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\begin{document}
\title{Sonsuzk\"u\c c\"uk Analiz}
\subtitle{\textit{Infinitesimal Analysis}}
\author{David Pierce}
\date{17 \c Subat 2020}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{david.pierce@msgsu.edu.tr}}

\maketitle

\addchap{Okura Not}

Bu metinde,
ger\c cel analizin temel kavramlar\i n\i\ ve teoremlerini
``standart olmayan,'' \emph{sonsuzk\"u\c c\"uk} y\"ontemi ile
en basit \c sekilde
a\c c\i klamaya \c cal\i\c s\i yorum.

%\"Orne\u gin
B\"ol\"um \numarada{ch:continuity}
Spivak'\i n ``\"u\c c zor teoremi''
\cite[sayfa 108]{Spivak-Calc},
%a\c sa\u g\i daki
Teoremler \ref{thm:ivt}, \ref{thm:icc}, ve \ref{thm:max-min} olur.

\"O\u grencilerden
baz\i s\i\
``standart,'' \emph{epsilon-delta} y\"ontemi,
baz\i s\i\ sonsuzk\"u\c c\"uk y\"ontemi
tercih edebilir.
Bundan dolay\i\ bir \"o\u gretmen,
y\"ontemlerin ikisini bilmek zorundad\i r.

Epsilon-delta ve sonsuzk\"u\c c\"uk y\"ontemlerin denkli\u gini,
B\"ol\"um \numarada{ch:intro}n sonra,
sadece B\"ol\"um \numarada{ch:app} bahsedece\u gim.

Metinde bir teoremin kan\i t\i\ yoksa,
teoremi kan\i tlamay\i\ size b\i rakm\i\c st\i m.

Teknik terimler, \emph{italik} olabildi\u gi halde
tan\i mlan\i nca \textbf{siyaht\i r.}
\.Ingilizceden gelen terimler, \textsl{e\u gimlidir.}

Abraham Robinson
sonsuzk\"u\c c\"uk y\"ontemi
sert (\eng{rigorous}) bir \c sekilde meydana koydu \cite{MR1373196}.
Keisler \cite{Keisler-calculus,Keisler-foundations}
ve Henle ile Kleinberg \cite{MR1999278}
taraf\i ndan
yaz\i lm\i\c s
sonsuzk\"u\c c\"uk y\"ontemi kullanan ders kitaplar\i\ vard\i r.

\tableofcontents
%\listoffigures

\chapter{Giri\c s}\label{ch:intro}

Epsilon--delta y\"ontemini kullanmadan
limitleri tan\i mlay\i p
kalk\"ul\"us\"un temel teoremlerini kan\i tlayaca\u g\i z.

