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\begin{document}
\title{Analiz k\i sa s\i nav\i\ 6 \emph{ve \c c\"oz\"umleri}}
\author{David Pierce, MSGS\"U}
\date{9 Nisan 2012}
\maketitle\thispagestyle{empty}
%{\large \textbf{Analiz k\i sa s\i nav\i} (David Pierce), 20 \c Subat 2012}

\begin{problem}
Her Hausdorff topolojik uzayda, her tek noktal\i\ k\"umenin kapal\i\ oldu\u gunu g\"osterin.
\end{problem}

\begin{solution}
  $(X,\tau)$, Hausdorff topolojik bir uzay olsun, ve $a\in X$ olsun.
  $(X,\tau)$ Hausdorff oldu\u gundan $X\setminus\{a\}$ k\"umesinin her
  $b$ eleman\i\ i\c cin \"oyle $a$'n\i n $U$ ve $b$'nin $V$ kom\c
  suluklar\i\ vard\i r ki 
  \begin{equation*}
    U\cap V=\emptyset.
  \end{equation*}
O zaman $V\included X\setminus\{a\}$.  \"Oyleyse $X\setminus\{a\}$,
$b$'nin kom\c sulu\u gudur.  Dolay\i s\i yla $X\setminus\{a\}$ a\c c\i
ktir, ve $\{a\}$, kapal\i d\i r.
\end{solution}

\begin{problem}
Sonsuz bir topolojik uzayda, her tek noktal\i\ k\"ume kapal\i ysa,  bu
uzay Hausdorff olmal\i\ m\i? 
\end{problem}

\begin{solution}
  Hay\i r, olmayabilir.  Uzay\i n topolojisi Fr\'echet topolojisiyse,
  her iki a\c c\i k k\"umenin kesi\c simi bo\c s de\u gildir.  Asl\i
  nda uzay $X$ olsun, ve $U$ ile $V$, $X$'in a\c c\i k altk\"umeleri
  olsun.  O zaman
  \begin{equation*}
    U\cap V=X\setminus((X\setminus U)\cup(Y\setminus V)),
  \end{equation*}
ve bu k\"ume bo\c s de\u gildir, \c c\"unk\"u $(X\setminus
U)\cup(Y\setminus V)$ sonlu, ve $X$ sonsuzdur.
\end{solution}

\end{document}
