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\chead{K\"umeler kuram\i\ \"ozeti}
\ihead{10 Mart 2014}

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\begin{document}

\title{Aksiyomatik K\"umeler Kuram\i\ \"Ozeti}
\author{David Pierce}
\date{10 Mart 2014}
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\thispagestyle{empty}
\begin{center}
\large
  Aksiyomatik K\"umeler Kuram\i\ \"Ozeti

David Pierce

10 Mart 2014
\end{center}

$a$, $b$, $c$, \dots, \textbf{k\"umedir.}

$x$, $y$, $z$, \dots, k\"umeler i\c cin \textbf{de\u gi\c skendir.}

$x\in y$ form\"ul\"une g\"ore $x$, $y$'nin \textbf{eleman\i d\i r,}
ve $y$, $x$'i \textbf{i\c cerir.}

\begin{definition}
  $a=b\liff\Forall x(x\in a\liff x\in b)$,
ve bu durumda $a$ ve $b$, birbirine \textbf{e\c sittir.}
\end{definition}

\begin{axiom}
  $a=b\lto\Forall x(a\in x\lto b\in x)$.
\end{axiom}

\begin{theorem}
Her $\phi(x)$ form\"ul\"u i\c cin
  $a=b\lto\phi(a)\lto\phi(b)$.
(Bu teoremi kan\i tlamad\i k.)
\end{theorem}


Her $\phi(x)$ form\"ul\"u, $\{x\colon\phi(x)\}$
\textbf{s\i n\i f\i n\i} tan\i mlar.
$\bm A$, $\bm B$, $\bm C$, s\i n\i ft\i r.

\begin{definition}
$\bm A\included\bm B\liff\Forall x(x\in\bm A\lto x\in\bm B)$, 
ve bu durumda $\bm B$, $\bm A$'y\i\ \textbf{kapsar.}
\end{definition}

\begin{definition}
$\begin{aligned}[t]
\bm A=\bm B&\liff\bm A\included\bm B\land\bm B\included\bm A,\\
a=\bm B&\liff\Forall x(x\in a\liff x\in\bm B),\\
\bm B=a&\liff a=\bm B.
  \end{aligned}$
\end{definition}

\begin{theorem}
  K\"umelerin ve s\i n\i flar\i n e\c sitli\u gi, 
ayn\i l\i k olarak d\"u\c s\"un\"ule\-bilir.
\end{theorem}

\begin{theorem}
Her k\"ume, bir s\i n\i ft\i r: $a=\{x\colon x\in a\}$.
\end{theorem}

\begin{theorem}[Russell Paradoksu]
  Baz\i\ s\i n\i flar k\"ume de\u gildir.
Asl\i nda
$\{x\colon x\notin x\}$
  s\i n\i f\i,
bir k\"ume de\u gildir.
\end{theorem}

\begin{definition}
  $
  \begin{aligned}[t]
    \emptyset&=\{x\colon x\neq x\},\\
a\cup\{b\}&=\{x\colon x\in a\lor x=b\}.
  \end{aligned}$
\end{definition}

\begin{axiom}[Bo\c s K\"ume]
  $\emptyset$ bir k\"umedir: $\Exists xx=\emptyset$.
\end{axiom}

\begin{axiom}[Biti\c stirme]
  $a\cup\{b\}$ bir k\"umedir.
\end{axiom}

\begin{definition}
  $
  \begin{gathered}[t]
  \begin{aligned}[t]
    \emptyset\cup\{a\}&=\{a\},\\
\{a\}\cup\{b\}&=\{a,b\},\\
\{a,b\}\cup\{c\}&=\{a,b,c\},\\
\{a,b,c\}\cup\{d\}&=\{a,b,c,d\},
  \end{aligned}
\\\makebox[4cm]{\dotfill}
  \end{gathered}$
ve
$\begin{gathered}[t]
\begin{aligned}[t]
0&=\emptyset,\\
1&=\{0\},\\
2&=\{0,1\},\\
3&=\{0,1,2\},\\
\end{aligned}
\\\makebox[2cm]{\dotfill}
  \end{gathered}
  $

$a'=a\cup\{a\}$, ve bu k\"ume, $a$'n\i n \textbf{ard\i l\i d\i r.}
\end{definition}

