& 1& 3&      7&                &                   &\delta+\gamma'\\
     2& 4& 5&      8&                &                   &\delta+\gamma''\\
     3& 9&  &       &                &                   &\vdots\\
\vphantom{|}&&&&&&\\
\alpha&  &  &       &                &\bm G(\alpha,\beta)&\\
\vphantom{|}&&&&&&\\
\gamma&\delta&\delta'&&&&\delta+\gamma\cdot2\\
\gamma'&\delta+(\gamma\cdot2)'&&&&&
    \end{array}
  \end{equation*}
  \caption{$\lambda=\lambda\cardprod\lambda$}\label{fig:cnxcn}
  
\end{figure}
\c Simdi
\begin{equation*}
  \bm F(\alpha)=\bm G(\alpha,0)
\end{equation*}
olsun.  O zaman $\bm F(\alpha)\approx\alpha\times\alpha$, dolay\i s\i yla
\begin{align*}
  \alpha\approx\beta
&\lto\alpha\times\alpha\approx\beta\times\beta
&&\text{[\Teorem{thm:approx-times}]}\\
&\lto\bm F(\alpha)\approx\bm F(\beta).&&
\end{align*}
Ayr\i ca $\bm F$ kesin artand\i r.  Son olarak $\alpha$ limit ise
\begin{align*}
  \bm F(\alpha)
&=\{\bm G(\xi,\eta)\colon\max(\xi,\eta)<\alpha\}\\
&=\bigcup_{\zeta<\alpha}\{\bm G(\xi,\eta)\colon\max(\xi,\eta)<\zeta\}\\
&=\sup_{\zeta<\alpha}\bm F(\zeta).
\end{align*}
B\"oylece $\bm F$ normaldir, 
ve son teoreme g\"ore $\kappa$ sonsuz ise $\bm F(\kappa)=\kappa$,
dolay\i s\i yla
\begin{equation*}
  \kappa=\kappa\cardprod\kappa.\qedhere
\end{equation*}
\end{proof}

\begin{theorem}\label{thm:card+.}
  T\"um $\alpha$ ve $\beta$ i\c cin
\begin{equation*}
\aleph_{\beta}\cardsum\aleph_{\gamma}=
\aleph_{\beta}\cardprod\aleph_{\gamma}=\aleph_{\max\{\alpha,\beta\}}.
\end{equation*}
\end{theorem}

\begin{proof}
  \Teorem{thm:2kl} ve \ref{thm:l=ll}.
\end{proof}

\section{Ordinaller Kuvvetlerinin kardinalleri}

Herhangi bir $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}
\powf{\bm A},
\end{equation*}%
\glossary{$\powf{\bm A}$}
$\bm A$ s\i n\i f\i n\i n \emph{sonlu} altk\"umelerinin s\i n\i f\i\ olsun.

\begin{theorem}
  $\alpha$ sonsuz ise
  \begin{equation*}
\powf{\alpha}\approx\alpha^{\upomega}\approx\alpha.
  \end{equation*}
\end{theorem}

\begin{proof}
E\u ger $\upomega\leq\alpha$ ise,
o zaman
\begin{align*}
\alpha
&\preccurlyeq\alpha^{\upomega}&&\\
&\approx\exp(\alpha,\upomega)&&\text{[\Teorem{thm:exp}]}\\
&\included\powf{\upomega\times\alpha}&&\\
&\approx\powf{\alpha}&&\text{[\Teorem{thm:card+.}]}.
\end{align*}
Ayr\i ca $b\in\powf{\alpha}$ ise,
\Teoreme{thm:s} g\"ore,
bir ve tek bir $f$ i\c cin
\begin{equation*}
  f\colon\bigl(\card(b),\in\bigr)\xrightarrow{\cong}(b,\in).
\end{equation*}
\"Ozel olarak $f\in{}^{\card(b)}\alpha$.
B\"oylece
\begin{equation*}
  \powf{\alpha}\preccurlyeq\bigcup_{n\in\upomega}{}^n\alpha.
\end{equation*}
\"Ozyineleme ile,
tan\i m k\"umesi $\upomega$ olan,
\begin{align*}
  g_0&\colon{}^0\alpha\xrightarrow{\preccurlyeq}\alpha,&
g_{n+1}&\colon{}^{n+1}\alpha\xrightarrow{\approx}\alpha
\end{align*}
ko\c sullar\i n\i\ sa\u glayan bir $n\mapsto g_n$ g\"ondermesi vard\i r.
Asl\i nda
\begin{align*}
  g_0(0)&=0,&
g_1(f)&=f(0),&
h&\colon\alpha\times\alpha\xrightarrow{\approx}\alpha
\end{align*}
olsun.
E\u ger $g_n\colon{}^n\alpha\xrightarrow{\approx}\alpha$ ise,
o zaman
\begin{equation*}
  g_{n+1}(f)=h\bigl(g_n(f\restriction n),f(n)\bigr)
\end{equation*}
olsun.  \c Simdi
\begin{equation*}
  f\mapsto\bigl(\card(f),g_{\card(f)}(f)\bigr)\colon
\bigcup_{n\in\upomega}{}^n\alpha\xrightarrow{\preccurlyeq}\upomega\times\alpha.
\end{equation*}
Son olarak tekrar $\upomega\times\alpha\approx\alpha$.
\end{proof}

\begin{theorem}
  $\alpha$ ile $\beta$ sonsuz ise
  \begin{equation*}
    \alpha^{\beta}\approx\max\{\alpha,\beta\}.
  \end{equation*}
\end{theorem}

