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\cehead{Kumeler Kuram\i}
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 \begin{document}
%\frontmatter
 \title{K\"umeler kuram\i}
 \author{David Pierce}
%\date{\today, \printtime}
\date{12 \c Subat 2014}
 \publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\uppertitleback{\centering
Bu eser\\
 Creative Commons Attribution--Gayriticari--Share-Alike\\
3.0 Unported Lisans\i\ ile lisansl\i d\i r.\\
Lisans\i n bir kopyas\i n\i\ g\"orebilmek i\c cin,\\
\url{http://creativecommons.org/licenses/by-nc-sa/3.0/deed.tr}\\
adresini ziyaret edin ya da a\c sa\u g\i daki adrese yaz\i n:\\
Creative Commons,
444 Castro Street,
Suite 900,\\
Mountain View,
California, 94041, USA.\\
\mbox{}\\
\cc \ccby David Austin Pierce \ccnc \ccsa}

\lowertitleback{\centering
%\chapter*{\"Ons\"oz}
Bu notlar\i, MAT 340 kodlu Aksiyomatik K\"umeler Kuram\i\ dersi i\c cin
yaz\i yorum.  L\"utfen hatalar\i\ bana bildirin.
}

 \maketitle

\tableofcontents

%\mainmatter

\chapter{Giri\c s}

\section{Sayma ve ordinaller}

Bir torbada birka\c c tane satran\c c ta\c s\i m\i z var, 
onlar\i\ teker teker \c cekiyoruz, ve ayn\i\ zamanda say\i lar
diyoruz:
\begin{compactenum}[1:]
\item
piyade (\eng{pawn});
\item
kale (\eng{rook});
\item
at (\eng{knight});
\item
f{}il (\eng{bishop});
\item
vezir (\eng{queen});
\item
\c sah (\eng{king}).
\end{compactenum}
Bu \c sekilde ta\c slar\i\ \emph{saym\i\c s olduk.}
Sonu\c c olarak 6 tane ta\c s\i m\i z var deriz.  
Ama ta\c slar\i\ belli bir \emph{s\i rada} \c cektik.  
Ba\c ska bir s\i ra m\"umk\"und\"u.  
Ta\c slar\i\ tekrar \c cantaya koyup \c cekiyoruz:
\begin{compactenum}[1:]
\item
piyade;
\item
at;
\item
vezir;
\item
kale;
\item
f{}il;
\item
\c sah.
\end{compactenum}
Son ta\c s\i\ \c cekince yine 6 numaras\i n\i\ diyoruz.  
Her zaman \"oyle olacak:  
her zaman ta\c slar\i\ say\i nca 6'ya kadar sayaca\u g\i z.  
Ama nas\i l biliyoruz?\label{nasil}

Saymak nedir?  Sayman\i n nesnesi, bir
\textbf{topluluktur}\index{topluluk}
(\eng{collection}).\footnote{\textbf{K\"umeler}\index{k\"ume}
  (\eng{sets}), \"ozel topluluk olacak.}  Bir toplulu\u gu say\i nca
asl\i nda onu 
\textbf{s\i ral\i yoruz}%
\index{s\i ra}\index{s\i ra!---lama}
(\eng{order}). 

$A$ bir topluluk olsun, ve $R$, onun bir 
\textbf{s\i ralamas\i}
(\eng{ordering}) olsun.  O zaman $A$ toplulu\u gunun \textbf{elemanlar\i}\index{eleman} (\eng{elements}) veya \textbf{\"o\u geleri}\index{\"o\u ge} (\eng{members}) vard\i r; ve bu toplulu\u gun t\"um $b$, $c$, ve $d$ elemanlar\i\ i\c cin
\begin{compactenum}[1)]
\item
$b\mathrel Rb$ de\u gil, yani
\begin{equation*}
\lnot\;b\mathrel Rb;
\end{equation*}
\item
$b\mathrel Rc$ ve $c\mathrel Rd$ ise $b\mathrel Rd$, yani
\begin{equation*}
b\mathrel Rc\land c\mathrel Rd\lto b\mathrel Rd;
\end{equation*}
\item
$b$ ve $c$ birbirinden farkl\i ysa ya $b\mathrel Rc$ ya da $c\mathrel Rb$, yani
\begin{equation*}
b=c\lor b\mathrel Rc\lor c\mathrel Rb.
\end{equation*}
\end{compactenum}
B\"oylece $R$,
\begin{compactenum}[1)] 
\item
\textbf{yans\i mas\i z} veya \textbf{d\"on\"u\c ss\"uz}
(\eng{irreflexive}),\footnote{I\c s\i k, bir aynadan yans\i r; ses,
  bir kayal\i ktan yans\i r.  \emph{Y\i kanmak} f{}iili, \emph{kendi
    kendini y\i kamak} \"obe\u ginin anlam\i na gelirse, d\"on\"u\c
  sl\"ud\"ur; \emph{y\i kan\i lma} f{}iilinin anlam\i na gelirse,
  edilgendir \cite{Lewis2,Ozkirimli-T}.} 
\item
\textbf{ge\c ci\c sli} veya \textbf{ge\c ci\c sken} (\eng{transitive}),\footnote{\emph{Kaynatmak} f{}iili ge\c ci\c slidir, \c c\"unk\"u bir nesne ister; \emph{kaynamak} ge\c ci\c ssizdir.} ve 
\item
\textbf{do\u grusal} (\eng{linear}) veya \textbf{tam} (\eng{total})
\end{compactenum}
bir ba\u g\i nt\i d\i r.  O zaman $(A,R)$ ikilisi (asl\i nda \emph{s\i
  ral\i} ikilisi), bir \textbf{s\i rad\i r.}\index{s\i ra}  Bu s\i ra,
$A$ \textbf{toplulu\u gunun bir s\i ras\i d\i r.} 

\c Simdi $A$, satran\c c ta\c slar\i\ torbam\i z olsun.  O zaman $A$
toplulu\u gunun t\"um s\i ralar\i, birbiriyle
\textbf{izomorftur}\index{izomorf} (\eng{isomorphic}).  Yani $R$ ile
$S$, $A$ toplulu\u gunun iki s\i ralamas\i ysa, o zaman $A$ toplulu\u
gundan kendisine giden \"oyle bir birebir ve \"orten $f$ g\"ondermesi
vard\i r---yani $A$ toplulu\u gunun \"oyle bir $f$
\textbf{perm\"utasyonu} (\eng{permutation}) veya \textbf{e\c sle\c
  smesi}\index{e\c sle\c sme} vard\i r---ki $A$ toplulu\u gunun t\"um
$b$ ile $c$ elemanlar\i\ i\c cin 
\begin{equation*}
b\mathrel Rc\liff f(b)\mathrel Sf(c)
\end{equation*}
denkli\u gi do\u grudur.  Ama bunu nas\i l biliyoruz?

\c Simdi $A$, pozitif \emph{tamsay\i lar}\index{say\i!tam---lar\i}
toplulu\u gu olsun.  Yani $A=\N$ 
olsun.  Bu toplulu\u gun al\i\c s\i lm\i \c s ``do\u gal'' $<$ s\i
ralamas\i\ vard\i r.  Ama ba\c ska s\i ralamalar\i\ da vard\i r.
Mesela $\N$ toplulu\u gunun \"oyle bir $R$ \textbf{ba\u g\i nt\i
  s\i}\index{ba\u g\i nt\i} (veya \textbf{ili\c skisi:}\index{ili\c
  ski} \eng{relation}) vard\i r ki toplulu\u gun t\"um $k$ ile $m$
elemanlar\i\ i\c cin 
\begin{equation*}
k\mathrel Rm\liff (1<k\land k<m)\lor(1=m\land m<k)
\end{equation*}
denkli\u gi do\u grudur.  
\"Oyleyse $R$ ba\u g\i nt\i s\i, $\N$ toplulu\u gunu s\i ral\i yor; 
asl\i nda $R$ s\i ralamas\i, $<$ s\i ras\i\ ile hemen hemen ayn\i d\i r, 
ancak $R$ s\i ras\i na g\"ore $1$ eleman\i, 
$\N$ toplulu\u gunun \emph{son} eleman\i d\i r.  
O zaman $(\N,<)$ ile $(\N,R)$, birbirine izomorf de\u gildir:
\begin{equation*}
\begin{array}{c|*5c}
<&1,&2,&3,&\dots;&?\\\hline
R&2,&3,&4,&\dots;&1
\end{array}
\end{equation*}
\c Simdi
\begin{equation*}
k\mathrel Sm\liff(2\divides k+m\land k<m)\lor(2\ndivides k\land 2\divides m)
\end{equation*}
olsun.  O zaman $k\mathrel Sm$ ancak ve ancak
\begin{compactenum}[1)]
\item
hem $k$ hem $m$ ya tek ya \c cift, ve $k<m$, veya
\item
$k$ tek ve $m$ \c cift.
\end{compactenum}
O zaman $S$ ba\u g\i nt\i s\i\ da, $\N$ toplulu\u gunu s\i ral\i yor, ama $(\N,<)$ ile $(\N,S)$ s\i ralar\i, birbirine izomorf de\u gildir:
\begin{equation*}
\begin{array}{c|*8c}
<&1,&2,&3,&\dots;& ?& ?& ?&\dots\\\hline
S&1,&3,&5,&\dots;&2,&4,&6,&\dots
\end{array}
\end{equation*}

$\N$ toplulu\u gu say\i labilir mi?  
Normalde, sayarken, say\i lar diyoruz.  
$R$ s\i ralamas\i na g\"ore $\N$ toplulu\u gunu say\i nca
$1$ i\c cin hangi say\i y\i\ diyebiliriz?  
Yani yukar\i daki ilk tablonun alt sat\i r\i ndaki 
$1$ numaras\i n\i n \"ust\"unde, 
soru i\c saretinin yerine hangi say\i y\i\ koyabiliriz?  
Bu say\i\, $\upomega+1$ olacak.%
%\glossary{$\upomega$}
Ondan sonra $\upomega+2$, $\upomega+3$, vesaire
say\i lar\i\ olacak; bunlardan sonra, $\upomega+\upomega$, yani
$\upomega\cdot2$, $\upomega\cdot2+1$, vesaire say\i lar\i\ olacak.
Ama $\N$ toplulu\u gunun sadece $\upomega$ tane eleman\i\ olacak. 

Asl\i nda k\"umeler kuramc\i lar\i\ olarak sayarken, 
$0$'dan ba\c slayaca\u g\i z:
\begin{equation*}
\begin{array}{c|*8c}
 &0,&1,&2,&\dots;&\upomega,&\upomega+1,&\upomega+2,&\dots\\\hline
S&1,&3,&5,&\dots;&       2,&         4,&         6,&\dots
\end{array}
\end{equation*}
Burada $0$, $1$, $2$, $3$, \dots; $\upomega$, $\upomega+1$,
$\upomega+2$, \dots; $\upomega\cdot2$, $\upomega\cdot2+1$, \dots
numaralar\i, \textbf{ordinal say\i lar}\index{ordinal} veya
\textbf{ordinallerdir.}  (Her ordinal, bu s\i rada bulunacak.)  Ayr\i
ca $0$, $1$, $2$, $3$, \dots, $\upomega$ numaralar\i,
\textbf{kardinal} (\eng{cardinal}) \textbf{say\i lar}\index{kardinal}
veya \textbf{kardinaldirler} (ba\c ska kardinaller olacak); ama
$\upomega+1$, bir kardinal de\u gildir. 

Her kardinal, bir ordinal olacak, ama her ordinal, bir kardinal olmayacak.

Her ordinal, bir \textbf{k\"ume}\index{k\"ume} olacak; ama
baz\i\ k\"umeler, ordinal olmayacak. 

Her \textbf{k\"ume,} bir topluluk olacak; ve her k\"umenin her
eleman\i, bir k\"ume olacak.  O zaman $a$ ile $b$ k\"umeyse, ya $a$
k\"umesi, $b$ k\"umesinin eleman\i d\i r, ya da eleman\i\ de\u gildir.
\.Ilk durumda $b$ k\"umesi, $a$ k\"umesini \textbf{i\c
  cerir}\index{i\c cerme} (\eng{contains}), yani $a$ k\"umesi, $b$
k\"umesi taraf\i ndan \textbf{i\c cerilir,} ve 
\begin{equation*}
a\in b
\end{equation*}
%\glossary{$\in$}%
ifadesini yazar\i z;\footnote{\label{epsilon}Buradaki $\in$ i\c
  sareti, Yunan \foreignlanguage{polutonikogreek}{e} (epsilon)
  harf{}inden t\"urer.  Bu harf,
  \foreignlanguage{polutonikogreek}{>est'i} kelimesinin ilk harf{}idir,
  ve $A$ \foreignlanguage{polutonikogreek}{>est'i} $B$ c\"umlesi,
  ``$A$, $B$'dir'' (\eng{$A$ is $B$}) anlam\i na gelir.  Epsilonun bu kullan\i\c s\i n\i, Peano \cite{Peano} ortaya koymu\c stur.} ikinci
durumda $b$ k\"umesi, $a$ k\"umesini \textbf{i\c cermez,} ve 
\begin{equation*}
a\notin b
\end{equation*}
ifadesini yazar\i z.  Genelde $C$ bir topluluk ise, ya $a\in C$ ya da
$a\notin C$. 

Bize g\"ore \textbf{bo\c s bir topluluk}---elemanlar\i\ olmayan bir
topluluk---vard\i r, ve bu topluluk, bir k\"umedir.  Bu varsay\i m, 
\textbf{Bo\c s K\"ume Aksiyomudur}%
\label{boskume}%
\index{aksiyom!Bo\c s K\"ume A---u}%
\index{bo\c s k\"ume}%
\index{k\"ume!bo\c s ---} 
(\eng{Empty Set Axiom}).  
Bo\c s k\"umenin i\c sareti, 
\begin{equation*}
\emptyset.
\end{equation*}%
%\glossary{$\emptyset$}
Ayr\i ca $a$ ile $b$ k\"umeyse, o zaman \"oyle bir k\"ume vard\i r ki
her eleman\i, ya $a$ k\"umesinin bir eleman\i, ya da $b$ k\"umesinin
kendisidir.  Bu yeni k\"umenin ifadesi, 
\begin{equation*}
a\cup\{b\}.
\end{equation*}
%\glossary{$a\cup\{b\}$}%
Bu toplulu\u gun k\"ume oldu\u gu, 
\textbf{Biti\c stirme Aksiyomudur}%
\label{bitistirme}%
\index{aksiyom!Biti\c stirme A---u} 
(\eng{Adjunction Axiom}).%
\footnote{Bu aksiyom, Tarski ve Givant \cite[p.~223, QIII]{MR920815}
  kayna\u g\i nda bulunur; \.Ingilizce ad\i, Boolos
  \cite[p.~100]{Boolos-again} kayna\u g\i nda bulunur.}  
Burada $a$
bo\c s ise, yeni $a\cup\{b\}$ k\"umesi, 
\begin{equation*}
\{b\}
\end{equation*}
olarak yaz\i l\i r.  O zaman a\c sa\u g\i daki gibi k\"umelerimiz vard\i r:
\begin{align*}
&\emptyset,&
&\{\emptyset\},&
&\{\emptyset\}\cup\bigl\{\{\emptyset\}\bigr\},&
&\Bigl(\{\emptyset\}\cup\bigl\{\{\emptyset\}\bigr\}\Bigr)\cup\Bigl\{\{\emptyset\}\cup\bigl\{\{\emptyset\}\bigr\}\Bigr\}.
%\end{align*}
\intertext{Bu ifadelerin yerine}
%\begin{align*}
&\emptyset,&
&\{\emptyset\},&
&\bigl\{\emptyset,\{\emptyset\}\bigr\},&
&\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\}
\end{align*}
ifadelerini yazabiliriz.  Asl\i nda $0$%
%\glossary{$0$} 
say\i s\i n\i\ $\emptyset$
olarak tan\i mlar\i z, yani 
\begin{equation*}
0=\emptyset.
\end{equation*}
Bu say\i, \textbf{ilk ordinaldir.}  Her $\alpha$ ordinali i\c cin bir
sonraki ordinal olacak, ve bu ordinal, $\alpha\cup\{\alpha\}$ olacak.
Mesela $0$'dan bir sonraki ordinal $\{0\}$ olacak; yani
\begin{equation*}
1=\{0\}
\end{equation*}
%\glossary{$1$}%
olacak.  Ayr\i ca her $\alpha$ ordinal i\c cin
\begin{equation*}
\alpha+1=\alpha\cup\{\alpha\}
\end{equation*}
olacak.  Ama bildi\u gimiz gibi
\begin{align*}
1+1&=2,&
2+1&=3,&
3+1&=4,
\end{align*}
%\glossary{$2$, $3$, $4$, \dots}%
vesaire.  O zaman
\begin{gather*}
2=1\cup\{1\}=\{0,1\},\\
3=2\cup\{2\}=\{0,1,2\},\\
4=3\cup\{3\}=\{0,1,2,3\},
\end{gather*}
vesaire.  B\"oyle tan\i mlanm\i\c s say\i lar, \textbf{von Neumann
  do\u gal say\i lar\i d\i r}%
\index{say\i!von Neumann do\u gal ---lar\i}%
\index{von Neumann do\u gal say\i lar\i}\label{vnn}
(\eng{von Neumann natural numbers} \cite{von-Neumann}).  Bu say\i lar, bir
toplulu\u gu olu\c sturacak, ve bu topluluk, $\upomega$ olacak.  Yani
$\upomega$, \"oyle bir topluluktur ki 
\begin{compactenum}[1)]
\item
$0\in\upomega$,
\item
$\alpha\in\upomega$ ise $\alpha+1\in\upomega$, ve
\item
$\upomega$ toplulu\u gunun ba\c ska eleman\i\ yoktur.
\end{compactenum}
\"Oyleyse $\upomega$ toplulu\u gunun tan\i m\i, 
\textbf{\"ozyineli}%
\index{\"ozyineli tan\i m} 
veya 
\textbf{rek\"ursiftir}%
\index{rek\"ursif tan\i m} 
(\eng{recursive}).

\section{Ordinaller Hesaplar\i}

\textbf{Sonsuzluk Aksiyomuna}%
\index{aksiyom!Sonsuzluk A---u}%
\label{sonsuzluk}%
\footnote{Veya \textbf{Sonsuz K\"ume Aksiyomu} \cite{Nesin-SKK}.} 
(\eng{Axiom of Infinity} \cite{Zermelo-invest})
g\"ore $\upomega$ toplulu\u gu, bir k\"ume olacak.  O zaman $\upomega$
bir ordinal olacak, ve bu ordinalin her $k$ eleman\i\ i\c cin
$\upomega+k$ k\"umesi, bir ordinal olacak. 

Asl\i nda t\"um $\alpha$ ile $\beta$ ordinaller i\c cin
\begin{align*}
&\alpha+\beta\text{ toplam\i n\i,}&
&\alpha\cdot\beta\text{ \c carp\i m\i n\i, ve}&
&\alpha^{\beta}\text{ kuvvetini}
\end{align*}
tan\i mlayaca\u g\i z.  O zaman
\begin{gather*}
1+\upomega=\upomega<\upomega+1,\\
2\cdot\upomega=\upomega<\upomega\cdot2,\\
(\upomega+1)^{\upomega}=\upomega^{\upomega}<\upomega^{\upomega+1}
\end{gather*}
olacak.  Asl\i nda:
\begin{asparaitem}
\item
$1+\upomega$ toplam\i,
\begin{equation*}
(0,0,1,2,3,\dots)
\end{equation*}
s\i ras\i n\i n ordinalidir, ama $\upomega+1$,
\begin{equation*}
(0,1,2,3,\dots,0)
\end{equation*}
s\i ras\i n\i n ordinalidir.
\item
$2\cdot\upomega$ \c carp\i m\i,
\begin{equation*}
(0,1,0,1,0,1,\dots)
\end{equation*}
s\i ras\i n\i n ordinalidir, ama $\upomega\cdot2$,
\begin{equation*}
(0,1,2,3,\dots,0,1,2,3,\dots)
\end{equation*}
s\i ras\i n\i n ordinalidir; ayr\i ca
\begin{gather*}
	2\cdot\upomega=2+2+2+\dotsb,\\
	\upomega\cdot2=\upomega+\upomega=\upomega+1+1+1+\dotsb
\end{gather*}
(\pageref{o+o} numaral\i\ sayfaya bak\i n).
\item
$(\upomega+1)^{\upomega}$ kuvveti,
\begin{equation*}
((\upomega+1)^2,(\upomega+1)^3,(\upomega+1)^4,\dots)
\end{equation*}
dizisinin \textbf{limitidir,}\index{limit} ve
\begin{gather*}\allowdisplaybreaks
\begin{aligned}
(\upomega+1)^2
&=(\upomega+1)\cdot(\upomega+1)\\
&=(\upomega+1)\cdot\upomega+(\upomega+1)\cdot1\\
&=(\upomega+1+\upomega+1+\upomega+1+\dotsb)+\upomega+1\\
&=(\upomega+\upomega+\upomega+\dotsb)+\upomega+1\\
&=\upomega^2+\upomega+1,
\end{aligned}\\
\begin{aligned}
(\upomega+1)^3
&=(\upomega+1)^2\cdot(\upomega+1)\\
&=(\upomega^2+\upomega+1)\cdot(\upomega+1)\\
&=(\upomega^2+\upomega+1)\cdot\upomega+\upomega^2+\upomega+1\\
&=(\upomega^2+\upomega+1+\upomega^2+\upomega+1+\upomega^2+\dotsb) +\upomega^2+\upomega+1\\
&=(\upomega^2+\upomega^2+\dotsb)+\upomega^2+\upomega+1\\
&=\upomega^3+\upomega^2+\upomega+1,
\end{aligned}
\end{gather*}
ve genelde
\begin{equation*}
(\upomega+1)^n=\upomega^n+\upomega^{n-1}+\dots+\upomega+1.
\end{equation*}
\end{asparaitem}
Ayr\i ca her pozitif $\alpha$ ordinali i\c cin \"oyle bir $\ell$ do\u
gal say\i s\i, ve $\alpha_0$, \dots, $\alpha_{\ell}$ ordinalleri, ve
$a_0$, \dots, $a_{\ell}$ pozitif do\u gal say\i lar\i\ vard\i r ki 
\begin{align*}
\alpha_0&>\dots>\alpha_{\ell},&
\alpha&=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_{\ell}}\cdot a_{\ell}.
\end{align*}
Burada $\upomega^{\alpha_0}\cdot
a_0+\dots+\upomega^{\alpha_{\ell}}\cdot a_{\ell}$ ifadesi, $\alpha$
ordinalinin \textbf{Cantor normal bi\c cimidir}%
\index{Cantor normal bi\c cimi}%
\label{Cnf}%
\index{normal!Cantor --- bi\c cimi}
(\eng{Cantor normal form}).  
Her pozitif ordinalin tek
bir Cantor normal bi\c cimi vard\i r.  Bundan hesaplama
kurallar\i\ t\"ureyebilir. 

\section{K\"umeler ve S\i n\i flar}

Her topluluk, bir k\"ume de\u gildir.\label{Russell}  \"Orne\u gin
\"oyle bir $R$ toplulu\u gu vard\i r ki her eleman\i\ bir k\"ume, ama
bu k\"ume, kendisinin eleman\i\ de\u gildir.  Yani 
\begin{equation*}
R=\{x\colon x\notin x\}.
\end{equation*}
%\glossary{$x$}%
Burada $x$ de\u gi\c skeni her zaman bir k\"ume olacak.  \c
Simdi $a$ bir k\"ume olsun.  E\u ger $a\in a$ ise, o zaman $a\notin
R$, dolay\i s\i yla $a\neq R$.  E\u ger $a\notin a$ ise, o zaman $a\in
R$, dolay\i s\i yla $a\neq R$.  Her durumda $R$ toplulu\u gu,
$a$ k\"umesi de\u gildir.  Yani $R$, bir k\"ume de\u gildir.  Bu
teoreme 
\textbf{Russell Paradoksu}%
\index{paradoks!Russell P---u}%
\index{teorem!Russell Paradoksu}
denir \cite{Russell-letter}. 

Elemanlar\i\ k\"ume olan baz\i\ topluluklar, 
\textbf{s\i n\i f}%
\index{s\i n\i f}
olacak.  Her k\"ume, bir s\i n\i ft\i r, ancak baz\i\ s\i n\i flar,
k\"ume de\u gildir.  Mesela yukar\i daki gibi $\{x\colon x\notin x\}$
toplulu\u gu, bir s\i n\i ft\i r, ama g\"osterdi\u gimiz gibi k\"ume
de\u gildir.  Tan\i ma g\"ore her s\i n\i f, 
\begin{equation*}
\{x\colon\phi(x)\}
\end{equation*}
%\glossary{$\{x\colon\phi(x)\}$}%
bi\c ciminde yaz\i labilir.  Burada $\phi(x)$, k\"umeler kuram\i n\i n
mant\i\u g\i nda bir \textbf{form\"uld\"ur.}\index{form\"ul}  E\u ger
$a$ bir k\"umeyse, o zaman $\phi(a)$ ifadesi, bir
\textbf{c\"umledir.}\index{c\"umle}  Her c\"umle, ya do\u gru ya
yanl\i\c st\i r.  Bir $\{x\colon\phi(x)\}$ s\i n\i f\i n\i n
elemanlar\i, $\phi(a)$ c\"umesini do\u gru yapan $a$ k\"umeleridir.
Bu s\i n\i f, $\phi(x)$ form\"ul\"u taraf\i ndan  
\textbf{tan\i mlan\i r.}%
\index{tan\i mlama}

Bir $\phi(x)$ form\"ul\"un\"un bir ve tek bir \textbf{serbest de\u
  gi\c skeni}\index{ de\u gi\c sken!serbest} vard\i r, ve bu de\u gi\c
sken, $x$ olur.  Ancak bir form\"ul\"un birden fazla serbest de\u gi\c
skeni olabilir.  \"Orne\u gin 
\begin{equation*}
\Forall z(z\in x\liff z\in y)
\end{equation*}
ifadesi, bir form\"uld\"ur, ve serbest de\u gi\c skenleri, $x$ ile $y$
olur.  Bu form\"ulde $z$, \textbf{ba\u glant\i l\i\ de\u gi\c
  skendir.}\index{de\u gi\c sken!ba\u glant\i l\i}  Form\"ul,
k\"umelerin \textbf{e\c sitlik}\index{e\c sitlik} ba\u g\i nt\i s\i
n\i\ tan\i mlar.  Yani $a$ ile $b$ k\"umeleri birbirine e\c sittir,
ancak ve ancak 
\begin{equation*}
\Forall z(z\in a\liff z\in b),
\end{equation*}
yani elemanlar\i\ ayn\i d\i r.  K\"ume olmayan bir s\i n\i f\i n
oldu\u gunu kan\i tlarken, bu kural\i\ kulland\i k.  Yukar\i daki
$\Forall z(z\in x\liff z\in y)$ form\"ul\"un\"un yerine 
\begin{equation*}\label{=}
x=y
\end{equation*}
%\glossary{$=$}%
ifadesini yazar\i z.  O halde bir $\{x\colon x=x\}$ s\i n\i
f\i\ vard\i r, ve bu s\i n\i f, t\"um k\"umelerin s\i n\i f\i d\i r.
Bu s\i n\i f, 
\textbf{evrensel s\i n\i ft\i r}%
\index{evrensel s\i n\i f}
(\eng{universal class}),
ve i\c sareti,
\begin{equation*}\label{universe}
\universe
\end{equation*}
%\glossary{$\universe$}%
olacak.  Ayr\i ca $a$ bir k\"umeyse, o zaman bir $\{x\colon x\in a\}$
s\i n\i f\i\ vard\i r, ama bu s\i n\i f, $a$ k\"umenin kendisidir,
yani 
\begin{equation*}
a=\{x\colon x\in a\}.
\end{equation*}
\"Oyleyse, dedi\u gimiz gibi, her k\"ume, bir s\i n\i ft\i r.

Sonsuzluk Aksiyomunu kullanmadan $\upomega$ toplulu\u gunun s\i n\i f
oldu\u gu apa\c c\i k de\u gildir, ama s\i n\i f olacakt\i r.  Ondan sonra
Sonsuzluk Aksiyomu, $\Exists xx=\upomega$ bi\c ciminde olabilecektir.

Asl\i nda $\upomega$ s\i n\i f\i\ bir k\"ume oldu\u gundan,
\textbf{Yerle\c stirme Aksiyomuna}%
\index{aksiyom!Yerle\c stirme A---u}%
\index{Yerle\c stirme Aksiyomu}%
\label{yerlestirme}
(\eng{Replacement Axiom})\footnote{Skolem
  \cite{Skolem-some-remarks}, 1922 y\i l\i nda bu aksiyomu tavsiye
  etti; ayn\i\ y\i lda Fraenkel, benzer bir aksiyomu tavsiye etmi\c s.
  Ayr\i ca Cantor'a \cite[p.~114]{Cantor-letter} bak\i n.}
g\"ore $\{y\colon\Exists x(x\in\upomega\land y=\upomega+x)\}$ s\i n\i
f\i, bir k\"ume olacakt\i r.
Bu k\"ume
\begin{equation*}
\{\upomega+x\colon x\in\upomega\}
\end{equation*}
olarak yaz\i labilir. 
\textbf{Bile\c sim Aksiyomuna}%
\label{bilesim}%
\index{aksiyom!Bile\c sim A---u}%
\index{bile\c sim!B--- Aksiyomu}
(\eng{Union Axiom} \cite{Zermelo-invest})
g\"ore bu k\"umenin
\begin{align*}
&\bigcup\{\upomega+x\colon x\in\upomega\}&
&\text{ veya }&
&\bigcup_{x\in\upomega}(\upomega+x)
\end{align*}
bile\c simi de bir k\"umedir; tan\i ma g\"ore bu bile\c sim,
$\upomega+\upomega$%
\label{o+o} 
toplam\i d\i r.

K\"umelerden olu\c sturulmu\c s baz\i\ topluluklar, 
s\i n\i f de\u gildir.  
Bu sonu\c c, 
\textbf{G\"odel'in Eksiklik Teoremi}%
\index{teorem!G\"odel Eksiklik T---i} 
(\eng{G\"odel's Incompleteness Theorem}\linebreak 
\cite{Goedel-incompl}%\nocite{MR1890980}
)
veya 
\textbf{Tarski'nin Do\u grulu\u gun Tan\i mlanamamas\i\ Teoremi}%
\index{teorem!Tarski Do\u grulu\u gun Tan\i mlanamamas\i\ T---i} 
(\eng{Tarski's Theorem on the Indefinability of Truth}
\cite{Tarski-truth}%\nocite{MR736686}
)
gibidir.  
Bu teoremlerin as\i l bi\c cimleri,
$\N$ toplulu\u gu hakk\i ndad\i r, 
ve bu bi\c cimde teoremlerini kan\i tlamak zordur.  
Fakat bu teoremler,  
$\universe$ hakk\i nda yaz\i labilir; 
ve bu bi\c cimde onlar\i\ kan\i tlamak daha kolayd\i r. 

T\"um ordinallerin toplulu\u gu, bir s\i n\i f olacak, 
ve bu s\i n\i f\i n i\c sareti
\begin{equation*}
\on
\end{equation*}
%\glossary{$\on$}%
olacak.  Asl\i nda bu s\i n\i f, bir $a$ k\"umesiyse, o zaman
$a\in\on$ olurdu, yani $a\in a$ olurdu; ama bir ordinal i\c cin bu i\c
cerme imk\^ans\i zd\i r.  Sonu\c c olarak $\on$, bir k\"ume de\u
gildir.  Bu teorem, \textbf{Burali-Forti Paradoksu}%
\index{paradoks!Burali-Forti P---u}%
\index{teorem!Burali-Forti Paradoksu}\label{BFP}
\cite{Burali-Forti}
olarak bilinir.

\section{Kardinaller}

$\on$ s\i n\i f\i n\i n bir s\i ralamas\i\ vard\i r, ve bu s\i ralama,
\emph{i\c cerilmedir,} yani $\in$ ile g\"osterilen s\i ralamad\i r.  
\textbf{Se\c cim Aksiyomuna}%
\label{secim}%
\index{aksiyom!Se\c cim A---u} 
(\eng{Axiom of Choice} \cite{Zermelo-invest}) 
g\"ore, her $a$ k\"umesinden bir $\beta$
ordinaline giden bir \textbf{e\c sleme}\index{e\c sleme}\index{e\c slenik} (yani bir birebir \"orten g\"onderme) vard\i r.  O halde
\begin{equation*}
a\approx\beta
\end{equation*}
%\glossary{$\approx$}%
ifadesini yazal\i m, ve $a$ ile $\beta$ k\"umelerine \textbf{e\c
  slenik} densin \cite[s.~82]{Nesin-SKK}.  E\u ger $a$ verilirse, ve
$a\approx\beta$ ko\c sulunu sa\u glayan $\beta$ ordinallerinin en
k\"u\c c\"u\u g\"u $\kappa$ (``kappa'') ise, o zaman $\kappa$, $a$ k\"umesinin 
\textbf{kardinalidir.}%
\index{kardinal}  
T\"um kardinallerden olu\c sturulmu\c s topluluk, bir s\i n\i f
olacak, ve bu s\i n\i f\i n i\c sareti 
\begin{equation*}
\cn
\end{equation*}
%\glossary{$\cn$}%
olacak.  En k\"u\c c\"uk \emph{sonsuz} kardinal, $\upomega$ olur.
$\on$ s\i n\i f\i ndan $\cn$ s\i n\i f\i na giden bir
\begin{equation*}
\xi\mapsto\aleph_{\xi}
\end{equation*}
%\glossary{$\aleph_{\xi}$}%
g\"ondermesi vard\i r.  Burada
\begin{align*}
\aleph_0&=\upomega&
&\text{ ve }&
\alpha<\beta\liff\aleph_{\alpha}<\aleph_{\beta},
\end{align*}
ve her sonsuz kardinal, bir $\alpha$ ordinali i\c cin,
$\aleph_{\alpha}$ bi\c cimindedir.  \.Iki kardinalin \emph{kardinal}
toplam\i\ ve \emph{kardinal} \c carp\i m\i\ vard\i r, ama
\begin{equation*}
\aleph_{\alpha}\cardsum\aleph_{\beta}=
\aleph_{\alpha}\cardprod\aleph_{\beta}=\aleph_{\max(\alpha,\beta)}
\end{equation*}
Ayr\i ca $1\leq k<\upomega$ ise 
$\aleph_{\alpha}\cardsum k=k\cardsum \aleph_{\alpha}=\aleph_{\alpha}\cardprod
k=k\cardprod\aleph_{\alpha}=\aleph_{\alpha}$.