B\"oylece $1=0'$, $2=1'$, $3=2'$, \dots

\begin{definition}
  $
  \begin{aligned}[t]
    \universe&=\{x\colon x=x\},\\
\bigcap\bm A&=\{x\colon\Forall y(y\in\bm A\lto x\in y)\},\\
\bigcup\bm A&=\{x\colon\Exists y(y\in\bm A\land x\in y)\}.
  \end{aligned}$
\end{definition}

\begin{theorem}
  $\bigcap\emptyset=\universe$.
\end{theorem}

\begin{axiom}[Ay\i rma]
  $\bm A\included b$ ise $\bm A$ bir k\"umedir.
\end{axiom}

\begin{theorem}
  $\universe$ bir k\"ume de\u gildir.
\end{theorem}

\begin{theorem}
  $b\in\bm A$ ise $\bigcap\bm A\included b\included\bigcup\bm A$.
\"Ozel olarak $\bm A$ bo\c s de\u gilse $\bigcap\bm A$ bir k\"umedir.
\end{theorem}

\begin{definition}
$\Omega=\{x\colon 0\in x\land\Forall y(y\in x\lto y'\in x)\}$.
\end{definition}

\begin{axiom}[Sonsuzluk]
  $\Omega\neq\emptyset$.
\end{axiom}

\begin{definition}
  $\upomega=\bigcap\Omega$.  
Bu bir k\"umedir, ve elemanlar\i,
\textbf{(von Neumann) do\u gal say\i lar\i d\i r.}
\end{definition}

\begin{theorem}\label{thm:ind}
\begin{minipage}[t]{0.6\linewidth}
\begin{compactenum}
    \item
$0\in\upomega$.
\item
$k\in\upomega\lto k'\in\upomega$.
\item
\textbf{T\"umevar\i m:}
$A\included\upomega$ olsun, ve
  \begin{compactitem}
    \item
$0\in A$, ve
\item
$\Forall x(x\in A\lto x'\in A)$
  \end{compactitem}
varsay\i ls\i n.  O zaman $A=\upomega$.
  \end{compactenum}
\end{minipage}
\end{theorem}

\begin{definition}
  $\Forall x(x\in\bm A\lto x\included\bm A)$ ise 
$\bm A$ \textbf{ge\c ci\c slidir.}
\end{definition}

\begin{theorem}\label{thm:nat-trans}
  Her do\u gal say\i\ ge\c ci\c slidir.
\end{theorem}

\begin{proof}
  T\"umevar\i m.
\end{proof}

\begin{theorem}\label{thm:nat-irr}
$\Forall x(x\in\upomega\lto x\notin x)$.
\end{theorem}

\begin{proof}
  T\"umevar\i m ve \ref{thm:nat-trans} numaral\i\ teorem.
\end{proof}

\begin{theorem}\label{thm:inj}
    $\Forall x\Forall y(x\in\upomega\land y\in\upomega\land x'=y'\lto x=y)$.
\end{theorem}

\begin{proof}
  \ref{thm:nat-irr} numaral\i\ teorem.
\end{proof}

\ref{thm:ind} ve \ref{thm:inj} numaral\i\ teoremler 
ve $\Forall xx'\neq0$ teoremi, 
\textbf{Dedekind--Peano Aksiyomlar\i d\i r.}
Bizim i\c cin aksiyom de\u gil, teoremdir.

\begin{definition}
 $(a,b)=\bigl\{\{a\},\{a,b\}\bigr\}$.
\end{definition}

\begin{theorem}
  $(a,b)=(c,d)\liff a=c\land b=d$.
\end{theorem}

\begin{definition}
  $
  \begin{aligned}[t]
\{(x,y)\colon\phi(x,y)\}
&=\bigl\{z\colon\Exists x\Exists y\bigl(z=(x,y)\land\phi(x,y)\bigr)\bigr\},\\
\bm A\times\bm B&=\{(x,y)\colon x\in\bm A\land y\in\bm B\}.
  \end{aligned}$
\end{definition}

\begin{definition}
  $\bm R\included\universe\times\universe$ ise 
$\bm R$ bir \textbf{ikili ba\u g\i nt\i d\i r.}
Bu durumda
\begin{equation*}
  a\mathrel{\bm R}b
\liff(a,b)\in\bm R.
\end{equation*}
\end{definition}

\begin{definition}
E\u ger $\bm R$ bir ikili ba\u g\i nt\i\ ve
$\bm A$'n\i n t\"um $a$, $b$, ve $c$ elemanlar\i\ i\c cin
\begin{align*}
&\lnot(a\mathrel{\bm R}a),&
&(a\mathrel{\bm R}b\land b\mathrel{\bm R}c \lto a\mathrel{\bm R}c)
\end{align*}
\begin{comment}
  