Geleneksel, ``standart'' tan\i ma g\"ore
bir $a$ noktas\i nda bir $f$ fonksiyonunun limitinin $L$ olmas\i\ i\c cin,
yani
\begin{equation*}%\label{eqn:lim}
  \lim_af=L
\end{equation*}
i\c cin,
  \begin{multline*}
    \Forall{\epsilon}\Bigl(\epsilon>0\lto\Exists{\delta}
    \bigl(\delta>0\land\Forall x\\
    (0<\abs{x-a}<\delta\lto\abs{f(x)-L}<\epsilon)\bigr)\Bigr)
  \end{multline*}
  ko\c sulu yeter ve gerektir.
  \emph{\"Onekli yasal bi\c cimde} ko\c sul
  \begin{multline}\label{eqn:ed}
    \Forall{\epsilon}\Exists{\delta}\Forall x
    \bigl(\epsilon>0\lto\delta>0
    \land{}\\
    (0<\abs{x-a}<\delta\lto\abs{f(x)-L}<\epsilon)\bigr)
  \end{multline}
  olur.
  Burada \"u\c c niceleyici vard\i r:
  $\forall{\epsilon}$, $\exists{\delta}$, ve $\forall x$.
  Ko\c sul \esitligin{eqn:ed} yerine
  niceleyicilerden ikisini \c c\i kararak
  \begin{equation}\label{eqn:ns}
    \Forall{[\seq x]}([\seq x]\approx a\land[\seq x]\neq a\lto
    \stf[\seq x]\approx L)
  \end{equation}
  ko\c sulunu kullanaca\u g\i z.
  Burada
  \begin{itemize}
  \item
    $\seq x$, bir $(x_k\colon k\in\N)$ dizisidir;
  \item
    bir $\denk$ denklik ba\u g\i nt\i s\i na g\"ore
    $[\seq x]$,
    $\seq x$'in denklik s\i n\i f\i d\i r;
  \item
    $\seq x\denk\seq y$ i\c cin
    $\{k\in\N\colon x_k\neq y_k\}$ k\"umesinin sonlu olmas\i\
    yeter;
  \item
    $a$, $(a,a,a,\dots)$ sabit dizisinin
    $[a,a,a,\dots]$ denklik s\i n\i f\i\ olarak say\i l\i r;
  \item
    $\stf[\seq x]=[f(x_k)\colon k\in\N]$;
  \item
    $\approx$,
        \emph{sonsuzyak\i nl\i k}
        denklik ba\u g\i nt\i s\i d\i r;
      \item
        $[\seq x]\approx[\seq y]$ i\c cin
        $\{k\in\N\colon\abs{x_k-y_k}\geq1/n\}$
        k\"umesinin her $n$ sayma say\i s\i\ i\c cin sonlu olmas\i\
    yeter.
  \end{itemize}
  Yukar\i da
  $\seq x\denk\seq y$ ve $[\seq x]\approx[\seq y]$ i\c cin
   verilen
  yeter ko\c sullar gerekli de olarak al\i n\i rsa,
  o zaman ko\c sullar \eqref{eqn:ed} ve \eqref{eqn:ns} denktir.
  Asl\i nda kullanaca\u g\i m\i z tan\i ma g\"ore
  \begin{itemize}
  \item 
    $\seq x\denk\seq y$ i\c cin
    $\{k\in\N\colon x_k= y_k\}$ k\"umesinin \emph{b\"uy\"uk} olmas\i\
    gereklidir;
      \item
        $[\seq x]\approx[\seq y]$ i\c cin
        $\{k\in\N\colon\abs{x_k-y_k}<1/n\}$
        k\"umesinin her $n$ sayma say\i s\i\ i\c cin b\"uy\"uk olmas\i\
    gereklidir.
  \end{itemize}
    Burada $\N$'nin altk\"umelerinden
    \begin{enumerate}[1)]
    \item
      hi\c c sonlu k\"ume b\"uy\"uk de\u gildir;
    \item
      b\"uy\"uk olmayan bir k\"umenin t\"umleyeni b\"uy\"ukt\"ur;
    \item
      iki b\"uy\"uk k\"umenin kesi\c simi de b\"uy\"ukt\"ur.
    \end{enumerate}
    Bu tan\i m alt\i nda $[\seq x]$ s\i n\i flar\i\
    $\R$'yi kapsayan,
    $\R$'nin baz\i\ \"ozelliklerini payla\c san,
    \emph{do\u grusal s\i ralanm\i\c s bir cismi} olu\c sturur.

