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\theoremstyle{definition}
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\begin{document}
\title{Analiz k\i sa s\i nav\i\ 7}
\author{David Pierce, MSGS\"U}
\date{16 Nisan 2012}
\maketitle\thispagestyle{empty}
%{\large \textbf{Analiz k\i sa s\i nav\i} (David Pierce), 20 \c Subat 2012}


\begin{problem}
Her topolojik uzayda, iki k\"umenin bile\c siminin kapan\i\c s\i, o
k\"umelerin kapan\i\c slar\i n\i n bile\c simine e\c sit midir? 
\end{problem}

\begin{solution}
  Evet.  $\overline X\subseteq\overline{X\cup Y}$
  ve  $\overline Y\subseteq\overline{X\cup Y}$, dolay\i s\i yla
  \begin{equation*}
    \overline X\cup\overline Y\subseteq\overline{X\cup Y}.
  \end{equation*}
Ayr\i ca $a\in\overline{X\cup Y}\smallsetminus\overline X$ olsun.  O
zaman $a$'n\i n bir $V$ kom\c sulu\u gu i\c cin $V\cap X=\emptyset$, ama
$a$'n\i n her $U$ kom\c sulu\u gu i\c cin $U\cap(X\cup
Y)\neq\emptyset$.  O halde $(U\cap V)\cap Y\neq\emptyset$ (\c
c\"unk\"u $U\cap V$, $a$'n\i n bir kom\c sulu\u gudur).  \"Ozel
olarak $U\cap Y\neq\emptyset$.  \"Oyleyse $a\in\overline Y$.
\end{solution}


\begin{problem}
$M$ bir k\"ume, ve $f\colon M\times M\to \R$ olsun.  E\u ger $M$'nin her $a$, $b$, ve $c$ elemanlar\i
\begin{gather*}
	f(a,b)=0\iff a=b,\\
	f(a,b)\leq f(a,c)+f(b,c)
\end{gather*}
ko\c sullar\i n\i\ sa\u glarsa
$f$, $M$ \"uzerinde bir metrik midir?
\end{problem}

\begin{solution}
  Evet: $c=a$ ise  $f(a,b)\leq f(a,a)+f(b,a)$, yani $f(a,b)\leq
  f(b,a)$.  Ayn\i\ \c sekilde $f(b,a)\leq f(a,b)$.  Sonu\c c olarak
  $f(a,b)=f(b,a)$.  Ayr\i ca $b=a$ ise
  \begin{equation*}
    f(a,a)\leq f(a,c)+f(a,c),
  \end{equation*}
yani $0\leq 2f(a,c)$, dolay\i s\i yla $0\leq f(a,c)$.
\end{solution}

\end{document}