\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*}


\end{comment}
ise $\bm R$, $\bm A$'da bir \textbf{s\i ralamad\i r.}
\end{definition}

\begin{definition}
  E\u ger $\bm R$, $\bm A$'da bir s\i ralama ve
  \begin{equation*}
    \Forall x\Forall y(x\in\bm A\land y\in\bm A\lto 
x\mathrel{\bm R}y\lor x=y\lor x\mathrel{\bm R}y)
  \end{equation*}
ise $\bm R$, $\bm A$'da bir \textbf{do\u grusal s\i ralamad\i r.}
\end{definition}

\begin{theorem}\label{thm:nat-lin}
  $\in$, her do\u gal say\i da do\u grusal bir s\i ralamad\i r.
\end{theorem}

\begin{proof}
  T\"umevar\i m, 
ve \ref{thm:nat-irr} ve \ref{thm:nat-trans} numaral\i\ teoremler.
\end{proof}

\begin{definition}
  $\bm R$, $\bm A$'da do\u grusal bir s\i ralama olsun.
E\u ger $B\included\bm A$ ve $c\in B$ ve
\begin{equation*}
  \Forall x(x\in B\lto c\mathrel{\bm R}x)
\end{equation*}
ise, o zaman $c$, 
$B$'nin \textbf{en k\"u\c c\"uk} veya \textbf{minimum} eleman\i d\i r.
E\u ger $\bm A$'n\i n bo\c s olmayan 
her altk\"umesinin en k\"u\c c\"uk eleman\i\ varsa,
$\bm A$, $\bm R$ taraf\i ndan \textbf{iyi s\i ralan\i r.}
\end{definition}

\begin{theorem}\label{thm:nat-wo}
  Her do\u gal say\i, $\in$ taraf\i ndan iyi s\i ralan\i r.
\end{theorem}

\begin{proof}
\ref{thm:nat-lin} numaral\i\ teorem ve t\"umevar\i m.
\end{proof}

\begin{definition}
  $\bm A\setminus\bm B=\{x\colon x\in\bm A\land x\notin\bm B\}$.
\end{definition}

\begin{definition}
  $\bm A\pincluded\bm B\liff\bm A\included\bm B\land\bm A\neq\bm B$.
\end{definition}

\begin{theorem}\label{thm:omega-in-psub}
  $\upomega$'da $\in$ ve $\pincluded$ ayn\i\ ba\u g\i nt\i d\i r.
\end{theorem}

\begin{proof}
  $k\in m$ ise 
\ref{thm:nat-trans} ve \ref{thm:nat-irr} numaral\i\ teoremlerle 
$k\pincluded m$.

$k\pincluded m$ ise 
\ref{thm:nat-wo} ve \ref{thm:nat-trans} numaral\i\ teoremleri kullanarak 
\begin{equation*}
 k=\min(m\setminus k). \qedhere
\end{equation*}
\end{proof}

\begin{theorem}
  $\upomega$, $\in$ taraf\i ndan iyi s\i ralan\i r.
\end{theorem}

\begin{proof}
  \ref{thm:nat-trans} ve \ref{thm:nat-irr} numaral\i\ teoremlere g\"ore
$\in$, $\upomega$'da bir s\i ralamad\i r.

$m\notin k$ ve $m\neq k$ ise 
\ref{thm:nat-wo} ve \ref{thm:nat-trans} numaral\i\ teoremleri kullanarak 
\begin{equation*}
 \min(m\setminus k)\included k. 
\end{equation*}
\ref{thm:omega-in-psub} numaral\i\ teoremi kullararak $\min(m\setminus k)=k$.

\ref{thm:nat-wo} numaral\i\ teoremi kullanarak 
$\upomega$ iyi s\i ralanm\i\c st\i r.
\end{proof}

\begin{theorem}
  $\upomega$ ge\c ci\c slidir.
\end{theorem}

\begin{proof}
  T\"umevar\i m.
\end{proof}

\begin{definition}
  Ge\c ci\c sli ve $\in$ taraf\i ndan iyi s\i ralanm\i\c s bir k\"ume, 
bir \textbf{ordinaldir.}
Ordinaller, $\on$
s\i n\i f\i n\i\ olu\c stururlar.
\end{definition}

O zaman $\upomega\included\on$ ve $\upomega\in\on$.

\begin{theorem}[Burali-Forti Paradoksu]
  $\on$ ge\c ci\c sli ve $\in$ taraf\i ndan iyi s\i ralanm\i\c st\i r,
dolay\i s\i yla bir k\"ume olamaz.
\end{theorem}

\end{document}