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\begin{document}
%\frontmatter
\title{Ordinal Analiz}
\subtitle{Aksiyomatik K\"umeler Kuram\i\ Dersi}
\author{David Pierce}
\date{7 Kas\i m 2019 tasla\u g\i}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{mat.msgsu.edu.tr/~dpierce/}}

\maketitle

%\frontmatter

%\selectlanguage{english}

%\mainmatter

%\selectlanguage{turkish}

\tableofcontents

\listoffigures

%\mainmatter

\chapter{K\"umeler kuram\i\ olarak matematik}\label{ch:intro}


Matemati\u gi yaparak zaten k\"umelerin ne oldu\u gunu biliyoruz.
\"Orne\u gin
\begin{compactitem}
  \item
    ger\c cel say\i lar bir $\R$,
    \item
      kesirli say\i lar bir $\Q$,
      \item
        tamsay\i lar bir $\Z$,
        \item
          sayma say\i lar bir $\N$
\end{compactitem}
 k\"umesini
olu\c sturur.

\c Simdi $\R$'den $\R$'ye giden bir $f$ g\"ondermesi verilsin.
\"Orne\u gin
\begin{equation*}
  f(x)=\sin x+x^2-2
\end{equation*}
olsun.  O zaman $f$'nin kendisi,
\begin{equation*}
  \{(x,y)\in\R\times\R\colon y=\sin x+x^2-2\}
\end{equation*}
k\"umesidir.
Burada tan\i ma g\"ore
\begin{gather*}
  (a,b)=\bigl\{\{a\},\{a,b\}\bigr\},\\
  A\times B=\{(x,y)\colon x\in A\And y\in B\}.
\end{gather*}

\begin{theorem}
  T\"um $a$, $b$, $c$, ve $d$ i\c cin
    \begin{equation*}
    (a,b)=(c,d)\iff a=c\And b=d.
  \end{equation*}
\end{theorem}

\begin{proof}
  ($\Rightarrow$).
  $(a,b)=(c,d)$ olsun.  O zaman
  \begin{equation*}
    \bigl\{\{a\},\{a,b\}\bigr\}
    =\bigl\{\{c\},\{c,d\}\bigr\}.
  \end{equation*}
  O halde
  \begin{equation*}
    \{a\}\in\bigl\{\{c\},\{c,d\}\bigr\},
  \end{equation*}
  \.Iki durum vard\i r.
  \begin{enumerate}
  \item
    $\{a\}=\{c\}$ ise $a=c$.
  \item
    $\{a\}=\{c,d\}$ ise $c\in\{a\}$, dolay\i s\i yla tekrar $a=c$.
  \end{enumerate}
  \c Simdi
  \begin{equation*}
    \{a,b\}=\{c,d\}
  \end{equation*}
  olmal\i d\i r,
  dolay\i s\i yla $a=c$ oldu\u gundan $b=d$.

  ($\Leftarrow$).
  A\c c\i kt\i r.
\end{proof}

Her $A$ k\"umesinin altk\"umeleri, $A$'n\i n
\begin{equation*}
  \pow A
\end{equation*}
\textbf{kuvvet k\"umesini} olu\c sturur.
\"Orne\u gin
\begin{align*}
  \{a\}&\included\{a,b\},&
  \{a,b\}&\included\{a,b\}
\end{align*}
oldu\u gundan
\begin{align*}
  \{a\}&\in\pow{\{a,b\}},&
  \{a,b\}&\in\pow{\{a,b\}},
\end{align*}
dolay\i s\i yla
\begin{equation*}
  \bigl\{\{a\},\{a,b\}\bigr\}\included\pow{\{a,b\}},
\end{equation*}
yani
\begin{equation*}
  (a,b)\in\pow{\pow{\{a,b\}}}.
\end{equation*}

\begin{theorem}\label{thm:x}
  $A$ ve $B$ k\"ume ise
  \begin{equation*}
    A\times B\included\pow{\pow{A\cup B}}.
  \end{equation*}
\end{theorem}

\ktk

\c Simdi $\R$'den $\pow{\R}$'ye giden,
\begin{equation}\label{eqn:g}
  g(a)=\{x\in\Q\colon x<a\}
\end{equation}
tan\i m\i n\i\ sa\u glayan bir $g$ g\"ondermesi vard\i r.
$\Q$ $\R$'de \emph{yo\u gun} oldu\u gundan
$g$ birebirdir.
Ayr\i ca her $g(a)$ k\"umesi,
$\R$'nin
\begin{gather}\label{eqn:0AR}
  \emptyset\pincluded A\pincluded\R,\\\label{eqn:A}
    b\in A\And c<b\implies c\in A
\end{gather}
ko\c sullar\i n\i\ sa\u glayan bir $A$ altk\"umesidir.

Tam tersine $A$,
elemanlar\i\ ger\c cel say\i\ olan,
verilen \eqref{eqn:0AR} ve \eqref{eqn:A}
ko\c sullar\i n\i\ sa\u glayan bir k\"ume olsun.
O zaman
\begin{compactitem}
\item
  \eqref{eqn:0AR} sayesinde
  $\R\setminus A$ fark\i n\i n bir $c$ eleman\i\ vard\i r, ve
\item
  \eqref{eqn:A} ko\c sulunun
  \begin{equation*}
    c\in\R\setminus A\And b\in A\implies b<c
  \end{equation*}
  kar\c s\i t tersine g\"ore
$c$ $A$'n\i n bir \"usts\i n\i r\i d\i r.
\end{compactitem}
O zaman
\begin{compactitem}
  \item
$A$ bo\c s olmad\i g\i ndan,
\item
$\R$ do\u grusal s\i ralanm\i\c s bir k\"ume olarak
\emph{tam} oldu\u gundan
\end{compactitem}
$A$'n\i n en k\"u\c c\"uk \"usts\i n\i r\i\ vard\i r.
\c Simdi ayr\i ca
\begin{equation}\label{eqn:open}
  \sup A\notin A
\end{equation}
olsun.
O zaman
\begin{equation*}
  g(\sup A)=A.
\end{equation*}
Genel bir teorem vard\i r.

\begin{theorem}
  $(A,<)$,
  u\c csuz yo\u gun do\u grusal s\i ralanm\i\c s bir k\"ume olsun,
  ve $A$'n\i n
  \begin{gather*}
    X\neq\emptyset,\qquad
    X\neq A,\\
    \Forall x\Forall y(y\in X\land x<y\lto x\in X),\\
    \Forall x\Exists y(x\in X\lto x<y\land y\in X)
  \end{gather*}
  c\"umlelerini sa\u glayan $X$ altk\"umeleri $A^*$ k\"umesini olu\c stursun.
  O zaman
  \begin{compactenum}[(a)]
    \item
  $(A^*,\pincluded)$ yap\i s\i\ da
      u\c csuz yo\u gun do\u grusal s\i ralanm\i\c s bir k\"umedir,
    \item
      oraya $(A,<)$
      \begin{equation*}
        x\mapsto\{y\in A\colon y<x\}
      \end{equation*}
      g\"ondermesi alt\i nda g\"om\"ul\"ur,
      \item
        $(A^*,\pincluded)$ tamd\i r,
      \item
        E\u ger $B$, $A^*$'\i n bo\c s olmayan,
        \"usts\i n\i r\i\ olan bir altk\"umesi ise,
        o zaman
        \begin{equation*}
          \sup B=\bigcup B.
        \end{equation*}
  \end{compactenum}
\end{theorem}

\ktk

\c Simdi
$\R$'nin
\eqref{eqn:0AR},
\eqref{eqn:A}, ve \eqref{eqn:open} ko\c sullar\i n\i\ sa\u glayan
$A$ k\"umeleri
 $\mathscr S$ k\"umesini olu\c stursun.
O zaman tan\i m\i\ \eqref{eqn:g} olan $g$ g\"ondermesi,
$\R$'den $\mathscr S$'ye giden bir e\c slemedir.
Ayr\i ca $\R$'yi kullan\i lmadan $\mathscr S$ tan\i mland\i\u g\i ndan
$\R$, $\mathscr S$ olarak tan\i mlanabilir.
K\i saca her ger\c cel say\i,
elemanlar\i\ kesirli say\i\ olan bir k\"ume olarak tan\i mlanabilir.
\"Ozel olarak
\begin{equation*}
  \R\included\pow{\Q}.
\end{equation*}

Ayr\i ca tan\i ma g\"ore
\begin{equation*}
  \Q=\left\{\frac xy\colon(x,y)\in\Z\times(\Z\setminus\{0\})\right\},
\end{equation*}
ve burada
\begin{equation*}
  \frac ab=\frac cd\iff ad=bc.
\end{equation*}
Bir $a/b$ kesirli say\i s\i n\i n iyitan\i mlanmas\i\ i\c cin
$\Z\times(\Z\setminus\{0\})$ \c carp\i m\i nda
\begin{equation*}
  (x,y)\mathrel B(z,w)\iff xw=yz
\end{equation*}
kural\i\
taraf\i ndan tan\i mlanm\i\c s $B$ ba\u g\i nt\i s\i,
bir denklik ba\u g\i nt\i s\i\ olmal\i d\i r.
Bu durumda
\begin{equation*}
  \frac ab=\{(x,y)\in\Z\times(\Z\setminus\{0\})\colon(a,b)\mathrel B(x,y)\},
\end{equation*}
dolay\i s\i yla \Teorem{thm:x} sayesinde
\begin{gather*}
  \frac ab\included\pow{\pow{\Z}},\\
  \Q\included\pow{\pow{\pow{\Z}}}.
\end{gather*}
Benzer \c sekilde
\begin{equation*}
  a-b=\{(x,y)\in\N\times\N\colon a+y=b+x\}
\end{equation*}
olmak \"uzere
\begin{equation*}
  \Z=\{x-y\colon(x,y)\in\N\times\N\},
\end{equation*}
dolay\i s\i yla
\begin{gather*}
  \Z\included\pow{\pow{\pow{\N}}},\\
  \Q\included\pow{\pow{\pow{\pow{\pow{\pow{\N}}}}}},\\
  \R\included\pow{\pow{\pow{\pow{\pow{\pow{\pow{\N}}}}}}}.
\end{gather*}
K\"ume olarak $\N$'nin elemanlar\i\ nas\i l anla\c s\i labilir?


\chapter{Do\u gal say\i lar}\label{ch:ax} % check label


\section{S\i n\i flar ve k\"umeler}\label{sect:sets-classes}

Her \emph{k\"ume} bir \emph{s\i n\i ft\i r,}
ama her s\i n\i f bir k\"ume de\u gildir.

Tan\i ma g\"ore bir \textbf{s\i n\i f,}
\emph{tek serbest de\u gi\c skenli bir form\"ul} taraf\i ndan
tan\i mlan\i r.
Bir s\i n\i f\i n \textbf{elemanlar\i} vard\i r,
ve bunlar,
tan\i mlayan form\"ul\"un sa\u glayanlar\i d\i r.

Baz\i\ s\i n\i flar, k\"umedir.
K\"umelerin \"ozellikleri
bir tan\i m de\u gil,
aksiyomlar taraf\i ndan verilecektir.
\c Simdilik, e\u ger $a$ bir k\"ume ise,
o zaman
\begin{equation*}
  x\in a
\end{equation*}
ifadesi,
serbest de\u gi\c skeni $x$ olan bir form\"uld\"ur,
ve bu form\"ul, s\i n\i f olarak $a$'y\i\ tan\i mlar.

E\u ger $\phi$
\label{phi}%
serbest de\u gi\c skeni $x$ olan bir form\"ul ise,
o zaman $\phi$'nin tan\i mlad\i\u g\i\ s\i n\i f\i,
\begin{equation*}
\{x\colon\phi\}
\end{equation*}
olarak yaz\i l\i r.
E\u ger bu s\i n\i f i\c cin $\bm A$ harf\/ini yazarsak,
o zaman her $b$ k\"umesi i\c cin
$\phi(b)$ c\"umlesinin yerine $b\in\bm A$ ifadesini yazabiliriz:
k\i saca
\begin{equation*}
\phi(b)\iff  b\in\bm A.
\end{equation*}
Her taraf\i na g\"ore $b$, $\bm A$'n\i n eleman\i d\i r.

Form\"uller ve onlar\i n serbest de\u gi\c skenleri,
\ref{mantik} Eki'nde tan\i mlan\i r.
S\i n\i flar b\"uy\"uk siyah harfler ile g\"osterece\u giz;
k\"u\c c\"uk harfler
her zaman k\"ume olacakt\i r.
B\"uy\"uk normal harfler de k\"ume olabilir.

Bizim i\c cin her s\i n\i f\i n her eleman\i\ bir k\"ume olacakt\i r.
K\"ume olmayan bir s\i n\i f, bir s\i n\i f\i n eleman\i\ olamaz.
Bundan dolay\i, $x\in a$ veya $x\in\bm B$ gibi bir form\"ulde,
$x$ de\u gi\c skeninin yerine $\bm C$ gibi b\"uy\"uk siyah bir harf konulamaz.

Elemanlar\i\ ayn\i\ olan s\i n\i flar da ayn\i d\i r,
ama bu ayn\i l\i k, \textbf{e\c sitlik} olarak yaz\i l\i r.
(\"Oklid'de e\c sitlik, ayn\i l\i k de\u gildir.
\"Orne\u gin
ikizkenar bir \"u\c cgenin iki e\c sit kenar\i\ vard\i r,
ama bu kenarlar birbiriyle ayn\i\ de\u gildir.)
B\"oylece
\begin{equation*}
\bm A=\bm B\iff\Forall x(x\in\bm A\liff x\in\bm B).  
\end{equation*}
\"Ozel olarak
\begin{equation*}
  a=\{x\colon x\in a\}.
\end{equation*}

\begin{theorem}[Russell Paradoksu]\label{thm:Russell}
$\{x\colon x\notin x\}$
s\i n\i f\i\
bir k\"ume de\u gildir.
\end{theorem}

\begin{proof}
  $x\notin x$ form\"ul\"u $\phi$ olarak yaz\i ls\i n.
  M\"umk\"unse $\{x\colon\phi\}=a$ olsun.
O zaman her $b$ k\"umesi i\c cin
\begin{equation*}
b\in a\liff\phi(b).
\end{equation*}
\"Ozel olarak
$a\in a\liff\phi(a)$,
yani
\begin{equation*}
a\in a\liff a\notin a;
\end{equation*}
ama bu bir \c celi\c skidir.
Bu \c sekilde $\{x\colon x\notin x\}$ s\i n\i f\i,
bir $a$ k\"umesine e\c sit olamaz.
\end{proof}

Tan\i ma g\"ore
\begin{gather*}
  \universe=\{x\colon x\in x\lor x\notin x\},\\
  \emptyset=\{x\colon x\in x\land x\notin x\},\\
  \bm A\cup\bm B=\{x\colon x\in\bm A\lor x\in\bm B\},\\
  \bm A\cap\bm B=\{x\colon x\in\bm A\land x\in\bm B\},\\
  \bm A\setminus\bm B=\{x\colon x\in\bm A\land x\notin\bm B\}.
\end{gather*}
O zaman
\begin{align*}
  \bm B\cup\bm A&=\bm A\cup\bm B,&
  \bm A\cup\emptyset&=\bm A,&
  \bm A\cup\universe&=\universe,\\
  \bm B\cap\bm A&=\bm A\cap\bm B,&
  \bm A\cap\emptyset&=\emptyset,&
  \bm A\cap\universe&=\bm A,
\end{align*}
ve ayr\i ca
\begin{align*}
  \bm A\setminus\emptyset&=\bm A,&
  \bm A\setminus\universe&=\emptyset.
\end{align*}
Her $a$ k\"umesi i\c cin
tan\i ma g\"ore
\begin{equation*}
  \{a\}=\{x\colon x=a\};
\end{equation*}
burada $x=a$ ifadesi,
$\Forall y(y\in x\liff y\in a)$
form\"ul\"u i\c cin bir k\i saltmad\i r.
Benzer \c sekilde
\begin{gather*}
  \{a,b\}=\{x\colon x=a\lor x=b\},\\
  \{a,b,c\}=\{x\colon x=a\lor x=b\lor x=c\},\\
  \makebox[7cm]{\dotfill}
\end{gather*}


\begin{axiom}[Bo\c s K\"ume]
  $\emptyset$ s\i n\i f\i\
bir k\"umedir.
\end{axiom}

Bo\c s k\"ume $0$ olarak da yaz\i l\i r.

\begin{axiom}[Biti\c stirme]
  T\"um $a$ ve $b$ k\"umeleri i\c cin
  \begin{equation*}
    a\cup\{b\}
  \end{equation*}
  s\i n\i f\i\
  bir k\"umedir.
\end{axiom}

Sonu\c c olarak her $\{a\}$, $\{a,b\}$,
$\{a,b,c\}$, veya \lips s\i n\i f\i\ bir k\"umedir.
\"Ozellikle
\begin{align*}
  &\{0\},&
  &\bigl\{0,\{0\}\bigr\},&
  &\Bigl\{0,\{0\},\bigl\{0,\{0\}\bigr\}\Bigr\},&
  &\dots
\end{align*}
s\i n\i flar\i n\i n her biri, bir k\"umedir.
Bu k\"umeler, s\i ras\i yla $1$, $2$, $3$, \lips,
olarak yaz\i l\i r.