\begin{proof}
\.Ilk olarak
\begin{align*}
\alpha=\alpha^1&\leq\alpha^{\beta},&&\text{[\Teorem{thm:powers}]}\\
       \beta&\leq\alpha^{\beta}.&&\text{[\Teorem{thm:powers} ve \ref{thm:inc}]}
\end{align*}
  E\u ger $\bigl\{(\gamma_0,\delta_0),
  \dots,(\gamma_{n-1},\delta_{n-1})\bigr\}\in\exp(\alpha,\beta)$ ve
  $\gamma_0<\dots<\gamma_{n-1}$ ise, o zaman
  \begin{equation*}
    f\Bigl(\bigl\{(\gamma_0,\delta_0),
  \dots,(\gamma_{n-1},\delta_{n-1})\bigr\}\Bigr)
=\bigl(\{\gamma_0,\dots,\gamma_{n-1}\},(\delta_0,\dots,\delta_{n-1})\bigr)
  \end{equation*}
olsun.  O halde
$f\colon\exp(\alpha,\beta)
\xrightarrow{\preccurlyeq}\powf{\beta}\times\exp(\alpha,\upomega)$,
dolay\i s\i yla 
\begin{equation*}
\alpha^{\beta} \approx\exp(\alpha,\beta)
\preccurlyeq\powf{\beta}\times\exp(\alpha,\upomega)
\approx\beta\times\alpha \approx\max\{\alpha,\beta\}.\qedhere
\end{equation*}
\end{proof}

\section{Kontin\"u Hipotezi}

Cantor'un Teoreminden
\begin{equation*}
 \aleph_0=\upomega\prec\pow{\upomega}. 
\end{equation*}
\textbf{Kontin\"u Hipotezine}%
\index{Kontin\"u Hipotezi}
(\eng{Continuum Hypothesis}) g\"ore,
$\pow{\upomega}$ k\"umesi iyis\i ralanabilir ve ayr\i ca
\begin{equation*}
\pow{\upomega}\approx\aleph_1.
\end{equation*}
Bu bir aksiyom de\u gil, sadece bir c\"umledir.
Bu c\"umleye $\ch$%
\glossary{$\ch$}
denir.

\textbf{Genelle\c stirilmi\c s Kontin\"u Hipotezi}
(\eng{Generalized Continuum Hypothesis}) veya $\gch$,%
\glossary{$\gch$}
\begin{equation*}
\Forall x\Forall y(\upomega\preccurlyeq x\prec y\preccurlyeq\pow x
\lto y\approx\pow x)
\end{equation*}
c\"umlesidir.  
\c Simdilik $\gch\lto\ch$ gerektirmesinin do\u grulu\u gu, 
apa\c c\i k de\u gildir;
ama bu do\u grulu\u gu g\"osterece\u giz.

Ondan \"once $\pow{\upomega}\approx\R$ e\c slenikli\u gini g\"osterece\u giz.
\begin{equation*}
  \N=\{x\in\upomega\colon x\geq1\}
\end{equation*}
\glossary{$\N$}% 
olsun, ve $\N\times\N$ \c carp\i m\i nda $\sim$, 
\begin{equation*}
  (a,b)\sim(c,d)\liff a\cdot d=b\cdot c
\end{equation*}
ko\c sulunu sa\u glayan ikili ba\u g\i nt\i\ olsun.  

\begin{theorem}
$\sim$ bir denklik ba\u g\i nt\i s\i d\i r.  
\end{theorem}

\ktk

O zaman
$(a,b)\in\N\times\N$ ise
\begin{equation*}
  a/b=\frac ab=\{(x,y)\in\N\times\N\colon(x,y)\sim(a,b)\}
\end{equation*}
olsun, ve
\begin{align*}
  \Qp&=\{x/y\colon(x,y)\in\N\times\N\},&
\Q&=(\Qp\sqcup\Qp)\cup\{0\}
\end{align*}
\glossary{$\Qp$, $\Q$}%
olsun.
$\Q$ k\"umesinin elemanlar\i,
\textbf{kesirli say\i lar}%
\index{kesirli say\i lar}
(\eng{rational numbers})
olarak anla\c s\i labilir.

\begin{theorem}
  $\Q\approx\upomega$.
\end{theorem}

\ktk

\begin{theorem}
$x\mapsto x/1\colon\N\xrightarrow{\preccurlyeq}\Qp$.
Ayr\i ca
$\Qp$ k\"umesinde
\begin{align*}
  \frac ab+\frac cd&=\frac{a\cdot d+b\cdot c}{b\cdot d},&
\frac ab\cdot\frac cd&=\frac{ac}{bd},&
\frac{a/b}{c/d}&=\frac{a\cdot d}{b\cdot c}
\end{align*}
ve
\begin{equation*}
  \frac ab<\frac cd\liff a\cdot d<b\cdot c
\end{equation*}
tan\i mlar\i\ yap\i labilir.  
O zaman $<$, $\Qp$ k\"umesinin do\u grusal s\i ralamas\i d\i r, 
ve $a<b$ ise
\begin{equation*}
a<\frac{2a+b}3<\frac{a+2b}3<b.
\end{equation*}
\end{theorem}