\chapter{Ger\c cel say\i lar}

\section{Ger\c cel say\i lar}

\emph{Tam do\u grusal s\i ralanm\i\c s bir $\R$ cisminin}
var oldu\u gunu varsay\i yoruz;
$\R$'nin elemanlar\i, \textbf{ger\c cel say\i lard\i r}
(veya \emph{reel say\i lard\i r}).
\begin{enumerate}
\item 
\textbf{Do\u grusal s\i ralanm\i\c s} oldu\u gundan
$\R$'nin elemanlar\i, bir do\u gru olu\c sturur.
E\u ger $a$, $b$'nin solunda ise,
$a<b$ veya $b>a$ yaz\i l\i r.
Bu durumda $a$, $b$'den \textbf{k\"u\c c\"ukt\"ur}
ve $b$, $a$'dan \textbf{b\"uy\"ukt\"ur.}
Her durumda
  \begin{gather*}
    a\not<a,\\
    a<b\land b<c\lto a<c,\\
    a<b\lor a=b\lor a>b.
  \end{gather*}
\item
Do\u gru olarak $\R$ \textbf{tam} oldu\u gundan,
e\u ger
\Sekilde{fig:completeness}ki gibi
\begin{figure}[t]
  \centering
  \begin{pspicture}(-5,0)(5,1)
    \psset{dotsize=4pt 2}
    \psline{o->}(-1,1)(-5,1)
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    \psline{*->}(-1,0)(-5,0)
  \end{pspicture}
  \caption{Ger\c cel say\i lar\i n taml\i\u g\i}
  \label{fig:completeness}
\end{figure}
ger\c cel say\i lar do\u grusu iki par\c caya k\i r\i l\i rsa,
o zaman par\c calar\i n birinin u\c c noktas\i\ vard\i r.
Demek $\R$'nin her bo\c s olmayan, \"usts\i n\i r\i\ olan
$A$ altk\"umesinin
en k\"u\c c\"uk \"usts\i n\i r\i\
veya \textbf{supremumu} vard\i r, yani
\begin{multline*}
  \Exists xx\in A\land
  \Exists x\Forall y(y\in A\lto y\leq x)\lto{}\\
  \Exists x\Bigl(\Forall y(y\in A\lto y\leq x)\land{}\\
  \Forall y\bigl(\Forall z(z\in A\lto z\leq y)\lto x\leq y\bigr)\Bigr)
\end{multline*}
veya
\begin{multline*}
  \Exists{x_0}x_0\in A\land
  \Exists{x_1}\Forall{x_2}(x_2\in A\lto x_2\leq x_1)\lto{}\\
  \Exists{x_3}\Bigl(\Forall{x_4}(x_4\in A\lto x_4\leq x_3)\land{}\\
  \Forall{x_5}\bigl(\Forall{x_6}(x_6\in A\lto x_6\leq x_5)\lto x_3\leq x_5\bigr)\Bigr)
\end{multline*}
veya
\begin{multline*}
  \Forall{(x_0,x_1)}\Exists{(x_2,x_3)}\Forall{(x_4,x_5)}\Exists{x_6}\\
  \Bigl(x_0\in A\land(x_2\in A\lto x_2\leq x_1)\lto{}\\
  (x_4\in A\lto x_4\leq x_3)\land{}\\
  \bigl((x_6\in A\lto x_6\leq x_5)\lto x_3\leq x_5\bigr)\Bigr).
\end{multline*}
%Bir k\"umenin en k\"u\c c\"uk bir \"usts\i n\i r, k\"umenin \textbf{supremumudur.}
$\R$'nin taml\i\u g\i n\i\
\Teoremde{thm:stp} kullanaca\u g\i z.
\item
$\R$ bir \textbf{cisim} oldu\u gundan
  toplama ve \c carpma i\c slemi vard\i r,
  ve ayr\i ca
\begin{enumerate}
\item
  toplama alt\i nda $\R$ \textbf{abelyan bir gruptur,}
  yani her durumda
  \begin{gather*}
    a+b=b+a,\\a+(b+c)=(a+b)+c,\\a+0=a,\\a+(-a)=0;
  \end{gather*}
\item
  \c carpma alt\i nda $\R\setminus\{0\}$ abelyan bir gruptur,
  dolay\i s\i yla her durumda
  \begin{gather*}
    ab=ba,\\
    a(bc)=(ab)c,\\
    a\cdot1=a,\\
    a\cdot a\inv=1;
  \end{gather*}
\item
  toplama \"uzerinde \c carpma da\u g\i l\i r, yani
  \begin{equation*}
    a\cdot(b+c)=ab+ac.
  \end{equation*}
\end{enumerate}
\item
  $0$'dan b\"uy\"uk olan ger\c cer say\i lar \textbf{pozitiftir;}
  k\"u\c c\"uk olan, \textbf{negatiftir.}
  \item
  $\R$ \textbf{do\u grusal s\i ralanm\i\c s bir cisim} oldu\u gundan
  iki pozitif elemanlar\i n toplam\i\ ve \c carp\i m\i\ da pozitiftir,
  yani
  \begin{equation*}
    a>0\land b>0\lto a+b>0\land ab>0.
  \end{equation*}
\end{enumerate}