Genelde siyah harfler, s\i n\i flar\i\ g\"osterecek.  
\c Simdi $\bm A$ ile $\bm B$, s\i n\i f olsun.%
%\glossary{$\bm A$, $\bm B$, \dots}
E\u ger $\bm A$ s\i n\i f\i n\i n
her eleman\i, $\bm B$ s\i n\i f\i n\i n eleman\i ysa, o zaman $\bm A$ s\i
n\i f\i na $\bm B$ 
\textbf{s\i n\i f\i n\i n alts\i n\i f\i}%
\index{alts\i n\i f}
(\eng{subclass of the class $B$}) denir, ve
\begin{equation*}
\bm A\included\bm B
\end{equation*}
%\glossary{$\included$}%
ifadesi yaz\i l\i r.  Bu durumda $\bm B$ s\i n\i f\i, $\bm A$ s\i n\i
f\i n\i\ 
\textbf{kapsar}%
\index{kapsama}  
(\eng{includes}).
\.I\c cerilme ($\in$) ve kapsanma ($\included$) ili\c skileri,
birbirinden tamamen farkl\i d\i r.

\textbf{Ay\i rma Aksiyomuna}%
\label{ayirma}%
\index{aksiyom!Ay\i rma A---u} 
(\eng{Separation Axiom} \cite{Zermelo-invest}) 
g\"ore, her \emph{k\"umenin} her alts\i n\i
f\i, bir k\"umedir.  \c Simdi, e\u ger $\phi(x)$ bir form\"ul ise, ve
$a$ bir k\"umeyse, o zaman \"oyle bir s\i n\i f vard\i r ki her
eleman\i, hem $a$ k\"umesinin eleman\i d\i r, hem de $\phi(x)$
form\"ul\"un\"u sa\u glar.  Bu s\i n\i f, 
\begin{equation*}
\{x\in a\colon\phi(x)\}
\end{equation*}
%\glossary{$\{x\in a\colon\phi(x)\}$}%
olarak yaz\i l\i r.  Ay\i rma Aksiyomuna g\"ore, bu s\i n\i f, bir
k\"umedir.  O zaman bu k\"ume, 
\textbf{$a$ k\"umesinin bir altk\"umesidir}%
\index{altk\"ume}
(\eng{a subset of the set $a$}).

Bir $a$ k\"umesinin t\"um altk\"umeleri, bir s\i n\i f olu\c sturur.  Bu s\i n\i f, $a$ k\"umesinin \textbf{kuvvet s\i n\i f\i d\i r} (\eng{power class}), ve
\begin{equation*}
\pow a
\end{equation*}
%\glossary{$\pow a$}%
olarak yaz\i l\i r.  
\textbf{Kuvvet K\"umesi Aksiyomuna}%
\label{kuvvetkumesi}%
\index{aksiyom!Kuvvet K\"umesi A---u} 
(\eng{Power Set Axiom} \cite{Zermelo-invest}) 
g\"ore, bu s\i n\i f, her zaman bir
k\"umedir.  
\textbf{Cantor'un Teoremine}%
\index{teorem!Cantor'un T---i}%
\index{Cantor'un Teoremi}%
\footnote{Levy'ye \cite{MR1924429} g\"ore Cantor, bu teoremi 1892 y\i
  l\i nda yay\i mlad\i.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
g\"ore,
her k\"umenin kuvvet k\"umesi, k\"umeden kesinlikle daha
b\"uy\"ukt\"ur, yani kardinali daha b\"uy\"ukt\"ur.  Bu teorem, 
\begin{equation*}
a\prec\pow a
\end{equation*}
%\glossary{$\prec$}
ifadesiyle s\"oylenir.

E\u ger $a$ ile $b$, iki k\"umeyse, 
o zaman $a$ k\"umesinden $b$ k\"umesine giden 
g\"ondermeler toplulu\u gu, bir k\"umedir, ve bu k\"ume
\begin{equation*}
{}^ab
\end{equation*}
%\glossary{${}^ab$}%
olarak yaz\i labilir.  O zaman
${}^a2\approx\pow a$.
E\u ger $\kappa$ ile $\lambda$, iki kardinal ise, tan\i ma g\"ore
\begin{equation*}
\kappa^{\lambda}
\end{equation*}
kuvveti, ${}^{\lambda}\kappa$ k\"umesinin kardinalidir.  
E\u ger $2\leq\kappa\leq\lambda$ ise, o zaman
\begin{equation*}
2^{\lambda}\leq\kappa^{\lambda}\leq(2^{\kappa})^{\lambda}=2^{\kappa\cdot\lambda}=2^{\lambda};
\end{equation*}
\"ozel olarak $\kappa^{\lambda}=2^{\lambda}$.

\c Simdi $\Z$,%
%\glossary{$\Z$}
\textbf{tamsay\i lar}\index{say\i!tam---lar\i} toplulu\u gu olsun.  
O zaman
$\Z\approx\upomega$,
\c c\"unk\"u tamsay\i lar, sonsuz bir
\begin{equation*}
0,1,-1,2,-2,3,-3,4,\dots
\end{equation*}
listesinde yaz\i labilir.  Ayr\i ca her tamsay\i, $\upomega$ k\"umesinin
elemanlar\i\ gibi, bir k\"ume olarak d\"u\c s\"un\"ulebilir.  Bunu
g\"ostermek i\c cin, e\u ger $a$ ile $b$, herhangi iki k\"umeyse, o zaman
\begin{equation*}
  (a,b)
\end{equation*}
%\glossary{$(a,b)$}%
 \textbf{s\i ral\i\ ikilisi}%
\index{ikili}%
\index{s\i ra!---l\i\ ikili}
(\eng{ordered pair}),
$\big\{\{a\},\{a,b\}\bigr\}$
k\"umesi olarak tan\i mlan\i r.\footnote{\label{Kuratowski}Bu,
  Kuratowski'nin tan\i m\i d\i r \cite{Kuratowski}.  Daha \"once, Wiener 
  \cite{Wiener} daha karma\c s\i k bir tan\i m verdi.}  O zaman
$n\in\upomega$ ve $n>0$ ise, o zaman $-n$ tamsay\i s\i, $(0,n)$ olarak
tan\i mlanabilir.

Ba\c ska y\"ontemle $\Z$ toplulu\u gunun her $r$ eleman\i n\i,
\begin{equation*}
\{(x,y)\in\upomega\times\upomega\colon x=y+r\}
\end{equation*}
olarak tan\i
mlanabiliriz.  Bu tan\i ma g\"ore $\Z$ toplulu\u gunun her eleman\i,
bir 
\textbf{denklik s\i n\i f\i d\i r.}%
\index{s\i n\i f!denklik ---\i}%
\index{denk!---lik s\i n\i f\i, ba\u g\i nt\i s\i}
Asl\i nda $\upomega\times\upomega$ \c carp\i m\i nda \"oyle bir $E$
\textbf{denklik ba\u g\i nt\i s\i}%
\index{ba\u g\i nt\i!denklik ---s\i} 
vard\i r ki
\begin{equation*}
  (a,b)\mathrel E(c,d)\liff a+d=b+c,
\end{equation*}
ve $\Z$ toplulu\u gu, $(\upomega\times\upomega)/E$ b\"ol\"um\"u olarak
tan\i mlanabilir.

  \"Oyleyse
$\Z$ toplulu\u gu, bir s\i n\i ft\i r.  O zaman 
Yerle\c stirme Aksiyomuna
g\"ore $\Z$, bir k\"ume
olmal\i, \c c\"unk\"u $\Z\approx\upomega$. 

Benzer \c sekilde $\Q$%
%\glossary{$\Q$}
\textbf{kesirli say\i lar}\index{say\i!kesirli
  ---lar} toplulu\u gu, \"oyle bir
$(\Z\times\Z)/F$ b\"ol\"um\"ud\"ur ki 
\begin{equation*}
  (a,b)\mathrel F(c,d)\liff ad=bc.
\end{equation*}
Asl\i nda $\Q\approx\upomega$,
\c c\"unk\"u kesirli say\i lar, 
\ref{fig:SB} numaral\i\ \c sekildeki
%a\c sa\u g\i daki 
``Stern--Brocot a\u
gac\i'' olarak, ve ondan sonra bir liste
olarak, yaz\i labilir. 
\begin{figure}[ht]
\centering
\newcommand{\SB}[1]{\TR{\psframebox[linestyle=none]{\ensuremath{#1}}}}
\pstree[treemode=D,treesep=0mm%,showbbox=true,treenodesize=10pt
]{\SB{0}}
{\pstree{\SB{-1}}
       {\pstree{\SB{-2}}
               {\pstree{\SB{-3}}
                       {\SB{-4}
                        \SB{-\frac52}}
                \pstree{\SB{-\frac32}}
                       {\SB{-\frac53}
                        \SB{-\frac43}}}
        \pstree{\SB{-\frac12}}
               {\pstree{\SB{-\frac23}}
                       {\SB{-\frac34}
                        \SB{-\frac35}}
                \pstree{\SB{-\frac13}}
                       {\SB{-\frac25}
                        \SB{-\frac14}}}}
\pstree{\SB{1}}
       {\pstree{\SB{\frac12}}
               {\pstree{\SB{\frac13}}
                       {\SB{\frac14}
                        \SB{\frac25}}
                \pstree{\SB{\frac23}}
                       {\SB{\frac35}
                        \SB{\frac34}}}
        \pstree{\SB{2}}
               {\pstree{\SB{\frac32}}
                       {\SB{\frac43}
                        \SB{\frac53}}
                \pstree{\SB{3}}
                       {\SB{\frac52}
                        \SB{4}}}}}
\caption{Stern--Brocot A\u gac\i}\label{fig:SB}
\end{figure}

\c Simdi $\R$,%
%\glossary{$\R$}
\textbf{ger\c cel say\i lar}\index{say\i!gercel ---lar}
toplulu\u gu olsun.  Her 
kesirli say\i, ger\c cel say\i\ olarak d\"u\c s\"un\"ulebilir.  Ayr\i ca
her iki farkl\i\ ger\c cel say\i n\i n aras\i nda bir kesirli
say\i\ vard\i r.  O zaman $\R$ toplulu\u gundan $\pow{\Q}$ kuvvet
k\"umesine giden \"oyle bir $h$ g\"ondermesi vard\i r ki her $a$ ger\c
cel say\i s\i\ i\c cin 
\begin{equation*}
h(a)=\{x\in\Q\colon x<a\},
\end{equation*}
ve bu g\"onderme, birebirdir.  \"Oyleyse $a$ say\i s\i, $h(a)$
k\"umesi olarak d\"u\c s\"un\"ulebilir, ve $\R$, bir k\"umedir.  Ayr\i
ca 
\begin{align*}
\R\preccurlyeq\pow{\Q}&\approx\pow{\upomega}&
&\text{ ve }&
\pow{\upomega}&\preccurlyeq\R.
\end{align*}
\"Orne\u gin
\begin{equation*}
\pow{\upomega}\approx{}^{\upomega}2
\end{equation*}
\c c\"unk\"u ${}^{\upomega}2$ k\"umesinden $\pow{\upomega}$ k\"umesine giden bir
\begin{equation*}
f\mapsto\{x\in\upomega\colon f(x)=1\}
\end{equation*}
e\c slemesi vard\i r, ve ayr\i ca ${}^{\upomega}2$ k\"umesinden $\R$ k\"umesine giden bir birebir
\begin{equation*}
f\mapsto\sum_{k=0}^{\infty}\frac{2\cdot f(k)}{3^{k+1}}
\end{equation*}
g\"ondermesi vard\i r.  Sonu\c c olarak, 
\textbf{Schr\"oder--Bernstein Teoremine}%
\index{teorem!Schr\"oder--Bernstein T---i}\label{SBT}
g\"ore
\begin{equation*}
\R\approx\pow{\upomega},
\end{equation*}
\c c\"unk\"u bu teoreme g\"ore t\"um $a$ ile $b$ k\"umeleri i\c cin
\begin{equation*}
a\preccurlyeq b\preccurlyeq a\lto a\approx b.
\end{equation*}

\c Simdi Cantor'un Teoreminden
$\upomega\prec\R$.
\"Ozel olarak \"oyle bir $\alpha$ olacak ki $\alpha>0$ ve
$\R\approx\aleph_{\alpha}$.
Ama $\alpha$ ordinalinin $1$ olup olmad\i\u g\i n\i\ bilmiyoruz.
\textbf{Kontin\"u Hipotezine}%
\index{Kontin\"u Hipotezi}
(\eng{Continuum Hypothesis})
g\"ore $\alpha=1$, yani $\upomega\preccurlyeq
a\prec\pow{\upomega}$ ise $a\approx\upomega$.
\textbf{Genelle\c stirilmi\c s Kontin\"u Hipotezine}
(\eng{Generalized Continuum Hypothesis})
g\"ore her sonsuz $b$ k\"umesi i\c cin $b\preccurlyeq c\prec\pow{b}$
ise $b\approx c$.

Se\c cim Aksiyomu hari\c c k\"umeler kuram\i n\i n kullanaca\u g\i m\i
z aksiyomlar\i, 
\emph{Zermelo--Fraenkel Aksiyomlar\i d\i r.}%
\index{aksiyom!Zermelo--Fraenkel A---lar\i}%
\index{aksiyom!Se\c cim A---u}
Asl\i nda Zermelo'nun verdi\u gi aksiyomlar \cite{Zermelo-invest}, a\c
sa\u g\i dad\i r.
\begin{compactenum}[I.]
  \item
\emph{Uzama}%
\index{aksiyom!Uzama A---u}
 (\pageref{uzama} numaral\i\ sayfada).
\item
\textbf{Temel K\"umeler}%
\index{aksiyom!Temel K\"umeler A---u}
(\eng{Elementary Sets}):
 $\emptyset$, $\{a\}$, ve $\{a,b\}$
topluluklar\i, k\"umedir.
\item
Ay\i rma%
\index{aksiyom!Ay\i rma A---u}
 (\pageref{ayirma} numaral\i\ sayfada).
\item
Kuvvet K\"umesi (\pageref{kuvvetkumesi} numaral\i\ sayfada).
\item
Bile\c sim (\pageref{bilesim} numaral\i\ sayfada).
\item
Se\c cim%
\index{aksiyom!Se\c cim A---u}
  (\pageref{secim} numaral\i\ sayfada).
\item
Sonsuzluk%
\index{aksiyom!Sonsuzluk A---u}
 (\pageref{sonsuzluk} numaral\i\ sayfada).
\end{compactenum}
(\pageref{bitistirme} numaral\i\ sayfadaki Biti\c stirme Aksiyomumuz,%
\index{aksiyom!Biti\c stirme A---u}
Zermelo'nun II.\ ve V.\ aksiyomlar\i\ taraf\i ndan gerektirilir.  Ters
olarak Biti\c stirme ve Bo\c s K\"ume Aksiyomlar\i m\i z, Zermelo'nun
II.\ aksiyomunu gerektirir.)  Sonra iki aksiyom daha verildi:
\begin{compactenum}[I.]
\setcounter{enumi}7
  \item
Yerle\c stirme%
\index{aksiyom!Yerle\c stirme A---u}
 (\pageref{yerlestirme} numaral\i\ sayfada).
\item
\textbf{Temellendirme}%
\label{temellendirme}%
\index{aksiyom!Temellendirme A---u}
(\eng{Foundation} \cite{Skolem-some-remarks}):  Her bo\c s olmayan $a$ k\"umesinin \"oyle bir $b$
eleman\i\ vard\i r ki $a\cap b=\emptyset$ (\pageref{kesisim}
numaral\i\ sayfaya bak\i n).
\end{compactenum}
I--V ile VII--IX numaral\i\ aksiyomlar, 
\textbf{Zermelo--Fraenkel Aksiyomlar\i d\i r.} 

Birka\c c tane k\i saltmalar kullan\i l\i r:
\begin{align*}
\ac&=\text{Se\c cim Aksiyomu,}\\
\zf&=\text{Zermelo--Fraenkel Aksiyomlar\i,}\\
\zfc&=\text{Zermelo--Fraenkel Aksiyomlar\i yla Se\c cim Aksiyomu,}\\
\ch&=\text{Kontin\"u Hipotezi,}\\
\gch&=\text{Genelle\c stirilmi\c s Kontin\"u Hipotezi.}
\end{align*}
%\glossary{\ac, \zf, \zfc, \ch, \gch}%
O zaman
\begin{equation*}
\zfc=\zf+\ac.
\end{equation*}
G\"odel'in kan\i tlad\i\u g\i\ teoreme g\"ore $\zf$ \textbf{tutarl\i
  ysa}\index{tutarl\i} (yani ondan bir \c celi\c ski \c c\i kmazsa), o
zaman $\zfc$ aksiyomlar\i\ da tutarl\i d\i r, ve ayr\i ca $\zfc$
aksiyomlar\i yla $\gch$ tutarl\i d\i r
\cite{Goedel-pnas-1,Goedel-pnas-2}.  
Sierpi\'nski \cite{MR0020121},
\begin{equation*}
  \zf+\gch\lto\ac
\end{equation*}
gerektirmesinin g\"osterdi.\footnote{Sierpi\'nski'ye g\"ore 1926 y\i
  l\i nda Lindenbaum ve Tarski, bu gerektirmesini ilan ettiler, ama
  kan\i t\i n\i\ vermediler.}
Cohen'in \cite{MR0232676}
kan\i tlad\i\u g\i\ teoreme g\"ore $\zf$ tutarl\i ysa, o zaman
$\zf+\lnot\ac$ aksiyomlar\i\ da tutarl\i d\i r, ve ayr\i ca
$\zfc+\lnot\ch$ tutarl\i d\i r.  Sierpi\'nski'nin teoremi, a\c sa\u
g\i daki \ref{thm:Sierpinski} numaral\i\ teorem olacakt\i r;
G\"odel'in ve Cohen'in teoremlerini 
kan\i tlamayaca\u g\i z.

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

\section{Form\"uller}

Form\"ullerde kullanaca\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}\footnote{Bilinen de\u gerler i\c
  cin Latin alfabesinin ba\c slang\i c\i ndan harflerin kullan\i l\i\c
  s\i, ve bilinmeyen de\u gerler i\c cin Latin alfabesinin sonundan
  harflerin kullan\i l\i\c s\i, Descartes'te \cite{Descartes-Geometry}
  g\"or\"un\"ur.} 
\item
\textbf{ikili 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.  Bunlar\i\ kalemle yazmak daha kolayd\i r.  Ama bu notlarda, $\bm F\colon\bm A\to\bm B$ ifadesi, $\bm F$ g\"ondermesinin $\bm A$ s\i n\i f\i ndan $\bm B$ s\i n\i f\i na gitti\u ginin anlam\i na gelecek.  A\c sa\u g\i daki \pageref{to} numaral\i\ sayfaya bak\i n.}
\item
bir \textbf{birli 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).\footnote{Yukar\i daki \pageref{epsilon} numaral\i\ sayfadaki dipnota bak\i n.}
\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 ikili 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}).  \"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 $\Forall x(\phi\lto\psi)$ ile $\phi\lto\Forall x\psi$ denktir.
\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 grudur.  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}

\section{E\c sitlik}

Yukar\i daki \pageref{=} numaral\i\ sayfada dedi\u gimiz gibi
\begin{equation*}
  t=u
\end{equation*}%
\glossary{$=$}
ifadesi, $\Forall x(x\in t\liff
x\in u)$ form\"ul\"un\"un k\i saltmas\i\ olarak kullan\i labilir.
Burada $x$, herhangi bir de\u gi\c sken olabilir, ama $t$ ile $u$
terimlerinden farkl\i\ olmal\i d\i r.  \"Orne\u gin $x=y$ ifadesi,
$\Forall z(z\in x\liff z\in y)$ form\"ul\"un\"un k\i saltmas\i d\i r,
ama $\Forall x(x\in x\liff x\in y)$ form\"ul\"un\"un k\i
saltmas\i\ de\u gildir.

Tan\i ma g\"ore 
\begin{equation*}
t=u\denk\Forall x(x\in t\liff x\in u).
\end{equation*}
O zaman
\begin{equation}\label{eqn:=}
\Forall x\Forall y(x=y\liff\Forall z(z\in x\liff z\in y))
\end{equation}
c\"umlesi do\u grudur.  Yani t\"um $a$ ile $b$ k\"umeleri i\c cin
\begin{equation*}
a=b\liff\Forall x(x\in a\liff x\in b)
\end{equation*}
c\"umlesi do\u grudur.  Bu c\"umle, $\liff$ simgesinin tan\i m\i na
g\"ore, iki c\"umlenin birle\c smesine denktir, ve bu c\"umleler,
\begin{align*}
a=b&\lto\Forall x(x\in a\liff x\in b),&
\Forall x(x\in a\liff x\in b)&\lto a=b.
\end{align*}
O zaman t\"um $a$ ile $b$ k\"umeleri i\c cin, hem
\begin{equation*}
\Forall x(x\in a\liff x\in b)\lto a=b
\end{equation*}
do\u grudur, hem de, \ref{thm:lto} numaral\i\ teoreme g\"ore, her $c$ k\"umesi i\c cin,
\begin{equation*}
a=b\land c\in a\lto c\in b
\end{equation*}
do\u grudur.

Bizim i\c cin, \eqref{eqn:=} c\"umlesinin do\u grulu\u gu, bir tan\i
md\i r.  Yani, simgesi $\in$ olan \textbf{i\c cerilme}\index{i\c
  cerilme} ba\u g\i nt\i s\i, temel bir ba\u g\i nt\i d\i r, ama
\textbf{e\c sitlik}\index{e\c sitlik} ba\u g\i nt\i s\i, yukar\i daki
\eqref{eqn:=} c\"umlesini sa\u glayan bir $=$ ba\u g\i nt\i s\i d\i r. 

\begin{theorem}\label{thm:=-equiv}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{align*}
&a=a,&
a=b&\lto b=a,&
a=b\land b=c&\lto a=c
\end{align*}
c\"umleleri do\u grudur.
\end{theorem}

\ktk 

Teoreme g\"ore e\c sitlik ba\u g\i nt\i s\i, \textbf{d\"on\"u\c
  sl\"u} (\eng{reflexive}), \textbf{simetrik} (\eng{symmetric}), ve
\textbf{ge\c ci\c sli} (\eng{transitive}) bir ba\u g\i nt\i d\i r,
yani bir 
\textbf{denklik ba\u g\i nt\i s\i d\i r}%
\index{denklik}\label{denklik}
(\eng{equivalence relation}). 

Teoremin dolay\i s\i yla $a=b\land b=c$ c\"umlesinin k\i saltmas\i\ olarak $a=b=c$ ifadesi yaz\i l\i r; yani
\begin{equation*}
a=b=c\denk a=b\land b=c.
\end{equation*}

\.Ilk resmi aksiyomumuz \c su:

\begin{axiom}[E\c sitlik]\index{aksiyom!E\c sitlik A---u}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{equation*}
a=b\land a\in c\lto b\in c
\end{equation*}
c\"umlesi do\u grudur.  Yani
\begin{equation*}
  \Forall x\Forall y\Forall z(x=y\land x\in z\lto y\in z)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

Bu aksiyomun ba\c ska bi\c cimleri vard\i r, mesela:
\begin{compactenum}
\item
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin $a=b\lto a\in c\lto b\in c$.
\item
T\"um $a$ ile $b$ k\"umeleri i\c cin $\Forall x(a=b\lto a\in x\lto b\in x)$.
\item
T\"um $a$ ile $b$ k\"umeleri i\c cin $\Forall x(a=b\land a\in x\lto b\in x)$.
\item
T\"um $a$ ile $b$ k\"umeleri i\c cin $a=b\lto\Forall x(a\in x\lto b\in x)$.
\item
$\Forall x\Forall y(x=y\lto\Forall z(x\in z\lto y\in z))$.
\item
$\Forall x\Forall y\Forall z(x=y\lto x\in z\lto y\in z)$.
\end{compactenum}

\begin{xca}
$a=b\land\Forall x(a\in x\lto b\in x)$ c\"umlesi, E\c sitlik Aksiyomundan kan\i tlanabilir mi?
\end{xca}

\begin{theorem}
Her $\phi(x)$ tek serbest de\u gi\c skenli form\"ul\"u i\c cin
\begin{equation}\label{eqn:a=b}
a=b\land\phi(a)\lto\phi(b)
\end{equation}
c\"umlesi do\u grudur.
\end{theorem}

\begin{proof}
Form\"ullerin \"ozyineli tan\i m\i\ nedeni ile, \emph{t\"umevar\i m} kullanabiliriz.
\begin{asparaenum}
\item
\.Ilk olarak $\phi$ b\"ol\"unemesin.  Yani $\phi(x)$, ya $c\in x$ veya $x\in c$ bi\c ciminde olsun.  O zaman \eqref{eqn:a=b} c\"umlesi, ya e\c sitli\u gin tan\i m\i ndan, ya da E\c sitlik Aksiyomundan, do\u grudur.
\item
E\u ger $\phi$, ya $\psi$ ya da $\chi$ ise, \eqref{eqn:a=b} do\u gru
olsun.  \c Simdi $a=b\land(\psi(a)\land\chi(a))$ do\u gru olsun.  O
zaman hem $a=b\land\psi(a)$ hem $a=b\land\chi(a)$ do\u gru olmal\i.
Sonu\c c olarak varsay\i m\i m\i zdan hem $\psi(b)$ hem $\chi(b)$ do\u
gru olmal\i, yani $\psi(b)\land\chi(b)$ do\u gru olmal\i.  \"Oyleyse
$\phi$, $\psi\land\chi$ ise \eqref{eqn:a=b} do\u grudur. 
\item
Son olarak, t\"um $c$ i\c cin $\phi(x)$, $\psi(x,c)$ ise, \eqref{eqn:a=b} do\u gru olsun.  \c Simdi $a=b\land\Exists y\phi(a,y)$ do\u gru olsun.  O zaman bir $c$ i\c cin $a=b\land\phi(a,c)$ do\u gru olmal\i, dolay\i s\i yla $\phi(b,c)$ do\u gru olmal\i.  Sonu\c c olarak $\Exists y\phi(b,y)$ do\u grudur.  \"Oyleyse $\phi(x)$, $\Exists y\phi(x,y)$ ise \eqref{eqn:a=b} do\u grudur.\qedhere
\end{asparaenum}
\end{proof}

Kitaplar\i n \c co\u gunda hem $\in$ hem $=$, temel ba\u g\i nt\i d\i
r, ve yukar\i daki \pageref{eqn:=} numaral\i\ sayfadaki \eqref{eqn:=} c\"umlesi, tan\i m de\u gil, 
\textbf{Uzama Aksiyomudur}%
\label{uzama}%
\footnote{Veya \textbf{K\"ume E\c sitli\u gi Aksiyomu}
\cite{Nesin-SKK}.}%
\index{aksiyom!Uzama A---u} 
(\eng{Axiom of Extensionality} \cite{Zermelo-invest}). 
Bu kitaplarda her $\phi(x)$ tek serbest de\u gi\c skenli form\"ul\"u i\c cin \eqref{eqn:a=b} c\"umlesi, bir
\textbf{mant\i ksal aksiyomd\i r.}%
\index{aksiyom!mant\i ksal ---}%
\index{mant\i ksal aksiyom} 

\section{S\i n\i flar}

Bir $\phi(x)$ form\"ul\"u ve bir $a$ k\"umesi i\c cin $\phi(a)$ c\"umlesi do\u gruysa $a$ k\"umesi, $\phi(x)$ form\"ul\"un\"u 
\textbf{sa\u glar}%
\index{sa\u glamak}
(\eng{satisfies}).
  O zaman $\phi$ form\"ul\"un\"u sa\u glayan k\"umeler toplulu\u gu vard\i r.  Bu topluluk
\begin{equation*}
\{x\colon\phi(x)\}
\end{equation*}
olarak yaz\i l\i r, ve ona 
\textbf{$\phi$ taraf\i ndan tan\i mlanm\i\c s s\i n\i f}%
\index{s\i n\i f}%
\index{tan\i mlama}
(\eng{class defined by $\phi$}) denir.