\ktk

\c Simdi $\Rp$,%
\glossary{$\Rp$, $\R$}
\begin{gather*}
  0\pincluded a\pincluded\Qp,\\
  x<y\land y\in a\lto x\in a,\\
  y\in a\lto\Exists z(y<z\land z\in a)
\end{gather*}
ko\c sullar\i n\i\ sa\u glayan $a$ k\"umelerinin k\"umesi olsun,
ve
\begin{equation*}
\R=(\Rp\sqcup\Rp)\cup\{0\}
\end{equation*}
olsun.
$\R$ k\"umesinin elemanlar\i,
\textbf{ger\c cel say\i lar}%
\index{ger\c cel say\i lar}
(\eng{real numbers})
olarak anla\c s\i labilir.
Ayr\i ca $\R$ k\"umesine \textbf{kontin\"u}%
\index{kontin\"}
(\eng{continuum})
denebilir.

\begin{theorem}
  $\R\approx\pow{\upomega}$.
\end{theorem}

\begin{proof}
$n\in\upomega$ ve $\sigma\in{}^n2$ ise
\begin{equation*}
  f(\sigma)=1+\sum_{\xi<n}\frac{2\sigma(\xi)}{3^{\xi}}
\end{equation*}
olsun.  
Yani \"ozyinelemeyi kullanarak 
$f(0)=1$ olsun, ve $\sigma\in{}^m2$ ise
\begin{align*}
f\Bigl(\sigma\cup\bigl\{(m,0)\bigr\}\Bigr)&=f(\sigma),&
f\Bigl(\sigma\cup\bigl\{(m,1)\bigr\}\Bigr)&=f(\sigma)+2/3^m
\end{align*}
olsun.  \c Simdi $\sigma\in{}^{\upomega}2$ ise
\begin{equation*}
  g(\sigma)=\Bigl\{x\in\Qp\colon\Exists{\eta}\bigl(\eta<\upomega\land
  x<f(\sigma\restriction\eta)\bigr)\Bigr\}
\end{equation*}
olsun.  O zaman $g\colon{}^{\upomega}2\to\Rp$.  Asl\i nda $g$ bir
g\"ommedir, \c c\"unk\"u $\sigma\restriction n=\tau\restriction n$ ama
$\sigma(n)=0$ ve $\tau(n)=1$ ise, o zaman
\begin{equation*}
  f(\sigma)+1/3^n\in g(\tau)\setminus g(\sigma).
\end{equation*}
Ayr\i ca $\Qp\approx\upomega$ ve $\Rp\subseteq\pow{\Qp}$, dolay\i s\i yla
\begin{align*}
  {}^{\upomega}2\preccurlyeq\Rp\preccurlyeq\pow{\Qp}\approx\pow{\upomega}\approx{}^{\upomega}2.
\end{align*}
Schr\"oder--Bernstein Teoremine g\"ore ${}^{\upomega}2\approx\Rp$.
\end{proof}


\begin{theorem}\label{thm:Sierpinski}
$\gch$ do\u gruysa, her k\"ume iyi\-s\i ralanabilir.
\end{theorem}

\begin{proof}
Her $a$ k\"umesi, $\pow{a\sqcup\upomega}$ k\"umesine g\"om\"ulebilir,
\c c\"unk\"u
\begin{equation*}
x\mapsto\{(x,0)\}\colon a\xrightarrow{\preccurlyeq}\pow{a\sqcup\upomega}.
\end{equation*}
Bu nedenle $\pow{a\sqcup\upomega}$ iyi\-s\i ralanabilirse, 
$a$ k\"umesi de iyi\-s\i ralanabilir.  

Genelde $c\included b$ ve $\card(b\setminus c)=1$ ise
\begin{equation*}
  \pow b\approx\pow c\sqcup\pow c.
\end{equation*}
Ayr\i ca $a\sqcup\upomega\approx a\sqcup\upomega'$, dolay\i s\i yla
\begin{equation*}
\pow{a\sqcup\upomega}
\approx\pow{a\sqcup\upomega'}
\approx\pow{a\sqcup\upomega}\sqcup\pow{a\sqcup\upomega}.
\end{equation*}
O zaman dedi\u gimize g\"ore,
$a$ k\"umesini iyis\i ralamak i\c cin,
$a\approx a\sqcup a$ e\c slenikli\u gini varsayabiliriz.  
\"Ozel olarak $a$ k\"umesi sonsuzdur.
O zaman Schr\"oder--Bernstein Teoremi
ve t\"umevar\i mdan,
her $n$ do\u gal say\i s\i\ i\c cin
\begin{align*}
\pow[n]a&\approx\pow[n]a\sqcup\pow[n]a,&
  \pow[n]a&\approx\pow[n]a\sqcup\{0\}.
\end{align*}
\c Simdi $b\preccurlyeq\pow a$ olsun.  O zaman
\begin{equation*}
a\preccurlyeq b\sqcup a\preccurlyeq\pow a\sqcup\pow a\approx\pow a.
\end{equation*}
$\gch$ varsay\i m\i m\i za g\"ore 
$a\approx b\sqcup a$ veya $b\sqcup a\approx\pow a$.  
Birinci durumda $b\preccurlyeq a$.  \.Ikinci durumda
\begin{equation*}
b\sqcup a\approx\pow{a\sqcup a}\approx\pow a\times\pow a.
\end{equation*}
\c Simdi
\begin{align*}
  f&\colon b\sqcup a\xrightarrow{\approx}\pow a\times\pow a,&
\pi(x,y)&=x
\end{align*}
olsun.  
Cantor'un Teoremine g\"ore,
$a\approx f\bigl[a\times\{1\}\bigr]$ oldu\u gundan
\begin{equation*}
  \pi\Bigl[f\bigl[a\times\{1\}\bigr]\Bigr]\pincluded\pow a,
\end{equation*}
dolay\i s\i yla $\pow a\setminus\pi\bigl[f[a\times\{1\}]\bigr]$ fark\i n\i n 
$c$ eleman\i\ vard\i r.  E\u ger
\begin{equation*}
d=\Bigl\{x\in b\colon\pi\bigl(f(x,0)\bigr)=c\Bigr\}
\end{equation*}
ise, o zaman $f[d]=\{c\}\times\pow a$, dolay\i s\i yla
\begin{align*}
d&\approx\pow a,&
\pow a&\preccurlyeq b,&
b\approx\pow a.
\end{align*}
K\i saca, g\"osterdi\u gimiz
\begin{align*}
a\sqcup a&\approx a&
&\land&
b&\preccurlyeq\pow a&
&\lto&
b&\approx\pow a&
&\lor&
b&\preccurlyeq a.  
\end{align*}
Hartogs'un Teoremine (yani \Teoreme{thm:Hartogs}) g\"ore 
bir $\beta$ ordinali i\c cin $\beta\not\preccurlyeq a$, 
ama $\beta\preccurlyeq\pow[4]a$.  
O zaman $\beta$, 
ya $\pow[4]a$, 
ya $\pow[3]a$, 
ya $\pow[2]a$, 
ya da $\pow a$ ile e\c sleniktir.  
Her durumda $a\preccurlyeq\beta$, dolay\i s\i yla $a$ iyi\-s\i ralanabilir.
\end{proof}