Tan\i ma g\"ore her $a$ k\"umesi i\c cin
\begin{equation*}
  a'=a\cup\{a\}
\end{equation*}
olsun.
O zaman Biti\c stirme Aksiyomu'na g\"ore $a'$ s\i n\i f\i,
bir k\"umedir.
Bu k\"umeye $a$'n\i n \textbf{ard\i l\i} densin.
\c Simdi
\begin{align*}
  1&=0',&
  2&=1',&
  3&=2',&
  &\dots
\end{align*}
$0$'\i\ i\c ceren,
her eleman\i n\i n ard\i l\i n\i\ da i\c ceren bir s\i n\i fa
\textbf{t\"umevar\i ml\i} densin.
\"Oyleyse bir $\bm A$ s\i n\i f\i\ t\"umevar\i ml\i d\i r
ancak ve ancak
  \begin{compactenum}[1)]
  \item
    $0\in\bm A$,
  \item
    $\Forall x(x\in\bm A\lto x'\in\bm A)$.
  \end{compactenum}
O zaman $\universe$ t\"umevar\i ml\i d\i r.
  
\begin{axiom}[Sonsuzluk]
T\"umevar\i ml\i\ bir k\"ume vard\i r.
\end{axiom}



\section{Alts\i n\i flar}

E\u ger bir $\bm A$ s\i n\i f\i n\i n her eleman\i\
bir $\bm B$ s\i n\i f\i n\i n bir eleman\i\ ise,
o zaman $\bm A$ $\bm B$'nin bir \textbf{alts\i n\i f\i d\i r}
ve
\begin{equation*}
  \bm A\included\bm B
\end{equation*}
ifadesini yazar\i z.
Bu ifade,
\begin{equation*}
  \Forall x(x\in\bm A\lto x\in\bm B)
\end{equation*}
c\"umlesinin bir k\i saltmas\i d\i r.
O zaman
\begin{equation}\label{eqn:A=B}
  \bm A=\bm B\liff\bm A\included\bm B\land\bm B\included\bm A.
\end{equation}
Ayr\i ca
\begin{equation*}
  \bm A\included\bm B\land
  \bm B\included\bm C\lto
  \bm A\included\bm C,
\end{equation*}
dolay\i s\i yla tan\i ma g\"ore
\begin{equation*}
  \bm A\included\bm B\included\bm C\iff
  \bm A\included\bm B\land
  \bm B\included\bm C.
\end{equation*}
S\"ozc\"uklerde bir s\i n\i f
\begin{compactitem}
\item
  elemanlar\i n\i\ \textbf{i\c cerir,}
\item
  alts\i n\i flar\i n\i\ \textbf{kapsar.}
\end{compactitem}
O zaman $a'$ $a$'y\i\ hem i\c cerir hem de kapsar:
\begin{align*}
  a&\in a',&
  a&\included a'.
\end{align*}
E\u ger bir s\i n\i f\i n bir alts\i n\i f\i\ bir k\"ume ise,
bu k\"ume s\i n\i f\i n bir \textbf{altk\"umesidir.}



\begin{axiom}[Ay\i rma]
Her k\"umenin her alts\i n\i f\i\ bir k\"umedir.
\end{axiom}

Bu \c sekilde her $a$ k\"umesi ve $\{x\colon\phi(x)\}$ s\i n\i f\i\ i\c cin
\begin{equation*}
\{x\colon x\in a\land\phi(x)\}
\end{equation*}
s\i n\i f\i\ bir k\"umedir.
Bu k\"ume
\begin{equation*}
\{x\in a\colon\phi(x)\}
\end{equation*}
olarak yaz\i labilir.

Her s\i n\i f $\universe$'nin bir alts\i n\i f\i\ oldu\u gundan
\Teorem{thm:Russell} sayesinde $\universe$ bir k\"ume de\u gildir.

Tan\i ma g\"ore her $\bm A$ s\i n\i f\i\ i\c cin
\begin{gather*}
  \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{gather*}
Burada
\begin{compactitem}
  \item
    $\bigcap\bm A$ $\bm A$'n\i n \textbf{kesi\c simidir,}
    \item
  $\bigcup\bm A$ $\bm A$'n\i n \textbf{bile\c simidir.}
\end{compactitem}
\"Orne\u gin
\begin{align*}
  \bigcap\{a,b\}&=a\cap b,&\bigcup\{a,b\}&=a\cup b,\\
  \bigcap\{a\}&=a,&\bigcup\{a\}&=a,\\
  \bigcap\emptyset&=\universe,&\bigcup\emptyset&=\emptyset.
\end{align*}
Ayr\i ca
\begin{gather}\notag
  \bm A\included\bm B\lto\bigcap\bm B\included\bigcap\bm A
  \land\bigcup\bm A\included\bigcup\bm B,\\\label{eqn:a-in-B}
  a\in\bm B\lto\bigcap\bm B\included a\included\bigcup\bm B.
\end{gather}
Ay\i rma Aksiyomu'ndan $\bm A$ bo\c s de\u gilse $\bigcap\bm A$ bir k\"umedir.

Tan\i ma g\"ore t\"umevar\i ml\i\ k\"umelerin olu\c sturdu\u gu
s\i n\i f\i n kesi\c simi,
\textbf{do\u gal say\i lar\i} i\c cerir,
ve bu kesi\c simi i\c cin $\upomega$ (omega) harf\/ini yazar\i z.
B\"oylece
\begin{equation*}
  \upomega=\bigcap\{x\colon 0\in x\land\Forall y(y\in x\lto y'\in x)\}.
\end{equation*}
Sonsuzluk ve Ay\i rma aksiyomlar\i\ sayesinde $\upomega$ bir k\"umedir.

\begin{theorem}[T\"umevar\i m]\label{thm:ind}
  $A$, $\upomega$'n\i n \"oyle altk\"umesi olsun ki
  \begin{enumerate}[1)]
    \item
    $0\in A$ olsun ve
  \item
    her $n$ do\u gal say\i s\i\ i\c cin
      $n\in A\lto n'\in A$ olsun.
  \end{enumerate}
  O zaman $A=\upomega$.
  K\i saca
  $\upomega$'n\i n tek t\"umevar\i ml\i\ altk\"umesi, kendisidir.
\end{theorem}

\begin{proof}
  Varsay\i ma g\"ore $A$,
  $\upomega$'n\i n t\"umevar\i ml\i\ bir altk\"umesidir.
  \"Oyleyse
  \eqref{eqn:a-in-B} sayesinde
  \begin{align*}
    A&\included\upomega,&
    \upomega&\included A,
  \end{align*}
  dolay\i s\i yla \eqref{eqn:A=B} sayesinde $A=\upomega$.
\end{proof}

\begin{theorem}\label{thm:0-n'}
  Her do\u gal say\i n\i n ya $0$
  ya da do\u gal bir say\i n\i n ard\i l\i\ oldu\u gunu g\"osterin.
\end{theorem}

\ktk

\begin{theorem}\label{thm:o-trans}
  Her do\u gal say\i n\i n her eleman\i,
  do\u gal say\i d\i r.
\end{theorem}

\ktk

Her k\"ume, kendinin altk\"umesidir; k\"umenin di\u ger altk\"umeleri,
k\"umenin \textbf{\"ozaltk\"umeleridir.}
K\i saca
\begin{equation*}
  a\pincluded b\iff a\included b\land a\neq b.
\end{equation*}
O zaman
$\pincluded$ ba\u g\i nt\i s\i\
\textbf{yans\i mas\i z} ve
\textbf{ge\c ci\c slidir,} yani
\begin{gather*}
  a\npincluded a,\\
  a\pincluded b\land b\pincluded c\lto a\pincluded c.
\end{gather*}
K\i saca $\pincluded$, bir \textbf{s\i ralamad\i r}.


\begin{theorem}\label{thm:n-trans}
  Her do\u gal say\i n\i n her eleman\i,
  say\i n\i n bir \"ozaltk\"umesidir:
  E\u ger $n\in\upomega$ ise, o zaman
  \begin{equation*}
    k\in n\lto k\pincluded n.
  \end{equation*}
\end{theorem}

\begin{proof}
  T\"umevar\i m kullanaca\u g\i z.
  \begin{asparaenum}
    \item
      $0$'n\i n eleman\i\ olmad\i\u g\i ndan
      $0$'\i n her eleman\i\ bir \"ozaltk\"umedir.
    \item
      Bir $m$ do\u gal say\i s\i\ i\c cin
      t\"umevar\i m hipotezi olarak
      $m$'nin her eleman\i\
      bir \"ozaltk\"ume olsun.
      Bu durumda $m\npincluded m$ oldu\u gundan
        $m\notin m$,
      dolay\i s\i yla
      \begin{equation}\label{eqn:m-m'}
        m\pincluded m'.
      \end{equation}
      \c Simdi $k\in m'$ olsun.
      O zaman
      \begin{equation*}
        k\in m\lor k=m.
      \end{equation*}
      \begin{compactenum}
      \item
        E\u ger $k\in m$ ise hipotezden $k\pincluded m$.
      \item
        E\u ger $k=m$ ise $k\included m$.
      \end{compactenum}
      Her durumda \eqref{eqn:m-m'} sayesinde
      \begin{equation*}
        k\pincluded m'.
      \end{equation*}
  B\"oylece t\"umevar\i m tamamd\i r.\qedhere
  \end{asparaenum}
\end{proof}

\begin{corollary}
  Ard\i llar\i\ ayn\i\ olan do\u gal say\i lar da ayn\i d\i r:
  $\upomega$'da
  \begin{equation*}
    k'=n'\lto k=n.
  \end{equation*}
\end{corollary}

\ksk

\begin{compactenum}[1)]
\item
  \Teorem{thm:ind},
\item
  \Teoremin{thm:n-trans} sonucu,
  ve
\item
  $0$'\i n ard\i l olmad\i\u g\i na
\end{compactenum}
\textbf{Peano Aksiyomlar\i} denir.
Bunlardan do\u gal say\i lar\i n
matematikte kullan\i lan t\"um \"ozellikleri kan\i tlanabilir.
Do\u gal say\i lar\i n k\"umeler kuram\i ndan gelen tan\i m\i n\i\ kullanarak
\"ozellikleri kan\i tlamak daha kolayd\i r.

\begin{theorem}\label{thm:inc-in}
  E\u ger $n\in\upomega$ ise,
  o zaman her $k$ do\u gal say\i s\i\ i\c cin
  \begin{equation*}
    k\pincluded n\lto k\in n.
  \end{equation*}
\end{theorem}

\begin{proof}
  T\"umevar\i m kullanaca\u g\i z.
  \begin{asparaenum}
    \item
      Her $k$ do\u gal say\i s\i\ i\c cin $k\npincluded 0$,
      dolay\i s\i yla
      \begin{equation*}
        k\pincluded0\lto k\in0.
      \end{equation*}
    \item
      Bir $m$ do\u gal say\i s\i\ i\c cin
      t\"umevar\i m hipotezi olarak
   her $k$ do\u gal say\i s\i\ i\c cin
  \begin{equation*}
    k\pincluded m\liff k\in m
  \end{equation*}
  olsun.
    E\u ger $k\pincluded m'$ ise, o zaman $k\included m$ veya $m\in k$.
    Son durum imk\^ans\i zd\i r \c c\"unk\"u $m\in k$ ise,
    \Teoremde{thm:n-trans}n
    \begin{equation*}
      m\pincluded k\pincluded m',
    \end{equation*}
    ki bu imk\^ans\i zd\i r.
    Sonu\c c olarak $k\included m$.
    \begin{compactitem}
      \item
        E\u ger $k=m$ ise, o zaman $k\in m'$.
      \item
        E\u ger $k\pincluded m$ ise, o zaman $k\in m$,
        dolay\i s\i yla $k\in m'$.
    \end{compactitem}
    T\"umevar\i m tamamd\i r.\qedhere
  \end{asparaenum}
\end{proof}

Sonu\c c olarak $\upomega$'da $\in$ ve $\pincluded$,
ayn\i\ ba\u g\i nt\i d\i r.
\"Ozel olarak $\upomega$'da $\in$ ba\u g\i nt\i s\i\
bir s\i ralamad\i r.

  
\begin{theorem}
  T\"um $k$ ve $n$ do\u gal say\i lar\i\ i\c cin
  \begin{equation*}
    k\included n\lor n\pincluded k.
  \end{equation*}
\end{theorem}

\begin{proof}
  \begin{asparaenum}
    \item
    Her $k$ i\c cin $0\included k$, dolay\i s\i yla
    \begin{equation*}
      k\included 0\lor 0\pincluded k.
    \end{equation*}
    \item
    $\upomega$'da bir $m$ i\c cin,
    her $k$ i\c cin,
    \begin{equation*}
      k\included m\lor m\pincluded k
    \end{equation*}
    olsun, ama bir $k$ i\c cin $k\nincluded m'$ olsun.
    O zaman
    \begin{align*}
      k&\neq m',&
      k&\nincluded m,
    \end{align*}
    dolay\i s\i yla, hipotez sayesinde, $m\pincluded k$.
    Ayr\i ca \Teoremde{thm:inc-in}n $m\in k$, dolay\i s\i yla
    \begin{equation*}
      m'\pincluded k.\qedhere
    \end{equation*}
  \end{asparaenum}
\end{proof}

Sonu\c c olarak $\upomega$'da $\pincluded$ veya $\in$ s\i ralamas\i,
\textbf{do\u grusal} bir s\i ralamad\i r.

\begin{theorem}\label{thm:n-wo}
  $\in$ s\i ralamas\i na g\"ore
  her do\u gal say\i n\i n bo\c s olmayan her altk\"umesinin
  en k\"u\c c\"uk eleman\i\ vard\i r,
  ve bu eleman, altk\"umenin kesi\c simidir.
\end{theorem}

\begin{proof}
  \begin{asparaenum}
  \item
    $0$'\i n bo\c s olmayan hi\c c altk\"umesi olmad\i\u g\i ndan
    iddia $0$ i\c cin a\c sik\^ar bir \c sekilde do\u grudur.
  \item
    Bir $m$ i\c cin iddia do\u gru olsun, ve
    \begin{equation*}
      0\pincluded A\included m'
    \end{equation*}
    olsun.
    \.Iki durum vard\i r.
    \begin{compactenum}
      \item
      E\u ger $A=\{m\}$ ise,
      o zaman $m$,
      a\c sik\^ar bir \c sekilde
      $A$'n\i n di\u ger elemanlar\i n\i n eleman\i d\i r,
      \c c\"unk\"u ba\c ska eleman yoktur.
      \item
        Di\u ger durumda hipoteze g\"ore
        $\bigcap(A\setminus\{m\})$ kesi\c simi,
      $A\setminus\{m\}$ fark\i n\i n en k\"u\c c\"uk eleman\i d\i r.
      Ayr\i ca
      \begin{equation*}
        A\setminus\{m\}\included m,
      \end{equation*}
      dolay\i s\i yla
        $\bigcap(A\setminus\{m\})$
      $A$'n\i n en k\"u\c c\"uk eleman\i d\i r.
      \qedhere
    \end{compactenum}
  \end{asparaenum}
\end{proof}

\begin{corollary}%\label{cor:o-wo}
  $\in$ s\i ralamas\i na g\"ore
  $\upomega$'n\i n bo\c s olmayan her altk\"umesinin
  en k\"u\c c\"uk eleman\i\ vard\i r,
  ve bu eleman, altk\"umenin kesi\c simidir.  
\end{corollary}

\ksk

K\i saca $\upomega$'da $\in$ veya $\pincluded$
do\u grusal s\i ralamas\i,
\textbf{iyis\i ralamad\i r.}
Ayr\i ca her do\u gal say\i da da $\in$ bir iyis\i ralamad\i r.

\begin{xca}[G\"u\c cl\"u T\"umevar\i m]
  $A$, $\upomega$'n\i n \"oyle altk\"umesi olsun ki
  \begin{itemize}
  \item
    her $n$ do\u gal say\i s\i\ i\c cin
    \begin{equation*}
      n\included A\lto n\in A
    \end{equation*}
    olsun.
    (K\i saca $A$, kapsad\i\u g\i\ her do\u gal say\i y\i\ i\c cersin.)
  \end{itemize}
  $A$'n\i n $\upomega$ oldu\u gunu kan\i tlay\i n.
\emph{\.Ipucu}:
  \"Once $\bm A$'n\i n her do\u gal say\i y\i\
  kapsad\i\u g\i n\i\ kan\i tlay\i n.
\end{xca}

\section{Ba\u g\i nt\i lar ve G\"ondermeler}

Bildi\u gimiz gibi
\begin{compactitem}
\item
  $=$ yans\i mal\i, simetrik, ve ge\c ci\c sli bir ba\u g\i nt\i d\i r;
\item
  $\included$ yans\i mal\i\ ve ge\c ci\c sli bir ba\u g\i nt\i d\i r;
\item
  $\pincluded$ yans\i mas\i z ve ge\c ci\c sli bir ba\u g\i nt\i d\i r,
  dolay\i s\i yla bir s\i ralamad\i r;
\item
  $\upomega$'da $\pincluded$ do\u grusal bir s\i ralamad\i r
  ve $\in$ ile ayn\i d\i r.
\end{compactitem}
Genelde bir \textbf{ba\u g\i nt\i,}
iki serbest de\u gi\c skeni olan bir form\"ul taraf\i ndan tan\i mlan\i r.
\"Orne\u gin $=$, $\included$, $\pincluded$, ve $\in$,
s\i ras\i yla
\begin{gather*}
  \Forall z(z\in x\liff z\in y),\\
  \Forall z(z\in x\lto z\in y),\\
  \Forall z(z\in x\lto z\in y)\land\Exists z(z\notin x\land z\in y),\\
  x\in y
\end{gather*}
form\"ulleri taraf\i ndan tan\i mlan\i r.
E\u ger bir $\phi$ form\"ul\"u bir $\bm R$ ba\u g\i nt\i s\i n\i\ tan\i mlarsa,
o zaman $\phi(a,b)$ c\"umlesini
\begin{equation*}
  a\mathrel{\bm R}b
\end{equation*}
olarak yazabiliriz.