Yukar\i daki \pageref{terim} numaral\i\ sayfadaki tan\i ma g\"ore bir de\u gi\c sken veya sabit, bir \emph{terimdir.}  Daha kesinlikle bir
\textbf{k\"ume terimidir}%
\index{k\"ume!--- terimi}\index{terim!k\"ume ---i}
(\eng{set term}).  \c Simdi, e\u ger $x$ de\u gi\c skeni, $\phi$ form\"ul\"un\"un serbest bir de\u gi\c skeniyse, $\phi$ form\"ul\"un\"u
\begin{equation*}
\phi(\dots x\dots)
\end{equation*}
olarak yazar\i z.  O zaman
\begin{equation*}
\{x\colon\phi(\dots x\dots)\}
\end{equation*}
ifadesi, bir
\textbf{s\i n\i f terimi}%
\index{s\i n\i f!--- terimi}\index{terim!s\i n\i f ---i}
(\eng{class term})
olacak.
S\i n\i f terimlerini form\"ullerde kullanabiliriz, ama \c simdilik, sadece $\in$ i\c saretinin sa\u g\i nda.  Bir $x$ de\u gi\c skeninin bir $\phi(\dots y\dots)$ form\"ul\"undeki serbest ge\c ci\c si, bir
\begin{equation*}
t\in\{y\colon\phi(\dots y\dots)\}
\end{equation*}
form\"ul\"unde (h\^al\^a) serbesttir.
E\u ger $x$ de\u gi\c skeninin $\phi(\dots x\dots)$ form\"ul\"un\"undeki her serbest ge\c ci\c sinin yerine $a$ sabitini koyarsak $\phi(\dots a\dots)$ form\"ul\"u \c c\i kar.  \c Simdi tan\i ma g\"ore
\begin{equation*}
a\in\{x\colon\phi(\dots x\dots)\}\denk\phi(\dots a\dots).
\end{equation*}

Bir sabit veya bir $\{x\colon\phi(x)\}$ s\i n\i f terimi,
\textbf{kapal\i}%
\index{kapal\i}\index{terim!kapal\i\ ---}
(\eng{closed})
bir terimdir.  Kapal\i\ bir terim, bir k\"umenin veya bir s\i n\i f\i
n ad\i d\i r.  $\bm A$, $\bm B$, $\bm C$%
\glossary{$\bm A$, $\bm B$, $\bm C$, \dots}
gibi b\"uy\"uk siyah harfleri kapal\i\ s\i n\i f terimleri olarak
kullanaca\u g\i z.  O zaman \pageref{eqn:=} numaral\i\ sayfadaki tan\i
ma g\"ore
\begin{gather*}
\bm A=\bm B\denk\Forall x(x\in\bm A\liff x\in\bm B),\\
a=\bm B\denk\Forall x(x\in a\liff x\in\bm B),\\
\bm B=a\denk a=\bm B.
\end{gather*}
Sonu\c c olarak
\begin{equation*}
a=\{x\colon x\in a\}.
\end{equation*}
Yani her k\"ume, bir s\i n\i fa e\c sitt\i r.  Ama tersi yanl\i\c st\i r;
bildi\u gimiz gibi baz\i\ s\i n\i flar hi\c cbir k\"umeye e\c sit
de\u gildir: 

\begin{theorem}[Russell Paradoksu \cite{Russell-letter}]%
\index{paradoks!Russell P---u}%
\index{teorem!Russell Paradoksu}
$\{x\colon x\notin x\}$ s\i n\i f\i, hi\c cbir k\"umeye e\c sit de\u gildir.
\end{theorem}

\begin{proof}
Bu teoremi zaten \pageref{Russell} numaral\i\ sayfada kan\i tlad\i k.  \c
Simdi bir kan\i t daha verece\u giz.
$x\notin x$ form\"ul\"u taraf\i ndan tan\i mlanm\i\c s s\i n\i f, $\bm
  A$ olsun.  O zaman her $b$ k\"umesi i\c cin
  \begin{equation*}
    b\in\bm A\liff b\notin b
  \end{equation*}
do\u grudur.  O zaman $\Forall x(x\in\bm A\liff x\in b)$ c\"umlesi yanl\i\c st\i r.  E\c sitli\u gin tan\i m\i na g\"ore $b\neq\bm A$.
\end{proof}

\c Simdi s\i n\i f terimlerini $\in$ i\c saretinin solunda
kullanabiliriz, ama \c c\i kan c\"umle do\u gru olaca\u g\i\ i\c cin
s\i n\i f terimi bir k\"umeyi adland\i rmal\i: 
\begin{equation*}
\bm A\in\bm B\denk\Exists x(x=\bm A\land x\in\bm B).
\end{equation*}

E\u ger $\Forall x(x\in\bm A\lto x\in\bm B)$ do\u gruysa, o
zaman $\bm A$, $\bm B$ s\i n\i f\i n\i n \textbf{alts\i n\i f\i d\i
  r}\index{alts\i n\i f}
(\eng{subclass}), ve $\bm A\included\bm B$ ifadesini yazar\i z.  Yani
\begin{equation*}
\bm A\included\bm B\denk\Forall x(x\in\bm A\lto x\in\bm B).
\end{equation*}

\begin{theorem}\label{thm:=}\mbox{}
\begin{compactenum}
\item
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{equation*}
\bm A=\bm B\denk\bm A\included\bm B\land\bm B\included\bm A.
\end{equation*}
\item
T\"um $\bm A$, $\bm B$, ve $\bm C$ s\i n\i flar\i\ i\c cin
\begin{equation*}
\bm A\included\bm B\land\bm B\included\bm C\lto\bm A\included\bm C
\end{equation*}
c\"umlesi (mant\i\u ga g\"ore) do\u grudur.
\end{compactenum}
\end{theorem}

\ktk

\section{\.I\c slemler}

S\i n\i flarla birka\c c tane ikili i\c slem vard\i r.  \"Once
\begin{gather*}\label{kesisim}
  \bm A\cap\bm B=\{x\colon x\in\bm A\land x\in\bm B\},\\
  \bm A\cup\bm B=\{x\colon x\in\bm A\lor x\in\bm B\}.
\end{gather*}%
\glossary{$\bm A\cap\bm B$}%
\glossary{$\bm A\cup\bm B$}
Bunlar s\i ras\i yla $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
\textbf{kesi\c simi}%
\index{kesi\c sim}
(\eng{intersection}) ve
\textbf{bile\c simi}%
\index{bile\c sim}
(\eng{union}).

\begin{theorem}
  T\"um $\bm A$, $\bm B$, ve $\bm C$ s\i n\i flar\i\ i\c cin
  \begin{gather*}
\bm A\cap\bm B=\bm B\cap\bm A,\\      
\bm A\cup\bm B=\bm B\cup\bm A,\\
      \bm A\cap(\bm B\cap\bm C)=(\bm A\cap\bm B)\cap\bm C,\\
      \bm A\cup(\bm B\cup\bm C)=(\bm A\cup\bm B)\cup\bm C,\\
    \bm A\cap(\bm B\cup\bm C)=(\bm A\cap\bm B)\cup(\bm A\cap\bm C),\\
    \bm A\cup(\bm B\cap\bm C)=(\bm A\cup\bm B)\cap(\bm A\cup\bm C).
  \end{gather*}
\end{theorem}

\begin{proof}
  $x\in\bm A\land x\in\bm B\denk x\in\bm B\land x\in\bm A$, vesaire.
\end{proof}

Ondan sonra
\begin{equation*}
  \bm A\setminus\bm B=\{x\colon x\in\bm A\land x\notin\bm B\};
\end{equation*}
bu s\i n\i f, $\bm A$ s\i n\i f\i n\i n $\bm B$ s\i n\i f\i ndan
\textbf{fark\i d\i r}%
\index{fark}
(\eng{difference}).  O zaman
\begin{equation*}
    \bm A\symdiff\bm B=(\bm A\setminus\bm A)\cup(\bm B\setminus\bm A);  
\end{equation*}
bu s\i n\i f, $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
\textbf{simetrik fark\i d\i r} (\eng{symmetric difference}).

\begin{comment}
  




\begin{theorem}
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{align*}
\bm A\symdiff\bm B
&=(\bm A\setminus\bm B)\cup(\bm B\setminus\bm A)\\
&=(\bm A\cup\bm B)\setminus(\bm A\cap\bm B).
\end{align*}
\end{theorem}

\ktk


\end{comment}

\ref{thm:=} numaral\i\ teorem sayesinde bir $\bm A\included\bm
B\land\bm B\included\bm C$ c\"umlesinin yerine
\begin{equation*}
\bm A\included\bm B\included\bm C
\end{equation*}
ifadesini yazabiliriz.  \"Orne\u gin sonraki teoremi yazabiliriz.

\begin{theorem}
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{align*}
\bm A\cap\bm B&\included\bm A\included\bm A\cup\bm B,&
\bm A\cap\bm B&\included\bm B\included\bm A\cup\bm B.
\end{align*}
\end{theorem}

\ktk

S\i n\i flarda bir \emph{birli} i\c slem vard\i r:
\begin{equation*}
\bm A\comp=\{x\colon x\notin\bm A\};
\end{equation*}
bu s\i n\i f, $\bm A$ s\i n\i f\i n\i n \textbf{t\"umleyenidir}%
\index{t\"umleyen}
(\eng{complement}).

\begin{theorem}[De Morgan Kurallar\i\protect{\footnote{Asl\i nda bu kurallar\i,
    Augustus De Morgan'\i n (1806--71) eserlerinde bulamad\i m, ama
    Venedikli Paulus'un ($\sim$1369--1429) eserlerinde
    \cite[31.35]{MR0120142} buldum.}}]%
\index{teorem!De Morgan Kurallar\i}%
\index{De Morgan Kurallar\i}
T\"um $\bm A$ ile $\bm B$ s\i n\i flar\i\ i\c cin
\begin{align*}
	(\bm A\cap\bm B)\comp&=\bm A\comp\cup\bm B\comp,&
	(\bm A\cup\bm B)\comp&=\bm A\comp\cap\bm B\comp.
\end{align*}
\end{theorem}



\ktk

\.I\c cerilme ba\u g\i nt\i s\i n\i\ kullanarak birka\c c tane birli
i\c slemi daha tan\i mlayabiliriz:
\begin{gather*}
  \bigcap\bm A=\{x\colon\Forall y(y\in\bm A\lto x\in y)\},\\
\bigcup\bm A=\{x\colon\Exists y(x\in y\land y\in\bm A)\},\\
\begin{aligned}
\pow{\bm A}
&=\{x\colon\Forall y(y\in x\lto y\in\bm A)\}\\
&=\{x\colon x\included\bm A\};
\end{aligned}
\end{gather*}%
\glossary{$\bigcap\bm A$}%
\glossary{$\bigcup\bm A$}%
\glossary{$\pow{\bm A}$}
bunlar s\i ras\i yla $\bm A$ s\i n\i f\i n\i n \textbf{kesi\c simi}
(\eng{intersection}), \textbf{bile\c simi} (\eng{union}), ve
\textbf{kuvvet s\i n\i f\i d\i r} (\eng{power class}).

\begin{theorem}\label{thm:cap-cup}
E\u ger $a\in\bm B$ ise
\begin{equation*}
\bigcap\bm B\included a\included\bigcup\bm B.
\end{equation*}
\end{theorem}

\ktk

Son olarak \pageref{universe} numaral\i\ sayfadaki gibi
\begin{equation*}
\universe=\{x\colon x=x\},
\end{equation*}%
\glossary{$\universe$}
ve
\begin{gather*}
  \emptyset=\{x\colon x\neq x\},\\
\{a\}=\{x\colon x=a\},\\
\{a,b\}=\{x\colon x=a\lor x=b\},\\
\{a,b,c\}=\{x\colon x=a\lor x=b\lor x=c\},\\
\parbox{7cm}{\dotfill}
\end{gather*}
Buradaki $\emptyset$ s\i n\i f\i, \textbf{bo\c s s\i n\i ft\i r.}%
\index{s\i n\i f!bo\c s ---}%
\index{bo\c s s\i n\i f}
%Ancak birka\c c aksiyom daha verilmeden bunlar sadece s\i n\i ft\i rlar.

\begin{theorem}\label{thm:cap-empty}
  \begin{align*}
    \bigcap\emptyset&=\universe,&
\bigcup\emptyset&=\emptyset.
  \end{align*}
\end{theorem}

\ktk[\footnote{Baz\i\ kitaplarda $\bm A$ bo\c s ise $\bigcap\bm A$ kesi\c simi tan\i
mlanmaz.  \"Orne\u gin \cite[s.~51 \&\ 285]{Nesin-SKK} kayna\u g\i na
bak\i n.}]

Bu altb\"ol\"um\"un
\begin{align*}
  &\begin{gathered}
\bm A\cap\bm B,\\
\bm A\cup\bm B,\\
\bm A\symdiff\bm B,\\
\bm A\setminus\bm B,
   \end{gathered}&
&\begin{gathered}
    \bm A\comp,\\
\bigcap\bm A,\\
\bigcup\bm A,\\
\pow{\bm A},
  \end{gathered}&
&\universe,&
&\begin{gathered}
 \emptyset,\\
\{a\},\\
\{a,b\},\\
\{a,b,c\}   
  \end{gathered}
\end{align*}
ifadeleri, 
\emph{s\i n\i f} terimidir.%
\index{terim!s\i n\i f ---i}
Her $\bm A$ veya $\bm B$
teriminin yerine ba\c ska bir terimi koyabiliriz.  Zaten bu \c sekilde $(\bm
A\setminus\bm B)\cup(\bm B\setminus\bm A)$ gibi ifadeleri yazd\i k.
Fakat \c simdilik k\"u\c c\"uk harfler hari\c c, k\"ume terimlerimiz
yoktur.  Bu durum hemen de\u gi\c secek.

%\begin{comment}
  


\chapter{Do\u gal Say\i lar}

\section{Do\u gal say\i lar k\"umesi}

Do\u grulu\u gun \pageref{truth} numaral\i\ sayfadaki tan\i m\i na
g\"ore $\Exists xx=a$
c\"umlesi do\u gru mudur?  Yani $\Exists x\Forall y(y\in x\liff y\in a)$
c\"umlesi do\u gru mudur?  E\u ger bir $b$ k\"umesi i\c cin $b=a$
c\"umlesi, yani $\Forall y(y\in b\liff y\in a)$ c\"umlesi, do\u
gruysa, o zaman $\Exists xx=a$ c\"umlesi de do\u grudur.  Asl\i nda
\ref{thm:=-equiv} numaral\i\ teoreme g\"ore $a=a$ c\"umlesi do\u gru,
de\u gil mi?  O halde $\Exists xx=a$ c\"umlesi do\u gru olmal\i. 

Ama bu iddia pek do\u gru de\u gildir.  Bir $a$ k\"umesi varsa, o zaman
$\Exists xx=a$ c\"umlesi do\u grudur.  Bir k\"ume varsa, bu k\"umeye
$a$ denilebilir, ve sonu\c c olarak $\Exists xx=a$ c\"umlesi do\u
gru oluyor.  Bu ana kadar hi\c c kesin bir k\"umemiz olmad\i.  Ama k\"umeler olmal\i, ve birini zaten biliyoruz: 

\begin{axiom}[Bo\c s K\"ume]\index{aksiyom!Bo\c s K\"ume A---u}
  $\emptyset$ bo\c s s\i n\i f, bir k\"umedir:
  \begin{equation*}
    \Exists x\Forall y(y\notin x)
  \end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

Bu aksiyom sayesinde $\emptyset$ i\c sareti, bir 
k\"ume terimidir.%
\index{terim!k\"ume ---i}
Bu y\"uzden
$\{\emptyset\}$ ve $\{\emptyset,a\}$ gibi s\i n\i f terimlerini
yazabiliriz.  Bu terimler de, k\"ume terimi olacak.  Bo\c s k\"ume
gibi bilinen k\"umelerden yeni k\"umeler olu\c sturulabilir:

\begin{axiom}[Biti\c stirme]\index{aksiyom!Biti\c stirme A---u}
T\"um $a$ ile $b$ k\"umeleri i\c cin $a\cup\{b\}$ s\i n\i f\i, bir
k\"umedir:
\begin{equation*}
 \Forall x\Forall y\Exists z\Forall w(w\in z\liff w\in x\lor w=y)
\end{equation*}
c\"umlesi do\u grudur.  
\end{axiom}

\begin{theorem}[Temel K\"umeler]
  T\"um $a$ ile $b$ k\"umeleri i\c cin $\{a\}$ ile $\{a,b\}$ s\i n\i flar\i, 
k\"umedir:
\begin{gather*}
  \Forall x\Exists y\Forall z(z\in y\liff z=x),\\
\Forall x\Forall y\Exists z\Forall w(w\in z\liff w=x\lor w=y)
\end{gather*}
c\"umleleri do\u grudur.
\end{theorem}

\begin{proof}
Bo\c s K\"ume ile Biti\c stirme Aksiyomlar\i na g\"ore
  $\{a\}$ s\i n\i f\i, $\emptyset\cup\{a\}$ k\"umesine e\c sittir, ve
$\{a,b\}$ s\i n\i f\i, $\{a\}\cup\{b\}$ k\"umesine e\c sittir. 
\end{proof}

\"Ozel olarak her $a$ k\"umesi i\c cin $a\cup\{a\}$ bir k\"umedir.  Bu
son k\"ume, $a'$ olsun.  Yani her $a$ k\"umesi i\c cin
\begin{equation*}
a'=a\cup\{a\}
\end{equation*}%
\glossary{$a'$}
olsun.  $a'$ k\"umesi, $a$ k\"umesinin
\textbf{ard\i l\i d\i r}%
\index{ard\i l}
(\eng{successor}).
S\i k s\i k ard\i llar\i\ alarak\label{0123}
\begin{align*}
&\emptyset,&
&\emptyset',&
&\emptyset'',&
&\emptyset''',&
&\dots
\\
%\end{align*}
\intertext{
k\"ume dizisini olu\c
sturabiliriz.  Bu dizi,
}
%\begin{align*}
  &\emptyset,&
  &\{\emptyset\},&
  &\bigl\{\emptyset,\{\emptyset\}\bigr\},&
  &\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\},&
&\dots
\\
%\end{align*}
\intertext{Yukar\i daki \pageref{vnn} numaral\i\ sayfadaki gibi bu k\"umeler,
}
%\begin{align*}
&0,&
&1,&
&2,&
&3,&
&\dots
\end{align*}%
\glossary{$0$}
do\u gal say\i lar\i\ olacak.
Elemanlar\i\ \emph{t\"um} do\u gal say\i lar olan bir s\i n\i f var
m\i d\i r? 

Do\u gal say\i lar\i n \emph{toplulu\u gunun} iki \"ozelli\u gi vard\i
r:
\begin{compactenum}
  \item
$0$, bu topluluktad\i r.
\item
E\u ger $a$, bu topluluktaysa, $a\cup\{a\}$ k\"umesi de, bu
topluluktad\i r.
\end{compactenum}
Bu \"ozellikleri olan \emph{k\"umeler,} bir s\i n\i f olu\c sturur.  Yani
\begin{equation*}
\bm{\Omega}=\{x\colon 0\in x\land\Forall y(y\in x\lto y\cup\{y\}\in x)\}
\end{equation*}%
\glossary{$\bm{\Omega}$}
e\c sitli\u gini sa\u glayan bir $\bm{\Omega}$ s\i n\i f\i\ vard\i r.

\begin{theorem}\label{thm:Omega}
\mbox{}
  \begin{compactenum}
    \item
$0\in\bigcap\bm{\Omega}$.
\item
E\u ger $a\in\bigcap\bm{\Omega}$ ise, o zaman
$a\cup\{a\}\in\bigcap\bm{\Omega}$ .
\item
E\u ger $a\included\bigcap\bm{\Omega}$ ise, ve $a$,
\begin{align*}
  0&\in a,&
\Forall x(x\in a&\lto x\cup\{x\}\in a)
\end{align*}
\"ozelliklerini sa\u glarsa, o zaman $a=\bigcap\bm{\Omega}$.
  \end{compactenum}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
E\u ger $a\in\bm{\Omega}$ ise, o zaman $0\in a$.  Sonu\c c olarak $0\in\bigcap\bm{\Omega}$.
\item
$a\in\bigcap\bm{\Omega}$ olsun.  O zaman $\bm{\Omega}$ s\i n\i f\i n\i n her $b$ eleman\i\ i\c cin $a\in b$.  Ayr\i ca $b\in\bm{\Omega}$ y\"uz\"unden $\Forall y(y\in b\lto y\cup\{y\}\in b)$ c\"umlesi do\u grudur.  O zaman $a\cup\{a\}\in b$ olmal\i.  Sonu\c c olarak $a\cup\{a\}\in\bigcap\bm{\Omega}$.
\item
$0\in a$ ve
$\Forall x(x\in a\lto x\cup\{x\}\in a)$ do\u gru olsun.  O zaman $a\in\bm{\Omega}$.  Bu y\"uzden \ref{thm:cap-cup} numaral\i\ teoreme g\"ore $\bigcap\bm{\Omega}\included a$ olmal\i.  E\u ger ayr\i ca
$a\included\bigcap\bm{\Omega}$ ise, o zaman \ref{thm:=} numaral\i\ teoreme g\"ore $a=\bigcap\bm{\Omega}$.\qedhere
\end{asparaenum}
\end{proof}

Bu teoreme ra\u gmen e\u ger
\begin{align}\label{eqn:induction}
\bm A&\included\bigcap\bm{\Omega},&
0&\in\bm A,&
\Forall x(x\in\bm A&\lto x\cup\{x\}\in\bm A)
\end{align}
ise $\bm A=\bigcap\bm{\Omega}$ c\"umlesini sonu\c cland\i ram\i yoruz.
Neden?  \c C\"unk\"u \ref{thm:cap-empty} numaral\i\ teoreme g\"ore 
\begin{equation*}
\bigcap0=\universe
\end{equation*}
(yani $\bigcap\emptyset=\universe$), ve $\bm{\Omega}$ s\i n\i f\i n\i
n bo\c s olmad\i\u g\i n\i\ \c simdilik bilmiyoruz.  Bu durumu hemen
de\u gi\c stirebiliriz: 

\begin{axiom}[Sonsuzluk]%
\index{aksiyom!Sonsuzluk A---u}
$\bm{\Omega}\neq0$, yani
\begin{equation*}
\Exists x\bigl(0\in x\land\Forall y(y\in x\lto y\cup\{y\}\in x)\bigr)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

H\^al\^a yukar\i daki \eqref{eqn:induction} sat\i r\i ndaki varsay\i lar\i ndan $\bm A=\bigcap\bm{\Omega}$ c\"umlesini sonu\c cland\i ram\i yoruz.  Neden?  Bir tane aksiyomu daha kullanarak bunu sonu\c cland\i rabiliriz:

\begin{axiom}[Ay\i rma]%
\index{aksiyom!Ay\i rma A---u}
Bir k\"umenin her alts\i n\i f\i, bir k\"umedir, yani her $\phi(x)$ form\"ul\"u i\c cin
\begin{equation*}
\Forall x\Exists y\Forall z\bigl(z\in y\liff z\in x\land\phi(z)\bigr)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

\c Simdi her $a$ k\"umesi ve $\phi(x)$ form\"ul\"u i\c cin $\{x\colon x\in a\land\phi(x)\}$ s\i n\i f\i, bir k\"umedir, ve bu k\"ume
\begin{equation*}
\{x\in a\colon\phi(x)\}
\end{equation*}
olarak yaz\i l\i r.

\begin{theorem}
Bir s\i n\i f bo\c s de\u gilse, kesi\c simi bir k\"umedir.
\end{theorem}

\begin{proof}
$a\in\bm B$ olsun.  \ref{thm:cap-cup} numaral\i\ teoreme g\"ore $\bigcap\bm B\included a$.  Ay\i rma Aksiyomuna g\"ore $\bigcap\bm B$ kesi\c simi, bir k\"ume olmal\i.
\end{proof}

\"Ozel olarak
\begin{equation*}
\upomega=\bigcap\bm{\Omega}
\end{equation*}%
\glossary{$\upomega$}
e\c sitli\u gini sa\u glayan bir $\upomega$ k\"umesi vard\i r.  Bu k\"umenin elemanlar\i, \textbf{von Neumann do\u gal say\i lar\i d\i r.}%
\index{say\i!von Neumann do\u gal ---lar\i}%
\index{von Neumann do\u gal say\i lar\i}
$\upomega$ i\c sareti, yeni bir k\"ume terimidir.  Bundan sonra $\bm{\Omega}$ s\i n\i f terimini kullanmayaca\u g\i z.

\c Simdi \ref{thm:Omega} numaral\i\ teoremi a\c sa\u g\i daki bi\c cimde yazabiliriz:
  \begin{compactenum}\label{induction}
    \item
$0\in\upomega$.
\item
E\u ger $a\in\upomega$ ise, o zaman $a'\in\upomega$.
\item
E\u ger $a\included\upomega$ ise, ve $a$,
\begin{align*}
  0&\in a,&
\Forall x(x\in a&\lto x'\in a)
\end{align*}
\"ozelliklerini sa\u glarsa, o zaman $a=\upomega$.
  \end{compactenum}
Ayr\i ca her k\"umeninki gibi $\upomega$ k\"umesinin de her alts\i n\i f\i, bir k\"umedir.  Sonu\c c olarak $\upomega$ k\"umesinin baz\i\ \"ozelliklerini 
\textbf{t\"umevar\i m}%
\index{t\"umevar\i m}
(\eng{induction})
y\"ontemiyle kan\i tlayabilece\u giz.

Asl\i nda bazen $\upomega$ k\"umesinin iki \"ozelli\u gininin
daha kullan\i lmas\i\ gerekecek.  $\Forall xx'\neq0$ apa\c c\i kt\i r.
Ama $k$ ile $m$, do\u gal say\i lar ise, ve $k'=m'$ ise, $k=m$ e\c
sitli\u gini elde etmek, biraz daha zor olacak. 

M\"umk\"unse $k'=m'$ ama $k\neq m$ olsun.  O zaman $k\in m$ ve $m\in k$
olmal\i.  Bundan $k\in k$ c\"umlesini sonu\c cland\i rmak istiyoruz.

E\u ger bir $\bm A$ s\i n\i f\i,
\begin{equation*}
  \Forall x\Forall y(x\in\bm A\land y\in x\lto y\in\bm A)
\end{equation*}
c\"umlesini sa\u glarsa, o zaman $\bm A$ s\i n\i f\i na 
\textbf{ge\c ci\c sli}%
\index{ge\c ci\c sli}
(\eng{transitive}) denir.  \"Oyleyse her ge\c ci\c sli s\i n\i f\i n
her eleman\i, s\i n\i f\i n bir altk\"umesidir de.

\begin{theorem}\label{thm:n-trans}
$\upomega$ k\"umesinin her eleman\i, ge\c ci\c slidir.
\end{theorem}

\begin{proof}
$a$, $\upomega$ k\"umesinin ge\c ci\c sli elemanlar\i\ k\"umesi olsun.  Yani
\begin{align*}
a
&=\{x\in\upomega\colon\Forall y\Forall z(y\in x\land z\in y\lto z\in x)\}\\
&=\{x\in\upomega\colon\Forall y(y\in x\lto y\included x)\}
\end{align*}
olsun.  O zaman $0\in a$.  \emph{T\"umevar\i m hipotezi} olarak $b\in
a$ olsun.  $b'\in a$ c\"umlesinin do\u grulu\u gunu g\"osterece\u giz.
$c\in b'$ olsun.  Ya $c\in b$ ya da $c=b$.  E\u ger $c\in b$ ise, o
zaman hipotezimize g\"ore $c\included b$.  Her durumda $b\included
b'$.  \"Oyleyse $c\included b'$.  Ama $c$, $b'$ k\"umesinin herhangi
bir eleman\i d\i r.  Sonu\c c olarak $b'\in a$.  T\"umevar\i mdan
(yani \ref{thm:Omega} numaral\i\ teoremin \pageref{induction}
numaral\i\ sayfadaki bi\c ciminden) $a=\upomega$. 
\end{proof}

\begin{theorem}\label{thm:w-trans}
$\upomega$ k\"umesi, ge\c ci\c slidir.
\end{theorem}

\ktk

\begin{xca}
$\bigl\{0,1,\{1\}\bigr\}$ k\"umesinin ge\c ci\c sli oldu\u gunu kan\i tlay\i n.
\end{xca}

\begin{theorem}\label{thm:irr}
$\upomega$ k\"umesinin hi\c cbir eleman\i, kendisini i\c cermez.
\end{theorem}

\begin{proof}
Tekrar t\"umevar\i m\i\ kullanaca\u g\i z.  \c C\"unk\"u bo\c s
k\"umenin hi\c cbir eleman\i\ yok, $0\notin0$.  \c Simdi
$a\in\upomega$ ve $a\notin a$ olsun.  E\u ger $a'\in a'$ ise, ya
$a'\in a$ ya da $a'=a$.  Her durumda, ge\c cen teoreme g\"ore,
$a'\included a$, dolay\i s\i yla $a\in a$ (\c c\"unk\"u $a\in a'$).
Bu sonu\c c, varsay\i m\i m\i zla \c celi\c sir.  O zaman $a'\notin
a'$ olmal\i.  T\"umevar\i mdan kan\i t\i m\i z bitti. 
\end{proof}

\begin{theorem}
$\upomega$ k\"umesinin t\"um $k$ ile $m$ elemanlar\i\ i\c cin $k'=m'$
  ise $k=m$. 
\end{theorem}

\begin{proof}
M\"umk\"unse $k'=m'$ ama $k\neq m$ olsun.  Dedi\u gimiz gibi $k\in m$ ve $m\in k$
olmal\i.  \ref{thm:n-trans} ile \ref{thm:irr} numaral\i\ teoremlere
g\"ore $k\in k$ ve $k\notin k$, bir \c celi\c skidir. 
\end{proof}

\c Simdi, \ref{thm:Omega} numaral\i\ teoremdekiler dahil, $\upomega$
k\"umesinin be\c s tane \"ozelli\u gi vard\i r: 
%\begin{minipage}{\textwidth}
\begin{compactenum}
\item
$0\in\upomega$.
\item
$\Forall x(x\in\upomega\lto x'\in\upomega)$.
\item
$\Forall x\bigl(x\included\upomega\land 0\in x\land\Forall y(y\in x\lto y'\in x)\lto x=\upomega\bigr)$.
\item
$\Forall x(x\in\upomega\lto x'\neq0)$.
\item
$\Forall x\Forall y(x\in\upomega\land y\in\upomega\land x'=y'\lto x=y)$.
\end{compactenum}
%\end{minipage}
Bu \"ozelliklerin \"onemi, 1887 y\i l\i nda Dedekind \cite[II, \P71]{MR0159773}
taraf\i ndan, ve 1889 y\i l\i nda Peano \cite{Peano} taraf\i ndan,
fark edilmi\c stir.  S\i k s\i k 
\textbf{Peano Aksiyomlar\i,}%
\index{aksiyom!Peano A---lar\i, Dedekind--Peano A---lar\i}\label{PA}
bu \"ozelliklere denir, ama
\textbf{Dedekind--Peano Aksiyomlar\i}
de kullan\i labilir.  Asl\i nda bizim i\c cin aksiyomlar de\u gil,
teoremdirler. 

Peano Aksiyomlar\i ndan do\u gal say\i lar\i n t\"um \"ozellikleri
elde edilebilir.  Mesela \emph{iyi s\i ralama} \"ozelli\u gi elde
edilebilir.  Asl\i nda $\upomega$, i\c cerilme ($\in$) ba\u g\i nt\i
s\i\ taraf\i ndan iyi s\i ralan\i r.  Ama bir ba\u g\i nt\i\ nedir? 

\section{Ba\u g\i nt\i lar}

Herhangi $a$ ile $b$ k\"umeleri i\c cin $\bigl\{\{a\},\{a,b\}\bigr\}$
k\"umesi $(a,b)$ \textbf{s\i ral\i\ ikilisi}%
\index{ikili}%
\index{s\i ra!---l\i\ ikili}
(\eng{ordered pair})
olarak yaz\i l\i r.  Yani\footnote{\ref{Kuratowski}
  numaral\i\ sayfadaki notta dedi\u gimiz gibi bu tan\i m,
  Kuratowski'nin \cite{Kuratowski} 1921 y\i l\i nda verdi\u gi tan\i
  md\i r.} 
\begin{equation*}
(a,b)=\bigl\{\{a\},\{a,b\}\bigr\}.
\end{equation*}%
\glossary{$(a,b)$}

\begin{theorem}
  T\"um $a$, $b$, $c$, ve $d$ k\"umeleri i\c cin
  \begin{equation*}
    (a,b)=(c,d)\liff a=c\land b=d
  \end{equation*}
c\"umlesi do\u grudur.
\end{theorem}

\ktk

\begin{xca}
$\bigl\{\{a,1\},\{b,2\}\bigr\}
=\bigl\{\{c,1\},\{d,2\}\bigr\}\liff
a=c\land b=d$ c\"umlesini kan\i tlay\i n.\footnote{Heijenoort'a
  \cite[s.~224]{MR1890980} g\"ore bu c\"umlede, Hausdorff'un 1914 y\i
  l\i nda verdi\u gi s\i ral\i\ ikili tan\i m bulunmu\c stur.} 
\end{xca}

\begin{xca}
$\Bigl\{\bigl\{\{a\},0\bigr\},\bigl\{\{b\}\bigr\}\Bigr\}
=\Bigl\{\bigl\{\{c\},0\bigr\},\bigl\{\{d\}\bigr\}\Bigr\}
\liff a=c\land b=d$ c\"umlesini kan\i tlay\i n.\footnote{Bu c\"umlede,
  Wiener'in \cite{Wiener} 1914 y\i l\i nda verdi\u gi s\i ral\i\ ikili
  tan\i m bulunmu\c stur.} 
\end{xca}

\c Simdi her ikili $\phi(x,y)$ form\"ul\"u i\c cin
\begin{equation*}
  \bigl\{z\colon\Exists x\Exists y\bigl(z=(x,y)\land\phi(x,y)\bigr)\bigr\}
\end{equation*}
s\i n\i f\i,
\begin{equation*}
  \{(x,y)\colon\phi(x,y)\}
\end{equation*}
olarak yaz\i labilir.  \"Oyle bir s\i n\i f, bir 
\textbf{ikili ba\u g\i nt\i d\i r}%
\index{ba\u g\i nt\i}
(\eng{binary relation}).

\"Orne\u gin:
\begin{asparaenum}
\item
\.I\c cerilme ba\u g\i nt\i s\i, $\{(x,y)\colon x\in y\}$ s\i n\i f\i d\i r.
\item
E\c sitlik ba\u g\i nt\i s\i, $\{(x,y)\colon x=y\}$ s\i n\i f\i d\i r.
\end{asparaenum}


Ayn\i\ \c sekilde, e\u ger $\bm R$, bir ikili ba\u g\i nt\i ysa, o
zaman $(x,y)\in\bm R$ form\"ul\"un\"un k\i saltmas\i\ olarak
$x\mathrel{\bm R}y$ ifadesini yazar\i z, yani 
\begin{equation*}
  x\mathrel{\bm R}y\denk(x,y)\in\bm R.
\end{equation*}
$\bm R$ ba\u g\i nt\i s\i n\i n \textbf{ters ba\u g\i nt\i s\i}%
\index{ba\u g\i nt\i!ters ---} 
veya
\textbf{tersi}% 
\index{ters}
(\eng{converse}),
\begin{equation*}
  \{(y,x)\colon x\mathrel{\bm R}y\}
\end{equation*}
ba\u g\i nt\i s\i d\i r.  Bu ba\u g\i nt\i, $\conv{\bm R}$ olarak
yaz\i l\i r; yani
\begin{equation*}
  x\mathrel{\conv{\bm R}}y\denk y\mathrel{\bm R}x.
\end{equation*}%
\glossary{$\conv{\bm R}$}
$\bm A$ ile $\bm B$, iki s\i n\i f ise, o zaman tan\i ma g\"ore
\begin{equation*}
  \bm A\times\bm B=\{(x,y)\colon x\in\bm A\land y\in\bm B\};
\end{equation*}%
\glossary{$\bm A\times\bm B$}
bu ba\u g\i nt\i, $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
\textbf{\c carp\i m\i d\i r}%
\index{\c carp\i m}
(\eng{product}).
E\u ger $\bm R\included\bm A\times\bm B$, o zaman $\bm R$, $\bm A$ s\i
n\i f\i ndan $\bm B$ s\i n\i f\i na giden bir ba\u g\i nt\i d\i r.

S\i n\i flar aras\i ndaki bir ba\u g\i nt\i n\i n kendisi,
bir s\i n\i ft\i r.  S\i ral\i\ ikililerin tan\i m\i, s\i n\i flarla
ba\u g\i nt\i lar\i\ birle\c stirir.  Benzer \c sekilde Newton'un A\u
g\i rl\i k Kanunu, Ay'\i n Yerin etraf\i nda d\"on\"u\c s\"u ile
nesnelerin yere d\"u\c s\"u\c s\"un\"u birle\c stirir. 