\begin{theorem}
 $\gch\lto\ch$.  
\end{theorem}

\begin{proof}
$\gch$ do\u gru ise,
son teoreme g\"ore her k\"ume iyi\-s\i ralanabilir.
Bu durumda her k\"umenin cardinali vard\i r.  
\"Ozel olarak $\pow{\upomega}$ k\"ume\-sinin 
$\aleph_{\beta}$ cardinali vard\i r, 
ve $\beta\neq0$, dolay\i s\i yla
\begin{equation*}
\upomega\prec\aleph_1\preccurlyeq\pow{\upomega}.\qedhere
\end{equation*}
\end{proof}

\section{Kardinaller kuvvetleri}

Genelle\c stirilmi\c s Kontin\"u Hipotezini kabul etmeyece\u giz, 
ama \Teoremin{thm:Sierpinski} sonucunu kabul edece\u giz.  
Se\c cim Aksiyomunun \c cok bi\c cimleri vard\i r; 
ama bizim i\c cin en uygun bi\c cimi, a\c sa\u g\i dad\i r.

\begin{axiom}[Se\c cim]\label{ax:choice}
  Her k\"ume iyi\-s\i ralanabilir.
\end{axiom}

Her \emph{say\i labilen} k\"ume zaten iyi\-s\i ralanabildi.  
\c Simdi her say\i lamaz sonsuzluktaki k\"ume iyi\-s\i ralanabilir, 
yani bundan $0$ adl\i\ bir eleman\i\ se\c cilebilir, 
ve ondan sonra $1$ adl\i\ eleman\i, vesaire.

\"Oyleyse her k\"umenin kardinali vard\i r.  
Kuvvet K\"umesi Aksiyomunun sayesinde ${}^{\beta}\alpha$ s\i n\i
f\i\ bir k\"umedir, \c c\"unk\"u
\begin{equation*}
  {}^{\beta}\alpha\included\pow{\beta\times\alpha}.
\end{equation*}
O zaman
\begin{equation*}
  \kappa^{\lambda}=\card({}^{\lambda}\kappa)
\end{equation*}%
\glossary{$\kappa^{\lambda}$}
tan\i m\i n\i\ yapabiliriz.  (Bu kuvvet, \emph{ordinaller} kuvveti de\u gildir.)

\begin{theorem}
T\"um $\kappa$, $\lambda$, $\mu$, ve $\nu$ kardinalleri i\c cin
\begin{align*}
&\begin{gathered}
	0<\lambda\lto 0^{\lambda}=0,\\
	\kappa^0=1,\\
	1^{\lambda}=1,\\
	\kappa^1=\kappa,
	\end{gathered}&
&\begin{gathered}
	\kappa^{\lambda\cardsum\mu}=\kappa^{\lambda}\cardprod\kappa^{\mu},\\
	\kappa^{\lambda\cardprod\mu}=(\kappa^{\lambda})^{\mu},\\
	\kappa\leq\mu\land\lambda\leq\nu\lto\kappa^{\lambda}\leq\mu^{\nu}.	
\end{gathered}
\end{align*}
\end{theorem}

\ktk

\begin{theorem}
Her $\kappa$ i\c cin
\begin{equation*}
\kappa<2^{\kappa}.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}
$2\leq\kappa$, $1\leq\lambda$, ve $\aleph_0\leq\max\{\kappa,\lambda\}$ olsun.  O zaman
\begin{gather}\label{eqn:kl}
\kappa\leq 2^{\lambda}\lto\kappa^{\lambda}=2^{\lambda},\\\label{eqn:lk}
\lambda\leq\kappa\lto\kappa\leq\kappa^{\lambda}\leq2^{\kappa}.
\end{gather}
\end{theorem}

\begin{proof}
Hipoteze g\"ore $\kappa\leq2^{\lambda}$ ise $2\leq\kappa\leq 2^{\lambda}$ ve $\lambda$ sonsuzdur, dolay\i s\i yla
\begin{equation*}
2^{\lambda}\leq\kappa^{\lambda}\leq(2^{\lambda})^{\lambda}=2^{\lambda\cardprod\lambda}=2^{\lambda}.
\end{equation*}
Ayr\i ca $\lambda\leq\kappa$ ise $\kappa$ sonsuzdur, dolay\i s\i yla
\begin{equation*}
\kappa\leq\kappa^{\lambda}
\leq(2^{\kappa})^{\lambda}
=2^{\kappa\cardprod\lambda}
=2^{\kappa}.\qedhere
\end{equation*}
\end{proof}