\c Simdi $\bm R$ bir ba\u g\i nt\i\ ve $\bm D$ bir s\i n\i f olsun.
E\u ger $\bm D$'nin her $a$ eleman\i\ i\c cin
ve b\"ut\"un $b$ ve $c$ k\"umeleri i\c cin
\begin{equation*}
  a\mathrel{\bm R}b
  \land
  a\mathrel{\bm R}c
  \lto
  b=c
\end{equation*}
ise, o zaman $\bm D$'de $\bm R$ bir \textbf{g\"ondermeyi} tan\i mlar.
Bu g\"ondermeye $\bm F$ densin;
o zaman $\bm F$'nin \textbf{tan\i m s\i n\i f\i} $\bm D$'dir.
E\u ger
\begin{equation*}
  a\in\bm D\land a\mathrel{\bm R}b
\end{equation*}
ise, o zaman
\begin{equation*}
  \bm F(a)=b
\end{equation*}
yaz\i labilir;
ayr\i ca $\bm F$'nin yerine
\begin{equation*}
  x\mapsto\bm F(x)
\end{equation*}
yaz\i labilir.
\"Orne\u gin $\universe$'de
  $x\mapsto x'$
g\"ondermesi vard\i r,
ve bu g\"ondermeyi
\begin{equation*}
  \Forall z\bigl(z\in y\liff z\in x\lor\Forall w(w\in z\liff w\in x)\bigr)
\end{equation*}
form\"ul\"u,
k\i saca
$\Forall z(z\in y\liff z\in x\lor z=x)$,
tan\i mlar.


\begin{axiom}[Yerle\c stirme]
  Her $\bm F$ g\"ondermesi i\c cin,
  $\bm F$'nin tan\i m s\i n\i f\i n\i n her $a$ altk\"umesi i\c cin
\begin{equation*}
\{y\colon\Exists x(\bm F(x)=y\land x\in a)\}
\end{equation*}
s\i n\i f\i\ bir k\"umedir.
\end{axiom}

Aksiyomda verilen k\"ume
\begin{align*}
  &\{\bm F(x)\colon x\in a\},&
  &\bm F[a]
\end{align*}
ifadelerinin biri ile g\"osterilebilir.


  \section{Do\u gal Say\i larda \"Ozyineleme}


  \begin{theorem}[\"Ozyineleme]\label{thm:rec}
    Bir $A$ k\"umesi i\c cin
    \begin{compactenum}[1)]
    \item
      $b\in A$,
    \item
      $f\colon A\to A$ 
    \end{compactenum}
    olsun.
    O zaman $\upomega$'dan $A$'ya giden bir ve tek bir $g$ g\"ondermesi i\c cin
    \begin{compactenum}[1)]
    \item
      $g(0)=b$,
    \item
      her $k$ do\u gal say\i s\i\ i\c cin $g(k+1)=f(g(k))$.
    \end{compactenum}
    \Sekle{fig:rec} bak\i n.
    \begin{figure}
      \relscale{1.1}
      \begin{equation*}
        \xymatrix@=2cm{\{0\}\ar[d]_g\ar[r]^{\included}&\upomega\ar[d]_g\ar[r]^{x\mapsto x+1}&\upomega\ar[d]^g\\
          \{b\}\ar[r]_{\included}&A\ar[r]_f&A}
      \end{equation*}
      \caption{\"Ozyineleme}\label{fig:rec}
    \end{figure}
  \end{theorem}

  \begin{proof}
    T\"umevar\i mdan
    istedi\u gimiz \"ozellikleri olan bir g\"onderme varsa,
    tek bir \"ornek vard\i r.

    \c Simdi elemanlar\i\ g\"onderme olan
    bir $\mathscr C$ k\"umesini tan\i mlayaca\u g\i z.\label{curly-C}
    $\mathscr C$'nin her $h$ eleman\i\ i\c cin,
    \begin{compactenum}[1)]
    \item
      $h$'nin tan\i m k\"umesi 
      $\upomega$'nin bir altk\"umesidir, ve
    \item
      herhangi $\ell$ do\u gal say\i s\i\ i\c cin,
      $h(\ell)$ tan\i mlan\i rsa, o zaman
      \begin{compactenum}
      \item 
        ya $\ell=0$ ve $h(\ell)=b$,
      \item
        ya da bir $k$ do\u gal say\i s\i\ i\c cin $\ell=k+1$,
        $h(k)$ tan\i mlan\i r, ve
        \begin{equation*}
          h(\ell)=f(h(k)).
        \end{equation*}
      \end{compactenum}
    \end{compactenum}
    \.Istedi\u gimiz gibi $g$ g\"ondermesi varsa
    $\mathscr C$'nin eleman\i d\i r.
    Her $k$ do\u gal say\i s\i\ i\c cin,
    $A$'n\i n bir ve tek bir $d$ eleman\i\ i\c cin,
    $\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k)=d$ g\"osterece\u giz.
    Bu \c sekilde $g(k)=d$ tan\i mlanabilir.

    Yukar\i daki \"ozelli\u gi olan $k$ do\u gal say\i lar\i,
    $E$ k\"umesini olu\c stursun.
    Tan\i m k\"umesi $\{0\}$ olan
    bir $h$ g\"ondermesi i\c cin $h(0)=b$.
    O zaman $h\in\mathscr C$.
    Ayr\i ca $\mathscr C$'nin herhangi $h$ eleman\i\ i\c cin $h(0)$ tan\i mlan\i rsa,
    o zaman $h(0)=b$ olmal\i d\i r, 
    \c c\"unk\"u hi\c c $k$ do\u gal say\i s\i\ i\c cin $k+1=0$ de\u gildir.
    Bu \c sekilde $0\in E$.

    \c Simdi $k\in E$ olsun.
    O zaman $A$'n\i n bir ve tek bir $d$ eleman\i\ i\c cin,
    $\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k)=d$.
    \begin{compactenum}
    \item
      E\u ger $h(k+1)$ tan\i mlan\i rsa,
      o zaman $\mathscr C$'nin tan\i m\i na g\"ore
      $h(k+1)=f(d)$, 
      \c c\"unk\"u $k+1\neq0$, ve ayr\i ca herhangi $\ell$ do\u gal say\i s\i\ i\c cin
      e\u ger $\ell+1=k+1$ ise, o zaman $\ell=k$.
    \item
      E\u ger $h(k+1)$ tan\i mlanmazsa, 
      o zaman yeni bir $h^*$ g\"ondermesi i\c cin
      \begin{equation*}
        h^*(x)=\begin{cases}
	h(x),&\text{ e\u ger $h(x)$ tan\i mlan\i rsa,}\\
	f(d),&\text{ e\u ger $x=k+1$.}
        \end{cases}
      \end{equation*}
      O zaman $h^*\in\mathscr C$ ve $h^*(k+1)=f(d)$.
    \end{compactenum}
    Bu \c sekilde, her durumda,
    %$A$'n\i n $f(d)$ eleman\i\ i\c cin,
    $\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k+1)=f(d)$.

    M\"umk\"umse $d^*\in A$, $d^*\neq f(d)$ olsun, 
    ama $\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k+1)=d^*$ olsun.
    O zaman $k+1\neq0$ oldu\u gundan bir $\ell$ do\u gal say\i s\i\ i\c cin
    $\ell+1=k+1$, $h(\ell)$ tan\i mlan\i r, ve $d^*=f(h(\ell))$.
    Ama bu durumda $\ell=k$, dolay\i s\i yla $h(\ell)=d$ ve $d^*=f(d)$.

    Sonu\c c olarak $k+1\in E$.  T\"umevar\i m ile $E=\upomega$.
  \end{proof}

  \begin{sloppypar}
    Yukar\i daki kan\i t, sadece $\upomega$'nin a\c sa\u g\i daki \"ozelliklerini kullan\i r:
    \begin{compactenum}
    \item
      $0\in\upomega$.
    \item
      $k\in\upomega$ ise $k+1\in\upomega$.
    \item
      T\"umevar\i m y\"ontemi ge\c cerlidir.
    \item
      Her $k$ do\u gal say\i s\i\ i\c cin
      $0\neq k+1$.
    \item
      T\"um $k$ ve $\ell$ do\u gal say\i lar\i\ i\c cin
      $k+1=\ell+1$ ise $k=\ell$.
    \end{compactenum}
    Bu \"ozelliklere \textbf{Peano Aksiyomlar\i} denir.
    Peano Aksiyomlar\i,
    $\upomega$'da iki-konumlu toplama i\c sleminin tan\i mland\i\u g\i n\i\ varsaymaz;
    sadece tek-konumlu $x\mapsto x+1$ i\c slemi vard\i r.
    Ama \"ozyineleme y\"ontemiyle 
    $\upomega$'da toplama ve \c carpma i\c slemlerini tan\i mlayabiliriz:
    \begin{align*}
      a+(b+1)&=(a+b)+1,&a\cdot 1&=a,&a\cdot(b+1)&=ab+a.
    \end{align*}
    T\"umevar\i m ve kalan Peano Aksiyomlar\i\ ile
    toplaman\i n ve \c carpman\i n \"ozelliklerini kan\i tlayabiliriz;
    ayr\i ca $\upomega$'n\i n s\i ralamas\i n\i\ tan\i mlay\i p
    \"ozelliklerini kan\i tlayabiliriz.
    Ondan sonra
    B\"ol\"um \numarada{ch:intro}ki gibi
    $\N$, $\Z$, $\Q$ ve $\R$ yap\i lar\i n\i\ elde edebiliriz.
  \end{sloppypar}

  \begin{sloppypar}
    Do\u gal say\i lar, \emph{sonlu ordinallerdir.}
    Sonsuz ordinaller de vard\i r.
    Ordinallerin aksiyomlar\i n\i\ kullanarak
    toplama ve \c carpma i\c slemlerini tan\i mlayay\i p
    \"ozelliklerini kan\i tlayaca\u g\i z.
    Ondan sonra
    k\"ume aksiyomlar\i n\i\ kullanarak
    ordinalleri in\c sa edece\u giz.
    Bu \c sekilde bildi\u gimiz t\"um matematik,
    k\"ume aksiyomlar\i\ taraf\i ndan gerektirilir.
  \end{sloppypar}



\chapter{Ordinal Say\i lar}\label{ch:ord}

\section{Ordinaller}

E\u ger bir s\i n\i f\i n her eleman\i\ bir altk\"ume ise,
o zaman s\i n\i f \textbf{ge\c ci\c slidir.}
E\u ger $\bm A$ ge\c ci\c sli ise, o zaman
\begin{equation*}
  c\in b\land b\in\bm A\lto c\in\bm A.
\end{equation*}
\"Orne\u gin
\begin{compactitem}
  \item
    \Teoreme{thm:o-trans} g\"ore $\upomega$ ge\c ci\c slidir;
    \item
      \Teoreme{thm:n-trans} g\"ore
      $\upomega$'n\i n her eleman\i\ da ge\c ci\c slidir.
\end{compactitem}
Tan\i ma g\"ore
\begin{compactenum}[1)]
  \item
    ge\c ci\c sli olan
  \item
    $\in$ taraf\i ndan iyis\i ralanm\i\c s olan
\end{compactenum}
bir k\"ume bir \textbf{ordinal say\i d\i r.}
\begin{compactitem}
  \item
    \Teorem{thm:n-wo} sayes\i nde $\upomega$'n\i n her eleman\i\ bir ordinaldir;
    \item
  teoremin sonucu sayes\i nde $\upomega$ bir ordinaldir.
\end{compactitem}

\begin{xca}
  Bulun
  \begin{compactenum}[(a)]
    \item
  $\in$ ba\u g\i nt\i s\i n\i n ge\c ci\c sli oldu\u gu,
      ge\c ci\c sli olmayan bir k\"ume;
      \item
  $\in$ ba\u g\i nt\i s\i n\i n ge\c ci\c sli olmad\i\u g\i,
  ge\c ci\c sli olan bir k\"ume.
  \end{compactenum}
\end{xca}

Ordinaller
\begin{equation*}
  \on
\end{equation*}
s\i n\i f\i n\i\ olu\c stururlar.
$\on$'nin elemanlar\i\ her zaman
k\"u\c c\"uk Yunan harfleri taraf\i ndan g\"osterilecektir.
\"Ozel olarak $\alpha$, $\beta$, $\gamma$, $\delta$,
ve $\theta$,\label{minus-gr}
sabit ordinaldirler,
ama $\xi$, $\eta$, ve $\zeta$, ordinal de\u gi\c skendirler.
\"Ornegin
\begin{equation*}
\{\xi\colon\phi(\xi)\}
%=\{\eta\colon\phi(\eta)\}
%=\{\zeta\colon\phi(\zeta)\}
=\{x\colon x\in\on\land\phi(x)\}.
\end{equation*}

\begin{theorem}\label{thm:on-trans}
  $\on$ ge\c ci\c slidir,
  dolay\i s\i yla
 her ordinalin her eleman\i\ bir ordinaldir.
\end{theorem}

\begin{proof}
$\alpha\in\on$ be $b\in\alpha$ olsun.
O zaman $b\included\alpha$, dolay\i s\i yla
$\alpha$ gibi $b$,
$\in$ taraf\i ndan iyis\i ralanm\i\c st\i r.

\c Simdi $c\in b$ olsun.
O zaman $c\in\alpha$, dolay\i s\i yla $c\included\alpha$.
\"Ozel olarak $d\in c$ ise $d\in\alpha$.
Bu durumda $d$, $c$, ve $b$, $\alpha$'n\i n eleman\i d\i rlar;
ayr\i ca $d\in c$ ve $c\in b$,
dolay\i s\i yla $d\in b$ \c c\"unk\"u $\alpha$'da $\in$ ge\c ci\c slidir.
Sonu\c c olarak $c\included b$.
O halde $b$ ge\c ci\c slidir.
\end{proof}

\begin{lemma}\label{lem:on-o}
$\on$, $\in$ taraf\i ndan s\i ralanm\i\c st\i r.
\end{lemma}

\begin{proof}
$\alpha\in\on$ olsun.  $\alpha$'da $\in$ ba\u g\i nt\i s\i\ yans\i mas\i z oldu\u gundan
$\alpha\notin\alpha$,
\c c\"unk\"u $\alpha\in\alpha$ ise $\alpha$'n\i n bir $\beta$ eleman\i\ i\c cin $\beta\in\beta$.

E\u ger $\beta\in\alpha$ ve $\gamma\in\beta$ ise, 
$\alpha$ ge\c ci\c sli oldu\u gundan $\gamma\in\alpha$.
\end{proof}

\begin{lemma}
$\on$'de $\in$ ve $\pincluded$ s\i ralamalar\i\ ayn\i d\i r.
\end{lemma}

\begin{proof}
Kan\i t\i n iki par\c cas\i\ vard\i r.
\begin{asparaenum}
  \item
\fbox{\mathversion{bold}$\alpha\in\beta\lto\alpha\pincluded\beta$:}
$\alpha\in\beta$ olsun.
$\beta$ ge\c ci\c sli oldu\u gundan $\alpha\included\beta$.
$\beta$'da $\in$ yans\i mas\i z oldu\u gundan $\alpha\neq\beta$.
Bu \c sekilde $\alpha\pincluded\beta$.
\item
\fbox{\mathversion{bold}$\alpha\pincluded\beta\lto\alpha\in\beta$:}
$\alpha\pincluded\beta$ olsun.
O zaman $\beta\setminus\alpha$ k\"umesi bo\c s de\u gildir.
$\gamma=\min(\beta\setminus\alpha)$ olsun.
O zaman $\gamma\in\beta$.
Biz \fbox{\mathversion{bold}$\gamma=\alpha$} kan\i tlayaca\u g\i z.
Bu kan\i t\i n iki par\c cas\i\ vard\i r.
\begin{compactenum}
  \item
\fbox{\mathversion{bold}$\gamma\included\alpha$:}
$\delta\in\gamma$ olsun.
O zaman $\beta$ ge\c ci\c sli oldu\u gundan $\delta\in\beta$.
Ayr\i ca $\delta\notin\beta\setminus\alpha$, \c c\"unk\"u $\delta\in\min(\beta\setminus\alpha)$.
O halde $\delta\in\alpha$.
B\"oylece $\gamma\included\alpha$.
\item
\fbox{\mathversion{bold}$\alpha\included\gamma$:}
$\delta\in\alpha$ olsun.
O zaman $\delta\in\beta$, \c c\"unk\"u $\alpha\pincluded\beta$,
dolay\i s\i yla $\delta\notin\beta\setminus\alpha$.
Ama $\delta\in\gamma$, $\delta=\gamma$, veya $\gamma\in\delta$;
ve son iki imk\^an olmaz.
Zira $\gamma\in\beta\setminus\alpha$ oldu\u gundan $\delta\neq\gamma$;
ve $\gamma\notin\alpha$ oldu\u gundan $\gamma\notin\delta$, \c c\"unk\"u $\delta\in\alpha$.
Bu \c sekilde $\alpha\included\gamma$.\qedhere
\end{compactenum}
\end{asparaenum}
\end{proof}

\begin{theorem}
Her ordinalde $\in$ ve $\pincluded$ s\i ralamalar\i\ ayn\i d\i r.  
\end{theorem}

\ktk

\c Simdi $\on$'nin ve her eleman\i n\i n
$\in$ veya $\pincluded$ s\i ralamas\i n\i\ $<$ olarak yazabiliriz.

\begin{lemma}
$\on$'nin $<$ s\i ralamas\i\ do\u grusald\i r.
\end{lemma}

\begin{proof}
$\alpha\not\leq\beta$ olsun.
\fbox{\mathversion{bold}$\beta<\alpha$} g\"osterece\u giz.

Varsay\i mdan $\alpha\not\included\beta$,
dolay\i s\i yla $\alpha\setminus\beta\neq\emptyset$.
$\gamma=\min(\alpha\setminus\beta)$ olsun.
O zaman $\gamma\in\alpha$, yani $\gamma<\alpha$.
\fbox{\mathversion{bold}$\gamma=\beta$} g\"osterece\u giz.

\fbox{\mathversion{bold}$\gamma\included\beta$:}
$\delta\in\gamma$ olsun.
O zaman $\delta<\min(\alpha\setminus\beta)$,
ama $\delta\in\alpha$,
dolay\i s\i yla $\delta\in\beta$.