\section{Sayma say\i lar\i}

$\R$'nin
\begin{enumerate}[1)]
\item
  $1\in A$,
\item
  $\Forall x(x\in A\lto x+1\in A)$
\end{enumerate}
ko\c sulunu sa\u glayan bir $A$ altk\"umesi \textbf{t\"umevar\i ml\i d\i r.}
$\R$'nin en k\"u\c c\"uk t\"umevar\i ml\i\ altk\"umesi, $\N$ olur,
ve onun elemanlar\i,
\textbf{sayma say\i lar\i d\i r.}
Tan\i mdan e\u ger $\N$'nin bir $A$ altk\"umesi t\"umevar\i ml\i\ ise,
o zaman $A=\N$ olur:
o kan\i tlama y\"ontemi, \textbf{t\"umevar\i md\i r.}

\begin{theorem}\label{thm:N1n+1}
  $\N=\{1\}\cup\{x+1\colon x\in\N\}$.
\end{theorem}

\begin{proof}
  $\{1\}\cup\{x+1\colon x\in\N\}$
  hem t\"umevar\i ml\i d\i r,
  hem de $\N$'nin bir altk\"umesidir.
\end{proof}

\begin{theorem}\label{thm:wo}
  $\N$ \textbf{iyis\i ral\i d\i r,}
  yani $\N$'nin her bo\c s olmayan altk\"umesinin
  en k\"u\c c\"uk eleman\i\ vard\i r.
\end{theorem}

\begin{theorem}\label{thm:kn}
  $\N$'de e\u ger $k<n$ ise, o zaman $k+1\leq n$.
\end{theorem}

\begin{proof}
  M\"umk\"unse $\{x\in\N\colon\Exists y(y\in\N\land x<y<x+1)\}$
  bo\c s olmas\i n.
  \Teoremde{thm:wo}n k\"umenin en k\"u\c c\"uk eleman\i\ $k$ olsun.
  O zaman $\{x\in\N\colon x\leq k\lor k\leq x\}$,
  t\"umevar\i ml\i d\i r.
\end{proof}

\begin{theorem}[\"Ozyineleme]\label{thm:rec}
  $A\included\R$, $b\in A$, ve $f\colon\N\times A\to A$ olsun.
  O zaman bir ve tek bir $g$ i\c cin
  \begin{itemize}
  \item
    $g\colon\N\to A$,
  \item
    $g(1)=b$,
  \item
    her $n$ sayma say\i s\i\ i\c cin
    \begin{equation*}
      g(n+1)=f(n,g(n)).
    \end{equation*}
  \end{itemize}
\end{theorem}

\begin{proof}
  \"Once $g$'nin tek oldu\u gunu g\"osterece\u giz.
  E\u ger $g_1$ ve $g_2$, $g$'nin \"ozelliklerini sa\u glarsa,
  o zaman
  \begin{itemize}
  \item
    $g_1$'in ve $g_2$'nin tan\i m k\"umesi $\N$'dir,
    \item
      $g_1(1)=b=g_2(1)$,
    \item
      her $n$ sayma say\i s\i\ i\c cin
      \begin{multline*}
    g_1(n)=g_2(n)\lto{}\\
    g_1(n+1)=f(n,g_1(n))=f(n,g_2(n))=g_2(n+1).
      \end{multline*}
  \end{itemize}
  B\"oylece $\{x\colon g_1(x)=g_2(x)\}$ k\"umesi,
  hem $\N$'nin bir altk\"umesidir,
  hem de t\"umevar\i ml\i d\i r.
  Sonu\c c olarak o k\"ume $\N$'dir,
  dolay\i s\i yla $g_1=g_2$ olur.