Teoremin \eqref{eqn:kl} ile \eqref{eqn:lk} gerektirmelerinde $\kappa\leq2^{\lambda}$ ile $\lambda\leq\kappa$ ko\c sullar\i, ayn\i\ anda do\u gru olabilir.
Bu durumda \eqref{eqn:kl} gerektirmesi, di\u gerinden daha \c cok
bilgi verir.  B\"oylece \eqref{eqn:lk} sat\i r\i ndaki c\"umlenin yerine 
\begin{equation*}
2^{\lambda}<\kappa\lto\kappa\leq\kappa^{\lambda}\leq2^{\kappa}
\end{equation*}
c\"umlesi konulabilir.  \"Orne\u gin
\begin{gather*}
	2\leq\kappa\leq2^{\aleph_0}\lto\kappa^{\aleph_0}=2^{\aleph_0},\\
2^{\aleph_0}<\kappa\lto\kappa\leq\kappa^{\aleph_0}\leq2^{\kappa}.
\end{gather*}


\c Simdi
\begin{gather*}
\beth_0=\upomega,\\
\beth_{\alpha+1}=2^{\beth_{\alpha}},\\
\beta\text{ limit}\lto\beth_{\beta}=\sup\{\beth_{\xi}\colon\xi<\beta\}
\end{gather*}%
\glossary{$\beth_{\alpha}$}
olsun.  (Burada $\beth$, \.Ibrani \emph{beth} harf\/idir.)

\begin{theorem}
$\xi\mapsto\beth_{\xi}$ g\"ondermesi, normal bir i\c slemidir.
$\gch$ do\u grudur ancak ve ancak her $\alpha$ ordinali i\c cin
\begin{equation*}
\aleph_{\alpha}=\beth_{\alpha}.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}
T\"um $\kappa$ ve $\lambda$ i\c cin
\begin{gather*}
2\leq\kappa\leq\beth_{\alpha+1}\lto\kappa^{\beth_{\alpha}}=\beth_{\alpha+1},\\
1\leq\lambda\leq\beth_{\alpha}\lto{\beth_{\alpha+1}}^{\lambda}=\beth_{\alpha+1}.
\end{gather*}
\end{theorem}

\ktk

\c Simdi $\lambda\leq\kappa$ (veya $2^{\lambda}<\kappa$) durumunu iki duruma b\"olece\u giz.

\subsection{Kof{}inall\i k}

Sonsuz bir $\kappa$ kardinali limit ordinali oldu\u gundan
\begin{equation*}
\kappa=\sup\{\xi\colon\xi<\kappa\}=\bigcup_{\xi<\kappa}\xi.
\end{equation*}
Bazen bir kardinal, 
kendisinden k\"u\c c\"uk bir altk\"umenin supremumu\-dur.  
\"Orne\u gin $\upomega<\aleph_{\upomega}$, ama
\begin{equation*}
\aleph_{\upomega}=\sup\{\aleph_x\colon x\in\upomega\}.
\end{equation*}
Genelde $\alpha$ limit, $b\included\alpha$, ve
\begin{equation*}
  \Forall{\xi}\bigl(\xi<\alpha\lto\Exists{\eta}(\eta\in b\land\xi<\eta)\bigr)
\end{equation*}
ise,
$b$ altk\"umesi, $\alpha$ ordinalinin \textbf{s\i n\i rs\i z}%
\index{s\i n\i rs\i z}
(\eng{unbounded})%%%%%
\footnote{\cite[\S13.7, s.~156]{Nesin-AKK} kayna\u g\i nda $b$, 
$\alpha$ ordinalinin bir \emph{kof{}inal} altk\"umesidir.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
altk\"umesidir.
Bu durumda
\begin{equation*}
  \alpha=\sup(b).
\end{equation*}
\"Orne\u gin her limit ordinali, kendisinde s\i n\i rs\i zd\i r.  
Ayr\i ca $\{\aleph_x\colon x\in\upomega\}$, 
$\aleph_{\upomega}$ ordinalinde s\i n\i rs\i zd\i r.
Bir limit ordinalinin s\i n\i rs\i z alt\-k\"umelerinin 
en k\"u\c c\"uk kardinaline, 
ordinalin
\textbf{kof\/inalli\u gi}%
\index{kof{}inallik}
(\eng{cofinality})
denir, ve bu kardinal,
$\cof{\alpha}$
\glossary{$\cof{\alpha}$}%
olarak yaz\i labilir.  Yani
\begin{equation*}
\cof{\alpha}=\min\{\card(x)\colon x\included\alpha\land\sup(x)=\alpha\}.
\end{equation*}
Ayr\i ca, tan\i ma g\"ore,
\begin{align*}
  \cof0&=0,&\cof{\alpha+1}&=1
\end{align*}
denebilir,
ama bu durumlar\i\ kullanmayaca\u g\i z.