\fbox{\mathversion{bold}$\gamma\not\pincluded\beta$:}
$\gamma\in\alpha\setminus\beta$ oldu\u gundan $\gamma\notin\beta$,
yani $\gamma\not\pincluded\beta$.
\end{proof}

\begin{theorem}\label{thm:on-wo}
$\on$'nin $<$ do\u grusal s\i ralamas\i\ bir iyis\i ralamad\i r.
Asl\i nda $\on$'nin bo\c s olmayan her \emph{alts\i n\i f\i n\i n} en k\"u\c c\"uk eleman\i\ vard\i r.
\end{theorem}

\begin{proof}
$\bm A\included\on$ ve $\alpha\in\bm A$ olsun.
\begin{compactitem}
\item
$\alpha\cap\bm A=\emptyset$ ise
$\alpha=\min(\bm A)$.
\item
$\alpha\cap\bm A\neq\emptyset$ ise
$\min(\alpha\cap\bm A)=\min(\bm A)$.\qedhere
\end{compactitem}
\end{proof}

\begin{theorem}[Burali-Forti Paradoksu]\label{thm:BFP}
$\on$ k\"ume de\u gildir.
\end{theorem}

\begin{proof}
  \c Simdi Teorem \ref{thm:on-trans} ve \numarada{thm:on-wo}n
  $\on$ hem ge\c ci\c sli hem $\in$ taraf\i ndan iyis\i ralanm\i\c st\i r.
  Tan\i ma g\"ore $\on$'nin elemanlar\i n\i n ayn\i\ \"ozellikleri vard\i r.
  Ama $\on$ $\in$ taraf\i ndan s\i ralanm\i\c s oldu\u gundan kendinin eleman\i\ olamaz.
  Bu \c sekilde $\on$ k\"ume olamaz.
\end{proof}

\begin{theorem}\mbox{}
  \begin{compactenum}
    \item
      $\emptyset\in\on$.
    \item
      $\alpha\in\on$ ise $\alpha'\in\on$
ve ayr\i ca her $\beta$ ordinali i\c cin
\begin{equation*}
\beta\leq\alpha\lor\alpha'\included\beta.
\end{equation*}
  \end{compactenum}
\end{theorem}

\ktk

Tan\i ma g\"ore
ne $0$ ne bir ard\i l olan bir ordinal bir \textbf{limittir.}
O zaman $\upomega$ bir limittir
ve (\Teorem{thm:0-n'} sayesinde) en k\"u\c c\"uk limittir.
Sonsuzluk Aksiyomu'nu kullanmadan $\upomega$,
ne limit olan ne limit i\c ceren ordinallerin olu\c sturdu\u gu s\i n\i f
olarak tan\i mlanabilir.


\begin{theorem}\label{thm:'}
S\i f\i r olmayan bir $\alpha$ ordinalinin limit olmas\i\ i\c cin 
gerek ve yeter ko\c sul,
\begin{equation*}
\beta<\alpha\lto\beta'<\alpha.
\end{equation*}
\end{theorem}

\ktk


\begin{axiom}[Bile\c sim]
Her k\"umenin bile\c simi bir k\"umedir.
\end{axiom}

\begin{theorem}
  $\on$'n\i n her altk\"umesinin en k\"u\c c\"uk \"usts\i n\i r\i\ vard\i r.
  Asl\i nda
$B\pincluded\on$ ise
\begin{equation*}
  \sup B=\bigcup B.
\end{equation*}
\end{theorem}

\ktk


\begin{sloppypar}
  \c Simdi Burali-Forti Paradoksu'nun
  (yani \Teoremin{thm:BFP})
  ba\c ska bir kan\i t\i\ vard\i r.
  Her ordinalin daha b\"uy\"u\u g\"u oldu\u gundan
  $\on$'nin en b\"uy\"uk eleman\i\ yoktur,
  dolay\i s\i yla
  $\on$'nin $\on$'de olan \"usts\i n\i r\i\ yoktur.
  $\on$'nin her altk\"umesinin \"usts\i n\i r\i\ oldu\u gundan
  $\on$'nin kendisi k\"ume olamaz.
\end{sloppypar}



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


\begin{theorem}[Ordinal T\"umevar\i m]
$\bm A\included\on$ olsun.
E\u ger
\begin{compactenum}[1)]
\item
$0\in\bm A$,
\item
Her $\beta$ i\c cin
\begin{equation*}
\beta\in\bm A\lto\beta'\in\bm A,
\end{equation*}
\item
her $\gamma$ limiti i\c cin
\begin{equation}\label{eqn:sub=>in}
\gamma\included\bm A\lto\gamma\in\bm A
\end{equation}
\end{compactenum}
ise, o zaman $\bm A=\on$.
\end{theorem}

\begin{proof}
Verilen ko\c sullar alt\i nda
$\on\setminus\bm A$ fark\i n\i n en k\"u\c c\"uk eleman\i\ olamaz.
Zira m\"umk\"umse $\alpha=\min(\on\setminus\bm A)$ olsun.
\begin{asparaenum}
\item
$\alpha=0$ ise $\alpha\in\bm A$.
\item
$\alpha=\beta'$ ise $\beta<\alpha$ oldu\u gundan $\beta\in\bm A$,
ama bu durumda $\beta'\in\bm A$, yani $\alpha\in\bm A$.
\item
Varsay\i m\i m\i za g\"ore
$\beta<\alpha$ ise $\beta\in\bm A$.
Bu \c sekilde
$\alpha\included\bm A$.
E\u ger ayr\i ca $\alpha$ bir limit ise,
o zaman \eqref{eqn:sub=>in} sayesinde
$\alpha$ da $\bm A$'n\i n eleman\i\ olmal\i d\i r.
\end{asparaenum}

Bu \c sekilde her ordinal ya $0$, ya bir ard\i l, ya da bir limit oldu\u gundan
$\alpha\in\bm A$,
ama $\alpha=\min(\on\setminus\bm A)$ varsay\i m\i na g\"ore $\alpha\notin\bm A$.
\"Oyleyse varsay\i m imk\^ans\i zd\i r.
$\on$'nin her bo\c s olmayan altk\"umesinin en k\"u\c c\"uk eleman\i\ var oldu\u gundan
$\on\setminus\bm A=\emptyset$,
dolay\i s\i yla $\bm A=\on$.
\end{proof}

Ordinal t\"umevar\i m ile
\Teoremi{thm:ord-rec}, 
\Teoremi{thm:norm-cond},
\Teoremi{thm:0+},
ve daha sonraki teoremler 
kan\i tlayaca\u g\i z.
Ordinal t\"umevar\i m kullan\i lan bir kan\i t\i n \"u\c c ad\i m\i\ vard\i r:
\begin{compactenum}[1)]
\item
s\i f\i r ad\i m\i,
\item
ard\i l ad\i m\i, ve
\item
limit ad\i m\i.
\end{compactenum}
Ayr\i ca kan\i tta iki t\"umevar\i m hipotezi vard\i r.
Ordinal T\"umevar\i m Teoremini yazarken kulland\i\u g\i m\i z harflerde,
\begin{compactitem}
\item
ard\i l ad\i m\i n\i n hipotezi, $\beta\in\bm A$;
\item
limit ad\i m\i n\i n hipotezi, $\gamma\included\bm A$,
yani
\begin{equation*}
\Forall{\xi}(\xi<\gamma\lto\xi\in\bm A).
\end{equation*}
\end{compactitem}


\begin{theorem}[Ordinal \"Ozyineleme]\label{thm:ord-rec}
%$\theta$ bir ordinal, $\bm F\colon\on\to\on$, ve $\bm G\colon\pow{\on}\to\on$ olsun.
Varsay\i mlar\i m\i z,
\begin{compactenum}[1)]
\item
$\theta\in\on$, 
\item
$\bm F\colon\on\to\on$.
%\item
%$\bm G\colon\pow{\on}\to\on$.
\end{compactenum}
O zaman bir ve tek bir $\bm H$ ordinal i\c slemi i\c cin
\begin{compactenum}[1)]
\item
$\bm H(0)=\theta$,
\item
her $\beta$ ordinali i\c cin $\bm H(\beta')=\bm F(\bm H(\beta))$,
\item
her $\gamma$ limiti i\c cin
$\bm H(\gamma)=\sup\{\bm H(\xi)\colon\xi<\gamma\}$.
\end{compactenum}
\end{theorem}

\begin{proof}
Her $\alpha$ i\c cin, 
tan\i m k\"umesi $\{\xi\colon\xi\leq\alpha\}$ olan
bir ve tek bir $h_{\alpha}$ g\"ondermesi i\c cin,
\begin{compactenum}[1)]
\item
$h_{\alpha}(0)=\theta$,
\item
$\beta<\alpha$ ise $h_{\alpha}(\beta')=\bm F(h_{\alpha}(\beta))$,
\item
$\gamma\leq\alpha$ ve limit ise $h_{\alpha}(\gamma)=\sup\{h_{\alpha}(\xi)\colon\xi<\gamma\}$.
\end{compactenum}
Bunu kan\i tlamak i\c cin, ordinal t\"umevar\i m kullanaca\u g\i z.
\begin{asparaenum}
\item
$h_0$, $h_0(0)=\theta$ ile tan\i mlanabilir ve tan\i mlanmal\i d\i r.
Yani $\alpha=0$ durumunda iddia do\u grudur.
\item
E\u ger $\alpha=\delta$ durumunda iddia do\u gru ise
$h_{\delta'}$,
\begin{equation*}
h_{\delta'}(\xi)=\begin{cases}
	h_{\delta}(\xi),&\text{ $\xi\leq\delta$ durumunda,}\\
	\bm F(h_{\delta}(\delta)),&\text{ $\xi=\delta'$ durumunda}
\end{cases}
\end{equation*}
kural\i\ taraf\i ndan tan\i mlanabilir.
Ayr\i ca $h_{\delta'}$ bu \c sekilde tan\i mlanmal\i d\i r,
\c c\"unk\"u hipoteze g\"ore
\begin{equation*}
h_{\delta'}\restriction\{\xi\colon\xi\leq\delta\}=h_{\delta}
\end{equation*}
olmal\i d\i r.
Bundan dolay\i\ $\alpha=\delta'$ durumunda iddia do\u grudur.
\item
Benzer \c sekilde bir $\delta$ i\c cin $\alpha<\delta$ durumlar\i nda iddia do\u gru ise,
o zaman $\alpha<\beta<\delta$ durumlar\i nda $h_{\alpha}(\alpha)=h_{\beta}(\alpha)$.
E\u ger ayr\i ca $\delta$ bir limit ise,
o zaman $h_{\delta}$,
\begin{equation*}
h_{\delta}(\xi)=\begin{cases}
	h_{\xi}(\xi),&\text{ $\xi<\delta$ durumunda,}\\
	\sup\{h_{\xi}(\xi)\colon\xi<\delta\},&\text{ $\xi=\delta$ durumunda}
\end{cases}
\end{equation*}
kural\i\ taraf\i ndan tan\i mlanabilir ve tan\i mlanmal\i d\i r,
ve bu \c sekilde $\alpha=\delta$ durumunda iddia do\u grudur.
\end{asparaenum}

Ordinal t\"umevar\i m\i m\i z bitti.
\c Simdi $\bm H(\xi)=h_{\xi}(\xi)$ tan\i mlanabilir ve tan\i mlanmal\i d\i r.
\end{proof}

B\"ol\"umler \ref{ch:add}, \ref{ch:mul}, ve \numarada{ch:exp}
ordinal \"ozyinelemeyle ordinal toplama, 
\c carpma, ve kuvvet alma i\c slemlerini tan\i mlayaca\u g\i z.

\section{Normal i\c slemler}

\c Simdi $\bm F$,
herhangi tek-konumlu ordinal i\c slem olsun.  
Ordinal aksiyomar\i na g\"ore $\{\bm F(\xi)\colon\xi<\alpha\}$ s\i n\i f\i\
her zaman bir k\"umedir,
ve bu k\"umenin \"usts\i n\i r\i\ vard\i r.
Ayr\i ca ordinaller iyis\i ralanm\i\c s oldu\u gundan
$\{\bm F(\xi)\colon\xi<\alpha\}$
k\"umesinin \"usts\i n\i rlar\i n\i n en k\"u\c c\"u\u g\"u vard\i r,
yani k\"umenin \textbf{supremumu} vard\i r.
Bu supremum, 
\begin{align*}
	&\sup\{\bm F(\xi)\colon\xi<\alpha\},&
	\sup_{\xi<\alpha}\bm F(\xi)
\end{align*}
\c sekillerinde yaz\i labilir.

\begin{theorem}\label{thm:sup-a}
  Her $\alpha$ ordinali i\c cin
  \begin{equation*}
    \sup\{\xi\colon\xi<\alpha\}=
    \begin{cases}
      0,&\text{ $\alpha=0$ durumunda},\\
\beta,&\text{ $\alpha=\beta'$ durumunda},\\
\alpha,&\text{ $\alpha$'n\i n limit oldu\u gu durumda.}
    \end{cases}
  \end{equation*}
\end{theorem}

\ktk

\begin{xca}
  $\{\xi'\colon\xi<\alpha\}$ k\"umesinin supremumunu hesaplay\i n.
\end{xca}

E\u ger
\begin{equation*}
\alpha\leq\beta\lto\bm F(\alpha)\leq\bm F(\beta)
\end{equation*}
ise, o zaman $\bm F$ \textbf{artand\i r.}
E\u ger
\begin{equation}\label{eqn:a+x}
\alpha<\beta\lto\bm F(\alpha)<\bm F(\beta)
\end{equation}
ise, o zaman $\bm F$ \textbf{kesin artand\i r.}
E\u ger
\begin{compactenum}[1)]
\item
$\bm F$ kesin artan ve 
\item
her $\alpha$ limiti i\c cin
$\bm F(\alpha)=\sup\{\bm F(\xi)\colon\xi<\alpha\}$
\end{compactenum}
ise, 
o zaman $\bm F$'ye \textbf{normal} densin.

\begin{xca}\label{ex:xx'}
$\xi\mapsto\xi'$ i\c sleminin kesin artan olup
normal olmad\i\u g\i n\i\ g\"osterin.
\end{xca}

\begin{xca}
Normal olan bir i\c slem \"orne\u gi verin.
\end{xca}

Sonraki teoremin ilk kullan\i m\i,
\Teoremin{thm:a+x} kan\i t\i nda olacakt\i r.

\begin{theorem}\label{thm:norm-cond}
$\bm F\colon\on\to\on$ olsun.
E\u ger
\begin{compactenum}[1)]
\item
her $\alpha$ i\c cin
$\bm F(\alpha)<\bm F(\alpha')$ ve 
\item
her $\alpha$ limiti i\c cin
$\bm F(\alpha)=\sup\{\bm F(\xi)\colon\xi<\alpha\}$
\end{compactenum}
ise, 
o zaman $\bm F$ normaldir.
\end{theorem}

\begin{proof}
$\bm F$'nin kesin artan oldu\u gunu g\"ostermek yeter.
\eqref{eqn:a+x} gerektirmesini
$\beta$ \"uzerinden t\"umevar\i m kullanarak
kan\i tlayaca\u g\i z.
  \begin{asparaenum}
    \item
    $\beta=0$ ise, \eqref{eqn:a+x} iddias\i\ do\u grudur, 
\c c\"unk\"u hi\c cbir zaman $\alpha<0$ de\u gildir.  
\item
$\beta=\gamma$ durumda \eqref{eqn:a+x} 
iddia do\u gru olsun.
E\u ger $\alpha<\gamma'$ ise, o zaman $\alpha\leq\gamma$, dolay\i s\i yla
\begin{align*}
      \bm F(\alpha)
&\leq\bm F(\gamma)&&\text{[t\"umevar\i m hipotezi]}\\
&<\bm F(\gamma').&&\text{[varsay\i m]}\\
\end{align*}
\item
$\gamma$ limit
ve $\alpha<\gamma$ ise, o zaman $\alpha<\alpha'<\gamma$, dolay\i s\i yla
\begin{align*}
  \bm F(\alpha)
&<\bm F(\alpha')&&\text{[varsay\i m]}\\
&\leq\sup\{\bm F(\xi)\colon\xi<\gamma\}&&\text{[supremumun tan\i m\i]}\\
&=\bm F(\gamma).&&\text{[varsay\i m]}
\end{align*}
(Bu ad\i mda bir t\"umevar\i m hipotezi kullan\i lm\i yor.)\qedhere
  \end{asparaenum}
\end{proof}

Sonraki teoremin ilk kullan\i m\i,
\Teoremin{thm:+assoc} kan\i t\i nda olacakt\i r.

\begin{theorem}\sloppy
$\bm F\colon\on\to\on$ ve normal olsun.
O zaman $\on$'nin bo\c s olmayan her $A$ altk\"umesi i\c cin
\begin{equation}\label{eqn:F(sup)}
\bm F(\sup(A))=\sup_{\xi\in A}\bm F(\xi).
\end{equation}
\end{theorem}

\begin{proof}
$A$ k\"umesinin supremumu $\alpha$ olsun.
$\bm F$ kesin artan oldu\u gundan
$\beta\in A$ ise $\bm F(\beta)\leq\bm F(\alpha)$.
Bundan dolay\i,
e\u ger $\alpha\in A$ ise, o zaman
\begin{equation*}
  \sup_{\xi\in A}\bm F(\xi)=F(\alpha),
\end{equation*}
yani \eqref{eqn:F(sup)} do\u grudur.
\c Simdi $\alpha\notin A$ olsun.
O zaman $\alpha$ ard\i l olamaz.
$A$ bo\c s olmad\i\u g\i ndan $\alpha=0$ olamaz,
dolay\i s\i yla $\alpha$ limittir.
Bu durumda $\bm F$ normal oldu\u gundan
\begin{equation}\label{eqn:F1}
  \bm F(\alpha)=\sup_{\xi<\alpha}\bm F(\xi).
\end{equation}
Ayr\i ca
\begin{equation}\label{eqn:F2}
\sup_{\xi\in A}\bm F(\xi)\leq\sup_{\xi<\alpha}\bm F(\xi),  
\end{equation}
\c c\"unk\"u $A\included\{\xi\colon\xi<\alpha\}$.
E\u ger $\beta<\alpha$ ise, 
$A$'n\i n bir $\gamma$ eleman\i\ i\c cin $\beta\leq\gamma<\alpha$,
dolay\i s\i yla $\bm F(\beta)\leq\bm F(\gamma)\leq\sup_{\xi\in A}\bm F(\xi)$.
Bu \c sekilde
\begin{equation}\label{eqn:F3}
  \sup_{\xi<\alpha}\bm F(\xi)\leq\sup_{\xi\in A}\bm F(\xi).
\end{equation}
Sonu\c c olarak
\eqref{eqn:F1}, \eqref{eqn:F2}, ve \eqref{eqn:F3} beraber
 \eqref{eqn:F(sup)} e\c sitli\u gini tekrar gerektirir.
\end{proof}

\section{S\"ureklilik}

Normallik kavram\i n\i n yerine
ger\c cel analizden gelen s\"ureklilik kavram\i n\i\ kullanabiliriz.
Ordinallerde,
kesin artan bir i\c slemin normal olmas\i\ i\c cin
gerek ve yeter bir ko\c sul,
i\c slemin s\"urekli olmas\i d\i r.
Bu sonu\c cu kurmak, bu altb\"ol\"um\"un i\c sidir.