E\u ger $\bm F$,
\begin{equation}\label{eqn:F}
\Forall x\Forall y\Forall z(x\mathrel{\bm F}y\land x\mathrel{\bm F}z\lto y=z)
\end{equation}
c\"umlesini sa\u glayan bir ikili ba\u g\i nt\i ysa, o zaman
\begin{compactenum}[(1)]
\item
 $\bm F$ ba\u g\i nt\i s\i na \textbf{g\"onderme}%
\index{g\"onderme}
denir; 
\item
$\{x\colon\Exists yx\mathrel{\bm F}y\}$ s\i n\i f\i na $\bm F$ g\"ondermesinin 
\textbf{tan\i m s\i n\i f\i}%
\index{tan\i m s\i n\i f\i}
(\eng{domain})
denir;
\item
$\{y\colon\Exists xx\mathrel{\bm F}y\}$ s\i n\i f\i na $\bm F$ g\"ondermesinin 
\textbf{de\u ger s\i n\i f\i}%
\index{de\u ger s\i n\i f\i}
(\eng{range})
denir.\footnote{Bu notlarda bir g\"onderme, sadece \eqref{eqn:F} c\"umlesini sa\u glayan bir $\bm F$ ikili ba\u g\i nt\i s\i d\i r.  Fakat baz\i\ kaynaklarda (\"orne\u gin \cite[s.~70]{Nesin-SKK} kayna\u g\i nda) bir g\"onderme veya fonksiyon,
\begin{inparaenum}[(1)]
\item
\eqref{eqn:F} c\"umlesini sa\u glayan bir $\bm F$ ikili ba\u g\i nt\i s\i,
\item
$\{y\colon\Exists xx\mathrel{\bm F}y\}$ s\i n\i f\i na e\c sit bir $\bm A$ s\i n\i f\i, ve
\item
$\{y\colon\Exists xx\mathrel{\bm F}y\}$ s\i n\i f\i n\i\ \emph{kapsayan} bir $\bm B$ s\i n\i f\i\
\end{inparaenum}
taraf\i ndan olu\c sturulmu\c s bir \"u\c cl\"ud\"ur.  O halde (a\c
sa\u g\i daki \pageref{to} numaral\i\ sayfadaki gibi) $\bm F\colon\bm
A\to\bm B$ ifadesi yaz\i l\i r.  Ayr\i ca, $\bm B$ s\i n\i f\i na
\emph{g\"ondermenin de\u ger s\i n\i f\i} (veya \emph{var\i\c s s\i
  n\i f\i}) denilebilir.  \.Ingilizcede \eng{codomain} kullan\i l\i r.
Ama buradaki $\bm B$ s\i n\i f\i, sadece $\bm F$ s\i n\i
f\i\ taraf\i ndan belirtilmez, ve buna hi\c cbir ad vermiyoruz.} 
\end{compactenum}
Bu durumda $x\mathrel{\bm F}y$ form\"ul\"un\"un yerine
\begin{equation*}
  y=\bm F(x)
\end{equation*}
ifadesini yazar\i z, \c c\"unk\"u $a\mathrel{\bm F}b$ do\u gruysa, o
zaman $b$ k\"umesi, $a$ k\"umesi taraf\i ndan belirtilir.  Buradaki $\bm F(x)$ ifadesi, yeni bir k\"ume terimidir.  O zaman $\bm F$, 
\begin{equation*}
x\mapsto\bm F(x)
\end{equation*}
\glossary{$x\mapsto\bm F(x)$}%
olarak yaz\i labilir; yani
\begin{equation*}
(x\mapsto\bm F(x))=\{(x,y)\colon y=\bm F(x)\}.
\end{equation*}
\"Orne\u gin:
\begin{asparaenum}
\item
Her $a$ k\"umesi i\c cin, $x\mapsto a$
\textbf{sabit g\"onderme}%
\index{sabit!--- g\"onderme}\index{g\"onderme!sabit ---}
(\eng{constant function}) vard\i r, \"ozel olarak $x\mapsto0$, $x\mapsto1$, \dots, $x\mapsto\upomega$, \dots
\item
$x\mapsto x$, \textbf{\"ozde\c slik g\"ondermesidir}%
\index{\"ozde\c slik g\"ondermesi}\index{g\"onderme!\"ozde\c slik ---si}
(\eng{identity function}).
\item
$x\mapsto x'$,
\textbf{ard\i l g\"ondermesi}
(\eng{successor function}) veya 
\textbf{ard\i llamad\i r}%
\index{g\"onderme!ard\i l ---si}%
\index{ard\i l}
(\eng{succession}).
\end{asparaenum}
E\u ger $\bm F$ g\"ondermesinin tan\i m s\i n\i f\i\ $\bm A$ ise, ve
de\u ger s\i n\i f\i n\i, bir $\bm B$ s\i n\i f\i\ taraf\i ndan
kapsan\i rsa, o zaman
\begin{equation*}\label{to}
  \bm F\colon\bm A\to\bm B
\end{equation*}
ifadesini yazar\i z.  Yani bu ifade,
\begin{multline*}
  \Forall x\Forall y(x\mathrel{\bm F}y\lto x\in\bm A\land
  y\in\bm B)\\
\land\Forall x\bigl(x\in\bm A\lto\Exists y(x\mathrel{\bm F}y)\bigr)\\
\land\Forall x\Forall y\Forall z(x\mathrel{\bm F}y\land x\mathrel{\bm
  F}z\lto y=z)
\end{multline*}
c\"umlesinin k\i saltmas\i d\i r.


\section{S\i ralamalar}

\textbf{S\i ralama}%
\index{s\i ralama}
(\eng{ordering}),
\begin{align*}
\Forall x\lnot\; x&\mathrel{\bm R}x,&
\Forall x\Forall y\Forall z(x\mathrel{\bm R}y\land y\mathrel{\bm R}z&\lto x\mathrel{\bm R}z)
\end{align*}
c\"umlelerini sa\u glayan bir $\bm R$ ikili ba\u g\i nt\i s\i d\i r.  \"Orne\u gin yukaradaki \pageref{SBT} numaral\i\ sayfada bahsedildi\u gi ve a\c sa\u g\i daki \pageref{thm:SB} numaral\i\ sayfada kan\i tlanacak Schr\"oder--Bernstein Teoremine g\"ore $\prec$ ba\u g\i nt\i s\i, bir s\i ralama olacakt\i r.  Ayr\i ca
\begin{equation*}
\bm A\pincluded\bm B\denk\bm A\included\bm B\land\bm A\neq\bm B
\end{equation*}
olsun; o zaman $\pincluded$ ba\u g\i nt\i s\i\ da, bir s\i ralamad\i r. 

Belki bir $\bm R$ ba\u g\i nt\i s\i, bir s\i ralama de\u gildir, ama bir $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}
\bm R\cap(\bm A\times\bm A)
\end{equation*}
kesi\c simi, bir s\i ralama olabilir.  O zaman $\bm A$, $\bm R$ taraf{}\i ndan s\i ralan\i r.  \"Orne\u gin $\in$, s\i ralama de\u gil; ama \ref{thm:n-trans} ile \ref{thm:irr} numaral\i\ teoremlere g\"ore $\in$ ba\u g\i nt\i s\i\ $\upomega$ k\"umesini s\i ralar.

E\u ger $\bm A$ s\i n\i f\i, $\bm R$ taraf{}\i ndan s\i ralan\i rsa, ve \"ustelik
\begin{equation*}
\Forall x\Forall y(x\in\bm A\land y\in\bm A\land x\neq y\lto x\mathrel{\bm R}y\lor y\mathrel{\bm R}x)
\end{equation*}
do\u gruysa, o zaman $\bm R$, $\bm A$ s\i n\i f\i n\i n bir
\textbf{do\u grusal}%
\index{do\u grusal}
(\eng{linear})
s\i ralamas\i d\i r.

\begin{theorem}
$\in$ ba\u g\i nt\i s\i, her do\u gal say\i n\i n do\u grusal s\i ralamas\i d\i r.
\end{theorem}

\ktk

E\u ger $\bm R$, $\bm A$ s\i n\i f\i n\i n do\u grusal s\i ralamas\i ysa, ve \"ustelik $\bm A$ s\i n\i f\i n\i n her bo\c s olmayan $b$ altk\"umesinin $\bm R$ s\i ralamas\i na g\"ore
\textbf{en k\"u\c c\"uk}%
\index{k\"u\c c\"uk}
(\eng{least})
eleman\i\ varsa, yani
\begin{equation*}
\Forall x\Bigl(x\included\bm A\land x\neq0\lto\Exists y\bigl(y\in x\land\Forall z(z\in x\setminus\{y\}\lto y\mathrel{\bm R}z)\bigr)\Bigr)
\end{equation*}
do\u gruysa, o zaman $\bm A$, $\bm R$ taraf{}\i ndan
\textbf{iyi s\i ralan\i r}%
\index{iyi s\i ralama}\index{s\i ra!iyi ---lama}
(\eng{well-ordered}).

\begin{theorem}\label{thm:n-wo}
$\in$ ba\u g\i nt\i s\i, her do\u gal say\i n\i n iyi s\i ralamas\i d\i r.
\end{theorem}

\ktk

\begin{theorem}\label{thm:in-pinc}
$\upomega$ k\"umesinde $\in$ ile $\pincluded$, ayn\i\ ba\u g\i nt\i d\i r, yani
\begin{equation*}
\Forall x\Forall y\bigl(x\in\upomega\land y\in\upomega\lto(x\in y\liff x\pincluded y)\bigr)
\end{equation*}
do\u grudur.
\end{theorem}

\begin{proof}
$k$ ile $m$, do\u gal say\i lar olsun.  \ref{thm:n-trans} ile \ref{thm:irr} numaral\i\ teoremlere g\"ore $k\in m$ ise $k\pincluded m$.  

\c Simdi $k\pincluded m$ olsun.  \"Onceki teoreme g\"ore $m\setminus
k$ fark\i n\i n en k\"u\c c\"uk $\ell$ eleman\i\ vard\i r.  O zaman
$\ell\in m$, dolay\i s\i yla $\ell\included m$.  Ayr\i ca $a\in\ell$
ise $a\in k$ olmal\i\ (\c c\"unk\"u $a\in m$, ama i\c cerilmeye g\"ore
$\ell$, $m\setminus k$ fark\i n\i n en k\"u\c c\"uk eleman\i d\i r).
\"Oyleyse $\ell\included k$.  Ama $b\in k$ ise $b\in m$, dolay\i
s\i yla $\ell\in b$ veya $\ell=b$ veya $b\in\ell$.  Ancak
$\ell\notin b$ ve $\ell\neq b$ (\c c\"unk\"u $b\included k$ ve
$\ell\notin k$).  \"Oyleyse $b\in\ell$.  Sonu\c c olarak
$k\included\ell$.  Fakat $\ell\included k$.  O zaman $k=\ell$, dolay\i
s\i yla $k\in m$. 
\end{proof}

\begin{theorem}\label{thm:w-wo}
$\upomega$, i\c cerilme taraf{}\i ndan iyi s\i ralan\i r.
\end{theorem}

\begin{proof}
$\upomega$ k\"umesinde $m\notin k$ ve $m\neq k$ olsun.  Yani (\"onceki teoremi kullanarak) $m\not\included k$ olsun.  O zaman $m\setminus k$ fark\i n\i n en k\"u\c c\"uk $\ell$ eleman\i\ vard\i r.  Ge\c cen kan\i ttaki gibi $\ell\included k$, yani $\ell\in k$ veya $\ell=k$.  Fakat $\ell\notin k$.  Sonu\c c olarak $\ell=k$, dolay\i s\i yla $k\in m$.  \"Oyleyse i\c cerilme, $\upomega$ k\"umesinin bir do\u grusal s\i ralamas\i d\i r.

Ayr\i ca $a\included\upomega$ ve $n\in a$ ise, ya $n$ $a$ k\"umesinin en k\"u\c c\"uk eleman\i d\i r, ya da $n\cap a$ kesi\c simi bo\c s de\u gildir.  Son durumda bu kesi\c simin en k\"u\c c\"uk eleman\i\ vard\i r, ve bu eleman, $a$ k\"umesinin en k\"u\c c\"uk eleman\i d\i r.
\end{proof}

\section{Ordinaller}

\"Onceki iki teoremin kan\i tlar\i, do\u gal say\i lar\i n sadece ge\c ci\c slilik ve iyi s\i ralama \"ozelliklerini kullanmaktad\i r.  Bir 
\textbf{ordinal,}%
\index{ordinal}
\begin{compactenum}[1)]
\item
ge\c ci\c sli ve
\item
$\in$ taraf\i ndan iyi s\i ralanm\i\c s
\end{compactenum}
bir k\"umedir.  Ordinaller,
\begin{equation*}
\on
\end{equation*}%
\glossary{$\on$}
s\i n\i f\i n\i\ olu\c sturur.  O zaman \ref{thm:n-trans} ve \ref{thm:n-wo} numaral\i\ teoremlere g\"ore
\begin{equation*}
\upomega\included\on.
\end{equation*}
\"Ustelik \ref{thm:w-trans} ve \ref{thm:w-wo} numaral\i\ teoremlere g\"ore
\begin{equation*}
\upomega\in\on.
\end{equation*}
Dolay\i s\i yla $\upomega'\in\on$.

\begin{theorem}
Her ordinalin ard\i l\i, bir ordinaldir.
\end{theorem}

\ktk

\begin{theorem}\label{thm:on-in-pinc}
$\on$ s\i n\i f\i nda $\in$ ve $\pincluded$, ayn\i\ ba\u g\i nt\i d\i r.
\end{theorem}

\begin{xca}
\ref{thm:in-pinc} numaral\i\ teoremin kan\i t\i n\i\ kullanarak bu teoremi kan\i tlay\i n.
\end{xca}

\begin{theorem}[Burali-Forti Paradoksu \cite{Burali-Forti}]%
\index{teorem!Burali-Forti Paradoksu}%
\index{paradoks!Burali-Forti P---u}
$\on$ ge\c ci\c slidir, ve $\in$ taraf\i ndan iyi s\i ralan\i r.
\end{theorem}

\begin{proof}
$\alpha$ bir ordinal olsun, ve $\beta\in\alpha$ olsun.  O zaman
  $\beta\included\alpha$.  Bu durumda $\beta$, $\in$ taraf\i ndan
  iyi s\i ralan\i r.  \c Simdi $\gamma\in\beta$ olsun.  O zaman
  $\gamma\in\alpha$, dolay\i s\i yla $\gamma\included\alpha$.  O
  zaman $\delta\in\gamma$ ise $\delta\in\alpha$.  $\alpha$, $\in$
  taraf\i ndan iyi s\i raland\i\u g\i ndan, $\delta\in\beta$, \c
  c\"unk\"u $\beta$, $\gamma$, ve $\delta$, hepsi $\alpha$
  k\"umesindedir, ve $\delta\in\gamma$, ve $\gamma\in\beta$.  K\i saca
  $\delta\in\gamma\lto\delta\in\beta$, yani $\gamma\included\beta$.
  Ama $\gamma$, $\beta$ k\"umesinin herhangi bir eleman\i d\i r.
  \"Oyleyse $\beta$, ge\c ci\c slidir.  Sonu\c c olarak $\beta$, bir
  ordinaldir.  Ama $\beta$, $\alpha$ ordinalinin herhangi bir eleman\i
  d\i r.  O zaman $\alpha\included\on$.  Ve $\alpha$, herhangi bir
  ordinaldir.  \"Oyleyse $\on$ ge\c ci\c slidir. 

Ordinaller s\i n\i f\i n\i n $\in$ taraf\i ndan iyi s\i raland\i\u
g\i\ kan\i t, \ref{thm:w-wo} numaral\i\ teoremin kan\i t\i\ ile ayn\i
d\i r. 
\end{proof}

\pageref{BFP} numaral\i\ sayfada dedi\u gimiz gibi $\on$ bir k\"ume olsayd\i, $\on\in\on$, ki bu sa\c cmad\i r (\c c\"unk\"u $\on$ s\i n\i f\i nda $\in$ d\"on\"u\c ss\"uzd\"ur).

$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$, ve $\iota$%
\glossary{$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$, $\iota$}
k\"u\c c\"uk Yunan 
harfleri, her zaman \emph{ordinal} sabit olacakt\i r.  Yani
\begin{equation*}
\alpha\in\on,
\end{equation*}
vesaire.  Ayr\i ca
\ref{thm:on-in-pinc} numaral\i\ teorem sayesinde $\alpha\in\beta$ veya
$\alpha\pincluded\beta$ form\"ul\"un\"un yerine 
\begin{equation*}
\alpha<\beta
\end{equation*}
ifadesini yazabiliriz.  $\xi$%
\glossary{$\xi$} Yunan harf{}i, ordinal de\u gi\c sken olacakt\i r.
\"Ozel olarak
\begin{equation*}
\{\xi\colon\phi(\xi)\}=\{x\colon x\in\on\land\phi(x)\}.
\end{equation*}

\begin{theorem}\label{thm:'}
$\alpha'=\min\{\xi\colon\alpha<\xi\}$, yani her ordinal i\c cin
  daha b\"uy\"uk ordinaller s\i n\i f\i n\i n en k\"u\c c\"uk
  eleman\i, ordinalin ard\i l\i d\i
  r.
\end{theorem}

\ktk

E\u ger $\alpha$ bo\c s veya ard\i l de\u gilse, ve $\beta\in\alpha$ ise, o zaman $\beta'<\alpha$ olmal\i d\i r.  Bu durumda, $\alpha$ ordinaline 
\textbf{limit}%
\index{limit}%
\index{ordinal!limit}
denir.  \"Orne\u gin $\upomega$, bir limittir.

\begin{theorem}
$\upomega$, hem limit olmayan hem limit i\c cermeyen ordinaller s\i n\i f\i d\i r.  Yani
\begin{equation}\label{eqn:omega}
\upomega=
\{\xi\colon(\xi=0\lor\Exists yy'=\xi)
\land\Forall z(z\in \xi\lto z=0\lor\Exists yy'=z)\}.
\end{equation}
\end{theorem}

\begin{proof}
T\"umevar\i mla her do\u gal say\i, ne limittir ne limit i\c cerir.
\"Ote yandan, e\u ger $\alpha'$ ard\i l\i, hi\c c limit i\c cermezse, o
zaman $\alpha$ ordinal de, hi\c c limit i\c cermez.  \"Oyleyse en
k\"u\c c\"uk limit olmayan, limit i\c cermeyen, do\u gal
say\i\ olmayan ordinal yoktur.  O zaman hi\c c \"oyle ordinaller
yoktur. 
\end{proof}

Bu teoremin kan\i t\i, Sonsuzluk Aksiyomunu kullanmaz, dolay\i s\i yla
\eqref{eqn:omega} e\c sitli\u gi, $\upomega$ s\i n\i f\i n\i n tan\i
m\i\ olarak kullan\i labilir.  O halde \pageref{PA}
numaral\i\ sayfadaki Peano Aksiyomlar\i\ yeniden kan\i tlanmal\i d\i r.

\section{\"Ozyineleme}

$\upomega$ k\"umesinde toplama, bir \textbf{ikili i\c slem}%
\index{i\c slem}
olacak, yani $\upomega\times\upomega$ \c carp\i m\i ndan $\upomega$ k\"umesine giden bir g\"onderme.  Bu i\c slem,
\begin{equation*}
(x,y)\mapsto x+y
\end{equation*}
olarak yaz\i l\i r.  O zaman her $k$ do\u gal say\i s\i\ i\c cin bir
$x\mapsto k+x$ \textbf{birli i\c slemi} olacakt\i r.  Bu i\c slemin
\"ozelliklerinden ikisi, 
\begin{align}\label{eqn:+}
k+0&=k,&\Forall x\bigl(x\in\upomega\lto k+x'&=(k+x)'\bigr)
\end{align}
olacakt\i r.  Asl\i nda $\upomega$ k\"umesindeki birli i\c slemlerden en \c cok birinin bu \"ozellikleri vard\i r.  \c C\"unk\"u $f\colon\upomega\to\upomega$, $f(0)=k$, ve $\Forall x\bigl(x\in\upomega\lto f(x')=f(x)'\bigr)$ olsun.  O zaman $f(0)=k+0$, ve $f(m)=k+m$ ise $f(m')=f(m)'=(k+m)'=k+m'$.  T\"umevar\i mla her $n$ do\u gal say\i s\i\ i\c cin $f(n)=k+n$.

Neden $\upomega$ k\"umesindeki birli i\c slemlerden \emph{en az} birinin \eqref{eqn:+} sat\i r\i ndaki \"ozellikleri vard\i r?  $k=0$ durumunda her $n$ i\c cin $k+n=n$ olsun.  O zaman $k+0=0$, ve $k+m'=m'=(k+m)'$.  \"Ustelik $k=\ell$ durumunda \eqref{eqn:+} sat\i r\i ndaki gibi $x\mapsto k+x$ i\c slemi varsa $\ell'+n=(\ell+n)'$ olsun.  O zaman $\ell'+0=(\ell+0)'=\ell'$, ve $\ell'+m'=(\ell+m')'=(\ell+m)''=(\ell'+m)'$.  Yani $k=\ell'$ durumunda \eqref{eqn:+} sat\i r\i ndaki gibi $x\mapsto k+x$ i\c slemi vard\i r.

T\"umevar\i mla $\upomega$ k\"umesindeki her $k$ i\c cin \eqref{eqn:+}
sat\i r\i ndaki gibi $x\mapsto k+x$ i\c sleminin oldu\u gu sonucuna
varabilir miyiz?  T\"umevar\i mla bir \emph{k\"umenin} $\upomega$
k\"umesine e\c sit oldu\u gu kan\i tlanabilir.  \c Simdiki durumda
hangi k\"ume, $\upomega$ k\"umesine e\c sit olmal\i d\i r?
M\"umk\"unse $a$, $\upomega$ k\"umesinin \"oyle $k$
elemanlar\i\ taraf\i ndan olu\c sturulsun ki \eqref{eqn:+} sat\i r\i
ndaki \"ozelliklerini sa\u glayan bir i\c slem olsun.  O halde
g\"osterdi\u gimiz gibi $a=\upomega$ olmal\i d\i r.  Ama \"oyle bir
$a$ k\"umesi var m\i d\i r?  Hangi form\"ul, bu k\"umeyi tan\i
mlayabilir? 

\pageref{yerlestirme} ve \pageref{Yer} numaral\i\ sayfalardaki Yerle\c stirme Aksiyomuna g\"ore bir k\"umede birli bir i\c slemin kendisi, bir k\"umedir.  O halde istedi\u gimiz $a$ k\"umesi tan\i mlanabilir, dolay\i s\i yla $\upomega$ k\"umesindeki toplaman\i n kendisi tan\i mlanabilir.  Asl\i nda \eqref{eqn:+} sat\i r\i ndaki \"ozellikleri, toplaman\i n 
\textbf{\"ozyineli tan\i m\i n\i}%
\index{\"ozyineli tan\i m}
(\eng{recursive definition})
sa\u glar.

Benzer \c sekilde her $k$ do\u gal say\i s\i\ i\c cin
\begin{align}\label{eqn:.}
k\cdot0&=0,&\Forall x(x\in\upomega\lto k\cdot x'&=k\cdot x+k)
\end{align}
\"ozellikleri olan $x\mapsto k\cdot x$ i\c slemi vard\i r.  (Burada tabii ki $k\cdot x+k=(k\cdot x)+k$.)  \c C\"unk\"u $0\cdot n=0$ ise $0\cdot0=0$ ve $0\cdot m'=0=0+0=0\cdot m+0$.  Ayr\i ca istedi\u gimiz gibi $x\mapsto\ell\cdot x$ varsa $\ell'\cdot n=\ell\cdot n+n$ olsun.  O halde
\begin{align*}
\ell'\cdot m'
&=\ell\cdot m'+m'\\
&=(\ell\cdot m+\ell)+m'\\
&=\ell\cdot m+(\ell+m')\\
&=\ell\cdot m+(\ell+m)'\\
&=\ell\cdot m+(m+\ell)'\\
&=\ell\cdot m+(m+\ell')\\
&=(\ell\cdot m+m)+\ell'\\
&=\ell'\cdot m+\ell'.
\end{align*}
Ama burada toplaman\i n birle\c sme ve de\u gi\c sme \"ozelliklerini kulland\i k; bunlar kan\i tlanmal\i d\i r.

Buraya kadar gelmek i\c cin t\"umevar\i m yeter.  Yani \pageref{PA} numaral\i\ sayfadaki ilk \"u\c c Peano Aksiyomu yeter.  Say\i lar teorisinde, her pozitif $n$ mod\"ul\"use g\"ore tamsay\i lar, bu aksiyomlar\i\ sa\u glar.  Yani e\u ger bir $a$ k\"umesinin elemanlar\i\, tamsay\i\ ise, ve
\begin{equation*}
x\equiv 0\pmod n
\end{equation*}
denkli\u ginin $a$ k\"umesinden \c c\"oz\"um\"u varsa, ve ayr\i ca her $\ell$ tamsay\i s\i\ i\c cin
\begin{equation*}
x\equiv \ell\lto y\equiv \ell'\pmod n
\end{equation*}
kara\c st\i rmas\i n\i n $a$ k\"umesinden \c c\"oz\"um\"u varsa, o zaman her $k$ tamsay\i s\i\ i\c cin
\begin{equation*}
x\equiv k\pmod n
\end{equation*}
denkli\u ginin $a$ k\"umesinden \c c\"oz\"um\"u vard\i r.  \"Orne\u gin $p$, bir asal say\i\ olsun.  O zaman
\begin{equation*}
0^p\equiv0\pmod p,
\end{equation*}
ve
\begin{equation*}
a^p\equiv a\lto(a+1)^p\equiv a+1\pmod p,
\end{equation*}
\c c\"unk\"u
\begin{align*}
(a+1)^p&\equiv a^p+pa^{p-1}+\binom p2a^{p-2}+\dots+\binom p{p-2}a^2+pa+1\\
       &\equiv a^p+1\pmod p.
\end{align*}
Sonu\c c olarak Fermat'n\i n Teoremi%
\index{teorem!Fermat'n\i n T---i}
do\u grudur, yani her $p$ asal say\i s\i\ i\c cin, her $a$ tamsay\i
s\i\ i\c cin\footnote{Gauss'a \cite[\P50]{Gauss} g\"ore verdi\u gimiz
  kan\i t, Euler'indir.} 
\begin{equation*}
a^p\equiv a\pmod p.
\end{equation*}
Ayn\i\ sebeple t\"um $a$, $b$, $c$, ve $d$ tamsay\i lar\i\ i\c cin, her pozitif $n$ say\i s\i\ i\c cin
\begin{equation*}
a\equiv b\land c\equiv d\lto a+c\equiv b+d\land a\cdot c\equiv b\cdot d\pmod n.
\end{equation*}

Sadece t\"umevar\i m\i\ kullanarak $(x,y)\mapsto x^y$ ikili \"ustel i\c slemi tan\i mlanabilir mi?  \"Ozyineli tan\i m varsa $\upomega$ k\"umesindeki her $k$ i\c cin
\begin{align}\label{eqn:exp}
k^0&=1,&\Forall x(x\in\upomega\lto k^{x'}=k^x\cdot k).
\end{align}
\"Ozel olarak $0^0=1$, ama $n>0$ ise $0^n=0$.  \"Oyleyse
$0\equiv n$ ama $0^0\not\equiv 0^n\pmod n$.  \"Ustel i\c slem i\c cin
t\"umevar\i m yetmez.\footnote{\cite{Pierce-IR} makalesine bak\i n.}

\begin{theorem}[\"Ozyineleme \protect{[\eng{Recursion}]}]%
\index{teorem!\"Ozyineleme T---i}\label{thm:rec}
$\bm A$, bir s\i n\i f olsun, ve $b\in\bm A$ ile $\bm F\colon\bm A\to\bm A$ olsun.  O zaman $\upomega$ k\"umesinden $\bm A$ s\i n\i f\i na giden
\begin{align*}
\bm G(0)&=b,&
\Forall x\Bigl(x\in\upomega\lto\bm G(x')=\bm F\bigl(\bm G(x)\bigr)\Bigr)
\end{align*}
\"ozellikleri olan bir ve tek bir $\bm G$ g\"ondermesi vard\i r.
\end{theorem}

\begin{proof}%[Kan\i t\i n fikri.]
T\"umevar\i mla en \c cok bir $\bm G$ g\"ondermesi vard\i r.  En az biri varsa, ba\u g\i nt\i\ olarak, $\upomega\times\bm A$ \c carp\i m\i n\i n alts\i n\i f\i d\i r, ve her $(\ell,d)$ eleman\i\ i\c cin,
\begin{compactitem}
\item
ya $(\ell,d)=(0,b)$,
\item
ya da bir $(k,c)$ eleman\i\ i\c cin, $k'=\ell$ ve $\bm F(c)=d$.
\end{compactitem}
Bu \"ozelli\u gi olan \emph{k\"umeler} vard\i r, mesela
\begin{align*}
&\{(0,b)\},&
&\bigl\{(0,b),\bigl(1,\bm F(b)\bigr)\bigr\},&
&\Bigl\{(0,b),\bigl(1,\bm F(b)\bigr),\Bigl(2,\bm F\bigl(\bm F(b)\bigr)\Bigr)\Bigr\},&
&\dots
\end{align*}
\"Ozelli\u gi olan k\"umelerin olu\c sturdu\u gu s\i n\i f, $\bm C$
olsun.  O zaman $\bigcup\bm C$, istedi\u gimiz $\bm G$ g\"ondermesi
olacakt\i r.  Bunu g\"ostermek i\c cin, t\"um Peano
Aksiyomlar\i\ kullan\i lmal\i d\i r.

Hemen $\{(0,b)\}\in\bm C$, dolay\i s\i yla $(0,b)\in\bigcup\bm C$.  \c Simdi $(k,c)\in\bigcup\bm C$ varsay\i ls\i n.  O zaman $\bm C$ s\i n\i f\i n\i n bir $a$ eleman\i\ i\c cin $(k,c)\in a$.  O halde $a\cup\{(k',\bm F(c))\}\in\bm C$.  B\"oylece $(k',\bm F(c))\in\bigcup\bm C$.  T\"umevar\i mla her $k$ do\u gal say\i s\i\ i\c cin $\bm A$ s\i n\i f\i n\i n $(k,c)\in\bigcup\bm C$ i\c cerilmesini sa\u glayan $c$ eleman\i\ vard\i r, yani
\begin{equation*}
\{x\colon\Exists y(x,y)\in\bigcup\bm C\}=\upomega.
\end{equation*}
\c Simdi
\begin{equation*}
\bigl\{x\colon\Forall y\Forall z\bigl((x,y)\in\bigcup\bm C\land(x,z)\in\bigcup\bm C\lto y=z)\bigr)\bigr\}=\upomega
\end{equation*}
e\c sitli\u gini kan\i tlayaca\u g\i z.  Soldaki k\"ume, $a_0$ olsun.  E\u ger $(0,e)\in\bigcup\bm C$ ise, o zaman $\bm C$ s\i n\i f\i n\i n bir $a$ eleman\i\ i\c cin $(0,e)\in a$, dolay\i s\i yla $e=b$ olmal\i d\i r, \c c\"unk\"u $0$, ard\i l de\u gildir.  \"Oyleyse $0\in a_0$.
\c Simdi $k\in a_0$ olsun.  G\"osterdi\u gimiz gibi $\bm A$ s\i n\i f\i n\i n bir $c$ eleman\i\ i\c cin $(k,c)\in\bigcup\bm C$ ve $(k',\bm F(c))\in\bigcup\bm C$.  $(k',d)\in\bigcup\bm C$ varsay\i ls\i n.  O zaman $\bm C$ s\i n\i f\i n\i n bir $a$ eleman\i\ i\c cin $(k',d)\in a$.  O halde $a$ k\"umesinin bir $(j,e)$ eleman\i\ i\c cin $j'=k'$ ve $\bm F(e)=d$.  Bu durumda $j=k$ olmal\i d\i r.  B\"oylece $(k,e)\in\bigcup\bm C$, dolay\i s\i yla $e=c$ ve $d=\bm F(c)$, \c c\"unk\"u $k\in a_0$ varsay\i l\i r.  \"Oyleyse $k'\in a_0$.  T\"umevar\i mla $a_0=\upomega$.

Sonu\c c olarak $\bigcup\bm C$, $\upomega$ k\"umesinden $\bm A$ s\i n\i f\i na giden bir $\bm G$ g\"ondermesidir.  $\bm C$ s\i n\i f\i n\i n tan\i m\i ndan $\bm G$ g\"ondermesinin istedi\u gimiz \"ozellikleri vard\i r.
\end{proof}

\c Simdi, do\u gal say\i larda, \eqref{eqn:+}, \eqref{eqn:.}, ve
\eqref{eqn:exp} sat\i rlar\i 
ndaki b\"ut\"un tan\i mlar ge\c cerlidir.  



\chapter{Ordinaller}


\section{\"Ozyineleme}

Do\u gal say\i larda, bir g\"ondermenin \"ozyineli tan\i m\i n\i n iki
tane par\c cas\i\ vard\i r, biri $0$ i\c cin, biri ard\i llar i\c
cin.  Ordinallerde \"u\c c\"unc\"u bir par\c ca gerekir, limitler
i\c cin.