\begin{theorem}
Her $\alpha$ limit ordinali i\c cin,
tan\i m k\"umesi $\cof{\alpha}$ olan,
de\u ger k\"umesi $\alpha$ ordinalinin s\i n\i rs\i z bir altk\"umesi olan,
kesin artan bir g\"onderme vard\i r.
\end{theorem}

\begin{proof}
$f\colon\cof{\alpha}\to\alpha$ olsun, 
ve $f[\alpha]$, $\alpha$ ordinalinin s\i n\i rs\i z bir altk\"umesi olsun.  
\"Ozyinelemeyle, tan\i m k\"umesi $\cof{\alpha}$ olan,
\begin{equation*}
g(\beta)=\max\Bigl(f(\beta),\sup\bigl(g[\beta]\bigr)\Bigr)
\end{equation*}
ko\c sulunu sa\u glayan bir $g$ g\"ondermesi vard\i r.
E\u ger $\beta<\cof{\alpha}$ ve $g[\beta]\included\alpha$ ise, 
o zaman $g[\beta]$, $\alpha$ ordinalinin s\i n\i rs\i z altk\"umesi de\u gil,
dolay\i s\i yla $g(\beta)\in\alpha$; ayr\i ca $f(\beta)\leq g(\beta)$.  
\"Oyleyse $g$, istedi\u gimiz gibidir.
\end{proof}

\begin{theorem}
$\alpha$ ve $\beta$ limit ordinalleri olsun.
E\u ger $f\colon\alpha\to\beta$ ve kesin artan ise,
ve $\beta=\bigcup f[\alpha]$ ise,
o zaman
\begin{equation*}
\cof{\alpha}=\cof{\beta}.
\end{equation*}
\end{theorem}

\begin{proof}
$\cof{\beta}\leq\cof{\alpha}$ ve $\cof{\alpha}\leq\cof{\beta}$
e\c sitsizliklerini kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$g\colon\cof{\alpha}\to\alpha$
ve $\bigcup g[\cof{\alpha}]=\alpha$ olsun.  
%$(f\circ g)[\gamma]$ g\"or\"unt\"us\"un\"un $\beta$ ordinalinde s\i n\i rs\i z oldu\u gunu kan\i tlayaca\u g\i z.  
$\delta<\beta$ ise,
hipoteze g\"ore $\alpha$ ordinalinin bir $\theta$ eleman\i\ i\c cin
\begin{equation*}
\delta<f(\theta).
\end{equation*}
O zaman $\cof{\alpha}$ kardinalinin bir $\iota$ eleman\i\ i\c cin
\begin{align*}
\theta&<g(\iota),&
\delta<f(\theta)&<f\bigl(g(\iota)\bigr).
\end{align*}
\"Oyleyse $\bigcup(f\circ g)[\cof{\alpha}]=\beta$,
dolay\i s\i yla $\cof{\beta}\leq\cof{\alpha}$.
\item
$h\colon\cof{\beta}\to\beta$ ve $\bigcup h[\cof{\beta}]=\beta$ olsun.
$\delta<\cof{\beta}$ ise
\begin{equation*}
k(\delta)=\min\{\xi\in\alpha\colon h(\delta)<f(\xi)\}
\end{equation*}
olsun.  O zaman $k\colon\cof{\beta}\to\alpha$.  
E\u ger $\theta\in\alpha$ ise, o zaman
$\cof{\beta}$ kardinalinin
\begin{equation*}
  f(\theta)<h(\delta)
\end{equation*}
ko\c sulunu sa\u glayan bir $\delta$ eleman\i\ vard\i r.
O zaman
\begin{equation*}
f(\theta)<h(\delta)<f(k(\delta)),  
\end{equation*}
dolay\i s\i yla $\theta<k(\delta)$,
\c c\"unk\"u $f$ kesin artand\i r.
\"Oyleyse $\bigcup k[\cof{\beta}]=\alpha$,
dolay\i s\i yla
$\cof{\alpha}\leq\cof{\beta}$ 
ve asl\i nda $\cof{\alpha}=\cof{\beta}$.\qedhere
\end{asparaenum}
\end{proof}

\"Ozel durum olarak $\bm F$ normal ve $\alpha$ limit ise
\begin{equation*}
  \cof{\bm F(\alpha)}=\cof{\alpha}.
\end{equation*}

\begin{theorem}
  $\alpha$ limit ise $\cof{\aleph_{\alpha}}=\cof{\alpha}$.
\end{theorem}

\begin{proof}
$\xi\mapsto\aleph_{\xi}$ normaldir.
\end{proof}

\begin{theorem}
  Cantor normal bi\c ciminde
\begin{equation*}
\alpha=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_n}\cdot a_n
\end{equation*}
ve $\alpha_n>0$ ise, o zaman
\begin{equation*}
\cof{\alpha}
=\begin{cases}
	\upomega,&\text{ e\u ger $\alpha_n$ bir ard\i lsa},\\
	\cof{\alpha_n},&\text{ e\u ger $\alpha_n$ bir limitse}.
\end{cases}
\end{equation*}
\end{theorem}

\begin{proof}
Son teoreme g\"ore
$\alpha$ limit, $\gamma\geq1$, ve $\delta\geq2$ ise
\begin{equation*}
  \cof{\alpha} 
=\cof{\beta+\alpha} 
=\cof{\gamma\cdot\alpha}
=\cof{\delta^{\alpha}}.\qedhere
\end{equation*}
\end{proof}