Tekrar $\bm F\colon\on\to\on$ olsun.  Varsa,
$\bm F$'nin bir noktadaki s\"ureklili\u gi
ger\c cel analizdeki gibi tan\i mlan\i r.
Asl\i nda e\u ger
\begin{equation*}
\beta<\bm F(\alpha)<\gamma
\end{equation*}
ko\c sulunu sa\u glayan herhangi $\beta$ ve $\gamma$ i\c cin,
baz\i\ $\delta$ ve $\theta$ i\c cin,
\begin{equation*}
\delta<\alpha<\theta
\land\Forall{\xi}(\delta<\xi<\theta\lto\beta<\bm F(\xi)<\gamma)
\end{equation*}
ise, o zaman $\bm F$, $\alpha$'da \textbf{s\"ureklidir.}
E\u ger $\bm F(\alpha)=0$ veya $\alpha=0$ ise, 
o zaman $\beta=-1$ veya $\delta=-1$ olabilir.
Benzer \c sekilde \textbf{soldan} ve \textbf{sa\u gdan} olan s\"ureklilik
tan\i mlanabilir.

\begin{lemma}
$\on$'de her tek-konumlu i\c slem,
limit olmayan her noktada s\"ureklidir
ve her noktada sa\u gdan s\"ureklidir.
\end{lemma}

\klk

Ger\c cel analizdeki gibi
$\bm F\colon\on\to\on$ ise
\begin{equation*}
\limsup_{\xi\to\alpha^-}\bm F(\xi)=\min\left\{\sup_{\eta<\xi<\alpha}\bm F(\xi)\colon\eta<\alpha\right\}
\end{equation*}
tan\i m\i n\i\ yapar\i z.

\begin{lemma}
$\bm F\colon\on\to\on$ ve artan ise	
\begin{equation*}
\limsup_{\xi\to\alpha^-}\bm F(\xi)=\sup_{\xi<\alpha}\bm F(\xi).
\end{equation*}
\end{lemma}

\klk

\begin{lemma}
$\bm F\colon\on\to\on$ olsun.
$\bm F$ bir $\alpha$ limitinde s\"ureklidir ancak ve ancak
\begin{equation*}
\limsup_{\xi\to\alpha^-}\bm F(\xi)=\bm F(\alpha).
\end{equation*}
\end{lemma}

\klk

\begin{theorem}
$\bm F\colon\on\to\on$ ve kesin artan olsun.
O zaman $\bm F$ normaldir ancak ve ancak 
her noktada s\"ureklidir.
\end{theorem}

\ktk

\begin{xca}
  S\"urekli olup normal olmayan bir i\c slem \"orne\u gi verin.
\end{xca}

\chapter{Ordinal toplama}\label{ch:add}

\section{Tan\i m ve \"ozellikler}

\"Ozyineli tan\i ma g\"ore her $\alpha$ ordinali i\c cin
\begin{gather}\label{eqn:+0}
  \alpha+0=\alpha,\\\label{eqn:+'}
\alpha+\beta'=(\alpha+\beta)',\\\label{eqn:+lim}
\gamma\text{ limit ise }\alpha+\gamma=\sup\{\alpha+\xi\colon\xi<\gamma\}.
\end{gather}
Ordinal toplaman\i n \"ozelliklerinin \c co\u gu,
t\"umevar\i m ile kan\i tlan\i r;
ama ilk teoremimiz, t\"umevar\i mdan de\u gildir.

\begin{theorem}
  $\alpha+1=\alpha'$.
\end{theorem}

\begin{proof}
\hfill
$\begin{aligned}[t]
  \alpha+1
&=\alpha+0'&&\\
&=(\alpha+0)'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=\alpha'.&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}
\end{aligned}$\hfill\vspace{-0.8\baselineskip}

\mbox{}
\end{proof}

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

\begin{proof}
Ordinal t\"umevar\i m kullanaca\u g\i z.
  \begin{asparaenum}
    \item
E\u ger $\alpha=0$ ise
\begin{align*}
0+\alpha
&=0+0&&\text{[varsay\i mdan]}\\
&=0&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}\\
&=\alpha.&&\text{[varsay\i mdan]}
\end{align*}
\item
E\u ger
\begin{equation}\label{eqn:0+b=b}
0+\beta=\beta
\end{equation}
ise, o zaman
  \begin{align*}
    0+\beta'
&=(0+\beta)'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=\beta'.&&\text{[\eqref{eqn:0+b=b} hipotezinden]}
  \end{align*}
\item
Bir $\alpha$ limiti i\c cin
\begin{equation}\label{eqn:0+b=b-lim}
  \Forall{\xi}(\xi<\alpha\lto0+\xi=\xi)
\end{equation}
ise, o zaman
\begin{align*}
  0+\alpha
&=\sup\{0+\xi\colon\xi<\alpha\}&&\text{[\eqref{eqn:+lim} tan\i m\i ndan]}\\
&=\sup\{\xi\colon\xi<\alpha\}&&\text{[\eqref{eqn:0+b=b-lim} hipotezinden]}\\
&=\alpha.&&\text{[\Teoremde{thm:sup-a}n]}\qedhere
\end{align*}
  \end{asparaenum}
\end{proof}

\begin{xca}
\mathversion{false}
A\c sa\u g\i daki kan\i t
nerede yanl\i\c st\i r?
\begin{falseproof}
Her $\alpha$ i\c cin $1+\alpha=\alpha'$ kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$1+0=1=0'$.
\item
$1+\beta=\beta'$ ise, o zaman
\begin{equation*}
1+\beta'=(1+\beta)'=(\beta')'.
\end{equation*}
\item
$\gamma$ limit ve $\Forall{\xi}(\xi<\gamma\lto1+\xi=\xi')$ ise
\begin{equation*}
1+\gamma=\sup_{\xi<\gamma}(1+\xi)=\sup_{\xi<\gamma}(\xi')=\gamma'.
\end{equation*}
\end{asparaenum}

B\"oylece her $\alpha$ i\c cin $1+\alpha=\alpha'$.
\end{falseproof}
\end{xca}

\begin{theorem}\label{thm:a+x}
Her $\alpha$ ordinali i\c cin $\xi\mapsto\alpha+\xi$ normaldir.
\end{theorem}

\begin{proof}
\Teoremde{thm:norm-cond}n $\alpha+\beta<\alpha+\beta'$ g\"ostermek yeter.
Ayr\i ca
\begin{align*}
      \alpha+\beta
&<(\alpha+\beta)'&&\\
&=\alpha+\beta'.&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\qedhere
\end{align*}
\end{proof}

\"Orne\u gin
\Sekle{fig:w+x} bak\i n.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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  \caption{$\eta=\upomega+\xi$ denkleminin grafi\u gi}\label{fig:w+x}  
\end{figure}
\c Sekilde
\begin{align*}
	\upomega\cdot2&=\upomega+\upomega,&
\upomega\cdot3&=\upomega\cdot2+\upomega,&
\upomega\cdot4&=\upomega\cdot3+\upomega,
\end{align*}
ve genelde,
\Teoremi{thm:rec} kullanan resmi \"ozyineli tan\i ma g\"ore,
\begin{align*}
\alpha\cdot0&=0,&
	\alpha\cdot1&=\alpha,&
	\alpha\cdot(k+1)&=\alpha\cdot k+\alpha.
\end{align*}
Bu \c sekilde $\alpha\cdot n$, ``$\alpha$'d\i r $n$ kere''
veya ``$\alpha$'n\i n $n$ kat\i d\i r.''
Ayr\i ca
\begin{equation*}
\upomega^2=\upomega\cdot\upomega=\sup_{x<\upomega}(\upomega\cdot x).
\end{equation*}

\begin{xca}
$\xi\mapsto\xi\cdot2$ g\"ondermesi kesin artan m\i d\i r?
S\"urekli midir?
\end{xca}

\begin{xca}
\mathversion{false}
A\c sa\u g\i daki kan\i t
nerede yanl\i\c st\i r?
\begin{falseproof}
Her $\alpha$ i\c cin, her $\beta$ i\c cin,
$\alpha+\beta=\beta+\alpha$ kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$\alpha+0=\alpha=0+\alpha$.
\item
E\u ger $\alpha+\beta=\beta+\alpha$ ise, o zaman
\begin{equation*}
\alpha+\beta'=(\alpha+\beta)'=(\beta+\alpha)'=\beta'+\alpha.
\end{equation*}
\item
E\u ger $\gamma$ limit ve $\Forall{\xi}(\xi<\gamma\lto\alpha+\xi=\xi+\alpha)$ ise,
o zaman
\begin{equation*}
\alpha+\gamma
=\sup_{\xi<\gamma}(\alpha+\xi)
=\sup_{\xi<\gamma}(\xi+\alpha)
=\gamma+\alpha.
\end{equation*}
\end{asparaenum}
Bu \c sekilde her durumda $\alpha+\beta=\beta+\alpha$.
\end{falseproof}
\end{xca}

\begin{theorem}\label{thm:+assoc}
Ordinal toplama birle\c smelidir.
\end{theorem}

\begin{proof}
Her $\gamma$ i\c cin, 
t\"umevar\i m kullanarak
t\"um $\alpha$ ve $\beta$ i\c cin
  \begin{equation*}
    \alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma
  \end{equation*}
	g\"osterece\u giz.
  \begin{asparaenum}
    \item
$		\begin{aligned}[t]
\alpha+(\beta+0)
&=\alpha+\beta&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}\\
&=(\alpha+\beta)+0.&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}
					\end{aligned}$
\item
E\u ger
\begin{equation}\label{eqn:+assoc'}
\alpha+(\beta+\delta)=(\alpha+\beta)+\delta
\end{equation}
ise, o zaman
  \begin{align*}
    \alpha+(\beta+\delta')
		&=\alpha+(\beta+\delta)'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=(\alpha+(\beta+\delta))'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=((\alpha+\beta)+\delta)'&&\text{[\eqref{eqn:+assoc'} hipotezinden]}\\
&=(\alpha+\beta)+\delta'.&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}
  \end{align*}
\item
$\delta$ limit olsun, ve
\begin{equation}\label{eqn:+assoc-lim}
\Forall{\xi}\bigl(\xi<\delta\lto
\alpha+(\beta+\xi)=(\alpha+\beta)+\xi\bigr)
\end{equation}
olsun.  
O zaman
\begin{align*}
&\phantom{{}={}}(\alpha+\beta)+\delta\\
&=\sup\{(\alpha+\beta)+\xi\colon\xi<\delta\}&&\text{[\eqref{eqn:+lim} tan\i m\i]}\\
&=\sup\{\alpha+(\beta+\xi)\colon\xi<\delta\}&&\text{[\eqref{eqn:+assoc-lim} hipotezi]}\\
&=\alpha+\sup\{\beta+\xi\colon\xi<\delta\}&&\text{[$\xi\mapsto\alpha+\xi$ normaldir]}\\
&=\alpha+(\beta+\delta).&&\text{[\eqref{eqn:+lim} tan\i m\i]}\qedhere 
\end{align*}
  \end{asparaenum}
\end{proof}

\c Simdi herhangi $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
\alpha\cdot n=\underbrace{\alpha+\dots+\alpha}_n
\end{equation*}
anla\c s\i labilir.

\begin{theorem}
$k<\upomega$ ve $\ell<\upomega$ ise
$\alpha\cdot(k+\ell)=\alpha\cdot k+\alpha\cdot\ell$.
\end{theorem}

\ktk[ (T\"umevar\i m kullan\i n.)]

\begin{theorem}\label{thm:x+a}
Her $\xi\mapsto\xi+\alpha$ g\"ondermesi artand\i r.
\end{theorem}

\begin{proof}
  $\beta\leq\gamma$ olsun.
$\alpha$ \"uzerinden t\"umevar\i m kullanarak  
\begin{equation*}
\beta+\alpha\leq\gamma+\alpha
\end{equation*}
kan\i tlayaca\u g\i z.
\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
E\u ger $\delta$ limit ise
\begin{equation*}
\Forall{\xi}(\xi<\delta\lto\beta+\xi<\gamma+\xi)
\end{equation*}
olsun.  O zaman
  \begin{equation*}
    \beta+\delta=\sup_{\xi<\delta}(\beta+\xi)
\leq\sup_{\xi<\delta}(\gamma+\xi)=\gamma+\delta.\qedhere
  \end{equation*}
  \end{asparaenum}
\end{proof}

\section{Hesaplamalar}

Bu altb\"ol\"um\"un teoremleri
t\"umevar\i m kullanmaz.

\begin{theorem}%\label{thm:k+w=w}
  $k<\upomega$ ise $k+\upomega=\upomega$.
(\Sekle{fig:x+w} bak\i n.)
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  \caption{$\eta=\xi+\upomega$ denkleminin grafi\u gi}\label{fig:x+w}  
\end{figure}
\end{theorem}

\begin{proof}
  $k+\upomega=\sup\{k+x\colon x<\upomega\}=\upomega$.
\end{proof}

\begin{corollary}
$k<\upomega$ ve $1\leq n<\upomega$ ise
\begin{equation*}
k+\upomega\cdot n=\upomega\cdot n.
\end{equation*}
\end{corollary}

\ksk

\begin{comment}

\begin{corollary}
  Ordinal toplama de\u gi\c smeli de\u gildir.
\end{corollary}

\ksk
\end{comment}

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

\begin{proof}
Denklemin \c c\"oz\"um\"u varsa,
\Teoreme{thm:a+x} g\"ore tek \c c\"oz\"um vard\i r.

Teoremler \ref{thm:0+} ve \numarada{thm:x+a}n $\alpha+\beta\geq\beta$,
dolay\i s\i yla $\{\xi\colon\alpha+\xi\leq\beta\}$ 
s\i n\i f\i n\i n $\beta'$ \"usts\i n\i r\i\ vard\i r.
\c Simdi $\gamma$, s\i n\i f\i n\i n supremumu olsun.
O zaman
\begin{gather*}
\begin{aligned}
	\alpha+\gamma
	&=\alpha+\sup\{\xi\colon\alpha+\xi\leq\beta\}\\
	&=\sup\{\alpha+\xi\colon\alpha+\xi\leq\beta\}
	\leq\beta,
\end{aligned}\\
(\alpha+\gamma)'
		=\alpha+\gamma'
		>\beta,
\end{gather*}
dolay\i s\i yla $\alpha+\gamma=\beta$.
\end{proof}

\begin{xca}
$\alpha\leq\beta$ varsay\i nca,
$\{\xi\colon a+\xi\geq\beta\}$ s\i n\i f\i n\i n bo\c s olmay\i p
s\i n\i f\i n en k\"u\c c\"uk eleman\i n\i n \eqref{eqn:a+x=b} denkleminin
\c c\"oz\"um\"u oldu\u gunu g\"osterin.
\end{xca}

\begin{theorem}
$\upomega+\alpha=\alpha$ ancak ve ancak $\upomega^2\leq\alpha$.
\end{theorem}

\begin{proof}
$\begin{aligned}[t]
	\upomega+\upomega^2
	&=\upomega+\sup_{x<\upomega}(\upomega\cdot x)\\
	&=\sup_{x<\upomega}(\upomega+\upomega\cdot x)\\
	&=\sup_{x<\upomega}\bigl(\upomega\cdot(1+x)\bigr)\\
	&=\upomega^2.
\end{aligned}$

E\u ger $\alpha\geq\upomega^2$ ise, o zaman bir $\beta$ i\c cin $\upomega^2+\beta=\alpha$,
dolay\i s\i yla
\begin{equation*}
\upomega+\alpha
=\upomega+\upomega^2+\beta
=\upomega^2+\beta
=\alpha.
\end{equation*}
\c Simdi $\alpha<\upomega^2$ olsun.
O zaman bir $k$ do\u gal say\i s\i\ i\c cin
\begin{gather*}%\label{eqn:wk-a-w(k+1)}
\upomega\cdot k\leq\alpha<\upomega\cdot(k+1),\\%\notag
\upomega\cdot(k+1)\leq\upomega+\alpha,	
\end{gather*}
dolay\i s\i yla $\alpha<\upomega+\alpha$.
\end{proof}

Teorem sayesinde $\upomega\leq\alpha<\upomega^2$ ise,
o zaman bir $\alpha_1$ i\c cin
\begin{align*}
	\upomega+\alpha_1&=\alpha,&
	\alpha_1&<\alpha.
\end{align*}
E\u ger $\upomega\leq\alpha_1$ ise,
o zaman bir $\alpha_2$ i\c cin
\begin{align*}
	\upomega+\alpha_2&=\alpha_1,&
	\alpha_2&<\alpha_1,
\end{align*}
ve saire.
O zaman bir $k$ i\c cin
\begin{equation*}
\alpha
={\underbrace{\upomega+\dots+\upomega}_k}+\alpha_k
=\upomega\cdot k+\alpha_k.
\end{equation*}
$\on$ iyis\i ral\i\ oldu\u gundan bir $k$ i\c cin
$\alpha_k<\upomega$.
Bu \c sekilde
\begin{equation*}
\{\xi\colon\xi<\upomega^2\}
\end{equation*}
k\"umesinin her eleman\i,
$\upomega\cdot k+\ell$ bi\c ciminde yaz\i labilir.
Verilen k\"ume, toplama alt\i nda kapal\i d\i r,
ve toplama kural\i,
\begin{equation*}
(\upomega\cdot k+\ell)+(\upomega\cdot m+n)
=\upomega\cdot(k+m)+n.
\end{equation*}

\begin{xca}
$\alpha=\upomega\cdot17+6$ ve $\beta=\upomega\cdot1000+5$ ise
$\alpha+\beta$ toplam\i n\i\ hesaplay\i n.
\end{xca}

\section{Kardinaller}

\c Simdi herhangi $A$ ve $B$ k\"umeleri i\c cin
\begin{equation*}
  A\sqcup B=(A\times\{0\})\cup(B\times\{1\})
\end{equation*}
olsun; bu bile\c sim, 
$A$ ve $B$'nin \textbf{ayr\i k bile\c simidir.}
B\"ol\"um \numarada{ch:ax}n
\begin{equation*}
\alpha=\{\xi\colon\xi<\alpha\}
\end{equation*}
anla\c smas\i n\i\ kullanaca\u g\i z.

\begin{theorem}\label{thm:a+b}
  $\alpha+\beta\approx\alpha\sqcup\beta$.
\end{theorem}

\begin{proof}
  \Teoremde{thm:subtraction}n
	\begin{equation*}
	\left\{
	\begin{aligned}
(\xi,0)&\mapsto\xi,\\
(\eta,1)&\mapsto\alpha+\eta
	\end{aligned}
	\right.
	\end{equation*}
kural\i,
$\alpha\sqcup\beta$
ayr\i k bile\c siminden
$\alpha+\beta$ k\"umesine
giden bir e\c sleme tan\i mlar.
\end{proof}

Bir $A$ k\"umesi bir ordinalle e\c slenik olsun.
Tan\i ma g\"ore
\begin{equation*}
\card A=\min\bigl\{\xi\colon\xi\approx A\bigr\};
\end{equation*}
bu ordinal, $A$'n\i n \textbf{kardinalidir.}
Kardinaller, $\kappa$, $\lambda$, $\mu$, ve $\nu$\label{card}
harfleri ile g\"osterilecektir.