T\"um $\alpha$ ordinalleri i\c cin
\begin{align}\label{eqn:++}
  \alpha+0&=\alpha,&
\alpha+\beta'&=(\alpha+\beta)'
\end{align}
olacak.  Ama $\beta$ limitse, $\alpha+\beta$ nedir?
Mesela $\alpha+\upomega$ nedir?  Asl\i nda \ref{thm:rec}
numaral\i\ teoreme g\"ore $\upomega$ k\"umesinden $\on$ s\i n\i f\i na
giden, \eqref{eqn:++} sat\i r\i ndaki \"ozellikleri olan
$x\mapsto\alpha+x$ g\"ondermesi vard\i r.
Her $n$ do\u gal
say\i s\i\ i\c cin, 
\begin{equation*}
  \alpha+n<\alpha+\upomega
\end{equation*}
e\c sitsizli\u gini isteriz.  Yani $\alpha+\upomega$,
$\{y\colon\Exists x(x\in\upomega\land y=\alpha+x)\}$ s\i n\i f\i n\i n
\"ust s\i n\i r\i\ olmal\i d\i r (s\i n\i f\i n \"ust s\i n\i
r\i\ varsa).  Bu s\i n\i f, $\{\alpha+x\colon x\in\upomega\}$ olarak
yaz\i labilir.  

Genelde $\bm F\colon\bm A\to\bm B$ ve $\bm C\included\bm A$ ise
\begin{equation*}
  \{y\colon\Exists x(x\in\bm C\land\bm F(x)=y)\}
\end{equation*}
s\i n\i f\i,
\begin{align*}
&\{\bm F(x)\colon x\in\bm C\},&
&\bm F[\bm C]
\end{align*}
ifadelerinin biri olarak yaz\i labilir.  Bu s\i n\i f, $\bm C$ s\i n\i f\i n\i n $\bm F$ alt\i nda \textbf{g\"or\"unt\"us\"ud\"ur.}%
\index{g\"or\"unt\"u}
Bu durumda, $\bm F$ g\"ondermesi $\bm C$ s\i n\i f\i nda
\textbf{tan\i mlan\i r,}% 
\index{tan\i mlama}
\c c\"unk\"u $\bm C$, $\bm F$ g\"ondermesinin tan\i m s\i n\i
f\i\ taraf\i ndan kapsan\i r.  E\u ger $\bm F$, $\bm C$ s\i n\i f\i
nda tan\i mlanmazsa, $\bm F[\bm C]$ ifadesini yazmayaca\u g\i z. 

$\upomega$ k\"umesinin $\{\alpha+x\colon x\in\upomega\}$ g\"or\"unt\"us\"un\"un \"ust s\i n\i
r\i\ varsa, en k\"u\c c\"uk \"ust s\i n\i r\i, yani \emph{supremumu,}
vard\i r (\c c\"unk\"u $\on$, iyi s\i ralan\i r).  \c Simdi
\eqref{eqn:++} sat\i r\i ndaki \"ozelliklere g\"ore  
\begin{equation*}
  \alpha\included\alpha+1\included\alpha+2\included\dotsb
\end{equation*}
E\u ger $\bigcup\{\alpha+x\colon x\in\upomega\}$ bir ordinalse,
$\{\alpha+x\colon x\in\upomega\}$ k\"umesinin \"ust s\i n\i r\i d\i r,
asl\i nda supremumudur.

\begin{theorem}
Elemanlar\i\ ordinal olan her s\i n\i f\i n bile\c simi, ya bir
ordinal, ya da ordinallerin s\i n\i f\i d\i r.
\end{theorem}

\begin{proof}
  $\bm A\included\on$ olsun.  $b\in\bigcup\bm A$ ise, $\bm A$ s\i n\i
  f\i n\i n bir $\alpha$ eleman\i\ i\c cin $b\in\alpha$, dolay\i s\i
  yla $b\included\alpha$ ve onun i\c cin $b\included\bigcup\bm A$.
  \"Oyleyse $\bigcup\bm A$ ge\c ci\c slidir.  Ayr\i ca, $\on$ s\i n\i
  f\i\ da ge\c ci\c sli oldu\u gundan, $\bigcup\bm A\included\on$,
  dolay\i s\i yla $\bigcup\bm A$, $\in$ taraf\i ndan iyi s\i ralan\i
  r.  \"Oyleyse $\bigcup\bm A$ ya bir ordinaldir, ya da k\"ume olmayan
  bir s\i n\i ft\i r.  \.Ikinci durumda $\bigcup\bm A$ bile\c siminin
  $\on$ oldu\u gunu g\"osterece\u giz.  E\u ger $\bigcup\bm
  A\pincluded\on$ ise $\beta\in\on\setminus\bigcup\bm A$ olsun.  O
  zaman $\bigcup\bm A\included\beta$, \c c\"unk\"u $\bigcup\bm A$ ge\c
  ci\c slidir (e\u ger $\gamma\in\bigcup\bm A$ ise
  $\gamma\included\bigcup\bm A$, dolay\i s\i yla $\beta\notin\gamma$
  ve $\gamma\leq\beta$).  Bu durumda $\bigcup\bm A$, bir k\"umedir,
  dolay\i s\i yla ordinaldir. 
\end{proof}

Sonu\c c olarak, e\u ger $\bm A\included\on$ ve $\bigcup\bm A$ bir
k\"umeyse, o zaman bir ordinaldir; de\u gilse, $\on$ s\i n\i f\i d\i
r.  \c Simdi $\bm A$ bir k\"umeyse, $\bigcup\bm A$ bile\c siminin bir
k\"ume oldu\u gunu isteriz:   

\begin{axiom}[Bile\c sim]%
\index{aksiyom!Bile\c sim A---u}
  Her k\"umenin bile\c simi, bir k\"umedir:
  \begin{equation*}
    \Forall x\Exists y\Forall z\bigl(z\in y\liff\Exists w(w\in x\land
    z\in w)\bigr). 
  \end{equation*}
\end{axiom}

\begin{theorem}
  Ordinallerin olu\c sturdu\u gu her k\"umenin bile\c simi, k\"umenin
  supremumudur. 
\end{theorem}

\begin{proof}
  $a\included\on$ olsun.  Son teorem ve Bile\c sim Aksiyomuna g\"ore $\bigcup a$, bir $\alpha$
  ordinalidir.  O zaman $\alpha$, $a$ k\"umesinin bir \"ust s\i n\i r\i
  d\i r.  E\u ger $\beta<\alpha$ ise, o zaman $\beta\in\alpha$,
  dolay\i s\i yla $a$ k\"umesinin bir $\gamma$ eleman\i\ i\c cin
  $\beta\in\gamma$, yani $\beta<\gamma$.  Sonu\c c olarak $\beta$, $a$
  k\"umesinin \"ust s\i n\i r\i\ de\u gildir.  \"Oyleyse $\alpha=\sup(a)$.
\end{proof}

\c Simdi $\{\alpha+x\colon x\in\upomega\}$ gibi g\"or\"unt\"uler, k\"ume
olsun:

\begin{axiom}[Yerle\c stirme]%
\index{aksiyom!Yerle\c stirme A---u}\label{Yer}
Her g\"ondermenin tan\i m s\i n\i f\i n\i n altk\"umesinin g\"onderme alt\i nda g\"or\"unt\"us\"u, bir k\"umedir.  Yani her ikili $\phi(x,y)$
form\"ul\"u i\c cin
\begin{multline*}
\Forall w\biggl(\Forall x\Forall y\Forall z\bigl(\phi(x,y)\land\phi(x,z)\land x\in w\lto y=z\bigr)\\
\lto\Exists z\Forall y\Bigl(y\in z\liff\Exists x\bigl(x\in w\land\phi(x,y)\bigr)\Bigr)\biggr).
\end{multline*}
\end{axiom}

\c Simdi $\beta$ ordinalinde $x\mapsto\alpha+x$ g\"ondermesi tan\i
mlan\i rsa $\{\alpha+x\colon x\in\beta\}$ g\"or\"unt\"us\"u, bir
k\"umedir.  $\beta$ limitse 
\begin{equation}\label{eqn:+lim}
  \alpha+\beta=\sup\{\alpha+\xi\colon\xi<\beta\}
\end{equation}
olsun.  Bu ko\c sul, \eqref{eqn:++} sat\i r\i ndaki ko\c sullarla, $\on$ s\i n\i f\i nda $x\mapsto\alpha+x$ i\c
slemesini tan\i mlayacakt\i r.

\begin{theorem}[T\"umevar\i m]%
\index{teorem!T\"umevar\i m T---i}
$\bm A\included\on$ olsun.  E\u ger
\begin{compactenum}[1)]
\item
$0\in\bm A$,
\item
her $\alpha$ i\c cin $\alpha\in\bm A\lto\alpha'\in\bm A$,
\item
her $\alpha$ limiti i\c cin $\alpha\included\bm A\lto\alpha\in\bm A$
\end{compactenum}
ise, o zaman $\bm A=\on$.
\end{theorem}

\begin{proof}
Hipotez alt\i nda $\on\setminus\bm A$ fark\i n\i n en k\"u\c c\"uk eleman\i\ olamaz.
\end{proof}

E\u ger $\bm F\colon\bm A\to\bm B$ ve $\bm C\included\bm A$ ise
\begin{equation*}
\bm F\cap(\bm C\times\bm B)=\bm F\restriction\bm C
\end{equation*}
olsun.  Bu $\bm F\restriction\bm C$ g\"ondermesi, $\bm F$ g\"ondermesinin $\bm C$ s\i n\i f\i na \textbf{s\i n\i rlamas\i d\i r}%
\index{s\i n\i rlama}
(\eng{restriction}).

\begin{theorem}[\"Ozyineleme]%
\index{teorem!\"Ozyineleme T---i}\label{thm:rec-ord}
$\bm A$, bir s\i n\i f olsun, ve $b\in\bm A$, $\bm F\colon\bm
A\to\bm A$, ve $\bm G\colon\pow{\bm A}\to\bm A$ olsun.  O zaman
$\on$ s\i n\i f\i ndan $\bm A$ s\i n\i f\i 
na giden 
\begin{gather*}
  \bm H(0)=b,\\
\bm H(\alpha')=\bm F(\bm H(\alpha)),\\
\alpha\text{ limit ise }\bm H(\alpha)=\bm G\bigl(\{\bm
H(\xi)\colon\xi<\alpha\}\bigr) 
\end{gather*}
\"ozellikleri olan bir ve tek bir $\bm H$ g\"ondermesi vard\i r.
\end{theorem}

\begin{proof}
T\"umevar\i mla en \c cok bir $\bm H$ g\"ondermesi vard\i r.  \c C\"unk\"u $\bm H_1$ g\"ondermesinin ve $\bm H$ g\"ondermesinin \"ozellikleri ayn\i\ olsun.  O zaman
\begin{compactenum}[1)]
\item
$\bm H_1(0)=b=\bm H(0)$;
\item
$\bm H_1(\alpha)=\bm H(\alpha)$ ise
\begin{equation*}
\bm H_1(\alpha')=\bm F(\bm H_1(\alpha))=\bm F(\bm H(\alpha))=\bm H(\alpha');
\end{equation*}
\item
$\alpha$ limit ise ve $\bm H_1\restriction\alpha=\bm H\restriction\alpha$ ise
\begin{equation*}
\bm H_1(\alpha)=\bm G(\bm H_1[\alpha])=\bm G(\bm H[\alpha])=\bm H(\alpha).
\end{equation*}
\end{compactenum}
 \ref{thm:rec} numaral\i\ teoremin kan\i t\i ndaki gibi bir $\bm C$ s\i n\i f\i\ i\c cin $\bm H=\bigcup\bm C$ olacakt\i r.  Bu s\i n\i f\i n tan\i m\i na g\"ore her $a$ eleman\i\ i\c cin $a\subseteq\on\times\bm A$, ve $a$ k\"umesinin her $(\alpha,d)$ eleman\i\ i\c cin,
\begin{compactitem}
\item
ya $(\alpha,d)=(0,b)$,
\item
ya da bir $(\beta,c)$ eleman\i\ i\c cin, $\beta'=\alpha$ ve $\bm F(c)=d$,
\item
ya da $\alpha$ limit, ve $a\cap(\alpha\times\bm A)$ kesi\c simi, tan\i m k\"umesi $\alpha$ olan bir $f$ g\"ondermesi, ve $\bm G(f[\alpha])=d$.
\end{compactitem}
E\u ger $\bigcup\bm C$ bile\c simi, tan\i m s\i n\i f\i\ $\on$ olan bir g\"onderme de\u gilse, bir \emph{en k\"u\c c\"uk} $\alpha$ i\c cin
\begin{equation*}
\{x\colon x\in\bm A\land(\alpha,x)\in\bigcup\bm C\}
\end{equation*}
s\i n\i f\i n\i n ya hi\c c eleman\i\ yoktur ya da en az iki
eleman\i\ vard\i r.  O zaman $\alpha\neq0$.  E\u ger
$\alpha=\beta'$ ise, o zaman bir $c$ i\c cin $(\beta,c)\in\bigcup\bm
C$, dolay\i s\i yla $(\alpha,\bm F(c))\in\bigcup\bm C$.  Bu
durumda e\u ger $(\alpha,d)\in\bigcup\bm C$ ise bir $e$ i\c cin $d=\bm
F(e)$ ve $(\beta,e)\in\bigcup\bm C$, dolay\i s\i yla $c=e$ ve $d=\bm
F(c)$ (\c c\"unk\"u $\alpha$ en k\"u\c c\"ukt\"ur).  \"Oyleyse
$\alpha$ ard\i l olamaz.  Benzer \c sekilde $\alpha$ limit olamaz.
Sonu\c c olarak $\bigcup\bm C$ bile\c simi, tan\i m s\i n\i f\i\ $\on$
olan bir g\"onderme olmal\i d\i r.  Bu g\"ondermenin, tan\i m\i ndan
dolay\i\ istedi\u gimiz \"ozellikleri vard\i r.
\end{proof}

\section{Toplama}

Son teoreme g\"ore her $\alpha$ i\c cin \eqref{eqn:++} ve
\eqref{eqn:+lim} sat\i rlar\i ndaki ko\c sullar $\on$ s\i n\i f\i nda
$x\mapsto\alpha+x$ i\c slemini tan\i mlar.  Yani
\begin{gather*}
  \alpha+0=\alpha,\\
\alpha+\beta'=(\alpha+\beta)',\\
\beta\text{ limit}\lto\alpha+\beta=\sup\{\alpha+\xi\colon\xi<\beta\}.
\end{gather*}%
\glossary{$\alpha+\beta$}
\"Ozel olarak
\begin{equation*}
  \alpha+1=\alpha'.
\end{equation*}
O zaman
\begin{equation*}
  1+\upomega=\sup\{1+x\colon x\in\upomega\}=\upomega<\upomega+1,
\end{equation*}
ve genelde $0<n<\upomega$ ise
\begin{equation*}
n+\upomega=\upomega<\upomega+n.
\end{equation*}
B\"oylece $\on$ s\i n\i f\i nda toplama de\u gi\c smeli de\u gildir.

\begin{theorem}\label{thm:<}
  $\beta<\gamma$ ise $\alpha+\beta<\alpha+\gamma$.
\end{theorem}

\begin{proof}
  $\gamma$ \"uzerinden t\"umevar\i m kullanaca\u g\i z.  
  \begin{asparaenum}
    \item
    $\gamma=0$ ise, iddia do\u grudur, \c c\"unk\"u hi\c cbir zaman
  $\beta<0$ de\u gildir.  
\item
$\gamma=\delta$ durumda iddian\i n do\u gru oldu\u gu varsay\i ls\i n.  E\u ger $\beta<\delta'$ ise, o zaman $\beta\leq\delta$, dolay\i s\i yla
  \begin{equation*}
    \alpha+\beta\leq\alpha+\delta<(\alpha+\delta)'=\alpha+\delta'.
  \end{equation*}
\item
$\delta$ limit, $\gamma<\delta$ durumunda iddia do\u gru, ve $\beta<\delta$ ise, o zaman $\beta<\beta'<\delta$, dolay\i s\i yla
\begin{equation*}
\alpha+\beta<\alpha+\beta'\leq\sup_{\xi<\delta}(\alpha+\xi)=\alpha+\delta.\qedhere
\end{equation*}
  \end{asparaenum}
\end{proof}

\begin{theorem}\label{thm:leq}
  $\beta\leq\gamma$ ise $\beta+\alpha\leq\gamma+\alpha$.
\end{theorem}

\begin{proof}
  \c Simdi $\alpha$ \"uzerinden t\"umevar\i m kullanaca\u g\i z.
  $\beta\leq\gamma$ olsun.
  \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, \ref{thm:'}
numaral\i\ teoreme g\"ore
\begin{equation*}
  \beta+\alpha'
  =(\beta+\alpha)'\leq\gamma+\alpha<(\gamma+\alpha)'=\gamma+\alpha'.  
\end{equation*}
\item
$\delta$ limit olsun, ve $\alpha<\delta$ ise,
  $\beta+\alpha\leq\gamma+\alpha$ olsun.  O zaman
  \begin{equation*}
    \beta+\delta=\sup_{\xi<\gamma}(\beta+\xi)
\leq\sup_{\xi<\gamma}(\gamma+\xi)=\gamma+\delta.\qedhere
  \end{equation*}
  \end{asparaenum}
\end{proof}

G\"ord\"u\u g\"um\"uz gibi ayn\i\ zamanda $\beta<\gamma$ ama
$\beta+\alpha=\gamma+\alpha$ olabilir, mesela $0<k<\ell<\upomega$ ise $k+\upomega=\ell+\upomega$.

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

\ktk

\begin{theorem}
  $\alpha\leq\beta$ ise
  \begin{equation*}
    \alpha+\xi=\beta
  \end{equation*}
denkleminin bir ve tek bir \c c\"oz\"um\"u vard\i r.  
Yani $\alpha+x=\beta$ denkleminin bir ve tek bir ordinal \c c\"oz\"um\"u vard\i r.
\end{theorem}

\begin{proof}
  \ref{thm:<} numaral\i\ teoreme g\"ore denklemin en \c cok bir \c
  c\"oz\"um\"u vard\i r.  \ref{thm:leq} ve \ref{thm:0+}
  numaral\i\ teoremlere g\"ore
  \begin{equation*}
    \alpha+\beta\geq0+\beta=\beta,
  \end{equation*}
dolay\i s\i yla $\{\xi\colon\beta\leq\alpha+\xi\}$ bo\c s de\u
gildir (\c c\"unk\"u $\beta$ eleman\i n\i\ i\c cerir).  En k\"u\c c\"uk eleman\i, $\delta$ olsun.  \"U\c c durum
vard\i r.
\begin{asparaenum}
  \item
$\delta=0$ ise
\begin{equation*}
  \beta\leq\alpha+\delta=\alpha,
\end{equation*}
dolay\i s\i yla $\beta=\alpha=\alpha+\delta$ (\c c\"unk\"u
$\alpha\leq\beta$ da sa\u glan\i r).
\item
$\delta=\gamma'$ ise $\alpha+\gamma<\beta$, dolay\i s\i yla
  \begin{align*}
    \alpha+\delta&=(\alpha+\gamma)'\leq\beta,&
\alpha+\delta&=\beta
  \end{align*}
(\c c\"unk\"u $\alpha+\delta\geq\beta$ da sa\u glan\i r).
\item
$\delta$ limit olsun.  E\u ger $\gamma<\delta$ ise
  $\alpha+\gamma<\beta$ olmal\i d\i r.  O zaman
  \begin{align*}
    \alpha+\delta&=\sup_{\xi<\delta}(\alpha+\xi)\leq\beta,&
\alpha+\delta&=\beta.\qedhere
  \end{align*}
\end{asparaenum}
\end{proof}

$\alpha\leq\beta$ durumunda $\alpha+\xi=\beta$ denkleminin \c c\"oz\"um\"u i\c cin
\begin{equation*}
\beta-\alpha
\end{equation*}
ifadesi yaz\i labilir.  \"Orne\u gin $\alpha'-1=\alpha$.

\begin{theorem}\label{thm:lim}
  $\beta$ limitse $\alpha+\beta$ toplam\i\ da limittir.
\end{theorem}

\ktk

\begin{theorem}
  T\"um $\alpha$, $\beta$, ve $\gamma$ i\c cin
  \begin{equation*}
    \alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma.
  \end{equation*}
\end{theorem}

\begin{proof}
  $\gamma$ \"uzerinden t\"umevar\i m kullanaca\u g\i z.
  \begin{asparaenum}
    \item
$\alpha+(\beta+0)=\alpha+\beta=(\alpha+\beta)+0$.
\item
$\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$ ise
  \begin{align*}
    \alpha+(\beta+\gamma')=\alpha+(\beta+\gamma)'
&=(\alpha+(\beta+\gamma))'\\
&=((\alpha+\beta)+\gamma)'
=(\alpha+\beta)+\gamma'.
  \end{align*}
\item
$\delta$ limit olsun, ve $\gamma<\delta$ ise
  $\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$ olsun.  Ama
  $\gamma<\delta$ ise
  \begin{align*}
    \beta+\gamma&<\beta+\delta,&
    \alpha+(\beta+\gamma)&<\alpha+(\beta+\delta).
  \end{align*}
\"Oyleyse
\begin{equation*}
  (\alpha+\beta)+\delta
=\sup_{\xi<\delta}\bigl((\alpha+\beta)+\xi\bigr)
=\sup_{\xi<\delta}\bigl(\alpha+(\beta+\xi)\bigr)
\leq\alpha+(\beta+\delta).
\end{equation*}
Ayr\i ca, \ref{thm:lim} numaral\i\ teoreme g\"ore
$\beta+\delta$ limit oldu\u gundan
\begin{equation*}
  \alpha+(\beta+\delta)=\sup_{\xi<\beta+\delta}(\alpha+\xi).
\end{equation*}
Dahas\i\ $\theta<\beta+\delta$ ise, bir $\gamma$ i\c cin
$\gamma<\delta$ ve $\theta<\beta+\gamma$ (\c c\"unk\"u $\beta+\delta=\sup\{\beta+\xi\colon\xi<\delta\}$).  O halde
\begin{equation*}
  \sup_{\xi<\beta+\delta}(\alpha+\xi)\leq\sup_{\xi<\delta}(\alpha+(\beta+\xi))
=(\alpha+\beta)+\delta.
\end{equation*}
Sonu\c c olarak $\alpha+(\beta+\delta)=(\alpha+\beta)+\delta$.\qedhere 
  \end{asparaenum}
\end{proof}

Asl\i nda son teoremin kan\i t\i, genel bir y\"ontemin \"orne\u
gidir.  $\on$ s\i n\i f\i nda $x\mapsto\alpha+x$ i\c slemi,
\begin{compactenum}[1)]
  \item
$\beta<\gamma\lto\bm F(\beta)<\bm F(\gamma)$,
\item
$\gamma$ limitse $\bm F(\gamma)=\sup\{\bm F(\xi)\colon\xi<\gamma\}$
\end{compactenum}
ko\c sullar\i n\i\ sa\u glayan bir $\bm F$ i\c slemidir.  Bu tip herhangi
bir i\c sleme \textbf{normal}%
\index{normal!--- i\c slem}
(\eng{normal}) denir.  \.Ilk ko\c sula g\"ore $\bm F$, \textbf{kesin
  artan}%
\index{artan g\"onderme}
(\eng{strictly increasing})
bir g\"ondermedir.  O zaman ikinci ko\c sula g\"ore $\bm F$,
\textbf{s\"ureklidir}%
\index{s\"urekli g\"onderme}
(\eng{continuous}).

\begin{theorem}
  E\u ger $0\pincluded a\pincluded\on$ ve $\sup(a)\notin a$, o zaman
  $\sup(a)$, bir limittir.
\end{theorem}

\ktk

\begin{theorem}
  $\bm F$, $\on$ s\i n\i f\i nda normal bir i\c slem olsun.  T\"um
  bo\c s olmayan ordinallerin olu\c sturdu\u gu $a$ k\"umeleri i\c cin
  \begin{equation*}
    \bm F(\sup(a))=\sup(\bm F[a]),
  \end{equation*}
  yani
  \begin{equation*}
\bm F(\sup_{x\in a}x)=\sup_{x\in a}\bm F(x).
\end{equation*}
\end{theorem}

\begin{proof}
  $\sup(a)=\beta$ olsun.  E\u ger $\beta\in a$ ise, $\bm F$ i\c
  slemi artan oldu\u gundan
  \begin{equation*}
    \bm F(\beta)=\sup_{x\in a}\bm F(x).
  \end{equation*}
\c Simdi $\beta\notin a$ olsun.  Son teoreme g\"ore $\beta$
bir limit olmal\i d\i r.  Ayr\i ca $a\included\beta$, dolay\i s\i yla
\begin{equation*}
  \sup(\bm F[a])\leq\sup(\bm F[\beta])=\bm F(\beta)
\end{equation*}
(\c c\"unk\"u $\bm F$ s\"ureklidir).  Dahas\i\ $\alpha<\beta$ ise, $a$
k\"umesinin bir $\gamma$ eleman\i\ i\c cin $\alpha<\gamma$, dolay\i s\i yla
\begin{equation*}
  \bm F(\beta)=\sup(\bm F[\beta])\leq\sup(\bm F[a])
\end{equation*}
(yine $\bm F$ s\"urekli oldu\u gu i\c cin).
\end{proof}

Bu teoremde $a=0$ ise $\sup(\bm F[a])=\sup(0)=0$, ama $\bm F$ normal olunca, $\bm F(0)>0$
olabilir. 

Bir $\delta$ limiti i\c cin, e\u ger
\begin{equation*}
  \gamma<\delta\lto\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma
\end{equation*}
ise, o zaman teoreme g\"ore
\begin{align*}
  \alpha+(\beta+\delta)
&=\alpha+\sup_{\xi<\delta}(\beta+\xi)\\
&=\sup_{\xi<\delta}(\alpha+(\beta+\xi))
=\sup_{\xi<\delta}((\alpha+\beta)+\xi)
=(\alpha+\beta)+\delta,
\end{align*}
\c c\"unk\"u $x\mapsto\alpha+x$, normaldir.

\section{\c Carpma}

Her $\alpha$ i\c cin $\on$ s\i n\i f\i nda $x\mapsto\alpha\cdot x$ i\c slemi, tan\i m\i na g\"ore,
\begin{gather*}
	\alpha\cdot0=0,\\
	\alpha\cdot\beta'=\alpha\cdot\beta+\alpha,\\
	\gamma\text{ limit}\lto\alpha\cdot\gamma=\sup\{\alpha\cdot\xi\colon\xi<\gamma\}
\end{gather*}%
\glossary{$\alpha\cdot\beta$}
ko\c sullar\i\ sa\u glar.  \"Ozel olarak
\begin{equation*}
\alpha\cdot1=\alpha.
\end{equation*}
\"Orne\u gin
\begin{gather*}
	\upomega\cdot2 =\upomega\cdot1+\upomega=\upomega+\upomega=\sup_{x\in\upomega}(\upomega+x),\\
	2\cdot\upomega=\sup_{x\in\upomega}(2\cdot x)=\upomega,
\end{gather*}
dolay\i s\i yla
\begin{equation*}
  	2\cdot\upomega<\upomega\cdot2.
\end{equation*}

\begin{theorem}
$0\cdot\alpha=0$ ve $1\cdot\alpha=\alpha$.
\end{theorem}

\ktk

$\alpha\geq1$ ise, $\xi\mapsto\alpha\cdot\xi$ i\c sleminin normal oldu\u gunu kan\i tlayaca\u g\i z.  Bu i\c slem kesin artan ise, tan\i mdan dolay\i\ s\"ureklidir, dolay\i s\i yla normaldir.  O zaman \ref{thm:<} numaral\i\ teorem gibi bir teorem yeterli olacakt\i r, \c c\"unk\"u\ \ref{thm:<} numaral\i\ teorem, genel bir y\"ontem g\"osterir:

\begin{theorem}
E\u ger $\bm F\colon\on\to\on$, ve t\"um $\alpha$ i\c cin
\begin{equation*}
\bm F(\alpha)<\bm F(\alpha'),
\end{equation*}
ve limit olan t\"um $\beta$ i\c cin
\begin{equation*}
\bm F(\beta)=\sup_{\xi<\beta}\bm F(\xi)
\end{equation*}
ise, o zaman $\bm F$ normaldir.
\end{theorem}

\ktk

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

\ktk

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

\begin{proof}
$\gamma=0$ durumunda kan\i t kolayd\i r.  $\gamma=\delta$ durumunda iddia do\u gruysa
\begin{align*}
\alpha\cdot(\beta+\delta')
&=\alpha\cdot(\beta+\delta)'\\
&=\alpha\cdot(\beta+\delta)+\alpha\\
&=(\alpha\cdot\beta+\alpha\cdot\delta)+\alpha\\
&=\alpha\cdot\beta+(\alpha\cdot\delta+\alpha)\\
&=\alpha\cdot\beta+\alpha\cdot\delta';
\end{align*}
b\"oylece $\gamma=\delta'$ durumunda iddia do\u grudur.  Son olarak $\delta$ limit, ve $\gamma<\delta$ durumunda iddia do\u gru olsun.  $\gamma=\delta$ durumunu kan\i tlayaca\u g\i z.  $\alpha=0$ ise iddia kolayd\i r.  $\alpha\geq1$ olsun.  O zaman $\xi\mapsto\alpha\cdot\xi$ ve $\xi\mapsto\alpha\cdot\beta+\xi$ i\c slemleri normal oldu\u gundan
\begin{align*}
\alpha\cdot(\beta+\delta)
&=\alpha\cdot\sup_{\xi<\delta}(\beta+\xi)\\
&=\sup_{\xi<\delta}(\alpha\cdot(\beta+\xi))\\
&=\sup_{\xi<\delta}(\alpha\cdot\beta+\alpha\cdot\xi)\\
&=\alpha\cdot\beta+\sup_{\xi<\delta}(\alpha\cdot\xi)
=\alpha\cdot\beta+\alpha\cdot\delta.\qedhere
\end{align*}
\end{proof}

G\"ord\"u\u g\"um\"uz gibi $2\cdot\upomega<\upomega+\upomega$,
dolay\i s\i yla
\begin{equation*}
(1+1)\cdot\upomega<1\cdot\upomega+1\cdot\upomega.  
\end{equation*}

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

\ktk

\begin{theorem}
$\beta\leq\gamma$ ise $\beta\cdot\alpha\leq\gamma\cdot\alpha$.
\end{theorem}

\ktk

\c Simdi $\xi$ gibi $\eta$,%
\glossary{$\eta$} 
ordinal de\u gi\c sken olsun.

\begin{theorem}
$1\leq\alpha$ ise
\begin{equation*}
\alpha\cdot\xi+\eta=\beta\land\eta<\alpha
\end{equation*}
sisteminin bir ve tek bir \c c\"oz\"um\"u vard\i r.
\end{theorem}

\begin{proof}
$\alpha\cdot\beta\geq1\cdot\beta=\beta$ ve $\xi\mapsto\alpha\cdot\xi$ artan oldu\u gundan
\begin{equation*}
\{\xi\colon\alpha\cdot\xi\leq\beta\}\included\beta',
\end{equation*}
dolay\i s\i yla $\bigcup\{\xi\colon\alpha\cdot\xi\leq\beta\}$, bir $\gamma$ ordinalidir.  B\"oylece
\begin{equation*}
\gamma=\sup\{\xi\colon\alpha\cdot\xi\leq\beta\}.
\end{equation*}
E\u ger $\gamma\in\{\xi\colon\alpha\cdot\xi\leq\beta\}$ ise $\alpha\cdot\gamma\leq\beta$.  De\u gilse $\gamma$ limit olmal\i d\i r,
dolay\i s\i yla
\begin{equation*}
\alpha\cdot\gamma=\sup_{\xi<\gamma}(\alpha\cdot\xi)\leq\beta.
\end{equation*}
\c Simdi $\alpha\cdot\gamma+\eta=\beta$ denkleminin $\delta$ \c c\"oz\"um\"u vard\i r.  E\u ger $\delta\geq\alpha$ ise $\alpha+\xi=\delta$ denkleminin bir $\theta$ \c c\"oz\"um\"u vard\i r, dolay\i s\i yla
\begin{gather*}
	\beta =\alpha\cdot\gamma+\delta =\alpha\cdot\gamma+\alpha+\theta =\alpha\cdot\gamma'+\theta,\\
	\gamma'\in\{\xi\colon\alpha\cdot\xi\leq\beta\},\\
	\gamma'\leq\gamma,
\end{gather*}
ki bu imk\^ans\i zd\i r.  \"Oyleyse $\delta<\alpha$, ve $(\gamma,\delta)$, istedi\u gimiz \c c\"oz\"umd\"ur.  Benzer \c sekilde ba\c ska \c c\"oz\"um yoktur, \c c\"unk\"u
\begin{align*}
\alpha\cdot\gamma+\delta&=\alpha\cdot\gamma_1+\delta_1,&
\gamma<\gamma_1
\end{align*}
ise $\gamma'\leq\gamma_1$, dolay\i s\i yla bir $\theta$ i\c cin
\begin{gather*}
	\gamma'+\theta=\gamma_1,\\
\alpha\cdot\gamma+\delta =\alpha\cdot\gamma_1+\delta_1 =\alpha\cdot\gamma+\alpha+\alpha\cdot\theta+\delta_1,\\
	\delta=\alpha+\alpha\cdot\theta+\delta_1\geq\alpha.\qedhere
\end{gather*}
\end{proof}