Bazen bu hesaplama bize yard\i m etmez.  Mesela $f(0)=0$ ve $f(n+1)=\upomega^{f(n)}$ ve $\alpha=\sup(f[\upomega])$ ise, yani
\begin{equation*}
\alpha =\sup\{0,1,\upomega,\upomega^{\upomega},\upomega^{\upomega^{\upomega}},\dots\}
\end{equation*}
ise, o zaman $\cof{\alpha}=\upomega$, ama $\alpha=\upomega^{\alpha}$.

\begin{theorem}\label{thm:succ-cof}
Her $\alpha$ ordinali i\c cin
\begin{equation*}
\cof{\aleph_{\alpha+1}}=\aleph_{\alpha+1}.  
\end{equation*}
\end{theorem}

\begin{proof}
$\beta<\aleph_{\alpha+1}$ ve $f\colon\beta\to\aleph_{\alpha+1}$ olsun.  O zaman
\begin{equation*}
\sup(f[\beta])=\bigcup_{\xi<\beta}f(\xi).
\end{equation*}
Bu bile\c simden $\aleph_{\alpha}\times\aleph_{\alpha}$ \c carp\i m\i na giden bir $h$ g\"ommesini tan\i mlayaca\u g\i z.
Se\c cim Aksiyomu sayesinde $\bigcup\{{}^{\xi}\aleph_{\alpha}\colon\xi<\aleph_{\alpha+1}\}$ k\"umesi iyi\-s\i ralanabilir.  Bu s\i ralamaya g\"ore $\delta<\aleph_{\alpha+1}$ ise  ${}^\delta\aleph_{\alpha}$ k\"umesinin en k\"u\c c\"uk \emph{g\"ommesi,} $g_{\delta}$ olsun.  O zaman $\gamma<\sup(f[\beta])$ ise
\begin{align*}
\delta&=\min\{z\in\beta\colon\gamma<f(z)\},& h(\gamma)&=\bigl(g_{\beta}(\delta),g_{\delta}(\gamma)\bigr)
\end{align*}
olsun.
B\"oylece
\begin{equation*}
\card\bigl(\sup(f[\beta])\bigr)\leq\card(\aleph_{\alpha}\times\aleph_{\alpha})=\aleph_{\alpha},
\end{equation*}
dolay\i s\i yla $\sup(f[\beta])<\aleph_{\alpha+1}$.  Sonu\c c olarak $\cof{\aleph_{\alpha+1}}=\aleph_{\alpha+1}$.
\end{proof}

\subsection{Hesapmalar}

\begin{theorem}
$2\leq\kappa$, $1\leq\lambda$, ve $\aleph_0\leq\max\{\kappa,\lambda\}$ olsun.  O zaman
\begin{align*}
	\lambda\geq\cof{\kappa}&\lto\kappa<\kappa^{\lambda},\\
	\gch\land\lambda<\cof{\kappa}&\lto\kappa=\kappa^{\lambda}.
\end{align*}
\end{theorem}

\begin{proof}
$\cof{\kappa}\leq\lambda$ ise ${}^{\lambda}\kappa$ k\"umesinin
\begin{equation*}
\kappa=\bigcup_{\xi<\lambda}f(\xi)
\end{equation*}
ko\c sulunu sa\u glayan bir $f$ eleman\i\ vard\i r.  
\c Simdi $\xi\mapsto g_{\xi}\colon\kappa\to{}^{\lambda}\kappa$ olsun.  
O zaman ${}^{\lambda}\kappa$ k\"umesinin 
$\{g_{\xi}\colon\xi<\kappa\}$ k\"umesinde olmayan bir
\begin{equation*}
\eta\mapsto\min\Bigl(\kappa\setminus\bigl\{g_{\xi}(\eta)\colon\xi<f(\eta)\bigr\}\Bigr)
\end{equation*}
eleman\i\ vard\i r.  

\c Simdi $\lambda<\cof{\kappa}$ olsun.  
O zaman \Teoremin{thm:succ-cof} kan\i t\i ndaki gibi
\begin{equation*}
{}^{\lambda}\kappa
=\bigcup_{\xi<\kappa}{}^{\lambda}\xi
=\bigcup_{\lambda\leq\xi<\kappa}{}^{\lambda}\xi
\preccurlyeq\bigcup_{\lambda\leq\xi<\kappa}{}^{\lambda}(\card{\xi})
=\bigcup_{\substack{\lambda\leq\xi<\kappa\\\xi\in\cn}}{}^{\lambda}\xi
\preccurlyeq\bigcup_{\substack{\lambda\leq\xi<\kappa\\\xi\in\cn}}{}^{\xi}2.
\end{equation*}
E\u ger $\gch$ do\u gruysa $\mu<\kappa\lto2^{\mu}\leq\kappa$, dolay\i s\i yla $\kappa^{\lambda}\leq\kappa$.
\end{proof}