  Benzer \c sekilde
  her $n$ sayma say\i s\i\ i\c cin,
  en \c cok bir $h_n$ i\c cin,
  \begin{itemize}
  \item
    $h_n\colon\{1,\dots,n\}\to\R$,
  \item
    $h_n(1)=b$,
  \item
    e\u ger $1\leq k<n$ ise, o zaman
    \begin{equation*}
      h_n(k+1)=f(h_n(k)).
    \end{equation*}
  \end{itemize}
  Ayr\i ca en az bir $h_n$'nin verilen \"ozellikleri vard\i r,
  \c c\"unk\"u
  \begin{enumerate}[1)]
  \item
    $h_1=\{(1,b)\}$ olabilir,
  \item
    e\u ger bir $m$ i\c cin $h_m$ varsa, o zaman
    Teorem \ref{thm:kn} sayesinde
    \begin{equation*}
      h_{m+1}=h_m\cup\biggl\{\Bigl(m+1,f\bigl(m,h_m(m)\bigr)\Bigr)\biggr\}
    \end{equation*}
    olabilir.
  \end{enumerate}
  \c Simdi her $n$ i\c cin
  \begin{equation*}
    g(n)=h_n(n)
  \end{equation*}
  olsun.
\end{proof}

\section{Ar\c simet \"ozelli\u gi}

Tan\i ma g\"ore
\begin{equation*}
  \Z=\N\cup\{0\}\cup\{-x\colon x\in\N\},
\end{equation*}
ve $\Z$'nin elemanlar\i, \textbf{tamsay\i lard\i r.}
Hem toplama hem de \c carpma alt\i nda kapal\i\ oldu\u gundan
ve $1$'i i\c cerdi\u ginden $\Z$ bir \textbf{halkad\i r.}
E\u ger $\R$'de $\Z$'nin $a$ \"usts\i n\i r\i\ varsa,
o zaman $a-1$ de $\Z$'nin $a$ \"usts\i n\i r\i d\i r.
Bundan dolay\i\ $\Z$'nin supremumu yoktur.
Sonu\c c olarak $\R$'de $\Z$, s\i n\i rl\i\ de\u gildir.
B\"oylece $\R$'nin \textbf{Ar\c simet \"ozelli\u gi} vard\i r.

\"Oklid d\"uzleminde
\"Oklid'in \emph{\"O\u geler}'indeki gibi
iki do\u gru par\c cas\i\ \emph{e\c sit} olabilir.
Bu e\c sitlikten iki \emph{y\"onlendirilmi\c s} do\u gru par\c cas\i n\i n
e\c sitli\u gini elde edebiliriz.
Asl\i nda e\u ger $ABCD$ bir paralelkenar ise,
o zaman $\overrightarrow{AB}=\overrightarrow{DC}$;
e\u ger ayr\i ca $E$ ve $F$, $AB$ do\u grusunda
ve $ABEF$ de bir paralelkenar ise,
o zaman $\overrightarrow{AB}=\overrightarrow{EF}$.
E\c sitli\u ge g\"ore
y\"onlendirilmi\c s bir do\u gru par\c cas\i n\i n denklik s\i n\i f\i,
bir \emph{vekt\"ord\"ur.}
\.Iki vekt\"or toplanabilir,
ve vekt\"orler abelyan bir grubu olu\c sturur.

iki paralel vekt\"or kar\c s\i la\c st\i r\i labilir.
Ar\c simet \"ozelli\u gine g\"ore
daha k\"u\c c\"u\u g\"un\"un bir kat\i,
di\u gerinden b\"uy\"ukt\"ur.