E\u ger $f\colon A\to B$ ve $C\included A$ ise
\begin{equation*}
f[C]=\{f(x)\colon x\in C\}
\end{equation*}
olsun.
E\u ger $f$ birebir ise, o zaman
$A$'n\i n $B$'ye bir \textbf{g\"ommesidir,}
ve
\begin{equation*}
A\approx f[A]\included B.
\end{equation*}
Bu durumda
\begin{equation*}
f\colon A\xrightarrow{\preccurlyeq}B
\end{equation*}
yazal\i m, ve \"oyle bir $f$ g\"ommesi varsa
\begin{equation*}
A\preccurlyeq B
\end{equation*}
yazal\i m.

\begin{theorem}[Schr\"oder--Bernstein]
$A\preccurlyeq B$ ve $B\preccurlyeq A$ ise
\begin{equation*}
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&=A,&A_{n+1}&=g[B_n],\\
B_0&=B,&B_{n+1}&=f[A_n]
\end{align*}
olsun.
O zaman
\begin{gather*}
	f[A_0\setminus A_1]=B_1\setminus B_2,\\
	g[B_0\setminus B_1]=A_1\setminus A_2,
\end{gather*}
dolay\i s\i yla $A_0\setminus A_2\approx B_0\setminus B_2$.
Benzer \c sekilde
\begin{equation*}
A_n\setminus A_{n+2}\approx B_n\setminus B_{n+2},
\end{equation*}
dolay\i s\i yla
\begin{equation*}
A\setminus\bigcap_{i<\upomega}A_i\approx B\setminus\bigcap_{i<\upomega}B_i.
\end{equation*}
Ayr\i ca
\begin{equation*}
f\left[\bigcap_{i<\upomega}A_i\right]=\bigcap_{i<\upomega}f[A_i]=\bigcap_{0<i<\upomega}B_i=\bigcap_{i<\upomega}B_i,
\end{equation*}
dolay\i s\i yla 
$\bigcap_{i<\upomega}A_i\approx\bigcap_{i<\upomega}B_i$,
ve sonu\c c olarak $A\approx B$.
\end{proof}

\begin{theorem}
$\xi\mapsto\card{\xi}$ artand\i r.
\end{theorem}

\begin{proof}
E\u ger $\alpha\leq\beta$ ama $\card{\beta}\leq\card{\alpha}$ ise,
o zaman
\begin{equation*}
\alpha\preccurlyeq\beta\approx\card{\beta}\preccurlyeq\card{\alpha}\approx\alpha,
\end{equation*}
dolay\i s\i yla $\alpha\approx\beta$.
\end{proof}

\begin{theorem}
$k<\upomega$ ise $\card k=k$.
\end{theorem}

\ktk

\begin{theorem}
$\card{\upomega+\upomega}=\upomega$.
\end{theorem}

\ktk

\begin{corollary}
$\{\xi\colon\upomega\leq\xi<\upomega^2\}$ k\"umesinin her eleman\i n\i n kardinali
$\upomega$'d\i r.
\end{corollary}

\ksk

\chapter{Ordinal \c carpma}\label{ch:mul}

\section{Tan\i m ve \"ozellikler}

\"Ozyineli tan\i ma g\"ore her $\alpha$ i\c cin
\begin{gather*}
	\alpha\cdot0=0,\\
	\alpha\cdot\beta'=\alpha\cdot\beta+\alpha,\\
	\gamma\text{ limit ise }\alpha\cdot\gamma=\sup\{\alpha\cdot\xi\colon\xi<\gamma\}.
\end{gather*}

Ordinal \c carpma hakk\i nda ilk teoremimizin bir \c s\i kk\i\ t\"umevar\i m kullanmaz;
kalanlar t\"umevar\i m kullan\i yor.

\begin{theorem}\mbox{}
\begin{compactenum}
\item
$\alpha\cdot1=\alpha$.
\item
$1\cdot\alpha=\alpha$.
\item
$0\cdot\alpha=0$.
\end{compactenum}
\end{theorem}

\ktk

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

\ktk
\"Orne\u gin
\Sekle{fig:wx} bak\i n.
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  \caption{$\eta=\upomega\cdot\xi$ denkleminin grafi\u gi}\label{fig:wx}  
\end{figure}
\c Sekilde
\begin{align*}
	\upomega^2&=\upomega\cdot\upomega,&
\upomega^3&=\upomega^2\cdot\upomega,&
\upomega^4&=\upomega^3\cdot\upomega,
\end{align*}
ve genelde,
\Teoremi{thm:rec} kullanan resmi \"ozyineli tan\i ma g\"ore,
\begin{align*}
\alpha^0&=1,&
	\alpha^1&=\alpha,&
	\alpha^{k+1}&=\alpha^k\cdot\alpha.
\end{align*}
Ayr\i ca
\begin{equation*}
\upomega^{\upomega}=\sup_{x<\upomega}(\upomega^x).
\end{equation*}

\begin{xca}
$\xi\mapsto\xi^2$ g\"ondermesi kesin artan m\i d\i r?
S\"urekli midir?
\end{xca}

\begin{theorem}
Ordinal \c carpma, toplama \"uzerine soldan da\u g\i l\i r.
\end{theorem}

\begin{proof}
Ordinal t\"umevar\i m ile
\begin{equation}\label{eqn:dist}
 \alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma
\end{equation}
kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$\alpha\cdot(\beta+0)
=\alpha\cdot\beta
=\alpha\cdot\beta+0
=\alpha\cdot\beta+\alpha\cdot0$.
\item
E\u ger \eqref{eqn:dist} do\u gru ise, o zaman
\begin{align*}
	\alpha\cdot(\beta+\gamma')
	&=\alpha\cdot(\beta+\gamma)'\\
	&=\alpha\cdot(\beta+\gamma)+\alpha\\
	&=(\alpha\cdot\beta+\alpha\cdot\gamma)+\alpha\\
	&=\alpha\cdot\beta+(\alpha\cdot\gamma+\alpha)\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma'.
\end{align*}
\item\c Simdi $\gamma$ limit ve
  \begin{equation*}
\Forall{\xi}(\xi<\gamma\lto\alpha\cdot(\beta+\xi)
=\alpha\cdot\beta+\alpha\cdot\xi)
  \end{equation*}
olsun.
E\u ger $\alpha=0$ ise, iddia apa\c c\i kt\i r,
dolay\i s\i yla $\alpha>0$ varsayaca\u g\i z.  
\begin{align*}
	\alpha\cdot(\beta+\gamma)
	&=\alpha\cdot\sup_{\xi<\gamma}(\beta+\xi)&&\text{[tan\i m]}\\
	&=\sup_{\xi<\gamma}\bigl(\alpha\cdot(\beta+\xi)\bigr)&&\text{[$\eta\mapsto\alpha\cdot\eta$ normaldir]}\\
	&=\sup_{\xi<\gamma}(\alpha\cdot\beta+\alpha\cdot\xi)&&\text{[t\"umevar\i m hipotezi]}\\
	&=\alpha\cdot\beta+\sup_{\xi<\gamma}(\alpha\cdot\xi)&&\text{[$\eta\mapsto\alpha\cdot\beta+\eta$ normaldir]}\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma.&&\text{[tan\i m]}\qedhere
\end{align*}
\end{asparaenum}
\end{proof}

\begin{xca}
A\c sa\u g\i daki kan\i t nerede yanl\i\c st\i r?
\begin{falseproof}\noindent
\begin{compactenum}
\item
$0\cdot(\beta+\gamma)=0=0+0=0\cdot\beta+0\cdot\gamma$.
\item
E\u ger \eqref{eqn:dist} do\u gru ise, o zaman
\begin{align*}
\alpha'\cdot(\beta+\gamma)
&=\alpha\cdot(\beta+\gamma)+(\beta+\gamma)\\
&=(\alpha\cdot\beta+\alpha\cdot\gamma)+(\beta+\gamma)\\
&=(\alpha\cdot\beta+\beta)+(\alpha\cdot\gamma+\gamma)\\
&=\alpha'\cdot\beta+\alpha'\cdot\gamma.
\end{align*}
\item
E\u ger $\alpha$ limit ve
$\Forall{\xi}\bigl(\xi<\alpha\lto\xi\cdot(\beta+\gamma)=\xi\cdot\beta+\xi\cdot\gamma\bigr)$ ise
\begin{align*}
	\alpha\cdot(\beta+\gamma)
	&=\sup_{\xi<\alpha}\bigl(\xi\cdot(\beta+\gamma)\bigr)\\
	&=\sup_{\xi<\alpha}\bigl(\xi\cdot\beta+\xi\cdot\gamma)\\
	&=\sup_{\xi<\alpha}(\xi\cdot\beta)+\sup_{\xi<\alpha}(\xi\cdot\gamma)\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma.
\end{align*}
\end{compactenum}
\end{falseproof}
\end{xca}

\begin{xca}
A\c sa\u g\i daki kan\i t nerede yanl\i\c st\i r?
\begin{falseproof}\noindent
\begin{compactenum}
\item
$(\alpha+\beta)\cdot0=0=0+0=\alpha\cdot0+\beta\cdot0$.
\item
E\u ger $(\alpha+\beta)\cdot\gamma=\alpha\cdot\gamma+\beta\cdot\gamma$ ise, o zaman
\begin{align*}
(\alpha+\beta)\cdot\gamma'
&=(\alpha+\beta)\cdot\gamma+(\alpha+\beta)\\
&=(\alpha\cdot\gamma+\beta\cdot\gamma)+(\alpha+\beta)\\
&=(\alpha\cdot\gamma+\alpha)+(\beta\cdot\gamma+\beta)\\
&=\alpha\cdot\gamma'+\beta\cdot\gamma'.
\end{align*}
\item
E\u ger $\gamma$ limit ve
$\Forall{\xi}\bigl(\xi<\gamma\lto(\alpha+\beta)\cdot\xi=\alpha\cdot\xi+\beta\cdot\xi\bigr)$ ise
\begin{align*}
(\alpha+\beta)\cdot\gamma
&=\sup_{\xi<\gamma}\bigl((\alpha+\beta)\cdot\xi\bigr)\\
&=\sup_{\xi<\gamma}(\alpha\cdot\xi+\beta\cdot\xi)\\
&=\sup_{\xi<\gamma}(\alpha\cdot\xi)+\sup_{\xi<\gamma}(\beta\cdot\xi)\\
&=\alpha\cdot\gamma+\beta\cdot\gamma.
\end{align*}
\end{compactenum}
\end{falseproof}
\end{xca}

\begin{theorem}\label{thm:.assoc}
Ordinal \c carpma birle\c smelidir.
\end{theorem}

\ktk

\c Simdi herhangi $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
\alpha^n=\underbrace{\alpha\cdots\alpha}_n
\end{equation*}
anla\c s\i labilir.

\begin{theorem}
$k<\upomega$ ve $\ell<\upomega$ ise
$\alpha^{k+\ell}=\alpha^k\cdot\alpha^{\ell}$.
\end{theorem}

\ktk[ (T\"umevar\i m kullan\i n.)]

\begin{theorem}
Her $\xi\mapsto\xi\cdot\alpha$ i\c slemi artand\i r.
\end{theorem}

\ktk

\section{Hesaplamalar}

\begin{lemma}
	$0<\ell<\upomega$ ise $1+\upomega^{\ell}=\upomega^{\ell}$.
\end{lemma}

\ktk

\begin{theorem}\label{thm:w^k+w^m=w^m}
$k<m<\upomega$ ise $\upomega^k+\upomega^m=\upomega^m$.
\end{theorem}

\begin{proof}
Bir $\ell$ i\c cin, $k+\ell=m$ ve $0<\ell<\upomega$,
dolay\i s\i yla
\begin{align*}
\upomega^k+\upomega^m
&=\upomega^k+\upomega^{k+\ell}\\
&=\upomega^k+\upomega^k\cdot\upomega^{\ell}\\
&=\upomega^k\cdot(1+\upomega^{\ell})\\
&=\upomega^k\cdot\upomega^{\ell}\\
&=\upomega^{k+\ell}\\
&=\upomega^m.\qedhere
\end{align*}
\end{proof}

\begin{theorem}
  $1\leq k<\upomega$ ise $k\cdot\upomega=\upomega$.
(\Sekle{fig:xw} bak\i n.)
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  \caption{$\eta=\xi\cdot\upomega$ denkleminin grafi\u gi}\label{fig:xw}  
\end{figure}

\end{theorem}

\ktk

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

\begin{xca}
  Teoremi kan\i tlay\i n.
\"Orne\u gin, a\c sa\u g\i daki iddialar\i\ g\"osterin.
\begin{asparaenum}
  \item
$\alpha>0$ ise $\alpha\cdot\beta\geq\beta$.
\item
$\{\eta\colon\alpha\cdot\eta\leq\beta\}$ k\"umesinin \"usts\i n\i r\i\ vard\i r.
\item
$\sup\{\eta\colon\alpha\cdot\eta\leq\beta\}=\delta$ olsun.
O zaman $\alpha\cdot\gamma+\xi=\beta$ denkleminin 
$\gamma$ \c c\"oz\"um\"u vard\i r, ve $\delta<\alpha$.
Ayr\i ca $(\gamma,\delta)$,
\eqref{eqn:ax+y=b} sisteminin tek \c c\"oz\"um\"u vard\i r.
\end{asparaenum}
\end{xca}

\begin{theorem}
$\upomega^{\upomega}$,
\begin{equation*}
\upomega\cdot\xi=\xi
\end{equation*}
denkleminin en k\"u\c c\"uk \c c\"oz\"um\"ud\"ur.
\end{theorem}

\begin{proof}
$\begin{aligned}[t]
\upomega\cdot\upomega^{\upomega}
&=\upomega\cdot\sup_{x<\upomega}(\upomega^x)\\
&=\sup_{x<\upomega}(\upomega\cdot\upomega^x)\\
&=\sup_{x<\upomega}(\upomega^{1+x})\\
&=\upomega^{\upomega}.\end{aligned}$

\c Simdi $\alpha<\upomega^{\upomega}$ olsun.
O zaman bir $k$ do\u gal say\i s\i\ i\c cin
\begin{gather*}
	\upomega^k\leq\alpha<\upomega^{k+1},\\
	\upomega^{k+1}\leq\upomega\cdot\alpha,
\end{gather*}
dolay\i s\i yla $\alpha<\upomega\cdot\alpha$.
\end{proof}

Teorem sayesinde $\alpha<\upomega^{\upomega}$ ise,
o zaman baz\i\ $\alpha_1$ ve $a_0$ i\c cin
\begin{align*}
	\upomega\cdot\alpha_1+a_0&=\alpha,&
	\alpha_1&<\alpha,&
	a_0&<\upomega.
\end{align*}
E\u ger $\alpha_1>0$ ise,
o zaman baz\i\ $\alpha_2$ ve $a_1$ i\c cin
\begin{align*}
	\upomega\cdot\alpha_2+a_1&=\alpha_1,&
	\alpha_2&<\alpha_1,&
	a_1&<\upomega,
\end{align*}
ve saire.
O zaman bir $k$ i\c cin
\begin{align*}
	\alpha_{k+1}&=0,\\
	\alpha_k&=a_k,\\
	\alpha_{k-1}&=\upomega\cdot a_k+a_{k-1},\\
	\alpha_{k-2}&=\upomega^2\cdot a_k+\upomega\cdot a_{k-1}+a_{k-2},\\
	&\dots\\
	\alpha_1&=\upomega^{k-1}\cdot a_k+\upomega^{k-2}\cdot a_{k-1}+\dots+\upomega\cdot a_2+a_1,\\
	\alpha&=\upomega^k\cdot a_k+\upomega^{k-1}\cdot a_{k-1}+\dots+\upomega^2\cdot a_2+\upomega\cdot a_1+a_0.
\end{align*}
Burada baz\i\ $a_i$ s\i f\i r olabilir.
S\i f\i r terimler silinirse,
o zaman bir $n$ i\c cin, 
\begin{equation*}
\upomega>b_0>b_1>\dots>b_{n-1}
\end{equation*}
ko\c sulunu sa\u glayan
baz\i\ $b_i$ i\c cin, ve baz\i\ $c_i$ sayma say\i lar\i\ i\c cin
\begin{equation*}
\alpha=\upomega^{b_0}\cdot c_0+\upomega^{b_1}\cdot c_1+\dots+\upomega^{b_{n-1}}\cdot c_{n-1}.
\end{equation*}
Bu ifadeye $\alpha$'n\i n \textbf{Cantor normal bi\c cimi} denir.
($0$'\i n Cantor normal bi\c cimi $0$'d\i r.)