\c Simdi, g\"osterdiklerimize g\"ore, e\u ger $\kappa+\lambda$
sonsuzsa, o zaman
\begin{align*}
2\leq\kappa\leq2^{\lambda}&\lto\kappa^{\lambda}=2^{\lambda},\\
\cof{\kappa}\leq\lambda\leq\kappa&\lto\kappa<\kappa^{\lambda}\leq2^{\kappa},\\
1\leq\lambda<\cof{\kappa}&\lto\kappa\leq\kappa^{\lambda}\leq2^{\kappa}.
\end{align*}
Ayr\i ca%, e\u ger $\gch$ do\u gruysa,
\begin{equation*}
\gch\lto
\kappa^{\lambda}=
\begin{cases}
  \lambda^+,&\text{ e\u ger $2\leq\kappa<\lambda$ ise},\\
\kappa^+,&\text{ e\u ger $\cof{\kappa}\leq\lambda\leq\kappa$ ise},\\
\kappa,&\text{ e\u ger $1\leq\lambda<\cof{\kappa}$ ise}.
\end{cases}
\end{equation*}
\"Ozel olarak% e\u ger $\gch$ do\u gruysa,
\begin{equation*}
\gch\lto
{\aleph_{\alpha}}^{\aleph_{\beta}}=
\begin{cases}
	\aleph_{\beta+1},&\text{ e\u ger $\alpha<\beta$ ise},\\
	\aleph_{\alpha+1},&\text{ e\u ger $\cof{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$ ise},\\
	\aleph_{\alpha},&\text{ e\u ger $\aleph_{\beta}<\cof{\alpha}$ ise}.	
\end{cases}
\end{equation*}

%\end{comment}

\appendix

\chapter{Schr\"oder--Bernstein Teoremi}\label{Z-SB}

Tekrar Schr\"oder--Bernstein Teoremine
(yani \Teoreme{thm:SB}) bakaca\u g\i z.

\begin{theorem*}[Schr\"oder--Bernstein]
T\"um $a$ ve $b$ k\"umeleri i\c cin
\begin{equation*}
a\preccurlyeq b\land b\preccurlyeq a\lto a\approx b.
\end{equation*}
\end{theorem*}

\begin{proof}[Kan\i t (Zermelo \cite{Zermelo-invest}).]
$f\colon a\xrightarrow{\preccurlyeq}b$ 
ve $g\colon b\xrightarrow{\preccurlyeq}a$ olsun.  Bu durumda
\begin{align*}
(g\circ f)[a]&\included g[b]\included a,&
g[b]&\approx b.
\end{align*}
Biz $a\approx g[b]$ e\c slenikli\u gini kan\i tlayaca\u g\i z.  Sonu\c c olarak $a\approx b$ olacakt\i r.

$a$ k\"umesinden $g[b]$ k\"umesine giden bir $h$ e\c slemesini tan\i mlayabilirsek, herhalde $a$ k\"umesinin bir $c$ altk\"umesi i\c cin
\begin{equation}\label{h}
h=\{(x,x)\colon x\in c\}\cup\{(x,(g\circ f)(x))\colon x\in a\setminus c\}
\end{equation}
olacakt\i r.  O halde
\begin{equation}\label{cup}
c\cup(g\circ f)[a\setminus c]=g[b]
\end{equation}
olmal\i d\i r, \c c\"unk\"u $h[a]=g[b]$ olacakt\i r.  Ayr\i ca
\begin{equation}\label{cap}
c\cap(g\circ f)[a\setminus c]=\emptyset
\end{equation}
olmal\i d\i r, \c c\"unk\"u $h$ bir g\"omme olacakt\i r.  O zaman
\begin{equation*}
c=g[b]\setminus(g\circ f)[a\setminus c]
\end{equation*}
olmal\i d\i r.  $g\circ f$ birebir oldu\u gundan
\begin{equation*}
(g\circ f)[a\setminus c]=(g\circ f)[a]\setminus(g\circ f)[c],
\end{equation*}
dolay\i s\i yla
\begin{equation*}
c=g[b]\setminus\bigl((g\circ f)[a]\setminus(g\circ f)[c]\bigr)
\end{equation*}
olmal\i d\i r.  $(g\circ f)[c]\included(g\circ f)[a]\included g[b]$ oldu\u gundan
\begin{equation}\label{c}
c=\bigl(g[b]\setminus(g\circ f)[a]\bigr)\cup(g\circ f)[c]
\end{equation}
olmal\i d\i r.  Ters olarak, e\u ger $c$, \eqref c sat\i r\i ndaki gibiyse, o zaman \eqref{cup} ile \eqref{cap} sat\i rlar\i\ do\u grudur, ve sonu\c c olarak, $g\circ f$ birebir oldu\u gundan, \eqref h sat\i r\i ndaki gibi $h$ g\"ondermesi, $a$ k\"umesinden $g[b]$ k\"umesine giden bir e\c slemedir.

\c Simdi \"oyle bir $c$ k\"umesini bulmal\i y\i z.  O zaman
\begin{equation*}
\bm A=\bigl\{x\colon \bigl(g[b]\setminus(g\circ f)[a]\bigr)\cup(g\circ f)[x]\included x\included a\big\}
\end{equation*}
olsun.  Bu durumda $a\in\bm A$, dolay\i s\i yla $\bigcap\bm A$ bir k\"ume olmal\i d\i r.  Bu k\"ume $c$ olsun.  O zaman $c\in\bm A$ olmal\i d\i r (neden?).  E\u ger \eqref c sat\i r\i\ yanl\i\c s ise, o zaman
\begin{equation*}
d\in c\setminus\Bigl(\bigl(g[b]\setminus(g\circ f)[a]\bigr)\cup(g\circ f)[c]\Bigr)
\end{equation*}
c\"umlesini sa\u glayan bir $d$ vard\i r.  Bu durumda
\begin{align*}
c\setminus\{d\}&\in\bm A,&
\bigcap\bm A&\included c\setminus\{d\},&
c&\included c\setminus\{d\},&
d&\notin c.
\end{align*}
Bu bir \c celi\c skidir.  O zaman \eqref c sat\i r\i\ do\u gru olmal\i d\i r, ve $a\approx g[b]$, dolay\i s\i yla $a\approx b$.
\end{proof}


\backmatter

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