\.Iki paralel vekt\"or\"un \emph{oran\i} vard\i r.
\emph{\"O\u geler}'in 5.\ ve 6.\ kitab\i nda bulunan
orant\i lar kuram\i na g\"ore
oranlar,
\emph{Thales Teoremi}'ni sa\u glar.
Bir vekt\"or\"u \emph{birim} olarak tan\i mlayarak,
birime paralel olan iki vekt\"or\"un \c carp\i m\i n\i\ tan\i mlayarak,
Descartes,
s\i ralanm\i\c s bir cismi elde etti
\cite{Descartes-Geometrie}.
Ar\c simet \"ozelli\u gini kullanmadan
Hilbert, ayn\i\ sonu\c cu elde etti \cite{Hilbert-10}.

\chapter{Ger\c cel\"ust\"u say\i lar}

\section{B\"uy\"uk k\"umeler}

Her eleman\i\ $\N$'nin bir altk\"umesi olan,
a\c sa\u g\i daki ko\c sullar\i\ sa\u glayan bir $\mathscr A$
k\"umesi,
$\N$ \"uzerinde bir \textbf{filtredir:}
\begin{enumerate}
\item
  $\N\in\mathscr A$.
\item
  $\emptyset\notin\mathscr A$.
\item
  $B\in\mathscr A\land B\included C\included\N\lto C\in\mathscr A$.
\item
  $B\in\mathscr A\land C\in\mathscr A\lto B\cap C\in\mathscr A$.
\end{enumerate}
\"Orne\u gin
$\N$'nin altk\"umesi olan,
t\"umleyeni sonlu olan k\"umeler,
$\N$ \"uzerinde \textbf{Fr\'echet filtresini} olu\c sturur.
\Sayfada{ch:algebra} bahsedilen
\emph{Zorn Lemmas\i}'na g\"ore
$\N$ \"uzerinde
Fr\'echet filtresini kapsayan,
\textbf{maksimal} (daha b\"uy\"u\u g\"u olmayan)
filtreler vard\i r.

\emph{\.Ilk ve son olarak $\N$ \"uzerinde
maksimal bir filtre se\c cilsin.}
Elemanlar\i na \textbf{b\"uy\"uk}
(veya \emph{\c co\u gunluk}) densin,
ve $\N$'nin di\u ger altk\"umelerine
\textbf{k\"u\c c\"uk} densin.
O zaman
\begin{itemize}
\item
  $\N$'nin her altk\"umesi
ya b\"uy\"uk ya da k\"u\c c\"ukt\"ur,
ama ikisi de\u gildir;
\item
  $\N$'nin b\"uy\"uk bir altk\"umesinin t\"umleyeni k\"u\c c\"ukt\"ur.
\end{itemize}

\section{Diziler ve denklikleri}

$\N$'den $\R$'ye giden bir fonksiyon,
bir \textbf{dizidir.}
Diziler
\begin{equation*}
  \R^{\N}
\end{equation*}
k\"umesini olu\c sturur.
Ayn\i\ dizi
\begin{itemize}
  \item
ya $(a_k\colon k\in\N)$,
\item
  ya da $(a_1,a_2,a_3,\dots)$,
  \item
    ya da k\i saca $\seq a$
\end{itemize}
olarak yaz\i l\i r.
Girdilerine g\"ore
iki dizinin toplam\i n\i\ ve \c carp\i m\i n\i
\begin{align*}
  \seq a+\seq b&=(a_k+b_k\colon k\in\N),&
    \seq a\seq b&=(a_kb_k\colon k\in\N)
\end{align*}
kurallar\i n\i\ kullanarak
tan\i mlar\i z.
\c Simdi tan\i ma g\"ore
\begin{equation}\label{eqn:D}
  \seq a\denk\seq b\liff\{k\in\N\colon a_k=b_k\}\text{ b\"uy\"ukt\"ur}
\end{equation}
olsun.

\begin{theorem}
  $\denk$ ba\u g\i nt\i s\i\ bir denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