\begin{theorem}\label{thm:a+w^m=w^m}
$0<m<\upomega$ ve $\alpha<\upomega^m$ ise $\alpha+\upomega^m=\upomega^m$.
\end{theorem}

\begin{proof}
$\alpha$'n\i n Cantor normal bi\c cimini yaz\i n
ve \Teoremi{thm:w^k+w^m=w^m} kullan\i n.
\end{proof}

\begin{corollary}
$m<\upomega$, $n\in\N$ ve $k\in\N$ ise
\begin{equation*}
(\upomega^m\cdot n+\alpha)\cdot k=\upomega^m\cdot n\cdot k+\alpha.
\end{equation*}
\end{corollary}

\ksk

\"Orne\u gin
\begin{equation*}
(\upomega^5\cdot10+\upomega^3\cdot8+\upomega)\cdot6
=\upomega^5\cdot60+\upomega^3\cdot8+\upomega.
\end{equation*}

Sonucun durumunda a\c sa\u g\i daki e\c sitlik \c c\i kar.
\begin{align*}
(\upomega^m\cdot n+\alpha)\cdot\upomega
&=\upomega^m\cdot n+\underbrace{\alpha+\upomega^m\cdot n}_{\upomega^m\cdot n}+\underbrace{\alpha+\upomega^m\cdot n}_{\upomega^m\cdot n}+\cdots\\
&=\upomega^m\cdot n\cdot\upomega\\
&=\upomega^{m+1}.
\end{align*}
Asl\i nda e\c sitli\u gin ger\c cek kan\i t\i n\i n
\Teoreme{thm:a+w^m=w^m} ihtiyac\i\ yoktur.

\begin{theorem}
$m<\upomega$, $n\in\N$ ve $\alpha<\upomega^m$ ise
\begin{equation*}
(\upomega^m\cdot n+\alpha)\cdot\upomega=\upomega^{m+1},
\end{equation*}
dolay\i s\i yla
$k\in\N$ ise
\begin{equation*}
(\upomega^m\cdot n+\alpha)\cdot\upomega^k=\upomega^{m+k}.
\end{equation*}
\end{theorem}

\begin{proof}
$(\upomega^m\cdot n+\alpha)\cdot k
<\upomega^m\cdot(n+1)\cdot k$
oldu\u gundan
\begin{align*}
	\upomega^{m+1}
	&\leq(\upomega^m\cdot n+\alpha)\cdot\upomega\\
  &=\sup_{x<\upomega}\bigl((\upomega^m\cdot n+\alpha)\cdot x\bigr)\\
  &\leq\sup_{x<\upomega}\bigl(\upomega^m\cdot(n+1)\cdot x\bigr)\\
	&=\upomega^{m+1}.\qedhere
\end{align*}
\end{proof}

\"Orne\u gin
\begin{align*}
	&\phantom{{}={}}(\upomega^3\cdot4+\upomega\cdot6)\cdot(\upomega^2\cdot3+8)\\
	&=(\upomega^3\cdot4+\upomega\cdot6)\cdot\upomega^2\cdot3+(\upomega^3\cdot4+\upomega\cdot6)\cdot8\\
	&=\upomega^5\cdot3+\upomega^3\cdot32+\upomega\cdot6.
\end{align*}

\begin{xca}
$(\upomega^9\cdot9+\upomega^2\cdot9+\upomega\cdot9+9)\cdot(\upomega^2\cdot9+\upomega\cdot9+9)$
\c carp\i m\i n\i n Cantor normal bi\c cimini hesaplay\i n.
\end{xca}

\section{Kardinaller}

\begin{theorem}\label{thm:ab=axb}
  $\alpha\cdot\beta
\approx\alpha\times\beta$.
\end{theorem}

\begin{proof}
  \Teoremde{thm:division}n
	\begin{equation*}
	(\xi,\eta)\mapsto\alpha\cdot\eta+\xi,
	\end{equation*}
$\alpha\times\beta$
kartezyan
\c carp\i m\i ndan
$\alpha\cdot\beta$ ordinal \c carp\i m\i na giden
bir e\c slemedir.
\end{proof}

\begin{theorem}
$\card{\upomega\cdot\upomega}=\upomega$.
\end{theorem}

\ktk

\begin{theorem}
$\{\xi\colon\upomega\leq\xi<\upomega^{\upomega}\}$ k\"umesinin her eleman\i n\i n kardinali
$\upomega$'d\i r.
\end{theorem}

\ktk

\begin{theorem}\label{thm:cntbl-u}
Her $k$ do\u gal say\i s\i\ i\c cin $f_k$
bir $A_k$ k\"umesini $\upomega$'ya g\"oms\"un.
O zaman
\begin{equation*}
\bigcup_{i<\upomega}A_i\preccurlyeq\upomega.
\end{equation*}
\end{theorem}

\begin{proof}
$\bigcup_{i<\upomega}A_i$ bile\c siminde
\begin{equation*}
g(x)=\min\{i\colon x\in A_i\}
\end{equation*}
olsun.  O zaman $x\mapsto\left(g(x),f_{g(x)}(x)\right)$ g\"ondermesi,
bile\c simin $\upomega\times\upomega$ \c carp\i m\i na bir g\"ommesidir.
\end{proof}

\begin{corollary}
$\upomega^{\upomega}\approx\upomega$.
\end{corollary}

\begin{proof}
Her $n$ i\c cin, $\upomega^{n+1}=\upomega^n\cdot\upomega$ oldu\u gundan,
\Teoremin{thm:ab=axb} kan\i t\i n\i ndan
kesin bir $f_n$ i\c cin
\begin{equation*}
f_n\colon\upomega^{n+1}\xrightarrow{\approx}\upomega^n\times\upomega.
\end{equation*}
\c Simdi $g\colon\upomega\times\upomega\xrightarrow{\approx}\upomega$ olsun.
O zaman
\begin{equation*}
  g\circ f_1\colon\upomega^2\xrightarrow{\approx}\upomega.
\end{equation*}
M\"umk\"unse
\begin{equation}\label{eqn:hm}
h_m\colon\upomega^m\xrightarrow{\approx}\upomega
\end{equation}
olsun.
O zaman bir ve tek bir $h_{m+1}$ i\c cin,
\begin{gather*}
h_{m+1}\colon\upomega^{m+1}\to\upomega,\\  
\Forall{\xi}\Forall{\eta}\Forall z
\Bigl(f_n(\xi)=(\eta,z)\lto h_{m+1}(\xi)=g\bigl(h_m(\eta),z\bigr)\Bigr);
\end{gather*}
ve bu durumda
\begin{equation*}
  h_{m+1}\colon\upomega^{m+1}\xrightarrow{\approx}\upomega.
\end{equation*}
T\"umevar\i m ve \"ozyinelemeyle her $m$ sayma say\i s\i\ i\c cin,
bir ve tek bir $h_m$ i\c cin, \eqref{eqn:hm} do\u grudur.
\c Simdi
\begin{equation*}
\upomega^{\upomega}=\sup_{0<x<\upomega}(\upomega^x)=\bigcup_{0<x<\upomega}\upomega^x
\end{equation*}
oldu\u gundan teoremi kullan\i labilir.
\end{proof}

\chapter{Ordinal kuvvet alma}\label{ch:exp}

\section{Tan\i m ve \"ozellikler}

\"Ozyineli tan\i ma g\"ore her $\alpha$ i\c cin
\begin{gather*}
	\alpha^0=1,\\
	\alpha^{\beta'}=\alpha^\beta\cdot\alpha,\\
	\gamma\text{ limit ise }\alpha^\gamma
=\sup_{0<\xi<\gamma}(\alpha^{\xi})
=\limsup_{\xi\to\gamma^-}(\alpha^{\xi}).
\end{gather*}

\begin{theorem}
  $\alpha^1=\alpha$, $1^{\alpha}=1$, ve
  \begin{equation*}
  0^{\alpha}=
  \begin{cases}
    1,&\text{ $\alpha=0$ durumunda},\\
0,&\text{ $\alpha>0$ durumunda}.
  \end{cases}
  \end{equation*}
\end{theorem}

\ktk

\begin{theorem}
$\alpha\geq2$ ise $\xi\mapsto\alpha^{\xi}$ i\c slemi, normaldir.
\end{theorem}

\ktk

\Sekle{fig:w^x} bak\i n.
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  \caption{$\eta=\upomega^{\xi}$ denkleminin grafi\u gi}\label{fig:w^x}  
\end{figure}
\c Sekilde
\begin{equation*}\label{epsilon}
\upvarepsilon_0=\sup\left\{\upomega,\upomega^{\upomega},\upomega^{\upomega^{\upomega}},\dots\right\}.
\end{equation*}

\begin{xca}
$\xi\mapsto\xi^{\xi}$ i\c slemi kesin artan m\i d\i r?
S\"urekli midir?
\end{xca}

\begin{theorem}
$\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma}$
ve
$\alpha^{\beta\cdot\gamma}=(\alpha^{\beta})^{\gamma}$.
\end{theorem}

\ktk

\begin{theorem}
$\alpha\geq1$ ise $\xi\mapsto\xi^{\alpha}$ artand\i r.
\end{theorem}

\ktk

\section{Hesaplamalar}

\begin{theorem}[Logaritma alma]
$2\leq\alpha$ ve $1\leq\beta$ ise $(\xi,\eta,\zeta)$ i\c cin
\begin{equation*}
\alpha^{\xi}\cdot\eta+\zeta=\beta\land0<\eta<\alpha\land\zeta<\alpha^{\xi}
\end{equation*}
sisteminin bir ve tek bir \c c\"oz\"um\"u vard\i r.
\end{theorem}

\ktk

Teorem sayesinde $1\leq\alpha$ ise,
baz\i\ $\alpha_0$, $a_0$, ve $\beta_1$ i\c cin
\begin{align*}
	\upomega^{\alpha_0}\cdot a_0+\beta_1&=\alpha,&
	0<a_0&<\upomega,&
	\beta_1&<\upomega^{\alpha_0}.
\end{align*}
E\u ger $1\leq\beta_1$ ise, o zaman
baz\i\ $\alpha_1$, $a_1$, ve $\beta_2$ i\c cin
\begin{align*}
	\upomega^{\alpha_1}\cdot a_1+\beta_2&=\beta_1,&
	0<a_1&<\upomega,&
	\beta_2&<\upomega^{\alpha_1},
\end{align*}
ve saire.  O zaman bir $k$ i\c cin
\begin{gather*}
\alpha_0>\alpha_1>\dots>\alpha_k,\\
\{a_0,\dots,a_k\}\included\N,\\
\alpha=\upomega^{\alpha_0}\cdot a_0+\upomega^{\alpha_1}\cdot a_1+\dots+\upomega^{\alpha_k}\cdot a_k.
\end{gather*}
Son ifade, $\alpha$'n\i n \textbf{Cantor normal bi\c cimidir.}

\begin{theorem}
$\alpha<\upomega^{\beta}$ ise $\alpha+\upomega^{\beta}=\upomega^{\beta}$.
\end{theorem}

\ktk[ (\Teoreme{thm:a+w^m=w^m} bak\i n.)]

\begin{corollary}
$\alpha<\upomega^{\beta}$, $n\in\N$, ve $k\in\N$ ise
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)\cdot k=\upomega^{\beta}\cdot n\cdot k.
\end{equation*}
\end{corollary}

\ksk

\begin{theorem}
$\alpha<\upomega^{\beta}$, $n\in\N$, ve $1\leq\gamma$ ise
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)\cdot\upomega^{\gamma}=\upomega^{\beta+\gamma}.
\end{equation*}
\end{theorem}

\ktk[ (Bir $\delta$ i\c cin $\gamma=1+\delta$ oldu\u gunu kullanabiliriz.)]

\c Simdi iki Cantor normal bi\c ciminin \c carp\i m\i n\i n 
Cantor normal bi\c cimini hesaplayabiliriz.

\begin{theorem}
  $0<k<\upomega$ ise
  \begin{equation*}
    k^{\upomega^{\xi}}=
    \begin{cases}
      k,&\text{ $\xi=0$ durumunda,}\\
\upomega^{\upomega^{\xi-1}},&\text{ $0<\xi<\upomega$ durumunda,}\\
\upomega^{\upomega^{\xi}},&\text{ $\upomega\leq\xi$ durumunda.}
    \end{cases}
  \end{equation*}
\end{theorem}

\ktk

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

\begin{theorem}
$\upvarepsilon_0$,
\begin{equation*}
\upomega^{\xi}=\xi
\end{equation*}
denkleminin en k\"u\c c\"uk \c c\"oz\"um\"ud\"ur.
\end{theorem}

\ktk

\section{Kardinaller}

Herhangi $\alpha$ ve $\beta$ ordinalleri i\c cin,
$\beta$'dan $\alpha$'ya giden g\"ondermeler,
\begin{equation*}
{}^{\beta}\alpha
\end{equation*}
s\i n\i f\i n\i\ olu\c stursun, ve
\begin{equation*}
\exp(\alpha,\beta)=\bigl\{f\colon f\in{}^{\beta}\alpha\land\{\xi\in\beta\colon f(\xi)>0\}\prec\upomega\bigr\}
\end{equation*}
olsun.

\begin{theorem}
$\alpha^{\beta}\approx\exp(\alpha,\beta)$.
\end{theorem}

\begin{proof}
$\exp(1,\beta)\approx1=1^{\beta}$; ayr\i ca
\begin{equation*}
\exp(0,\beta)\approx
\begin{cases}
	1,&\text{ $\beta=0$ durumunda,}\\
	0,&\text{ $\beta>0$ durumunda,}
\end{cases}
\end{equation*}
dolay\i s\i yla $\exp(0,\beta)\approx 0^{\beta}$.
\c Simdi $\alpha\geq2$ olsun.
E\u ger $\gamma<\alpha^\beta$ ise,
o zaman Cantor normal bi\c cimi gibi,
baz\i\ $n$ do\u gal say\i s\i\ i\c cin,
baz\i\ $\gamma_i$ ve $\delta_i$ i\c cin,
\begin{gather*}
	\beta>\gamma_0>\dots>\gamma_{n-1},\\
	\{\delta_i\colon i<n\}\included\alpha\setminus\{0\},\\
	\gamma=\alpha^{\gamma_0}\cdot c_0+\dots+\alpha^{\gamma_{n-1}}\cdot c_{n-1}.
\end{gather*}
\c Simdi tan\i ma g\"ore
\begin{equation*}
f_{\gamma}(\xi)=
\begin{cases}
	\delta_i,&\text{ $\xi=\gamma_i$ durumunda},\\
0,&\text{ $\xi\in\beta\setminus\{\gamma_i\colon i<n\}$ durumunda}
\end{cases}
\end{equation*}
olsun.
O zaman $f_{\gamma}\in\exp(\alpha,\beta)$.
As\i nda
\begin{equation*}
\xi\mapsto f_{\xi}\colon\alpha^{\beta}\xrightarrow{\approx}\exp(\alpha,\beta).\qedhere
\end{equation*}
\end{proof}

\begin{theorem}
$\upvarepsilon_0\approx\upomega$.
\end{theorem}

\ktk

\chapter{Kardinal kuvvetler}\label{ch:card}

\section{Say\i lamaz k\"umeler}

E\u ger $A\preccurlyeq\upomega$ ise,
o zaman $A$ \textbf{say\i labilir;}
di\u ger durumda $A$ \textbf{say\i lamaz.}
G\"ord\"u\u g\"um\"uz gibi
say\i labilir k\"umelerden
ordinal toplama, \c carpma, ve kuvvet alma ile
say\i lamaz k\"umeler elde edilemez.


Herhangi $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}\label{pow}
\pow{\bm A},
\end{equation*}
$\bm A$'n\i n \emph{altk\"umeleri} taraf\i ndan olu\c sturulmu\c s s\i n\i ft\i r.
Yani
\begin{equation*}
\pow{\bm A}=\{X\colon X\included\bm A\}.
\end{equation*}
Buradaki $X$ siyah olmad\i\u g\i ndan \emph{k\"ume} de\u gi\c skenidir.
K\"ume olmayan bir s\i n\i f, bir s\i n\i f\i n eleman\i\ olamaz.
E\u ger $\universe$, t\"um k\"umeler taraf\i ndan olu\c sturulmu\c s s\i n\i f ise,
o zaman
\begin{equation*}
\pow{\universe}=\universe.
\end{equation*}
Ama $n\in\upomega$ ise
\begin{equation*}
n<2^n=\card{\pow n}.
\end{equation*}

\begin{theorem}[Cantor]\label{thm:pow}
Her $A$ k\"umesi i\c cin
\begin{equation*}
A\prec\pow A.
\end{equation*}
\end{theorem}

\begin{proof}
$x\mapsto\{x\}\colon A\xrightarrow{\preccurlyeq}\pow A$.
\c Simdi $f\colon A\xrightarrow{\preccurlyeq}\pow A$ ise
\begin{equation*}
B=\{x\in A\colon x\notin f(x)\}
\end{equation*}
olsun.  O zaman $A$'n\i n her $c$ eleman\i\ i\c cin
\begin{equation*}
c\in B\liff c\notin f(c),
\end{equation*}
dolay\i s\i yla $B\neq f(c)$.
Bu \c sekilde $f$, e\c sleme olamaz.
\end{proof}

\begin{xca}
Cantor Teoreminin kan\i t\i\
$A$'n\i n k\"ume oldu\u gunu nas\i l kullan\i r?
\end{xca}

\begin{axiom}[Kuvvet K\"ume]
Her $A$ k\"umesi i\c cin $\pow A$ s\i n\i f\i\ bir k\"umedir.
\end{axiom}

Herhangi $a$ ve $b$ i\c cin
\begin{equation*}
(a,b)=\bigl\{\{a\},\{a,b\}\bigr\}
\end{equation*}
olsun.