\section{Kuvvet alma}

Her $\alpha$ i\c cin, $\alpha>0$ ise, $\on$ s\i n\i f\i nda $x\mapsto\alpha^x$ i\c slemi, tan\i m\i na g\"ore,
\begin{gather*}
	\alpha^0=1,\\
	\alpha^{\beta'}=\alpha^\beta\cdot\alpha,\\
	\gamma\text{ limit}\lto\alpha^\gamma=\sup\{\alpha^{\xi}\colon\xi<\gamma\}
\end{gather*}%
\glossary{$\alpha^{\beta}$}
ko\c sullar\i n\i\ sa\u glar.  \"Ozel olarak
\begin{equation*}
\alpha^1=\alpha.
\end{equation*}
Ayr\i ca, tan\i ma g\"ore,
\begin{align*}
0^0&=1,&\beta>0&\lto 0^{\beta}=0.
\end{align*}
\"Oyleyse $\gamma$ limit ise $0^{\gamma}$ kuvveti, $\sup\{0^{\xi}\colon\xi<\gamma\}$ de\u gildir, ama
\begin{equation*}
0^{\gamma}=\sup\{0^{\xi}\colon 0<\xi<\gamma\}.
\end{equation*}

\begin{theorem}
$1^{\alpha}=1$.
\end{theorem}

\ktk

\begin{theorem}
$\alpha\leq\beta$ ise $\alpha^{\gamma}\leq\beta^{\gamma}$.
\end{theorem}

\ktk

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

\ktk

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

\ktk

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

\ktk

\.Ilkokuldan bildi\u gimiz gibi, e\u ger $2\leq t<\upomega$ ve $1\leq a<\upomega$ ise, o zaman bir $n$ do\u gal say\i s\i\ i\c cin, $n+1$ tane $a_0$, $a_1$, \dots, $a_n$ do\u gal say\i s\i\ i\c cin
\begin{align*}
\{a_0,a_1,\dots,a_n\}&\included t,&
a_0&\neq0,
\end{align*}
ve
\begin{equation*}
a=t^n\cdot a_0+t^{n-1}\cdot a_1+\dots+t^0\cdot a_n.
\end{equation*}
O zaman $a$ say\i s\i,
\begin{align*}
&a_0a_1\dots a_n&
&\text{ veya }&
&(a_0a_1\dots a_n)_t
\end{align*}
olarak yaz\i labilir; bu ifade, $a$ say\i s\i n\i n 
$t$ \textbf{taban\i nda yaz\i l\i m\i d\i r}%
\index{taban}\index{yaz\i l\i m}
\cite[17.\ b\"ol.]{Nesin-AKK}
(\eng{base-$t$ numeral}).  $0$ olan $a_i$ 
rakamlar\i%
\index{rakam}
(\eng{digits}) 
\c c\i kart\i l\i rsa, $\upomega$ k\"umesinin bir $m$ eleman\i\ i\c cin, $\upomega$ k\"umesinin
\begin{align*}
b_0&>b_1>\dots>b_m,&
\{c_0,c_1,\dots,c_m\}&\included\{1,\dots,t-1\},
\end{align*}
ve
\begin{equation*}
a=t^{b_0}\cdot c_0+t^{b_1}\cdot c_1+\dots+t^{b_m}\cdot c_m
\end{equation*}
ko\c sullar\i n\i\ sa\u glayan $b_i$ ve $c_i$ elemanlar\i\ vard\i r.  B\"oylece $a$ say\i s\i,
\begin{equation*}
\{(b_0,c_0),(b_1,c_1),\dots,(b_m,c_m)\}
\end{equation*}
g\"ondermesini belirtir.  G\"osterece\u gimiz gibi her $0$ olmayan ordinal, b\"oyle bir g\"onderme belirtir.

\begin{theorem}\label{thm:inc}
E\u ger $\bm F$, $\on$ s\i n\i f\i nda kesin artan bir
i\c slemse, t\"um $\alpha$ i\c cin 
\begin{equation*}
\alpha\leq\bm F(\alpha).
\end{equation*}
\end{theorem}

\begin{proof}
$\alpha>\bm F(\alpha)$ ise, $\bm F$ kesin artan oldu\u gundan $\bm
  F(\alpha)>\bm F(\bm F(\alpha))$, dolay\i s\i yla $\{\xi\colon\xi>\bm
  F(\xi)\}$ s\i n\i f\i n\i n en k\"u\c c\"uk eleman\i\ yoktur.  $\on$
  iyi s\i ralanm\i\c s oldu\u gundan $\{\xi\colon\xi>\bm F(\xi)\}$ s\i
  n\i f\i\ bo\c s olmal\i d\i r. 
\end{proof}

\"Orne\u gin $\alpha>1$ ise $\alpha^{\beta}\geq\beta$, dolay\i s\i yla
sonraki teorem kan\i tlanabilir.  \c Simdi $\xi$ ve $\eta$ gibi
$\zeta$,%
\glossary{$\zeta$} 
ordinal de\u gi\c sken olsun. 

\begin{theorem}\label{thm:exp-sol}
$\alpha\geq2$ ise her $0$ olmayan $\beta$ i\c cin
\begin{equation*}
\alpha^{\xi}\cdot\eta+\zeta=\beta\land\eta<\alpha\land\zeta<\alpha^{\xi}
\end{equation*}
sisteminin bir ve tek bir $(\gamma,\delta,\theta)$ \c c\"oz\"um\"u vard\i r, ve ayr\i ca $\gamma\leq\beta$.
\end{theorem}

\ktk

Sonu\c c olarak $\alpha>1$ ise, her $0$ olmayan $\beta$ i\c cin
\begin{equation*}
\gamma_0>\gamma_1>\dotsb,
\end{equation*}
ve
\begin{align*}
0<\delta_0&<\alpha,&
0<\delta_1&<\alpha,&
&\dots,
\end{align*}
ve
\begin{equation*}
\beta=\alpha^{\gamma_0}\cdot\delta_0+\alpha^{\gamma_1}\cdot\delta_1+\dotsb
\end{equation*}
ko\c sullar\i n\i\ sa\u glayan $\gamma_i$ ve $\delta_i$ ordinalleri vard\i r.  Ayr\i ca, $\on$ iyi s\i ralanm\i\c s oldu\u gundan, kesin azalan $(\gamma_0,\gamma_1,\dots)$ dizisi sona ermelidir.  Yani bir $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
\beta=\alpha^{\gamma_0}\cdot\delta_0+\alpha^{\gamma_1}\cdot\delta_1+\dots+\alpha^{\gamma_n}\cdot\delta_n.
\end{equation*}
Buradaki $\{(\gamma_0,\delta_0),(\gamma_1,\delta_1),\dots,(\gamma_n,\delta_n)\}$ k\"umesine, sonraki teoremde
\begin{equation*}
\beta_{\alpha}
\end{equation*}%
\glossary{$\beta_{\alpha}$}
ad\i\ verilecek.\footnote{Bu $\beta_{\alpha}$ ifadesi, benimdir; ba\c
  ska kitaplarda g\"ormedim.}  Genel olarak bu k\"umeyi tan\i mlamak
i\c cin (yani $\xi\mapsto\xi_{\alpha}$ g\"ondermesini tan\i mlamak i\c
cin), \"ozyinelemeyi \ref{thm:rec-ord} numaral\i\ teoremden
farkl\i\ bir \c sekilde kullanaca\u g\i z. 

\begin{theorem}\label{thm:gonderme}
Tan\i m s\i n\i f\i\ bir k\"ume olan her g\"onderme bir k\"umedir.
\end{theorem}

\ktk

Herhangi $\bm A$ s\i n\i f\i\ ve $b$ k\"umesi i\c cin
\begin{equation*}
{}^b\bm A,
\end{equation*}%
\glossary{${}^b\bm A$}
$b$ k\"umesinden $\bm A$ s\i n\i f\i na giden g\"ondermelerin s\i n\i f\i\ olsun.  

\begin{xca}
$\pow a\approx{}^a2$ e\c slenikli\u gini kan\i tlay\i n.
\end{xca}

\begin{theorem}[\"Ozyineleme]\label{thm:rec-ord-again}
$\bm A$ bir s\i n\i f olsun, ve
\begin{equation*}
\bm F\colon\{x\colon\Exists{\eta}x\in{}^{\eta}\bm A\}\to\bm A
\end{equation*}
olsun.\footnote{$\{x\colon\Exists{\eta}x\in{}^{\eta}\bm A)\}$ s\i n\i f\i, $\bigcup_{\eta}{}^{\eta}\bm A$ olarak yaz\i labilir; ama ${}^{\beta}\bm A$ s\i n\i flar\i\, k\"ume olmayabilir.}   
O zaman $\on$ s\i n\i f\i ndan $\bm A$ s\i n\i f\i na giden ve her $\alpha$ ordinali i\c cin
\begin{equation*}
\bm G(\alpha)=\bm F(\bm G\restriction\alpha)
\end{equation*}
ko\c sulunu sa\u glayan bir ve tek bir $\bm G$ g\"ondermesi vard\i r.
\end{theorem}

\begin{proof}
Tan\i m k\"umesi bir ordinal olan, ve bu ordinalin t\"um $\alpha$ elemanlar\i\ i\c cin
\begin{equation*}
g(\alpha)=\bm F(g\restriction\alpha)
\end{equation*}
ko\c sulunu sa\u glayan $g$ g\"ondermelerinin olu\c sturdu\u gu s\i n\i f, $\bm B$ olsun.  E\u ger $\bm B$ s\i n\i f\i n\i n bir $g$ eleman\i n\i n tan\i m k\"umesi $\alpha$, bir $h$ eleman\i n\i n tan\i m k\"umesi $\beta$, ve $\alpha\leq\beta$ ise, o zaman
\begin{equation*}
g\included h
\end{equation*}
(neden?).  Sonu\c c olarak istedi\u gimiz g\"onderme, $\bigcup\bm B$ (neden?).
\end{proof}

\begin{theorem}\label{thm:base}
$\alpha\geq2$ ise bir ve tek bir $\xi\mapsto\xi_{\alpha}$ g\"ondermesi vard\i r \"oyle ki
\begin{align}\label{eqn:base}
\beta&=\alpha^{\gamma}\cdot\delta+\theta,&
\delta&<\alpha,&
\theta&<\alpha^{\gamma}
\end{align}
ise, o zaman
\begin{equation*}
\beta_{\alpha}=\{(\gamma,\delta)\}\cup\theta_{\alpha}.
\end{equation*}
\end{theorem}

\begin{proof}
\ref{thm:exp-sol} numaral\i\ teoreme g\"ore, \eqref{eqn:base} ko\c sullar\i n\i\ sa\u glayan bir ve tek bir $(\gamma,\delta,\theta)$ \"u\c cl\"us\"u vard\i r.  Ayr\i ca $\theta<\beta$.  O zaman \"oyle bir $\bm F$ g\"ondermesi vard\i r ki her $\beta$ i\c cin, e\u ger $g$, tan\i m k\"umesi $\beta$ olan (ve de\u ger k\"umesi herhangi bir k\"ume olan) bir g\"onderme, ve $(\gamma,\delta,\theta)$ \"u\c cl\"us\"u, \eqref{eqn:base} sat\i r\i ndaki gibiyse, o zaman
\begin{equation*}
\bm F(g)=\{(\gamma,\delta)\}\cup g(\beta).
\end{equation*}
\"Oyleyse istedi\u gimiz $\beta\mapsto\beta_{\alpha}$ g\"ondermesi, son teoreme g\"ore
\begin{equation*}
\bm G(\beta)=\bm F(\bm G\restriction\beta)
\end{equation*}
ko\c sulunu sa\u glayan $\bm G$ g\"ondermesidir.
\end{proof}

\section{Cantor normal bi\c cimi}

E\u ger $\beta_{\upomega}=\{(\alpha_0,k_0),\dots,(\alpha_n,k_n)\}$ ise, o zaman
\begin{equation*}
\beta=\upomega^{\alpha_0}\cdot k_0+\upomega^{\alpha_1}\cdot k_1+\dots+\upomega^{\alpha_n}\cdot k_n.
\end{equation*}
\pageref{Cnf} numaral\i\ sayfada dedi\u gimiz gibi bu toplam, $\beta$ ordinalinin
\textbf{Cantor normal bi\c cimidir}%
\index{Cantor!--- normal bi\c cimi}%
\index{normal!Cantor --- bi\c cimi}
(\eng{Cantor normal form}).  Buradaki $\alpha_0$ \"uss\"u, $\beta$ ordinalinin \textbf{derecesidir}%
\index{derece}
(\eng{degree}).  Bu derece
\begin{equation*}
\deg(\beta)
\end{equation*}%
\glossary{$\deg(\alpha)$}
olarak yaz\i ls\i n.

\begin{theorem}
$\alpha>0$ ise
\begin{equation*}
1+\upomega^{\alpha}=\upomega^{\alpha}.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}
$\deg(\alpha)<\deg(\beta)$ ise
\begin{equation*}
\alpha+\beta=\beta.
\end{equation*}
\end{theorem}

\begin{proof}
$\alpha<\beta$ ise $\upomega^{\alpha}+\upomega^{\beta}=\upomega^{\beta}$ e\c sitli\u gini kan\i tlayaca\u g\i z.  Bu durumda bir $\gamma$ i\c cin, $\beta=\alpha+\gamma$ ve $\gamma>0$, dolay\i s\i yla
\begin{equation*}
\upomega^{\alpha}+\upomega^{\beta}
=\upomega^{\alpha}+\upomega^{\alpha+\gamma}
=\upomega^{\alpha}\cdot(1+\upomega^{\gamma})
=\upomega^{\alpha}\cdot\upomega^{\gamma}
=\upomega^{\alpha+\gamma}
=\upomega^{\beta}.\qedhere
\end{equation*}
\end{proof}

\"Orne\u gin
\begin{multline*}
(\upomega^{\upomega^2}+\upomega^{\upomega\cdot4+7}\cdot5+\upomega^{\upomega\cdot2}\cdot 7+2)
+(\upomega^{\upomega\cdot4+7}\cdot 8+\upomega\cdot 3+16)\\
=\upomega^{\upomega^2}+\upomega^{\upomega\cdot4+7}\cdot13+\upomega\cdot3+16.
\end{multline*}

\begin{theorem}
$\alpha>\deg(\beta)$ ve $1\leq k<\upomega$ ve $1\leq n<\upomega$ ise
\begin{equation*}
(\upomega^{\alpha}\cdot k+\beta)\cdot n=\upomega^{\alpha}\cdot k\cdot n+\beta.
\end{equation*}
\end{theorem}

\begin{proof}
$n=1$ durumunda iddia do\u grudur.  $n=m$ durumunda do\u gruysa
\begin{align*}
(\upomega^{\alpha}\cdot k+\beta)\cdot(m+1)
&=(\upomega^{\alpha}\cdot k+\beta)\cdot m+\upomega^{\alpha}\cdot k+\beta\\
&=\upomega^{\alpha}\cdot k\cdot m+\beta+\upomega^{\alpha}\cdot k+\beta\\
&=\upomega^{\alpha}\cdot k\cdot m+\upomega^{\alpha}\cdot k+\beta\\
&=\upomega^{\alpha}\cdot k\cdot(m+1)+\beta,
\end{align*}
dolay\i s\i yla $n=m+1$ durumunda da do\u grudur.  T\"umevar\i mdan $1\leq n<\upomega$ ise iddia do\u grudur.
\end{proof}

\"Orne\u gin
\begin{equation*}
(\upomega^{\upomega}\cdot2+\upomega+5)\cdot7
=\upomega^{\upomega}\cdot14+\upomega+5.
\end{equation*}

\begin{theorem}
$1\leq\alpha$ ve $0<\beta$ ise
\begin{equation*}
\alpha\cdot\upomega^{\beta}=\upomega^{\deg(\alpha)+\beta}.
\end{equation*}
\end{theorem}

\begin{proof}
\"Once $\beta=1$ durumunda iddiay\i\ kan\i tlayaca\u g\i z.  $1\leq \alpha<\upomega$ ise $\deg(\alpha)=0$, dolay\i s\i yla
\begin{equation*}
\alpha\cdot\upomega^1
=\alpha\cdot\upomega
=\sup_{x\in\upomega}\alpha\cdot x=\upomega=\upomega^{\deg(\alpha)+1}.
\end{equation*}
\c Simdi $\upomega\leq\alpha$ olsun.  O zaman
\begin{align*}
\alpha&=\upomega^{\gamma}\cdot k+\delta,&
\alpha&>\deg(\gamma),&
1&\leq k<\upomega
\end{align*}
ko\c sullar\i n\i\ sa\u glayan $\gamma$, $k$, ve $\delta$ vard\i r.
O halde $1\leq n<\upomega$ ise
\begin{gather*}
	\upomega^{\gamma}\cdot k\leq\alpha\leq\alpha\cdot n
	=\upomega^{\gamma}\cdot k\cdot n+\delta
	<\upomega^{\gamma}\cdot(k\cdot n+1)<\upomega^{\gamma+1},\\
	\upomega^{\gamma+1}
	=\sup_{1\leq\xi<\upomega}(\upomega^{\gamma}\cdot\xi)
	\leq\sup_{1\leq\xi<\upomega}(\alpha\cdot\xi)
	\leq\upomega^{\gamma+1},\\
	\alpha\cdot\upomega=\upomega^{\gamma+1}=\upomega^{\deg(\alpha)+1}.
\end{gather*}
Genelde $\alpha\geq1$ ve $\beta\geq1$ ise, bir $\theta$ i\c cin $\beta=1+\theta$, dolay\i s\i yla
\begin{equation*}
\alpha\cdot\upomega^{\beta}=\alpha\cdot\upomega\cdot\upomega^{\theta}
=\upomega^{\deg(\alpha)+1+\theta}
=\upomega^{\deg(\alpha)+\beta}.\qedhere
\end{equation*}
\end{proof}

\"Orne\u gin
\begin{gather*}
5\cdot(\upomega^2\cdot 3+\upomega\cdot16+7)=\upomega^2\cdot 3+\upomega\cdot16+35,\\
(\upomega^{\upomega}\cdot2+\upomega+5)
\cdot(\upomega^2\cdot 3+\upomega\cdot16)
=\upomega^{\upomega+2}\cdot 3+\upomega^{\upomega+1}\cdot16,
\end{gather*}
ve
\begin{multline*}
(\upomega^{\upomega}\cdot2+\upomega+5)
\cdot(\upomega^2\cdot 3+\upomega\cdot16+7)\\
=\upomega^{\upomega+2}\cdot 3+\upomega^{\upomega+1}\cdot16+\upomega^{\upomega}\cdot14+\upomega+5.
\end{multline*}

\begin{theorem}
$1\leq k<\upomega$ ve $n<\upomega\leq\alpha$ ise
\begin{align*}
k^{\upomega^{n+1}}&=\upomega^{\upomega^n},&
k^{\upomega^{\alpha}}&=\upomega^{\upomega^{\alpha}}.
\end{align*}
\end{theorem}

\ktk[ \emph{\.Ipucu:}  $n+1=1+n$ ve $\alpha=1+\alpha$.]

\"Orne\u gin
\begin{equation*}
2^{\upomega^{\upomega}\cdot3+\upomega^5\cdot4+\upomega\cdot7+5}
=\upomega^{\upomega^{\upomega}\cdot3+\upomega^4\cdot4+7}\cdot32.
\end{equation*}

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

\ktk[ \emph{\.Ipucu:}  \"Once $\alpha^{\upomega}=\upomega^{\deg(\alpha)\cdot\upomega}$ denkli\u gini kan\i tlay\i n.]

\"Orne\u gin
\begin{multline*}
(\upomega^{\upomega+1}+\upomega^2+1)^{\upomega^2+\upomega\cdot3+2}\\
\begin{aligned}
&=\upomega^{(\upomega+1)\cdot(\upomega^2+\upomega\cdot3)} \cdot(\upomega^{\upomega+1}+\upomega^2+1)^2\\
&=\upomega^{\upomega^3+\upomega^2\cdot3}\cdot(\upomega^{\upomega+1}+\upomega^2+1)^2\\
&=\upomega^{\upomega^3+\upomega^2\cdot3} \cdot(\upomega^{\upomega+1+\upomega+1}+\upomega^{\upomega+1+2}+\upomega^{\upomega+1}+\upomega^2+1)\\
&=\upomega^{\upomega^3+\upomega^2\cdot3} 
\cdot(\upomega^{\upomega\cdot2+1}
+\upomega^{\upomega+3}
+\upomega^{\upomega+1}
+\upomega^2+1)\\
&=\upomega^{\upomega^3+\upomega^2\cdot3+\upomega\cdot2+1}
+\upomega^{\upomega^3+\upomega^2\cdot3+\upomega+3}
\end{aligned}\\
+\upomega^{\upomega^3+\upomega^2\cdot3+\upomega+1}
+\upomega^{\upomega^3+\upomega^2\cdot3+2}
+\upomega^{\upomega^3+\upomega^2\cdot3}.
\end{multline*}


\chapter{Kardinaller}

\section{E\c sleniklik}

E\u ger $\bm R$ ve $\bm S$, iki ba\u g\i nt\i ysa, o zaman tan\i ma g\"ore
\begin{equation*}
\bm R/\bm S=\{(x,z)\colon\Exists y(x\mathrel{\bm R}y\land y\mathrel{\bm S}z)\}.
\end{equation*}%
\glossary{$\bm R/\bm S$}
Bu yeni ba\u g\i nt\i, $\bm R$ ile $\bm S$ ba\u g\i nt\i lar\i n\i n 
\textbf{bile\c skesidir}%
\index{bile\c ske}
(\eng{composite}).
$\bm S/\bm R$ bile\c skesi, $\bm R/\bm S$ bile\c skesinden farkl\i\ olabilir.

\begin{theorem}
E\u ger $\bm F\colon\bm A\to\bm B$ ve $\bm G\colon\bm B\to\bm C$ ise, o zaman 
\begin{align*}
\bm F/\bm G&\colon\bm A\to\bm C,&
\Forall x\bigl(x\in\bm A\lto(\bm F/\bm G)(x)&=\bm G(\bm F(x))\bigr).
\end{align*}
\end{theorem}

\ktk

Teoremdeki durumda $\bm F/\bm G$ g\"ondermesi,
\begin{equation*}
\bm G\circ\bm F
\end{equation*}%
\glossary{$\bm B\circ\bm F$}
olarak yaz\i l\i r.

\c Simdi $\bm F\colon\bm A\to\bm B$ olsun.  E\u ger $\bm F$ ba\u g\i
nt\i s\i n\i n $\conv{\bm F}$ ters ba\u g\i nt\i s\i, $\bm B$ s\i n\i f\i
ndan $\bm A$ s\i n\i f\i na giden bir g\"ondermeyse, o zaman bu
g\"onderme, $\bm F$ g\"ondermesinin 
\textbf{ters g\"ondermesi}%
\index{g\"onderme!ters ---}
veya \textbf{tersidir}%
\index{ters}
(\eng{inverse}), ve
\begin{equation*}
\inv{\bm F}
\end{equation*}%
\glossary{$\inv{\bm F}$}
olarak yaz\i l\i r.  Bu durumda $\bm F$ g\"ondermesi, $\bm A$ s\i n\i f\i ndan
$\bm B$ s\i n\i f\i na giden bir \textbf{e\c slemedir}%
\index{e\c sleme}
(\eng{bijection}), ve $\bm A$ ile $\bm B$ s\i n\i flar\i n\i n
kendileri, birbiriyle \textbf{e\c sleniktir}%
\index{e\c slenik}
(\eng{equipollent}).

\begin{theorem}\label{thm:sinif}
Bir s\i n\i f, bir k\"umeyle e\c slenikse, s\i n\i f da bir k\"umedir.
\end{theorem}

\ktk

\ref{thm:sinif} ve \ref{thm:gonderme} numaral\i\ teoremler sayesinde
\emph{k\"umelerin} e\c slenikli\u gi, ikili bir ba\u g\i nt\i d\i r.  Bu ba\u
g\i nt\i n\i n i\c sareti
\begin{equation*}
\approx
\end{equation*}%
\glossary{$a\approx b$}
olsun.  O zaman $=$ gibi $\approx$, yeni bir y\"uklemdir.  Ayr\i ca
\begin{multline*}
a\approx b\denk\Exists w
\biggl(\Forall x\Exists y\Bigl(\bigl(x\in a\lto y\in b\land(x,y)\in w\bigr)\\
\land\bigl(x\in b\lto y\in a\land(y,x)\in w\bigr)\Bigr)\\
\land\Forall x\Forall y\Forall z
\Bigl(\bigl((x,y)\in w\land(x,z)\in w\land x\in a\lto y=z\land y\in b\bigr)\\
\land\bigl((x,z)\in w\land(y,z)\in w\land z\in b\lto x=y\land x\in a\bigr)\Bigr)\biggr).
\end{multline*}

E\c sitlik gibi e\c sleniklik, bir denklik ba\u g\i nt\i s\i d\i r
(\pageref{denklik} numaral\i\ sayfaya bak\i n):

\begin{theorem}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{align*}
&a\approx a,&
a\approx b&\lto b\approx a,&
a\approx b\land b\approx c&\lto a\approx c
\end{align*}
c\"umleleri do\u grudur.
\end{theorem}

\ktk

\begin{theorem}\label{thm:s}
$a$, bir $\bm S$ ba\u g\i nt\i s\i\ taraf\i ndan iyi s\i ralanm\i\c s bir k\"ume olsun.  O zaman bir ve tek bir $\beta$ i\c cin, $a$ k\"umesinden $\beta$ ordinaline giden
\begin{equation*}
x\mathrel{\bm S}y\liff f(x)<f(y)
\end{equation*}
ko\c sulunu sa\u glayan bir $f$ e\c slemesi vard\i r.
\end{theorem}

\begin{proof}
\ref{thm:rec-ord-again} numaral\i\ \"Ozyineleme Teoreminde $a$ k\"umesi, $\on$ s\i n\i f\i n\i n yerini alabilir.  Yani tan\i m k\"umesi $a$ olan ve
\begin{equation*}
f(b)=\{f(c)\colon c\in a\land c\mathrel{\bm S}b\}
\end{equation*}
ko\c sulunu sa\u glayan bir $f$ g\"ondermesi vard\i r.  O zaman $a$ k\"umesinin her $b$ eleman\i\ i\c cin $f(b)$ bir ordinaldir.  Zira de\u gilse $b$, $f(b)$ de\u geri ordinal olmayan $a$ k\"umesinin en k\"u\c c\"uk eleman\i\ olsun.  O zaman $f(b)\included\on$, ve ayr\i ca $f(b)$ ge\c ci\c slidir, dolay\i s\i yla $f(b)$ bir ordinal olmal\i d\i r.  Ayn\i\ \c sekilde $f[a]$ bir ordinaldir.  E\u ger $f_0$ ile $f_1$, ayn\i\ ko\c sulu sa\u glar, ve
\begin{equation*}
f_0\restriction\{c\in a\colon c\mathrel{\bm S}b\}=
f_1\restriction\{c\in a\colon c\mathrel{\bm S}b\}
\end{equation*}
ise, o zaman $f_0(b)=f_1(b)$; \"oyleyse $f_0=f_1$.
\end{proof}

\begin{theorem}
T\"um $a$ ile $b$ k\"umeleri i\c cin $a\times b$ \c carp\i m\i, bir k\"umedir.
\end{theorem}

\begin{proof}
Yerle\c stirme Aksiyomuna g\"ore her $c$ i\c cin $a\times\{c\}$ bir k\"umedir, dolay\i s\i yla bir $\bigl\{a\times\{x\}\colon x\in b\bigr\}$ s\i n\i f\i\ vard\i r, ve bu s\i n\i f da bir k\"umedir.  Ayr\i ca
\begin{equation*}
a\times b=\bigcup\bigl\{a\times\{x\}\colon x\in b\bigr\},
\end{equation*}
ve Bile\c sim Aksiyomuna g\"ore bu bile\c sim, bir k\"umedir.
\end{proof}

Tekrar $a$, $s$, ve $\beta$, \ref{thm:s} numaral\i\ teoremdeki gibi olsun.  $\bigl(a,\bm S\cap(a\times a)\bigr)$ s\i ral\i\ ikilisi,
\begin{equation*}
(a,\bm S)
\end{equation*}
olarak yaz\i labilir.  O zaman
 $\beta$, $(a,\bm S)$ ikilisinin \textbf{ordinalidir}%
\index{ordinal}
(\eng{$\beta$ is the ordinal of $(a,\bm S)$}), ve
\begin{equation*}
\beta=\ord(a,\bm S)
\end{equation*}%
\glossary{$\ord(a,\bm S)$}
yazabiliriz.  O halde
\begin{equation*}
a\approx\ord(a,\bm S).
\end{equation*}
Tan\i ma g\"ore
\begin{equation*}
\card(a)=\min\{\beta\colon a\approx\beta\};
\end{equation*}%
\glossary{$\card(a)$}
bu ordinal, $a$ k\"umesinin \textbf{kardinalidir}%
\index{kardinal}
(\eng{cardinal}).  \"Oyleyse her \emph{iyi s\i ralanabilir} k\"umenin kardinali vard\i r.  \"Ozel olarak her ordinalin kardinali vard\i r.
Kardinaller, bir
\begin{equation*}
\cn
\end{equation*}%
\glossary{$\cn$}
s\i n\i f\i n\i\ olu\c sturur.  O zaman $\cn\included\on$.  $\kappa$,
$\lambda$, $\mu$ ve $\nu$%
\glossary{$\kappa$, $\lambda$, $\mu$, $\nu$} 
k\"u\c c\"uk Yunan harfleri, her zaman kardinalleri g\"osterecektir.

Asl\i nda \pageref{secim} numaral\i\ sayfadaki Se\c cim Aksiyomuna g\"ore her k\"ume iyi s\i ralanabilir; ama \c su anda bu aksiyom, resmi askiyomlar\i m\i zdan biri de\u gildir.


\section{Sonlu k\"umeler}

Bir do\u gal say\i yla e\c slenik bir k\"ume,
\textbf{sonludur}%
\index{sonlu, sonsuz}
(\eng{finite}); sonlu olmayan bir s\i n\i f,
\textbf{sonsuzdur}
(\eng{infinite}).
O zaman her sonlu kardinal, bir do\u gal say\i d\i r.

Birka\c c tane von Neumann do\u gal say\i s\i n\i n tan\i m\i n\i,
\pageref{vnn} ve \pageref{0123} numaral\i\ sayfalardan hat\i rlayal\i m:
\begin{align*}
  0&=\emptyset,&
1&=\{0\},&
2&=\{0,1\},&
3&=\{0,1,2\},&
4&=\{0,1,2,3\}.
\end{align*}
Bir $a$ k\"umesinin
\begin{compactenum}[1)]
  \item
hi\c cbir eleman\i\ yoksa, o zaman $a\approx0$; asl\i
    nda $a=0$;
  \item
tek bir eleman\i\ varsa, o zaman $a\approx1$;
  \item
iki (ve sadece iki) eleman\i\ varsa, o zaman $a\approx2$;
  \item
\"u\c c (ve sadece \"u\c c) eleman\i\ varsa, o zaman
    $a\approx3$. 
\end{compactenum}
Ayr\i ca
\begin{align*}
  0&\not\approx1,&
  0&\not\approx2,&
  0&\not\approx3,&
  1&\not\approx2,&
  1&\not\approx3,&
  2&\not\approx3.
\end{align*}
Ancak herhangi iki e\c slenik do\u gal say\i\ e\c sit olmal\i\ m\i?  Bu soruyu, \pageref{nasil} numaral\i\ sayfada sormu\c stuk.

\begin{theorem}
Her do\u gal say\i, ya $0$ ya bir do\u gal say\i n\i n ard\i l\i d\i r.
\end{theorem}

\ktk

\begin{theorem}
\.Iki do\u gal say\i\ birbiriyle e\c slenikse, birbirine e\c sittir:
\begin{equation*}
\Forall x\Forall y(x\in\upomega\land y\in\upomega\land x\approx y\lto x=y).
\end{equation*}
Yani her do\u gal say\i, sonlu bir kardinaldir.
\end{theorem}

\begin{proof}
T\"umevar\i mla her $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
\Forall x(x\in\upomega\land x\approx n\lto x=n)
\end{equation*}
c\"umlesini kan\i tlayaca\u g\i z.  $n=0$ ise do\u grudur.  $n=m$ ise do\u gru olsun, ve bir $\ell$ do\u gal say\i s\i\ i\c cin $m'\approx\ell$ olsun.  O zaman $\ell$ bo\c s de\u gil.  Son teoreme g\"ore $\ell$ bir ard\i l olmal\i.  $\ell=k'$ olsun.  $m'$ say\i s\i ndan $k'$ say\i s\i na giden bir $f$ e\c slemesi vard\i r.  E\u ger $f(m)=k$, o zaman $f\setminus\{(m,k)\}$, $m$ say\i s\i ndan $k$ say\i s\i na giden bir e\c slemedir.  E\u ger $f(m)\neq k$, o zaman
\begin{equation*}
\bigl\{(x,y)\colon x\in m\setminus\{\inv f(k)\}\land y=f(x)\bigr\}\cup\bigl\{(\inv f(k),f(m))\bigr\}
\end{equation*}
ba\u g\i nt\i s\i, $m$ say\i s\i ndan $k$ say\i s\i na giden bir e\c slemedir.  \"Oyleyse her durumda $m\approx k$.  Hipotezimize g\"ore $m=k$ olmal\i, dolay\i s\i yla $m'=\ell$.  Kan\i t bitti.
\end{proof}

K\i saca $\upomega\included\cn$.

