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\theoremstyle{definition}
\newtheorem{problem}{Problem}
\newtheorem*{solution}{\c C\"oz\"um}

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%\let\endsolution=\endcomment

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\begin{document}

\title{Aksiyomatik K\"umeler Kuram\i\ (MAT 340)}
\date{12 Nisan 2013}
\author{David Pierce}
\maketitle
\thispagestyle{empty}

\begin{problem}
%\mbox{}
\begin{enumerate}[a)]
\item
\"Oyle bir $\phi(x)$ form\"ul\"u yaz\i n ki
\begin{center}
 $\phi(a)$ do\u grudur ancak ve ancak $a$ ge\c ci\c slidir 
 \end{center}
 ko\c sulu sa\u glans\i n.
\item
Ordinal olmayan ama ge\c ci\c sli bir k\"ume \"orne\u gi yaz\i n.
\item
Ordinal olmayan ama en az \"u\c c eleman\i\ olan ve i\c cerilme taraf\i ndan iyi s\i ralanm\i\c s bir k\"ume \"orne\u gi yaz\i n.
\end{enumerate}
\end{problem}

\begin{solution}
%\mbox{}
\begin{enumerate}[a)]
\item
$\Forall y(y\in x\lto y\included x)$.
\item
$\Bigl\{\bigl\{\{0\}\bigr\},\{0\},0\Bigr\}$.
\item
$\{1,2,3\}$.
\end{enumerate}
\end{solution}

\newpage

\begin{problem}
%\mbox{}
  \begin{enumerate}[a)]
  \item 
Yerle\c stirme Aksiyomu ne diyor?
\item
Yerle\c stirme ve Bo\c s K\"ume Aksiyomlar\i n\i\ kullanarak Ay\i rma
Aksiyomunu kan\i tlay\i n.  
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[a)]
  \item 
Her g\"onderme alt\i nda her k\"umenin g\"or\"unt\"us\"u, bir k\"umedir.  Yani, her $\bm F$ g\"ondermesi i\c cin, e\u ger $a$ k\"umesi, $\bm F$ g\"ondermesinin tan\i m s\i n\i f\i\ taraf\i ndan kapsan\i rsa, o zaman $\bm F[a]$ s\i n\i f\i, bir k\"umedir.   Burada 
\begin{equation*}
\bm F[a]=\{\bm F(x)\colon x\in a\}=\{y\colon\Exists x(x\in a\land\bm F(x)=y)\}.
\end{equation*}
\item
$\bm A\included b$ olsun.  E\u ger $\bm A=0$ ise, o zaman Bo\c s K\"ume Aksiyomuna g\"ore $\bm A$ bir k\"umedir.  $c\in\bm A$ olsun.  O zaman
\begin{equation*}
\{(x,x)\colon x\in\bm A\}\cup\{(x,c)\colon x\in b\setminus\bm A\}
\end{equation*}
ba\u g\i nt\i s\i, $b$ k\"umesinden ayn\i\ k\"umeye giden bir $\bm F$ g\"ondermesidir, ve $\bm F[b]=\bm A$.
  \end{enumerate}
\end{solution}

\newpage

\begin{problem}
\mbox{}
  \begin{enumerate}[a)]
\item
Ordinaller \"uzerinde toplaman\i n tan\i m\i n\i\ verin.
  \item 
  Her $\alpha$ ordinali i\c cin $0+\alpha=\alpha$ e\c sitli\u gini
  kan\i tlay\i n.
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[a)]
\item
$\begin{gathered}[t]
\alpha+0=\alpha,\\
\alpha+\beta'=(\alpha+\beta)',\\
\gamma \text{ limit}\implies\alpha+\gamma=\sup_{\beta<\gamma}(\alpha+\beta).
\end{gathered}$
  \item
  T\"umevar\i m kullanaca\u g\i z.
  \begin{enumerate}[i.]
  \item
  Tan\i mdan $0+0=0$.
  \item
  $0+\beta=\beta$ ise $0+\beta'=(0+\beta)'=\beta'$.
  \item
  $\gamma$ limit ve bunun b\"ut\"un $\beta$ elemanlar\i\ i\c cin $0+\beta=\beta$ ise, o zaman
  \begin{equation*}
0+\gamma=\sup_{\beta<\gamma}(0+\beta)=\sup_{\beta<\gamma}\beta=\gamma.
\end{equation*}
  \end{enumerate}
  \end{enumerate}
\end{solution}

\newpage

\begin{problem}
Cevaplar\i n\i z\i\ k\i saca a\c c\i klay\i n:
  \begin{enumerate}[a)]
  \item 
K\"ume olmayan bir s\i n\i f var m\i d\i r?
\item
S\i n\i f olmayan bir k\"ume var m\i d\i r?
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[a)]
  \item 
Var:  Russell Paradoksuna g\"ore $\{x\colon x\notin x\}$ s\i n\i f\i, k\"ume de\u gil.
\item
Yok:  Her $a$ k\"umesi, $\{x\colon x\in a\}$ s\i n\i f\i na e\c sittir.
  \end{enumerate}
\end{solution}

\begin{problem}
\c C\"oz\"un:
  \begin{enumerate}[a)]
\item
$x+\upomega+y=15+\upomega+16$.
  \item 
$x\cdot\upomega+y\cdot\upomega=(x+y)\cdot\upomega\land
    (x,y)\in\upomega\times\upomega$. 
  \end{enumerate}
\end{problem}

\begin{solution}
\mbox{}
  \begin{enumerate}[a)]
\item
$n\in\upomega$ ise $n+\upomega=\upomega$, ama
\begin{center}
 $\alpha\geq\upomega$ ise $\alpha+\upomega\geq\upomega+\upomega>\upomega+n$,
 \end{center}
dolay\i s\i yla denklemin \c c\"oz\"um k\"umesi, $\upomega\times\{16\}$.
  \item 
$0<n<\upomega$ ise $n\cdot\upomega=\upomega$, dolay\i s\i yla \c c\"oz\"um k\"umesi,
\begin{equation*}
(\upomega\times\{0\})\cup(\{0\}\times\upomega).
\end{equation*}
  \end{enumerate}
\end{solution}

\end{document}