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

\ktk

\c Simdi $A\times B=\{(x,y)\colon x\in A\land y\in B\}$
tan\i mlanabilir.

\begin{theorem}
$A\times B\included\pow{\pow{A\cup B}}$.
\end{theorem}

\ktk

\begin{theorem}[Hartogs]
Her kardinalin daha b\"uy\"u\u g\"u vard\i r.
\end{theorem}

\begin{proof}
$\bm A=\{\xi\colon\xi\preccurlyeq\kappa\}$ olsun.
O zaman $\bm A\included\on$, ve ayr\i ca $\bm A$ ge\c ci\c slidir,
dolay\i s\i yla $\bm A$ bir k\"umeyse, bir $\alpha$ ordinalidir.
Bu durumda $\alpha\notin\alpha$ oldu\u gundan $\alpha>\kappa$.

E\u ger $f$ bir $\beta$'y\i\ $\kappa$'ya g\"om\"urse,
o zaman bir
\begin{equation*}
\Bigl\{\bigl(f(\xi),f(\eta)\bigr)\colon\xi\leq\eta<\beta\Bigr\}
\end{equation*}
k\"umesi
elde edilebilir.
Bu \c sekilde elde edilen t\"um k\"umeler,
$\kappa\times\kappa$ \c carp\i m\i n\i n bir $B$ altk\"umesini olu\c sturur.
O halde $B\approx\bm A$ (neden?),
dolay\i s\i yla $\bm A$ da bir k\"umedir.
\end{proof}

Sonu\c c olarak
\begin{equation*}
\kappa^+=\min\{\xi\colon\kappa<\xi\}
\end{equation*}
tan\i mlanabilir, $\kappa^+$, $\kappa$'n\i n \emph{kardinal} ard\i l\i d\i r.

\c Simdi \"ozyineli tan\i ma g\"ore
\begin{gather*}
	\aleph_0=\upomega,\\
	\aleph_{\alpha'}=(\aleph_{\alpha})^+,\\
	\alpha\text{ limit ise }\aleph_{\alpha}=\sup_{\xi<\alpha}\aleph_{\xi}.
\end{gather*}
(Burada $\aleph$, \.Ibrani \emph{alef} harfidir.)

\begin{theorem}
$\xi\mapsto\aleph_{\xi}$ normaldir,
ve her sonsuz kardinal, bir $\alpha$ i\c cin, $\aleph_{\alpha}$'d\i r.
\end{theorem}

\ktk

\begin{lemma}
  Her sonsuz kardinal, $\upomega$'n\i n bir kuvvetidir.
\end{lemma}

\klk

Tan\i ma g\"ore
\begin{align*}
  \kappa\oplus\lambda&=\card{\kappa\sqcup\lambda}=\card{\kappa+\lambda},\\
  \kappa\otimes\lambda&=\card{\kappa\times\lambda}=\card{\kappa\cdot\lambda}
\end{align*}
olsun;
bunlar $\kappa$ ve $\lambda$'n\i n \textbf{kardinal toplam\i}
ve \textbf{kardinal \c carp\i m\i d\i r.}

\begin{theorem}
E\u ger $\kappa$ ve $\lambda$'n\i n biri sonsuz ise
  \begin{equation*}
    \kappa\oplus\lambda=\max(\kappa,\lambda).
  \end{equation*}
E\u ger
$\kappa$ ve $\lambda$'n\i n biri sonsuz ise ve di\u geri s\i f\i r de\u gilse
  \begin{equation*}
    \kappa\otimes\lambda=\max(\kappa,\lambda).
  \end{equation*}
\end{theorem}

\begin{proof}
$\kappa\leq\lambda$ olsun.  O zaman
  \begin{equation*}
   \lambda
\leq\kappa+\lambda
\leq\lambda+\lambda
\preccurlyeq2\cdot\lambda
\leq\lambda\cdot\lambda,
  \end{equation*}
ve $\kappa>0$ ise
\begin{equation*}
  \lambda\leq\kappa\cdot\lambda\leq\lambda\cdot\lambda,
\end{equation*}
dolay\i s\i yla $\lambda\approx\lambda^2$ kan\i tlamak yeter.

Lemmadan bir $\alpha$ i\c cin $\lambda=\upomega^{\alpha}$.
O zaman $\lambda\approx\exp(\upomega,\alpha)$.
\c Simdi $f\colon\upomega\times\upomega\xrightarrow{\approx}\upomega$ olsun.
E\u ger $g$ ve $h$, $\exp(\upomega,\alpha)$ k\"umesinin eleman\i\ ise $g*h$,
\begin{equation*}
  \xi\mapsto f(g(\xi),h(\xi))
\end{equation*}
eleman\i\ olsun.
O zaman
\begin{equation*}
  (g,h)\mapsto g*h\colon\exp(\upomega,\alpha)\xrightarrow{\approx}
\exp(\upomega,\alpha)\times\exp(\upomega,\alpha).\qedhere
\end{equation*}
\end{proof}

Sonu\c c olarak
\begin{equation*}
  \card{\aleph_{\alpha}+\aleph_{\beta}}=\aleph_{\max(\alpha,\beta)}
=\card{\aleph_{\alpha}\cdot\aleph_{\beta}}.
\end{equation*}

\c Simdi herhangi $A$ k\"umesi i\c cin
\begin{equation*}\label{powf}
  \powf A=\{X\in\pow A\colon\card X<\upomega\}
\end{equation*}
olsun.

\begin{theorem}
E\u ger $\kappa$ sonsuz ise $\powf{\kappa}\approx\kappa$.
\end{theorem}

\begin{proof}
Her $m$ i\c cin
  $\{\xi\in\kappa\colon\card{\xi}=m\}\preccurlyeq\kappa^m\approx\kappa$,
dolay\i s\i yla
\begin{equation*}
  \powf{\kappa}=\bigcup_{i\in\upomega}\{\xi\in\kappa\colon\card{\xi}=i\}
\preccurlyeq\upomega\times\kappa\approx\kappa.\qedhere
\end{equation*}
\end{proof}

\begin{theorem}
E\u ger $\beta$ sonsuz ve $2\leq\alpha\leq\beta$ ise
\begin{equation*}
  \card{\alpha^{\beta}}=\card{\beta}.
\end{equation*}
E\u ger $\alpha$ sonsuz ve $1\leq\beta\leq\alpha$ ise
\begin{equation*}
  \card{\alpha^{\beta}}=\card{\alpha}.
\end{equation*}
  \end{theorem}

\ktk

\section{Se\c cme}

\Teoremde{thm:cntbl-u}, 
$\bigcup_{i<\upomega}A_i$ bile\c siminin say\i labilir olmas\i\ i\c cin,
her $A_k$ k\"umesinin say\i labilir olmas\i\ yetmez,
ama $A_k$ k\"umesinin $\upomega$'ya kesin bir g\"ommesi bilinmelidir.
Her $k$ i\c cin,
$A_k$ k\"umesinin $\upomega$'ya g\"ommeleri,
bo\c s olmayan bir $\mathscr B_k$ k\"umesini olu\c sturabilirler;\label{curly-B}
ama g\"ord\"u\u g\"um\"uz aksiyomlarla
\begin{equation*}
\Forall x(x\in\upomega\lto f_x\in\mathscr B_x)
\end{equation*}
ko\c sulunu sa\u glayan 
$x\mapsto f_x$ g\"ondermesinin olup olmad\i\u g\i n\i\ bilmiyoruz.

G\"ord\"u\u g\"um\"uz aksiyomlar, 
\textbf{Zermelo--Fraenkel} veya \textbf{ZF} aksiyomlar\i d\i r.

Her $k$ i\c cin $\mathscr B_k$ k\"umesinden bir $f_k$ se\c cmek isteriz.
\emph{Se\c cim Aksiyomunun} 
bi\c cimlerinin birine g\"ore, bu se\c cme m\"umk\"und\"ur.
Bizim i\c cin, a\c sa\u g\i daki bi\c cim kullan\i l\i\c sl\i\ olacakt\i r.

\begin{axiom}[Se\c cim]
  Her k\"ume iyis\i ralanabilir.
\end{axiom}

\"Orne\u gin $\bigcup_{x\in\upomega}\mathscr B_x$ iyis\i ralan\i rsa,
o zaman istedi\u gimiz $x\mapsto f_x$ g\"ondermesi
\begin{equation*}
  x\mapsto\min\mathscr(B_x)
\end{equation*}
olabilir.

G\"odel'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZF aksiyomlar\i n\i n bir \emph{modelinde,}
Se\c cim Aksiyomu do\u grudur.
Cohen'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZF aksiyomlar\i n\i n bir modelinde,
Se\c cim Aksiyomu yanl\i\c st\i r.
K\i saca Se\c cim Aksiyomu, ZF'den ba\u g\i ms\i zd\i r.

\begin{sloppypar}
  Se\c cim Aksiyomunu varsay\i yoruz.
  Bununla ZF, \textbf{ZFC}'dir.
  \c Simdi her k\"umenin kardinali vard\i r.
  Tan\i ma g\"ore
  \begin{equation*}
    \kappa^{\lambda}=\card{{}^{\kappa}\lambda}.
  \end{equation*}
  Bu kuvvet, ordinal k\"uvvet de\u gil,
  \textbf{kardinal kuvvettir.}
  Orne\u gin
  $\aleph_0=\upomega$ oldu\u gu halde
  $2^{\aleph_0}$, kardinal kuvvet olarak anla\c s\i l\i r,
  ve bu kuvvet, $2^{\upomega}$ ordinal kuvvetinden farkl\i d\i r.
  Asl\i nda $2^{\upomega}=\upomega$, ama sonraki teoreme g\"ore
  $2^{\aleph_0}>\aleph_0$.
\end{sloppypar}

\begin{theorem}
  $2^{\kappa}=\card{\pow{\kappa}}$.
\end{theorem}

\ktk

A\c sa\u g\i daki kurallar kolayd\i r.
\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*}

\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}

\begin{sloppypar}
  Tekrar $\kappa$ ve $\lambda$'n\ in biri sonsuz olsun.
  E\u ger $\lambda\leq\kappa\leq2^{\lambda}$ ise, 
  o zaman \eqref{eqn:kl} gerektirmesine g\"ore
  \begin{equation*}
    \kappa^{\lambda}=2^{\lambda}\leq2^{\kappa};
  \end{equation*}
  burada \eqref{eqn:lk} gerekmez.
  Bir durumda, 
  e\u ger \eqref{eqn:kl} gerektirmesinin hipotezi do\u gru de\u gilse,
  o zaman $2^{\lambda}<\kappa$, dolay\i s\i yla $\lambda\leq\kappa$,
  ve \eqref{eqn:lk} kullan\i labilir.
  Bu \c sekilde teoremin yerine
  \begin{gather*}%\label{eqn:kl}
    \kappa\leq 2^{\lambda}\lto\kappa^{\lambda}=2^{\lambda},\\%\label{eqn:lk}
    2^{\lambda}<\kappa\lto
    %\kappa\leq
    \kappa^{\lambda}\leq2^{\kappa}.
  \end{gather*}
  kurallar\i\ kullan\i labilir.
  (Tekrar
  $2\leq\kappa$, $1\leq\lambda$, 
  ve $\aleph_0\leq\max\{\kappa,\lambda\}$ olmal\i d\i r.)
  \"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*}
\end{sloppypar}

\c Simdi a\c sa\u g\i daki tan\i m yapabiliriz.
\begin{gather*}
  \beth_0=\aleph_0,\\
\beth_{\alpha'}=\card{\pow{\beth_{\alpha}}}=2^{\beth_{\alpha}},\\
\alpha\text{ limit ise }\beth_{\alpha}=\sup_{\xi<\alpha}\beth_{\xi}.
\end{gather*}
(Burada $\beth$, \.Ibrani \emph{beth} harfidir.)
O zaman $\xi\mapsto\beth_{\xi}$ normaldir, ve
\begin{equation*}
  \aleph_{\alpha}\leq\beth_{\alpha}.
\end{equation*}

\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

\textbf{Kontinuum Hipotezi} veya \textbf{KH,}
$\aleph_1=\beth_1$ \"onermesidir.
\textbf{Genelle\c stirilmi\c s Kontinuum Hipotezi} veya \textbf{GKH,}
$\Forall{\xi}\aleph_{\xi}=\beth_{\xi}$ \"onermesidir.
G\"odel'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZFC aksiyomlar\i n\i n bir modelinde,
GKH do\u grudur.
Cohen'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZFC aksiyomlar\i n\i n bir modelinde,
KH yanl\i\c st\i r.
Bu \c sekilde KH, ZFC'den ba\u g\i ms\i zd\i r.

\appendix

\chapter{Harfler}\label{ch:let}

Metinde simge olarak kullan\i l\i rken harfler a\c sa\u g\i daki anlamlara gelir.

\begin{description}
\item[``Tahta siyah\i'' harfleri]\mbox{}
\begin{itemize}[]
\item
$\R$\letsep ger\c cel say\i lar k\"umesi \letterrefs{\sayfada{R}}
\item
$\Q$\letsep kesirli say\i lar k\"umesi \letterrefs{\sayfada{Q}}
\item
$\Z$\letsep tamsay\i lar k\"umesi \letterrefs{\sayfada{Z}}
\item
$\N$\letsep $\{1,2,3,\dots\}$ sayma say\i lar k\"umesi \letterrefs{\sayfada{N}}
\end{itemize}
\item[K\"u\c c\"uk Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$a$, $b$, $c$, $d$, $e$\letsep say\i lar veya k\"umeler
\item
$f$, $g$, $h$\letsep k\"umede tan\i mlanmi\c s g\"ondermeler
\item
$i$, $j$\letsep do\u gal say\i\ de\u gi\c skenler
\item
$k$, $\ell$, $m$, $n$\letsep do\u gal say\i lar
\item
$p$\letsep asal say\i\ \letterrefs{\ref{p} ve \sayfada{p2}}
\item
$u$, $x$, $y$, $z$\letsep say\i\ veya k\"ume de\u gi\c skenleri
\end{itemize}
\item[Dikey k\"u\c c\"uk Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\sup$\letsep supremum
\item
$\min$\letsep minimum (en k\"u\c c\"uk)
\item
$\max$\letsep maksimum (en b\"uy\"uk)
\end{itemize}
\item[B\"uy\"uk Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$A$, $B$, $C$, $D$\letsep k\"umeler \letterrefs{\sayfada{A}}
\item
$X$, $Y$, $Z$\letsep k\"ume de\u gi\c skenleri
\end{itemize}
\item[K\i v\i rc\i k Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\mathscr A$, $\mathscr B$, $\mathscr C$\letsep elemanlar\i\ k\"ume veya g\"onderme olan k\"umeler
\letterrefs{\pageref{curly-C} ve \sayfanumarada{curly-B}}
\item
$\pow A$\letsep$\{X\colon X\included A\}$ \letterrefs{\pageref{pow} ve \sayfanumarada{thm:pow}}
\item
$\powf A$\letsep$\{X\in\pow A\colon\card X<\upomega\}$ \letterrefs{\sayfada{powf}}
\end{itemize}
\item[B\"uy\"uk siyah Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\bm A$, $\bm B$, $\bm C$\letsep s\i n\i flar \letterrefs{\S \ref{sect:sets-classes}, \sayfada{sect:sets-classes}}
\item
$\bm F$, $\bm G$, $\bm H$\letsep s\i n\i fta tan\i mlanm\i\c s g\"ondermeler
\letterrefs{\Teorem{thm:ord-rec}, \sayfada{thm:ord-rec}}
\end{itemize}
\item[Dikey b\"uy\"uk siyah Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\universe$\letsep evrensel s\i n\i f
\item
$\on$\letsep ordinaller s\i n\i f\i
\item
$\cn$\letsep kardinaller s\i n\i f\i
\end{itemize}
\item[Yunan harfleri]\mbox{}
\begin{itemize}[]
\item
$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$\letsep ordinaller \letterrefs{\sayfada{minus-gr}}
\item
$\xi$, $\eta$, $\zeta$\letsep ordinal de\u gi\c skenler \letterrefs{\sayfada{minus-gr}}
\item
$\kappa$, $\lambda$, $\mu$, $\nu$\letsep kardinaller \letterrefs{\sayfada{card}}
\item
$\phi$, $\psi$, $\chi$\letsep form\"uller \letterrefs{\sayfada{phi}}
\end{itemize}
\item[Dikey Yunan harfi]\mbox{}
\begin{itemize}[]
\item
$\upvarepsilon_0$\letsep$\sup\{\upomega,\upomega^{\upomega},\upomega^{\upomega^{\upomega}},\dots\}$
\letterrefs{\sayfada{epsilon}}
\item
$\upomega$\letsep $\{0,1,2,\dots\}$ do\u gal say\i lar\i\ k\"umesi \letterrefs{\sayfada{eqn:omega=}}
\end{itemize}
\item[Harflerden t\"ureyen simgeler]\mbox{}
\begin{itemize}[]
\item
$\in$\letsep eleman olma ba\u g\i nt\i s\i\ 
(``$a$ \gr{>est`i} $B$'' demek ``$a$, bir $B$'dir'')
\item
$\forall$\letsep her \lips i\c cin (\emph{for \textbf All})
\item
$\exists$\letsep baz\i\lips i\c cin (\emph{there \textbf Exists})
\item
$\cup$, $\bigcup$\letsep bile\c sim (\emph{\textbf Union})
\end{itemize}
\end{description}

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

\section{Form\"uller}

Form\"ullerde kulland\i\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}
\item
\textbf{iki-konumlu 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.}
\item
bir \textbf{tek-konumlu 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).
\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 iki-konumlu 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}

\chapter{Kof\/inallik}\label{ch:cof}

\section{Tan\i m ve \"ozellikler}

\begin{sloppypar}
  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})
  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\{\cardinal(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.
\end{sloppypar}

\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*}\sloppy
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*}
\cardinal\bigl(\sup(f[\beta])\bigr)\leq\cardinal(\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}

\section{Hesaplamalar}

\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{multline*}
{}^{\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{multline*}
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{document}