\section{Say\i labilme}

Tekrar $\bm F\colon\bm A\to\bm B$ olsun.
\begin{compactenum}
  \item
E\u ger $\bm F[\bm A]=\bm B$ ise, o zaman $\bm F$, $\bm B$ s\i n\i f\i
n\i\ \textbf{\"orten}% 
\index{\"orten}
bir g\"ondermedir
(\eng{$\bm F$ is onto $\bm B$}), ve
\begin{equation*}
\bm F\colon\bm A\twoheadrightarrow\bm B
\end{equation*}%
\glossary{$\bm F\colon\bm A\twoheadrightarrow\bm B$}
ifadesini yazabiliriz.
\item
E\u ger
\begin{equation*}
\Forall x\Forall y(x\in\bm A\land y\in\bm A\land\bm F(x)=\bm F(y)\lto x=y)
\end{equation*}
ise, o zaman $\bm F$, \textbf{birebir}%
\index{g\"onderme!birebir ---}
(\eng{one-to-one}) veya
\textbf{injektif}
\index{g\"onderme!injektif ---}
(\eng{injective})
bir g\"ondermedir; ayr\i ca $\bm F$, bir \textbf{g\"ommedir}%
\index{g\"omme}
(\eng{embedding}).  Bu durumda
\begin{align*}
  \bm F\colon\bm A&\rightarrowtail\bm B&
  &\text{ veya }&
  \bm F\colon\bm A&\xrightarrow{\preccurlyeq}\bm B
\end{align*}
\glossary{$\bm F\colon\bm A\rightarrowtail\bm B$}
ifadesini yazabiliriz.  Bir $a$ \emph{k\"umesinden} $\bm B$ s\i n\i f\i na giden bir g\"omme varsa, bu g\"omme de bir k\"umedir, ve
\begin{equation*}
a\preccurlyeq\bm B
\end{equation*}%
\glossary{$a\preccurlyeq\bm B$}
ifadesini yazar\i z.
\item
E\u ger $\bm F$, $\bm B$ s\i n\i f\i n\i\ \"orten bir g\"ommeyse,
o zaman
\begin{equation*}
  \bm F\colon\bm A\xrightarrow{\approx}\bm B
\end{equation*}
ifadesini yazar\i z.\footnote{$\bm F\colon\bm A\twoheadrightarrowtail\bm B$ ifadesi de m\"umk\"und\"ur.}  Bu durumda $\bm F$, bir e\c slemedir.
\end{compactenum}

\begin{theorem}\label{thm:preccurlyeq}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{align*}
&a\preccurlyeq a,&
a\preccurlyeq b\land b\preccurlyeq c&\lto a\preccurlyeq c,&
a\approx b&\lto a\preccurlyeq b
\end{align*}
c\"umleleri do\u grudur.
\end{theorem}

\ktk

\begin{theorem}
Bir s\i n\i f, bir k\"umeye g\"om\"ulebilirse, bu s\i n\i f da bir k\"umedir.
\end{theorem}

\ktk

\begin{theorem}
$\upomega$, bir kardinaldir, yani
\begin{equation*}
\upomega\in\cn.
\end{equation*}
\end{theorem}

\begin{proof}
T\"umevar\i mla her $m$ do\u gal say\i s\i\ i\c cin $f\colon m\rightarrowtail\upomega$ ise $f[m]$ k\"umesinin en b\"uy\"uk $n$ eleman\i\ vard\i r, dolay\i s\i yla $n+1\in\upomega\setminus f[m]$; \"ozel olarak $f$, $\upomega$ k\"umesini \"orten de\u gildir.
\end{proof}

Teoremin sonucu olarak $\upomega$ k\"umesiyle e\c slenik her k\"ume, sonsuzdur.
\"Oyle bir k\"ume,
\textbf{say\i labilir sonsuzluktad\i r}
(\eng{countably infinite}).
Sonlu veya say\i labilir sonsuzluktaki bir k\"ume,
\textbf{say\i labilir}%
\index{say\i labilir, say\i lamaz}
(\eng{countable}).
Di\u ger k\"umeler ve s\i n\i flar,
\textbf{say\i lamaz sonsuzluktad\i r} 
(\eng{uncountably infinite})
veya \textbf{say\i lamaz}
(\eng{uncountable}).  Asl\i nda her k\"ume olmayan s\i n\i f, say\i lamaz.

\begin{theorem}
Bir $a$ k\"umesi i\c cin a\c sa\u g\i daki ko\c sullar, birbirine
  denktir.
  \begin{compactenum}
    \item
$a$ say\i labilir.
\item
$a\preccurlyeq\upomega$.
\item
Ya $a$ bo\c s, ya da $\upomega$ k\"umesinden $a$ k\"umesini \"orten bir
  g\"onderme vard\i r.
  \end{compactenum}
\end{theorem}

\begin{proof}
E\u ger $f\colon a\xrightarrow{\approx}\upomega$ ise, o zaman $f\colon a\xrightarrow{\preccurlyeq}\upomega$ ve $\inv{f}\colon\upomega\twoheadrightarrow a$.

E\u ger $n\in\upomega$ ve $f\colon a\xrightarrow{\approx}n$ ise, o zaman $f\colon a\xrightarrow{\preccurlyeq}\upomega$; ayr\i ca ya $n=0$ ya da
\begin{equation*}
\inv{f}\cup\{(x,0)\colon n\leq x<\upomega\}\colon\upomega\twoheadrightarrow a.
\end{equation*}

\c Simdi $f\colon a\xrightarrow{\preccurlyeq}\upomega$ olsun.  O zaman $f[a]$, $\in$ taraf\i ndan iyi s\i ralanm\i\c st\i r, dolay\i s\i yla \ref{thm:s} numaral\i\ teoreme g\"ore $f[a]$ k\"umesinden bir $\alpha$ ordinaline giden ve
\begin{equation*}
x\in y\liff g(x)\in g(y)
\end{equation*}
ko\c sulunu sa\u glayan bir $g$ e\c slemesi vard\i r.  Teoremin kan\i t\i ndaki gibi $f[a]$ k\"umesinin t\"um $n$ elemanlar\i\ i\c cin
\begin{equation*}
g(n)=\{g(k)\colon k\in f[a]\cap n\}.
\end{equation*}
E\u ger $f[a]\cap n$ k\"umesinin t\"um $k$ eleman\i\ i\c cin $g(k)\leq k$ ise, o zaman $g(n)\leq n$.  Sonu\c c olarak, $f[a]$ k\"umesi iyi s\i ralanm\i\c s oldu\u gundan her $n$ eleman\i\ i\c cin $g(n)\leq n$, dolay\i s\i yla $\alpha\leq\upomega$ ve $g\circ f\colon a\xrightarrow{\approx}\alpha$.

Son olarak $h\colon\upomega\twoheadrightarrow a$ olsun.  O zaman 
\begin{equation*}
x\mapsto\min\{y\in\upomega\colon h(y)=x\}\colon a\xrightarrow{\preccurlyeq}\upomega,
\end{equation*}
dolay\i s\i yla, g\"osterdi\u gimiz gibi, $a$ say\i labilir.
\end{proof}

Say\i lamaz sonsuzlukta bir k\"ume biliyor muyuz?

\subsection{Toplama}

Toplama veya ikili bile\c sim i\c slemiyle say\i lamaz sonsuzluktaki k\"umeler
olu\c sturulamaz. 

Tan\i ma g\"ore t\"um $a$ ile $b$ k\"umeleri i\c cin
\begin{equation*}
  a\sqcup b=(a\times\{0\})\cup(b\times\{1\}).
\end{equation*}%
\glossary{$a\sqcup b$}
Bu bile\c sim, $a$ ile $b$ k\"umelerinin \textbf{ayr\i k bile\c simidir}%
\index{ayr\i k bile\c sim}
(\eng{disjoint union}).

\begin{theorem}\label{thm:ord+}
T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin
\begin{equation*}
\alpha+\beta\approx\alpha\sqcup\beta.
%(\alpha\times\{0\})\cup(\beta\times\{1\}).
\end{equation*}
\end{theorem}

\begin{proof}
Bu kan\i t i\c cin t\"umevar\i m kullanmayaca\u g\i z.
$\alpha\leq\gamma$ ise $\bm F(\gamma)$,
\begin{equation*}
\alpha+x=\gamma
\end{equation*}
denkleminin tek \c c\"oz\"um\"u olsun. O zaman
\begin{equation*}
\bigl\{\bigl(x,(x,0)\bigr)\colon x<\alpha\bigr\}\cup\Bigl\{\Bigl(x,\bigl(\bm F(x),1\bigr)\Bigr)\colon\alpha\leq x<\alpha+\beta\Bigr\},
\end{equation*}
$\alpha+\beta$ toplam\i ndan $(\alpha\times\{0\})\cup(\beta\times\{1\})$ bile\c simine giden bir e\c slemedir.
\end{proof}

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

\begin{proof}
$(x,y)\mapsto 2x+y$ g\"ondermesi, $(\upomega\times\{0\})\cup(\upomega\times\{1\})$ bile\c siminden $\upomega$ k\"umesine giden bir e\c slemedir.
\end{proof}

\begin{theorem}
$\alpha\approx\beta$ ve $\gamma\approx\delta$ ise $\alpha+\gamma\approx\beta+\delta$.
\end{theorem}

\ktk

\begin{theorem}
$a$ ile $b$ say\i labilirse $a\cup b$ bile\c simi de say\i labilir.
\end{theorem}

\ktk

\subsection{\c Carpma}

\c Carpmayla say\i lamaz sonsuzluktaki k\"umeler olu\c sturulamaz.

\begin{theorem}\label{thm:ord.}
T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin
\begin{equation*}
\alpha\cdot\beta\approx\alpha\times\beta.
\end{equation*}
\end{theorem}

\begin{proof}
E\c sleme, $\bigl\{\bigl(x,(y,z)\bigr)\colon x<\alpha\cdot\beta\land y<\alpha\land x=\alpha\cdot z+y\bigr\}$. 
\end{proof}

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

\begin{proof}
$\bigl\{\bigl((x,y),z\bigr)\colon(x>y\lto z=x^2+y)\land(x\leq y\lto z=x^2+x+y)\bigr\}$ s\i n\i f\i, $\upomega\times\upomega$ \c carp\i m\i ndan $\upomega$ k\"umesine giden bir e\c sleme tan\i mlar.
\end{proof}

\begin{theorem}
$\alpha\approx\beta$ ve $\gamma\approx\delta$ ise $\alpha\cdot\gamma\approx\beta\cdot\delta$.
\end{theorem}

\ktk

\begin{theorem}
$a$ ile $b$ say\i labilirse $a\times b$ \c carp\i m\i\ da say\i labilir.
\end{theorem}

\ktk

\subsection{Kuvvet alma}

\emph{Ordinal} kuvvetleri alarak sonsuzluktaki k\"umeler olu\c sturulamaz.

T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin, tan\i m k\"umesi $\beta$ ordinalinin sonlu bir altk\"umesi olan ve de\u ger k\"umesi $\alpha$ ordinalinin bir altk\"umesi olan g\"ondermelerin s\i n\i f\i, 
\begin{equation*}
\exp(\alpha,\beta)
\end{equation*}%
\glossary{$\exp(\alpha,\beta)$}
olsun.%
\footnote{$\exp(\alpha,\beta)$ ifadesi, ve a\c sa\u g\i daki \ref{thm:exp} numaral\i\ teorem, Levy'nin \cite[IV.2.10]{MR1924429} kitab\i ndan al\i nm\i\c st\i r.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{theorem}\label{thm:exp}
T\"um $\alpha$ ile $\beta$ ordinalleri i\c cin
\begin{equation*}
\alpha^{\beta}\approx\exp(\alpha,\beta).
\end{equation*}
\end{theorem}

\begin{proof}
$\gamma_{\alpha}$, \ref{thm:base} numaral\i\ teoremdeki gibi olunca,
$\{(\gamma,\gamma_{\alpha})\colon\gamma<\alpha^{\beta}\}$ k\"umesi, $\alpha^{\beta}$ k\"umesinden $\exp(\alpha,\beta)$ k\"umesine giden bir e\c slemedir.
\end{proof}

\begin{theorem}
$\alpha\approx\beta$ ve $\gamma\approx\delta$ ise $\alpha^{\gamma}\approx\beta^{\delta}$.
\end{theorem}

\ktk

$n\in\upomega$ ise ${}^nb$ s\i n\i f\i n\i n bir eleman\i, $(a_0,\dots,a_{n-1})$ olarak yaz\i labilir.  O halde
\begin{equation*}
f=(a_0,\dots,a_{n-1})\lto f(i)=a_i.
\end{equation*}
Herhangi bir $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}
\powf{\bm A},
\end{equation*}%
\glossary{$\powf{\bm A}$}
$\bm A$ s\i n\i f\i n\i n \emph{sonlu} altk\"umelerinin s\i n\i f\i\ olsun.

\begin{theorem}\label{thm:ooo}
$\upomega^{\upomega}\approx\upomega$.
\end{theorem}

\begin{proof}
$\upomega^{\upomega}\approx\exp(\upomega,\upomega)$, ve
  $\exp(\upomega,\upomega)\included\powf{\upomega\times\upomega}$.
  Ayr\i ca $\upomega\times\upomega\approx\upomega$ oldu\u gundan
  $\powf{\upomega\times\upomega}\approx\powf{\upomega}$.  K\i saca
  \begin{equation*}
\upomega^{\upomega}
\approx\exp(\upomega,\upomega) 
\included\powf{\upomega\times\upomega}
\approx\powf{\upomega}.  
  \end{equation*}
\c Simdi $f\colon\upomega\times\upomega\xrightarrow{\approx}\upomega$
olsun.  \"Ozyinelemeyle $\upomega\setminus\{0\}$ k\"umesinde bir
$n\mapsto g_n$ g\"ondermesini tan\i mlayaca\u g\i z.  Asl\i nda
$g_n\colon{}^n\upomega\to\upomega$ olacakt\i r.  \"Ozyineli tan\i ma
g\"ore 
\begin{align*}
g_1\bigl((x)\bigr)&=x,&
g_{n+1}\bigl((x_0,\dots,x_n)\bigr)&=f\bigl((x_0,\dots,x_{n-1}),x_n\bigr).
\end{align*}
T\"umevar\i mdan $1\leq n<\upomega$ ise
$g_n\colon{}^n\upomega\xrightarrow{\approx}\upomega$, dolay\i s\i yla
${}^n\upomega$ ile $g_n$ s\i n\i flar\i, k\"umedir. 

$\powf{\upomega}$ s\i n\i f\i ndan $\bigcup\{{}^n\upomega\colon
n\in\upomega\}$ bile\c simine giden \"oyle bir $h$ g\"ondermesi vard\i
r ki $h(0)=0$, ve $a_0<a_1<\dots<a_n<\upomega$ ise 
\begin{equation*}
h\bigl(\{a_0,a_1,\dots,a_n\}\bigr)=(a_0,\dots,a_n).
\end{equation*}
\"Oyleyse $h$, bir g\"ommedir.  Ayr\i ca
\begin{equation*}
(b_0,\dots,b_{n-1})\mapsto f\Bigl(n,g_n\bigl((b_0,\dots,b_{n-1})\bigr)\Bigr)
\end{equation*}
g\"ondermesi, $\bigcup\{{}^n\upomega\colon n\in\upomega\}$ bile\c
siminden $\upomega$ k\"umesine giden bir g\"ommedir.  K\i saca
\begin{equation*}
  \powf{\upomega} 
\preccurlyeq\bigcup\{{}^n\upomega\colon n\in\upomega\} 
\preccurlyeq\upomega.
\end{equation*}
G\"osterdiklerimiz hep birlikte
$\upomega^{\upomega}$ k\"umesinden $\upomega$ k\"umesine giden bir
g\"omme oldu\u gunu kan\i tlar. 
\end{proof}

\section{B\"uy\"ukl\"uk}

Bir $a$ k\"umesinden bir $\bm B$ s\i n\i f\i na giden bir g\"omme varsa, ama e\c sleme yoksa,
\begin{equation*}
a\prec\bm B
\end{equation*}%
\glossary{$a\prec\bm B$}
ifadesini yazar\i z.  \"Oyleyse
\begin{equation*}
  a\prec\bm B\denk a\preccurlyeq\bm B\land a\not\approx\bm B.
\end{equation*}

O zaman bir $a$ k\"umesi sonludur ancak ve ancak $a\prec\upomega$.
\ref{thm:preccurlyeq} numaral\i\ teoremin \"ozel durumu vard\i r:
\begin{equation*}
a\prec b\land b\prec c\lto a\preccurlyeq c.  
\end{equation*}
Ama $a\prec b\land b\prec c$ ise $a\prec c$ sonucuna varabilir miyiz?

Bir $a$ k\"umesinin \textbf{b\"uy\"ukl\"u\u g\"u}%
\index{b\"uy\"ukl\"uk}
(\eng{size)},
$\{x\colon x\approx a\}$ denklik s\i n\i f\i\ olarak d\"u\c
s\"un\"ulebilir.  $a$ bo\c s de\u gilse $\{x\colon x\approx a\}$ s\i n\i f\i, k\"ume de\u gildir.  H\^al\^a b\"uy\"ukl\"uklerin s\i ralanabilir olup olmad\i\u g\i n\i\ sorabiliriz.  E\u ger $a\prec b$ ise, o zaman $\{x\colon x\approx a\}$ b\"uy\"ukl\"u\u g\"u, $\{x\colon x\approx b\}$ b\"uy\"ukl\"u\u g\"unden k\"u\c c\"uk gibidir; ama $\bm A$ k\"u\c c\"uk $\bm B$ ve $\bm B$ k\"u\c c\"uk $\bm C$ ise $\bm A$ k\"u\c c\"uk $\bm C$ do\u gru mudur?

Her k\"ume iyi s\i ralanabilirse, o zaman her b\"uy\"ukl\"uk, bir ve tek bir kardinal i\c cerir, dolay\i s\i yla b\"uy\"ukl\"ukler, i\c cerilen kardinallere g\"ore s\i ralanabilir.  Asl\i nda, a\c sa\u g\i da g\"orece\u gimiz gibi, Se\c cim Aksiyomunu kullanmadan b\"uy\"ukl\"ukler h\^al\^a s\i ralanabilir, ancak \emph{iyi} s\i ralanamaz.

\begin{theorem}[Schr\"oder--Bernstein]\label{thm:SB}
T\"um $a$ ve $b$ k\"umeleri i\c cin
\begin{equation*}
a\preccurlyeq b\land b\preccurlyeq a\lto a\approx b.
\end{equation*}
\end{theorem}

\begin{proof}[Kan\i t (Zermelo \cite{Zermelo-invest}).]
$f\colon a\rightarrowtail b$ ve $g\colon b\rightarrowtail a$ olsun.  Bu durumda
\begin{align*}
(g\circ f)[a]&\included g[b]\included a,&
g[b]&\approx b.
\end{align*}
Biz $a\approx g[b]$ e\c slenikli\u gini kan\i tlayaca\u g\i z.  Sonu\c c olarak $a\approx b$ olacakt\i r.

$a$ k\"umesinden $g[b]$ k\"umesine giden bir $h$ e\c slemesini tan\i mlayabilirsek, herhalde $a$ k\"umesinin bir $c$ altk\"umesi i\c cin
\begin{equation}\label{h}
h=\{(x,x)\colon x\in c\}\cup\{(x,(g\circ f)(x))\colon x\in a\setminus c\}
\end{equation}
olacakt\i r.  O halde
\begin{equation}\label{cup}
c\cup(g\circ f)[a\setminus c]=g[b]
\end{equation}
olmal\i d\i r, \c c\"unk\"u $h[a]=g[b]$ olacakt\i r.  Ayr\i ca
\begin{equation}\label{cap}
c\cap(g\circ f)[a\setminus c]=\emptyset
\end{equation}
olmal\i d\i r, \c c\"unk\"u $h$ bir g\"omme olacakt\i r.  O zaman
\begin{equation*}
c=g[b]\setminus(g\circ f)[a\setminus c]
\end{equation*}
olmal\i d\i r.  $g\circ f$ birebir oldu\u gundan
\begin{equation*}
(g\circ f)[a\setminus c]=(g\circ f)[a]\setminus(g\circ f)[c],
\end{equation*}
dolay\i s\i yla
\begin{equation*}
c=g[b]\setminus\bigl((g\circ f)[a]\setminus(g\circ f)[c]\bigr)
\end{equation*}
olmal\i d\i r.  $(g\circ f)[c]\included(g\circ f)[a]\included g[b]$ oldu\u gundan
\begin{equation}\label{c}
c=\bigl(g[b]\setminus(g\circ f)[a]\bigr)\cup(g\circ f)[c]
\end{equation}
olmal\i d\i r.  Ters olarak, e\u ger $c$, \eqref c sat\i r\i ndaki gibiyse, o zaman \eqref{cup} ile \eqref{cap} sat\i rlar\i\ do\u grudur, ve sonu\c c olarak, $g\circ f$ birebir oldu\u gundan, \eqref h sat\i r\i ndaki gibi $h$ g\"ondermesi, $a$ k\"umesinden $g[b]$ k\"umesine giden bir e\c slemedir.

\c Simdi \"oyle bir $c$ k\"umesini bulmal\i y\i z.  O zaman
\begin{equation*}
\bm A=\bigl\{x\colon \bigl(g[b]\setminus(g\circ f)[a]\bigr)\cup(g\circ f)[x]\included x\included a\big\}
\end{equation*}
olsun.  Bu durumda $a\in\bm A$, dolay\i s\i yla $\bigcap\bm A$ bir k\"ume olmal\i d\i r.  Bu k\"ume $c$ olsun.  O zaman $c\in\bm A$ olmal\i d\i r (neden?).  E\u ger \eqref c sat\i r\i\ yanl\i\c s ise, o zaman
\begin{equation*}
d\in c\setminus\Bigl(\bigl(g[b]\setminus(g\circ f)[a]\bigr)\cup(g\circ f)[c]\Bigr)
\end{equation*}
c\"umlesini sa\u glayan bir $d$ vard\i r.  Bu durumda
\begin{align*}
c\setminus\{d\}&\in\bm A,&
\bigcap\bm A&\included c\setminus\{d\},&
c&\included c\setminus\{d\},&
d&\notin c.
\end{align*}
Bu bir \c celi\c skidir.  O zaman \eqref c sat\i r\i\ do\u gru olmal\i d\i r, ve $a\approx g[b]$, dolay\i s\i yla $a\approx b$.
\end{proof}

\begin{theorem}
T\"um $a$, $b$, ve $c$ k\"umeleri i\c cin
\begin{equation*}
a\prec b\land b\prec c\lto a\prec c.
\end{equation*}
\end{theorem}

\ktk

\section{Say\i lamaz sonsuzluk}

\begin{theorem}[Cantor]%
\index{Cantor!---'un Teoremi}%
\index{teorem!Cantor'un T---i}%
  Her $a$ k\"umesi i\c cin
  \begin{equation*}
    a\prec\pow a.
  \end{equation*}
\end{theorem}

\begin{proof}
 $\bigl\{\bigl(x,\{x\}\bigr)\colon
  x\in a\bigr\}$ g\"ondermesinin sayesinde $a\preccurlyeq\pow
  a$.  \c Simdi $f\colon a\rightarrowtail\pow a$ ve
  \begin{equation*}
    b=\{x\in a\colon x\notin f(x)\}
  \end{equation*}
olsun.  O zaman $a$ k\"umesinin her $c$ eleman\i\ i\c cin
\begin{equation*}
  c\in b\liff c\notin f(c).
\end{equation*}
\"Oyleyse $b\neq f(c)$.  Dolay\i s\i yla $b\notin f[a]$, ve $f$, e\c
sleme de\u gildir.  O zaman $a\not\approx\pow a$.
\end{proof}

Cantor'un Teoremi, k\"ume olmayan s\i n\i flar i\c cin yanl\i\c st\i r.  Mesela
\begin{equation*}
\universe\approx\pow{\universe};
\end{equation*}
asl\i nda $\universe=\pow{\universe}$.  

\begin{xca}
Cantor'un Teoreminin kan\i t\i nda, $a$ k\"umesinin k\"ume oldu\u gunu nas\i l kulland\i k?
\end{xca}

Sonsuz bir k\"umenin kuvvet s\i n\i f\i\ bir k\"umeyse, bu k\"ume say\i lamaz.

\begin{axiom}[Kuvvet K\"umesi]%
\index{aksiyom!Kuvvet K\"umesi A---u}
Her k\"umenin kuvvet s\i n\i f\i, bir k\"umedir, yani
\begin{equation*}
\Forall x\Exists y\Forall z\bigl(z\in y\liff\Forall w(w\in z\lto w\in x)\bigr)
\end{equation*}
c\"umlesi do\u grudur.
\end{axiom}

\begin{theorem}[Hartogs]
Her k\"ume i\c cin, bu k\"umeye g\"om\"ulemeyen bir ordinal vard\i r.
\end{theorem}

\begin{proof}
$a$ bir k\"ume olsun, ve $b$, $\pow a\times\pow{a\times a}$ \c carp\i m\i n \"oyle bir altk\"umesi olsun ki \c carp\i m\i n her $(c,s)$ eleman\i\ i\c cin
\begin{equation*}
(c,s)\in b\liff\text{$s$, $c$ k\"umesini iyi s\i ralar.}
\end{equation*}
E\u ger $(c,s)\in b$ ve $\alpha<\ord(c,s)$ ise, $c$ k\"umesinin bir $d$ altk\"umesi i\c cin
\begin{equation*}
\ord(d,s)=\alpha.
\end{equation*}
\"Oyleyse $\{\ord(c,s)\colon(c,s)\in b\}$ ge\c ci\c slidir, dolay\i s\i yla bir $\beta$ ordinaline e\c sittir.  O zaman $\beta$, $a$ k\"umesine g\"om\"ulemez.  Asl\i nda $f\colon\beta\xrightarrow{\preccurlyeq}a$ ise
\begin{align*}
c&=f[\beta],&
s&=\bigl\{\bigl(f(x),f(y)\bigr)\colon x<y<\beta\bigr\}
\end{align*}
olsun.  O zaman $(c,s)\in b$ ve $\ord(c,s)=\beta$, dolay\i s\i yla $\beta\in\beta$, ki bu imk\^ans\i zd\i r.
\end{proof}

\c Simdi her $\kappa$ kardinali i\c cin, bu kardinale g\"om\"ulemeyen \emph{en k\"u\c c\"uk} ordinal vard\i r.  Bu ordinal, bir kardinal olmal\i d\i r.  Bu kardinale, $\kappa$ kardinalinin \textbf{ard\i l\i}%
\index{ard\i l}
(veya \emph{kardinal ard\i l\i})
denir, ve bu ard\i l
\begin{equation*}
\kappa^+
\end{equation*}%
\glossary{$\kappa^+$}
ifadesiyle g\"osterilir.  B\"oylece $\kappa<\kappa^+$, ve her $\lambda$ kardinali i\c cin ya $\lambda\leq\kappa$ ya da $\kappa^+\leq\lambda$.

\begin{theorem}
Elemanlar\i\ kardinal olan bir k\"umenin supremumu, bir kardinaldir.
\end{theorem}

\ktk

\c Simdi
\begin{gather*}
	\aleph_0=\upomega,\\
	\aleph_{\beta+1}=(\aleph_{\beta})^+,\\
	\gamma\text{ limit}\lto\aleph_{\gamma}=\sup\{\aleph_{\xi}\colon\xi<\gamma\}
\end{gather*}%
\glossary{$\aleph_{\alpha}$}
ko\c sullar\i, $\on$ s\i n\i f\i nda bir $x\mapsto\aleph_x$ i\c slemini tan\i mlar.  (Burada $\aleph$, \.Ibranice \emph{alef} harf{}idir.)

\begin{theorem}
$\xi\mapsto\aleph_{\xi}$ i\c slemi, $\cn\setminus\upomega$
  s\i n\i f\i n\i\ \"orten ve normal bir g\"ondermedir.
\end{theorem}

\ktk

\section{Toplama ve \c carpma}

Tan\i ma g\"ore
\begin{align*}
\kappa\cardsum\lambda&=\card(\kappa+\lambda),&
\kappa\cardprod\lambda&=\card(\kappa\cdot\lambda).
\end{align*}%
\glossary{$\kappa\cardsum\lambda$}%
\glossary{$\kappa\cardprod\lambda$}
O zaman \ref{thm:ord+} ve \ref{thm:ord.} numaral\i\ teoremler sayesinde
\begin{align*}
\kappa\cardsum\lambda&=\card(\kappa\sqcup\lambda),&
%\bigl((\kappa\times\{0\})\cup(\lambda\times\{1\})\bigr),&
\kappa\cardprod\lambda&=\card(\kappa\times\lambda).
\end{align*}

Sonraki teoremler sayesinde kardinal hesapmalar\i\ kolayd\i r.

\begin{theorem}
  T\"um $\kappa$ ile $\lambda$ kardinalleri i\c cin
  \begin{align*}
    \kappa\cardsum\lambda&=\lambda\cardsum\kappa,&
    \kappa\cardprod\lambda&=\lambda\cardprod\kappa.
  \end{align*}
\end{theorem}

\ktk

\begin{theorem}
$2\leq\kappa\leq\lambda$ ise
\begin{equation*}
\lambda\leq\kappa\cardsum\lambda\leq\kappa\cardprod\lambda\leq\lambda\cardprod\lambda.
\end{equation*}
\end{theorem}

\ktk

\begin{theorem}
  $\bm F$, ordinallerde bir normal i\c slem, ve
  \begin{equation}\label{eqn:F-approx}
    \alpha\approx\beta\lto\bm F(\alpha)\approx\bm F(\beta) 
  \end{equation}
ise, o zaman t\"um sonsuz $\kappa$ kardinali i\c cin
  \begin{equation*}
    \bm F(\kappa)=\kappa.
  \end{equation*}
\end{theorem}

\begin{proof}
  $\bm F$ kesin artan oldu\u gundan, \ref{thm:inc} numaral\i\ teorem
  sayesinde $\kappa\leq\bm F(\kappa)$.  M\"umk\"unse $\kappa<\bm
  F(\kappa)$ olsun.  $\kappa$ bir ordinal ard\i l\i\ (neden?) ve $\bm
  F$ normal oldu\u gundan bir $\alpha$ i\c cin $\kappa<\bm
  F(\alpha)$.  O halde $\lambda=\card(\alpha)$ ise, \eqref{eqn:F-approx}
  sat\i r\i ndaki ko\c sula g\"ore
  \begin{equation*}
    \lambda<\kappa\leq\bm F(\lambda).
  \end{equation*}
B\"oylece $\{x\colon x\in\cn\land x<\bm F(x)\}$ s\i n\i f\i n\i n en
k\"u\c c\"uk eleman\i\ yoktur, dolay\i s\i yla s\i n\i f bo\c stur.
\end{proof}

\begin{theorem}
$\lambda$ sonsuz ise
\begin{equation*}
\lambda=\lambda\cardprod\lambda.
\end{equation*}
\end{theorem}

\begin{proof}
$\on\times\on$ \c carp\i m\i, \"oyle bir $<$ ba\u g\i nt\i s\i\ taraf\i
  ndan iyi s\i ralan\i r ki 
\begin{equation*}
\max(\alpha,\beta)<\max(\gamma,\delta)\lto(\alpha,\beta)<(\gamma,\delta).
\end{equation*}
Mesela
\begin{multline*}
(\alpha,\beta)<(\gamma,\delta)\liff \max(\alpha,\beta)<\max(\gamma,\delta)
\lor\beta<\delta<\gamma=\alpha\\
\lor(\beta<\alpha=\delta\land\gamma\leq\delta)
\lor\alpha<\gamma\leq\delta=\beta
\end{multline*}
olsun, onun i\c cin
\begin{equation*}
  \alpha\times\alpha=
  \bigl\{(x,y)\colon(x,y)\in\on\times\on\land(x,y)<(\alpha,0)\bigr\}.
\end{equation*}
O zaman $\xi\mapsto \ord(\xi\times\xi,<)$ g\"ondermesi, son
teoremdeki gibi bir $\bm F$ i\c slemidir, dolay\i s\i yla her sonsuz
$\lambda$ i\c cin $\lambda=\bm F(\lambda)$, ve \"ozel olarak
$\lambda\approx\lambda\times\lambda$. 
\end{proof}

Sonu\c c olarak
\begin{equation*}
\aleph_{\beta}\cardsum\aleph_{\gamma}=
\aleph_{\beta}\cardprod\aleph_{\gamma}=\aleph_{\max(\alpha,\beta)}.
\end{equation*}

\section{Ordinaller Kuvvetlerinin kardinalleri}

\begin{theorem}
  $\alpha$ sonsuz ise
  \begin{equation*}
\powf{\alpha}\approx\alpha^{\upomega}\approx\alpha.
  \end{equation*}
\end{theorem}

\begin{proof}
E\u ger $\upomega\leq\alpha$ ise, \ref{thm:ooo} numaral\i\ teoremin kan\i
  t\i ndaki gibi
  \begin{equation*}
    \alpha\preccurlyeq\alpha^{\upomega} \approx\exp(\alpha,\upomega)
    \included\powf{\alpha\times\upomega}\approx\powf{\alpha}
    \preccurlyeq\bigcup\{{}^x\alpha\colon x\in\upomega\}
    \preccurlyeq\alpha
  \end{equation*}
\c c\"unk\"u $\alpha\times\upomega\approx\alpha$ ve
$\alpha\times\alpha\approx\alpha$. 
\end{proof}

\begin{theorem}
  $\alpha$ ile $\beta$ sonsuz ise
  \begin{equation*}
    \alpha^{\beta}\approx\max(\alpha,\beta).
  \end{equation*}
\end{theorem}

\begin{proof}
  E\u ger $\bigl\{(\gamma_0,\delta_0),
  \dots,(\gamma_{n-1},\delta_{n-1})\bigr\}\in\exp(\alpha,\beta)$ ve
  $\gamma_0<\dots<\gamma_{n-1}$ ise, o zaman
  \begin{equation*}
    f\Bigl(\bigl\{(\gamma_0,\delta_0),
  \dots,(\gamma_{n-1},\delta_{n-1})\bigr\}\Bigr)
=\bigl(\{\gamma_0,\dots,\gamma_{n-1}\},(\delta_0,\dots,\delta_{n-1})\bigr)
  \end{equation*}
olsun.  O halde
$f\colon\exp(\alpha,\beta)
\xrightarrow{\preccurlyeq}\powf{\beta}\times\exp(\alpha,\upomega)$,
dolay\i s\i yla 
\begin{equation*}
\alpha^{\beta} \approx\exp(\alpha,\beta)
\preccurlyeq\powf{\beta}\times\exp(\alpha,\upomega)
\approx\beta\times\alpha \approx\max(\alpha,\beta).\qedhere
\end{equation*}
\end{proof}

\section{Kontin\"u Hipotezi}

$\N=\{x\in\upomega\colon x\geq1\}$%
\glossary{$\N$} 
olsun, ve $\N\times\N$ \c carp\i
m\i nda $\sim$, 
\begin{equation*}
  (a,b)\sim(c,d)\liff a\cdot d=b\cdot c
\end{equation*}
ko\c sulunu sa\u glayan ikili bir ba\u g\i nt\i\ olsun.  O zaman
$\sim$ bir denklik ba\u g\i nt\i s\i d\i r.
$(a,b)\in\N\times\N$ ise
\begin{equation*}
  a/b=\frac ab=\{(x,y)\in\N\times\N\colon(x,y)\sim(a,b)\}
\end{equation*}
olsun, ve
\begin{equation*}
  \Qp =\{x/y\colon(x,y)\in\N\times\N\}
\end{equation*}%
\glossary{$\Qp$}
olsun.  O zaman $x\mapsto x/1$ g\"ondermesi, $\N$ k\"umesini $\Qp$
k\"umesine g\"omer.  \.Ilkokulda \"o\u grendi\u gimiz gibi $\Qp$ k\"umesinde
\begin{align*}
  \frac ab+\frac cd&=\frac{a\cdot d+b\cdot c}{b\cdot d},&
\frac ab\cdot\frac cd&=\frac{ac}{bd},&
\frac{a/b}{c/d}&=\frac{a\cdot d}{b\cdot c}
\end{align*}
ve
\begin{equation*}
  \frac ab<\frac cd\liff a\cdot d<b\cdot c
\end{equation*}
tan\i mlar\i\ yap\i labilir.  
O zaman $<$, $\Qp$ k\"umesinin do\u
grusal s\i ralamas\i d\i r, ve burada $a<b$ ise
\begin{equation*}
a<\frac{2a+b}3<\frac{a+2b}3<b.
\end{equation*}
\c Simdi $\Rp$,%
\glossary{$\Rp$}
\begin{gather*}
  0\pincluded a\pincluded\Qp,\\
  x<y\land y\in a\lto x\in a,\\
  y\in a\lto\Exists z(y<z\land z\in a)
\end{gather*}
ko\c sullar\i n\i\ sa\u glayan $a$ k\"umelerinin k\"umesi olsun.  Bu
$\Rp$ k\"umesine \textbf{kontin\"u}
(\eng{continuum})
denebilir.

\begin{theorem}
  $\Rp\approx\pow{\upomega}$.
\end{theorem}

\begin{proof}
$n\in\upomega$ ve $\sigma\in{}^n2$ ise
\begin{equation*}
  f(\sigma)=1+\sum_{\xi<n}\frac{2\sigma(\xi)}{3^{\xi}}
\end{equation*}
olsun.  Yani $f(0)=1$ olsun, ve $\sigma\in{}^m2$ ise
\begin{align*}
f\Bigl(\sigma\cup\bigl\{(m,0)\bigr\}\Bigr)&=a_{\sigma},&
f\Bigl(\sigma\cup\bigl\{(m,1)\bigr\}\Bigr)&=a_{\sigma}+2/3^m
\end{align*}
olsun.  \c Simdi $\sigma\in{}^{\upomega}2$ ise
\begin{equation*}
  g(\sigma)=\Bigl\{x\in\Qp\colon\Exists x\bigl(x\in\upomega\land
  x<f(\sigma\restriction x)\bigr)\Bigr\}
\end{equation*}
olsun.  O zaman $g\colon{}^{\upomega}2\to\Rp$.  Asl\i nda $g$ bir
g\"ommedir, \c c\"unk\"u $\sigma\restriction n=\tau\restriction n$ ama
$\sigma(n)=0$ ve $\tau(n)=1$ ise, o zaman
\begin{equation*}
  f(\sigma)+1/3^n\in g(\tau)\setminus g(\sigma).
\end{equation*}
Ayr\i ca $\Qp\approx\upomega$ ve $\Rp\subseteq\pow{\Qp}$, dolay\i s\i yla
\begin{align*}
  {}^{\upomega}2\preccurlyeq\Rp\preccurlyeq\pow{\Qp}\approx\pow{\upomega}\approx{}^{\upomega}2.
\end{align*}
Schr\"oder--Bernstein Teoremine g\"ore ${}^{\upomega}2\approx\Rp$.
\end{proof}


\textbf{Kontin\"u Hipotezi}%
\index{Kontin\"u Hipotezi}
(\eng{Continuum Hypothesis}) veya $\ch$,%
\glossary{$\ch$}
\begin{equation*}
\aleph_1\approx\pow{\upomega}
\end{equation*}
c\"umlesidir.  \textbf{Genelle\c stirilmi\c s Kontin\"u Hipotezi}
(\eng{Generalized Continuum Hypothesis}) veya $\gch$,%
\glossary{$\gch$}
\begin{equation*}
\Forall x\Forall y(\upomega\preccurlyeq x\prec y\preccurlyeq\pow x\lto y\approx\pow x)
\end{equation*}
c\"umlesidir.  $\gch\lto\ch$ gerektirmesinin do\u grulu\u gu, apa\c c\i k de\u gildir.

Sonraki teorem i\c cin
\begin{align*}
\pow[0]a&=a,&
\pow[n+1]a&=\pow{\pow[n]a}
\end{align*}
\"ozyineli tan\i m\i n\i\ yapar\i z.

\begin{theorem}\label{thm:Sierpinski}
$\gch$ do\u gruysa, her k\"ume iyi s\i ralanabilir.
\end{theorem}

\begin{proof}
Her $a$ k\"umesi, $\pow{a\sqcup\upomega}$ k\"umesine g\"om\"ulebilir.  Asl\i nda
\begin{equation*}
x\mapsto\{(x,0)\}\colon a\xrightarrow{\preccurlyeq}\pow{a\sqcup\upomega}.
\end{equation*}
Bu nedenle $\pow{a\sqcup\upomega}$ iyi s\i ralanabilirse, $a$ k\"umesi de iyi s\i ralanabilir.  Ayr\i ca $a\sqcup\upomega\approx a\sqcup\upomega'$, dolay\i s\i yla
\begin{equation*}
\pow{a\sqcup\upomega}\approx\pow{a\sqcup\upomega'}\approx\pow{a\sqcup\upomega}\sqcup\pow{a\sqcup\upomega}
\end{equation*}
(neden?).  \"Oyleyse $a\approx a\sqcup a$ e\c slenikli\u gini varsayabiliriz.  O zaman Schr\"oder--Bernstein Teoremine g\"ore $a\sqcup\{0\}\approx a$, dolay\i s\i yla her $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
\pow[n]a\approx\pow[n]a\sqcup\pow[n]a.
\end{equation*}
\c Simdi $b\preccurlyeq\pow a$ olsun.  O zaman
\begin{equation*}
a\preccurlyeq b\sqcup a\preccurlyeq\pow a\sqcup\pow a\approx\pow a.
\end{equation*}
$\gch$ varsay\i m\i m\i za g\"ore $a\approx b\sqcup a$ veya $b\sqcup a\approx\pow a$.  Birinci durumda $b\preccurlyeq a$.  \.Ikinci durumda
\begin{equation*}
b\sqcup a\approx\pow a\sqcup\pow a\approx\pow a\times\pow a.
\end{equation*}
\c Simdi $f\colon b\sqcup a\xrightarrow{\approx}\pow a\times\pow a$ olsun, ve $\pi$, $(x,y)\mapsto x$ g\"ondermesi olsun.  Cantor'un Teoremine g\"ore $\pi\bigl[f[a\times\{1\}]\bigr]=\pow a$ olamaz, dolay\i s\i yla $\pow a\setminus\pi\bigl[f[a\times\{1\}]\bigr]$ fark\i n\i n $c$ eleman\i\ vard\i r.  E\u ger
\begin{equation*}
d=\{x\in b\colon\pi\circ f(x)=c\}
\end{equation*}
ise, o zaman $f[d]=\{c\}\times\pow a$, dolay\i s\i yla
\begin{align*}
d&\approx\pow a,&
\pow a&\preccurlyeq b,&
b\approx\pow a.
\end{align*}
K\i saca $a\sqcup a\approx a$ ve $b\preccurlyeq\pow a$ ise, ya $b\approx\pow a$ ya da $b\preccurlyeq a$.

Hartogs'un Teoremine g\"ore \"oyle bir $\beta$ ordinali vard\i r ki $\beta\preccurlyeq\pow[4]a$ (neden?), ama $a$, $\beta$ ordinaline g\"om\"ulemez.  O zaman $\beta$, ya $\pow[4]a$, ya $\pow[3]a$, ya $\pow[2]a$, ya $\pow a$ ile e\c sleniktir.  Her durumda $a\preccurlyeq\beta$, dolay\i s\i yla $a$ iyi s\i ralanabilir.
\end{proof}

Her k\"ume iyi s\i ralanabilirse, her k\"umenin ordinali vard\i r.  \"Ozel olarak $\pow{\upomega}$ k\"umesinin $\aleph_{\beta}$ ordinali vard\i r, ve $\beta\neq0$, dolay\i s\i yla
\begin{equation*}
\upomega\prec\aleph_1\preccurlyeq\pow{\upomega}.
\end{equation*}
O zaman $\gch\lto\ch$ do\u grudur.

\section{Kardinaller kuvvetleri}

Genelle\c stirilmi\c s Kontin\"u Hipotezini kabul etmeyece\u giz, ama son teoremin sonucunu kabul edece\u giz.  Se\c cim Aksiyomunun \c cok bi\c cimleri vard\i r; ama bizim i\c cin en uygun bi\c cimi, a\c sa\u g\i dad\i r.

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

Her \emph{say\i labilen} zaten iyi s\i ralanabildi.  \c Simdi her say\i lamaz sonsuzluktaki k\"ume iyi s\i ralanabilir, yani bundan $0$ adl\i\ bir eleman\i\ se\c cilebilir, ve ondan sonra $1$ adl\i\ eleman\i, vesaire.

\"Oyleyse her k\"umenin kardinali vard\i r.  
Kuvvet K\"umesi Aksiyomunun sayesinde ${}^{\beta}\alpha$ s\i n\i
f\i\ bir k\"umedir, \c c\"unk\"u
\begin{equation*}
  {}^{\beta}\alpha\included\pow{\beta\times\alpha}.
\end{equation*}
O zaman
\begin{equation*}
  \kappa^{\lambda}=\card({}^{\lambda}\kappa)
\end{equation*}%
\glossary{$\kappa^{\lambda}$}
tan\i m\i n\i\ yapabiliriz.  (Bu kuvvet, \emph{ordinaller} kuvveti de\u gildir.)

\begin{theorem}
T\"um $\kappa$, $\lambda$, $\mu$, ve $\nu$ kardinalleri i\c cin
\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*}
\end{theorem}

\ktk

\begin{theorem}
Her $\kappa$ i\c cin
\begin{equation*}
\kappa<2^{\kappa}.
\end{equation*}
\end{theorem}

\ktk

\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}

Teoremin \eqref{eqn:kl} ile \eqref{eqn:lk} gerektirmelerinde $\kappa\leq2^{\lambda}$ ile $\lambda\leq\kappa$ ko\c sullar\i, ayn\i\ anda do\u gru olabilir.
Bu durumda \eqref{eqn:kl} gerektirmesi, di\u gerinden daha \c cok
bilgi verir.  B\"oylece \eqref{eqn:lk} sat\i r\i ndaki c\"umlenin yerine 
\begin{equation*}
2^{\lambda}<\kappa\lto\kappa\leq\kappa^{\lambda}\leq2^{\kappa}
\end{equation*}
c\"umlesi konulabilir.  \"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*}

\c Simdi $\lambda\leq\kappa$ (veya $2^{\lambda}<\kappa$) durumunu iki duruma b\"olece\u giz.

\subsection{Kof{}inall\i k}

Sonsuz bir $\kappa$ kardinali limit ordinal 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 supremumudur.  \"Orne\u gin $\upomega<\aleph_{\upomega}$, ama
\begin{equation*}
\aleph_{\upomega}=\sup\{\aleph_x\colon x\in\upomega\}.
\end{equation*}

\begin{theorem}
$b\included\alpha$ olsun.  A\c sa\u g\i daki ko\c sullar, birbirine denktir.
\begin{compactenum}
\item
$\alpha$ limit veya $0$ ise $\alpha=\sup(b)$, ve $\alpha=\gamma+1$ ise $\gamma\in b$.
\item
Her durumda
\begin{equation*}%\label{eqn:adb}
\alpha=\sup_{\xi\in b}(\xi+1)=\bigcup_{\xi\in b}\{\eta\in\alpha\colon\eta\leq\xi\}.
\end{equation*}
\item
$\alpha$ ordinalinin her $\gamma$ eleman\i\ i\c cin, $b$ k\"umesinin $\gamma\leq\delta$ ko\c sulunu sa\u glayan bir $\delta$ eleman\i\ vard\i r.
\end{compactenum}
\end{theorem}

\ktk

Teoremdeki ko\c sullar do\u gruysa $b$, $\alpha$ ordinalinde \textbf{s\i n\i rs\i zd\i r}
\index{s\i n\i rs\i z}
(\eng{unbounded}).%
\footnote{\cite[13.\ b\"ol.]{Nesin-AKK} kayna\u g\i nda $b$, $\alpha$ ordinalinin bir \emph{kof{}inal} altk\"umesidir.}
\"Orne\u gin bir ordinal, kendisinde s\i n\i rs\i zd\i r.  Bir ordinalin s\i n\i rs\i z altk\"umelerinin en k\"u\c c\"uk kardinaline, ordinalin
\textbf{kof{}inalli\u gi}%
\index{kof{}inallik}
(\eng{cofinality})
denir, ve bu kardinal,
\begin{equation*}
\cof{\alpha}
\end{equation*}%
\glossary{$\cof{\alpha}$}
olarak yaz\i labilir.  Yani
\begin{equation*}
\cof{\alpha}=\min\{\card(x)\colon x\included\alpha\land\sup_{\eta\in x}(\eta+1)=\alpha\}.
\end{equation*}
O zaman
\begin{align*}
\cof{\alpha}&\leq\alpha,&
	\cof{\alpha+1}&=1,
\end{align*}
ve ayr\i ca\footnote{Bu denklikte $f$, bir g\"onderme de\u gi\c skenidir.}
\begin{equation*}
\cof{\alpha} =\min\biggl\{\beta\colon\Exists f \Bigl(f\colon\beta\to\alpha \land\sup_{\xi<\beta}\bigl(f(\xi)+1\bigr)=\alpha\Bigr)\biggr\}.
\end{equation*}

\begin{theorem}
Her $\alpha$ ordinali i\c cin, kof{}inalli\u ginden giden \emph{kesin artan} ve de\u ger k\"umesi $\alpha$ ordinalinde s\i n\i rs\i z olan bir g\"onderme vard\i r.
\end{theorem}

\begin{proof}
$f\colon\cof{\alpha}\to\alpha$ olsun, ve de\u ger k\"umesi $\alpha$ ordinalinde s\i n\i rs\i z olsun.  O zaman $\cof{\alpha}$ k\"umesinde bir $g$ g\"ondermesinin
\begin{equation*}
g(\beta)=\max\Bigl(f(\beta),\sup_{\xi<\beta}\bigl(g(\xi)+1\bigr)\Bigr)
\end{equation*}
\"ozyineli tan\i m\i\ olsun.  E\u ger $\beta<\cof{\alpha}$ ve $g[\beta]\included\alpha$ ise, o zaman $g[\beta]$, $\alpha$ ordinalinde s\i n\i rs\i z olamaz, 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}
E\u ger 
%$\beta$ bir limit, 
$f\colon\alpha\to\beta$, $f$ \emph{kesin artan,} ve $f[\alpha]$, $\beta$ ordinalinde s\i n\i rs\i z ise, o zaman
\begin{equation*}
\cof{\alpha}=\cof{\beta}.
\end{equation*}
\end{theorem}

\begin{proof}
Bir $\gamma$ i\c cin $g\colon\gamma\to\alpha$ olsun, ve
$g[\gamma]$, $\alpha$ ordinalinde s\i n\i rs\i z 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.  $\delta<\beta$ olsun.
Hipoteze g\"ore $\alpha$ ordinalinin bir $\theta$ eleman\i\ i\c cin
\begin{equation*}
\delta\leq f(\theta).
\end{equation*}
O zaman $\gamma$ ordinalinin bir $\iota$ eleman\i\ i\c cin
\begin{align*}
\theta&\leq g(\iota),&
\delta\leq f(\theta)&\leq f\bigl(g(\iota)\bigr).
\end{align*}
\"Oyleyse $f\circ g$ g\"ondermesinin de\u ger k\"umesi, $\beta$ ordinalinde s\i n\i rs\i zd\i r.  Sonu\c c olarak
\begin{equation*}
\cof{\beta}\leq\cof{\alpha}.
\end{equation*}

\c Simdi $h\colon\gamma\to\beta$ olsun, ve $h[\gamma]$, $\beta$ ordinalinde s\i n\i rs\i z olsun.
$\delta<\gamma$ ise
\begin{equation*}
k(\delta)=\min\{\xi\in\alpha\colon h(\delta)\leq f(\xi)\}
\end{equation*}
olsun.  O zaman $k\colon\gamma\to\alpha$.  E\u ger $\theta\in\alpha$ ise, o zaman $\gamma$ ordinalinin \"oyle bir $\delta$ eleman\i\ vard\i r ki
\begin{align*}
f(\theta)&\leq h(\delta)\leq f(k(\delta)),&
\theta&\leq k(\delta).
\end{align*}
\"Oyleyse $k[\gamma]$, $\alpha$ ordinalinde s\i n\i rs\i zd\i r.  Sonu\c c olarak
$\cof{\alpha}\leq\cof{\beta}$ ve asl\i nda $\cof{\alpha}=\cof{\beta}$.
\end{proof}

\"Ozel olarak $\bm F$ normal ve $\alpha$ limitse
\begin{equation*}
  \cof{\bm F(\alpha)}=\cof{\alpha}.
\end{equation*}
\"Orne\u gin $\alpha$ limit, $\gamma\geq1$, ve $\delta\geq2$ ise
\begin{equation*}
  \cof{\alpha} 
=\cof{\beta+\alpha} 
=\cof{\gamma\cdot\alpha}
=\cof{\delta^{\alpha}}
=\cof{\aleph_{\alpha}}.  
\end{equation*}
B\"oylece e\u ger 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}
1,\text{ e\u ger $\alpha_n=0$ ise},&\\
\cof{\upomega^{\alpha_n}\cdot a_n}
=\cof{\upomega^{\alpha_n}}
=\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{cases}
\end{equation*}
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*}
\card\bigl(\sup(f[\beta])\bigr)\leq\card(\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}

\subsection{Hesapmalar}

\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*}
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 \ref{thm:succ-cof} numaral\i\ teoremin kan\i t\i ndaki gibi
\begin{equation*}
{}^{\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{equation*}
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*}

\c Simdi
\begin{gather*}
\beth_0=\upomega,\\
\beth_{\alpha+1}=2^{\beth_{\alpha}},\\
\beta\text{ limit}\lto\beth_{\beta}=\sup\{\beth_{\xi}\colon\xi<\beta\}
\end{gather*}%
\glossary{$\beth_{\alpha}$}
olsun.  (Burada $\beth$, \.Ibranice \emph{beth} harf{}idir.)

\begin{theorem}
T\"um $\kappa$ ve $\lambda$ i\c cin
\begin{gather*}
1\leq\lambda\leq\beth_{\alpha}\lto{\beth_{\alpha+1}}^{\lambda}=\beth_{\alpha+1},\\
2\leq\kappa\leq\beth_{\alpha+1}\lto\kappa^{\beth_{\alpha}}\leq\beth_{\alpha+1}.
\end{gather*}
\end{theorem}

\ktk

\begin{theorem}
$\gch$ do\u grudur ancak ve ancak her $\alpha$ ordinali i\c cin
\begin{equation*}
\aleph_{\alpha}=\beth_{\alpha}.
\end{equation*}
\end{theorem}

\ktk

%\end{comment}

\backmatter

%\bibliographystyle{plain}
%\bibliography{../../references}

%\bibliography{../references}

\def\rasp{\leavevmode\raise.45ex\hbox{$\rhook$}} \def\cprime{$'$}
  \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'$}
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\end{thebibliography}


\addchap{\.I\c saretler}

%\begin{multicols}{2}
\renewcommand{\arraystretch}{1.3}
\begin{longtable}{c r}
%\input{kumeler-kurami.glo}
\glossaryentry{$x$, $y$, $z$, \dots}{23}
\glossaryentry{$a$, $b$, $c$, \dots}{23}
\glossaryentry{$=$}{29}
\glossaryentry{$\bm A$, $\bm B$, $\bm C$, \dots}{33}
\glossaryentry{$\bm A\cap\bm B$}{35}
\glossaryentry{$\bm A\cup\bm B$}{35}
\glossaryentry{$\bigcap\bm A$}{37}
\glossaryentry{$\bigcup\bm A$}{37}
\glossaryentry{$\pow{\bm A}$}{37}
\glossaryentry{$\universe$}{37}
\glossaryentry{$a'$}{40}
\glossaryentry{$0$}{41}
\glossaryentry{$\bm{\Omega}$}{41}
\glossaryentry{$\upomega$}{43}
\glossaryentry{$(a,b)$}{45}
\glossaryentry{$\conv{\bm R}$}{47}
\glossaryentry{$\bm A\times\bm B$}{47}
\glossaryentry{$x\mapsto\bm F(x)$}{48}
\glossaryentry{$\on$}{51}
\glossaryentry{$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$, $\iota$}{52}
\glossaryentry{$\xi$}{52}
\glossaryentry{$\alpha+\beta$}{64}
\glossaryentry{$\alpha\cdot\beta$}{69}
\glossaryentry{$\eta$}{71}
\glossaryentry{$\alpha^{\beta}$}{73}
\glossaryentry{$\zeta$}{74}
\glossaryentry{$\beta_{\alpha}$}{75}
\glossaryentry{${}^b\bm A$}{76}
\glossaryentry{$\deg(\alpha)$}{77}
\glossaryentry{$\bm R/\bm S$}{81}
\glossaryentry{$\bm B\circ\bm F$}{81}
\glossaryentry{$\inv{\bm F}$}{81}
\glossaryentry{$a\approx b$}{82}
\glossaryentry{$\ord(a,\bm S)$}{83}
\glossaryentry{$\card(a)$}{84}
\glossaryentry{$\cn$}{84}
\glossaryentry{$\kappa$, $\lambda$, $\mu$, $\nu$}{84}
\glossaryentry{$\bm F\colon\bm A\twoheadrightarrow\bm B$}{86}
\glossaryentry{$\bm F\colon\bm A\rightarrowtail\bm B$}{86}
\glossaryentry{$a\preccurlyeq\bm B$}{86}
\glossaryentry{$a\sqcup b$}{88}
\glossaryentry{$\exp(\alpha,\beta)$}{90}
\glossaryentry{$\powf{\bm A}$}{91}
\glossaryentry{$a\prec\bm B$}{92}
\glossaryentry{$\kappa^+$}{96}
\glossaryentry{$\aleph_{\alpha}$}{96}
\glossaryentry{$\kappa\cardsum\lambda$}{97}
\glossaryentry{$\kappa\cardprod\lambda$}{97}
\glossaryentry{$\N$}{99}
\glossaryentry{$\Qp$}{100}
\glossaryentry{$\Rp$}{100}
\glossaryentry{$\ch$}{101}
\glossaryentry{$\gch$}{101}
\glossaryentry{$\kappa^{\lambda}$}{104}
\glossaryentry{$\cof{\alpha}$}{106}
\glossaryentry{$\beth_{\alpha}$}{110}
\end{longtable}
%\end{multicols}

%\printindex
\begin{theindex}
  \providecommand*\lettergroupDefault[1]{}
  \providecommand*\lettergroup[1]{%
      \par\textbf{#1}\par
      \nopagebreak
  }

  \lettergroup{A}
  \item aksiyom
    \subitem Ay\i rma A---u, 16, 20, 42
    \subitem Bile\c sim A---u, 15, 60
    \subitem Biti\c stirme A---u, 9, 21, 40
    \subitem Bo\c s K\"ume A---u, 9, 39
    \subitem E\c sitlik A---u, 31
    \subitem Kuvvet K\"umesi A---u, 17, 95
    \subitem mant\i ksal ---, 32
    \subitem Peano A---lar\i, Dedekind--Peano A---lar\i, 45
    \subitem Se\c cim A---u, 15, 20, 21
    \subitem Sonsuzluk A---u, 11, 21, 42
    \subitem Temel K\"umeler A---u, 20
    \subitem Temellendirme A---u, 21
    \subitem Uzama A---u, 20, 32
    \subitem Yerle\c stirme A---u, 14, 21, 61
    \subitem Zermelo--Fraenkel A---lar\i, 20
  \item altk\"ume, 17
  \item alts\i n\i f, 16, 34
  \item ard\i l, 40, 48, 96
  \item artan g\"onderme, 67
  \item ayra\c c, 23
  \item ayr\i k bile\c sim, 88
  \item ayr\i lma, 24

  \indexspace

  \lettergroup{B}
  \item ba\u glay\i c\i, 23
  \item ba\u g\i nt\i, 7, 46
    \subitem denklik ---s\i, 18
    \subitem ters ---, 46
  \item bile\c sim, 35
    \subitem B--- Aksiyomu, 15
  \item bile\c ske, 81
  \item birle\c sme, 24
  \item bo\c s k\"ume, 9
  \item bo\c s s\i n\i f, 37
  \item b\"uy\"ukl\"uk, 92

  \indexspace

  \lettergroup{C}
  \item Cantor
    \subitem --- normal bi\c cimi, 77
    \subitem ---'un Teoremi, 94
  \item Cantor normal bi\c cimi, 13
  \item Cantor'un Teoremi, 17
  \item c\"umle, 13, 25

  \indexspace

  \lettergroup{Ç}
  \item \c carp\i m, 47

  \indexspace

  \lettergroup{D}
  \item De Morgan Kurallar\i, 36
  \item de\u ger s\i n\i f\i, 47
  \item de\u gilleme, 24
  \item de\u gi\c sken, 23
    \subitem ba\u glant\i l\i, 14
    \subitem serbest, 13
  \item denk, 28
    \subitem ---lik s\i n\i f\i, ba\u g\i nt\i s\i, 18
  \item denklik, 24, 31
  \item derece, 77
  \item do\u gruluk, 26
  \item do\u grusal, 49

  \indexspace

  \lettergroup{E}
  \item eleman, 6
  \item e\c sitlik, 14, 30
  \item e\c sleme, 15, 81
  \item e\c slenik, 15, 81
  \item e\c sle\c sme, 7
  \item evetleme, 24
  \item evrensel s\i n\i f, 14

  \indexspace

  \lettergroup{F}
  \item fark, 36
  \item form\"ul, 13, 23

  \indexspace

  \lettergroup{G}
  \item ge\c cis, 25
  \item ge\c ci\c sli, 43
  \item genelle\c stirme, 24
  \item gerektirme, 24
  \item g\"omme, 86
  \item g\"onderme, 47
    \subitem ard\i l ---si, 48
    \subitem birebir ---, 86
    \subitem injektif ---, 86
    \subitem \"ozde\c slik ---si, 48
    \subitem sabit ---, 48
    \subitem ters ---, 81
  \item g\"or\"unt\"u, 60

  \indexspace

  \lettergroup{©}
  \item i\c   cerilme, 30
  \item i\c cerme, 8
  \item ikili, 18, 45
  \item ili\c   ski, 7
  \item i\c slem, 53
  \item iyi s\i ralama, 50
  \item izomorf, 6

  \indexspace

  \lettergroup{K}
  \item kapal\i, 33
  \item kapsama, 16
  \item kardinal, 8, 16, 84
  \item kesi\c sim, 35
  \item kof{}inallik, 106
  \item Kontin\"u Hipotezi, 20, 101
  \item k\"u\c c\"uk, 50
  \item k\"ume, 6, 8
    \subitem --- terimi, 33
    \subitem bo\c s ---, 9

  \indexspace

  \lettergroup{L}
  \item limit, 12, 53

  \indexspace

  \lettergroup{M}
  \item mant\i ksal aksiyom, 32

  \indexspace

  \lettergroup{N}
  \item niceleyici, 23
  \item normal
    \subitem --- i\c slem, 67
    \subitem Cantor --- bi\c cimi, 13, 77

  \indexspace

  \lettergroup{O}
  \item ordinal, 8, 51, 83
    \subitem limit, 53

  \indexspace

  \lettergroup{Ö}
  \item \"o\u ge, 6
  \item \"ornekleme, 24
  \item \"orten, 86
  \item \"ozde\c slik g\"ondermesi, 48
  \item \"ozyineli tan\i m, 10, 54

  \indexspace

  \lettergroup{P}
  \item paradoks
    \subitem Burali-Forti P---u, 15, 52
    \subitem Russell P---u, 13, 34

  \indexspace

  \lettergroup{R}
  \item rakam, 74
  \item rek\"ursif tan\i m, 10

  \indexspace

  \lettergroup{S}
  \item sabit, 23
    \subitem --- g\"onderme, 48
  \item sa\u glamak, 32
  \item say\i
    \subitem gercel ---lar, 18
    \subitem kesirli   ---lar, 18
    \subitem tam---lar\i, 7, 17
    \subitem von Neumann do\u gal ---lar\i, 10, 43
  \item say\i labilir, say\i lamaz, 87
  \item s\i n\i f, 13, 33
    \subitem --- terimi, 33
    \subitem bo\c s ---, 37
    \subitem denklik ---\i, 18
  \item s\i n\i rlama, 62
  \item s\i n\i rs\i z, 106
  \item sonlu, sonsuz, 84
  \item s\i ra, 6
    \subitem iyi ---lama, 50
    \subitem ---lama, 6
    \subitem ---l\i\ ikili, 18, 45
  \item s\i ralama, 49
  \item s\"urekli g\"onderme, 67

  \indexspace

  \lettergroup{T}
  \item taban, 74
  \item tan\i m s\i n\i f\i, 47
  \item tan\i mlama, 13, 33, 60
  \item teorem
    \subitem Burali-Forti Paradoksu, 15, 52
    \subitem Cantor'un T---i, 17, 94
    \subitem De Morgan Kurallar\i, 36
    \subitem Fermat'n\i n T---i, 56
    \subitem G\"odel Eksiklik T---i, 15
    \subitem \"Ozyineleme T---i, 56, 62
    \subitem Russell Paradoksu, 13, 34
    \subitem Schr\"oder--Bernstein T---i, 20
    \subitem Tarski Do\u grulu\u gun Tan\i mlanamamas\i\ T---i, 15
    \subitem T\"umevar\i m T---i, 61
  \item terim, 23
    \subitem kapal\i\ ---, 33
    \subitem k\"ume ---i, 33, 39
    \subitem s\i n\i f ---i, 33, 38
  \item ters, 46, 81
  \item topluluk, 6
  \item tutarl\i, 21
  \item t\"umel evetleme, 24
  \item t\"umevar\i m, 43
  \item t\"umleyen, 36

  \indexspace

  \lettergroup{V}
  \item von Neumann do\u gal say\i lar\i, 10, 43

  \indexspace

  \lettergroup{Y}
  \item yanl\i\c sl\i k, 26
  \item yaz\i l\i m, 74
  \item Yerle\c stirme Aksiyomu, 14
  \item y\"uklem, 23

\end{theindex}



\end{document}