Verilen $\denk$ ba\u g\i nt\i s\i na g\"ore
bir $\seq a$ dizisinin denklik s\i n\i f\i\
$[\seq a]$ olarak
yaz\i ls\i n.
B\"oylece
\begin{equation*}
  [\seq a]=\{\seq x\in\R^{\N}\colon\seq a\denk\seq x\}.
\end{equation*}
B\"ut\"un dizilerin denklik s\i n\i flar\i,
$\stR$ k\"umesini olu\c stursun.  B\"oylece
\begin{equation}\label{eqn:*R}
  \stR=\{[\seq x]\colon\seq x\in\R^{\N}\}.
\end{equation}

\begin{theorem}\label{thm:*R}
  $\denk$ ba\u g\i nt\i s\i,
  $\R^{\N}$ halkas\i n\i n i\c slemlerine sayg\i\ g\"osterir,
  yani
  \begin{equation*}
    \seq a\denk\seq c\land\seq b\denk\seq d
    \lto\seq a+\seq b\denk\seq c+\seq d\land\seq a\seq b\denk\seq c\seq d.
  \end{equation*}
  Bu \c sekilde $\stR$, iyitan\i mlanm\i\c s bir cisim olur.
  Ayr\i ca
  \begin{equation*}
    [\seq a]<[\seq b]\liff\{k\in\N\colon a_k<b_k\}\text{ b\"uy\"ukt\"ur}
  \end{equation*}
  kural\i na g\"ore $\stR$,
   iyitan\i mlanm\i\c s \emph{s\i ralanm\i\c s} bir cisim olur.
   Bu s\i ralanm\i\c s cisme
   \begin{equation*}
     x\mapsto[x,x,x,\dots]
   \end{equation*}
\textbf{k\"o\c segen g\"ommesi} ile
  $\R$ g\"om\"ul\"ur.
\end{theorem}

$\stR$ cisminin elemanlar\i na \textbf{ger\c cel\"ust\"u}
(\eng{hyperreal}) denebilir.
$\stR$'nin bir altk\"umesi olarak
$\R$'yi,
\Teoremi{thm:*R} kullanarak
d\"u\c s\"unece\u giz;
b\"oylece her ger\c cel say\i, ger\c cel\"ust\"ud\"ur.

\section{Sonsuz ve sonsuzk\"u\c c\"uk say\i lar}

Her ger\c cel say\i\ gibi
her $[\seq a]$ ger\c cel\"ust\"u say\i s\i n\i n,
\begin{equation*}
  \left.
  \begin{array}{rr}
    [\seq a]\geq0\text{ ise}&[\seq a]\\{}
    [\seq a]<0\text{ ise}&-[\seq a]
  \end{array}
  \right\}=
  \bigl|[\seq a]\bigr|
\end{equation*}
kural\i n\i\ sa\u glayan
$\bigl|[\seq a]\bigr|$ \textbf{mutlak de\u geri} vard\i r.
E\u ger ger\c cel\"ust\"u bir say\i n\i n mutlak de\u geri
\begin{itemize}
\item
  \emph{bir} pozitif ger\c cel say\i dan
k\"u\c c\"uk ise,
o zaman ger\c cel\"ust\"u say\i ya \textbf{sonlu} densin;
\item
  \emph{her} pozitif ger\c cel say\i dan
k\"u\c c\"uk ise,
o zaman ger\c cel\"ust\"u say\i ya \textbf{sonsuzk\"u\c c\"uk}
(\eng{infinitesimal}) densin.
\end{itemize}

\begin{example}
  $0$ sonsuzk\"u\c c\"ukt\"ur,
  ama hi\c c ba\c ska ger\c cel say\i\
  sonsuzk\"u\c c\"uk de\u gildir.  
\end{example}

Sonlu olmayan ger\c cel\"ust\"u bir say\i, \textbf{sonsuzdur.}

\begin{example}
  $[k\colon k\in\N]$,
  sonsuz ger\c cel\"ust\"u bir say\i d\i r.
  Sonu\c c olarak $\stR$'nin Ar\c simet \"ozelli\u gi yoktur.
\end{example}

\begin{example}\label{ex:k-inv}
  $[k\inv\colon k\in\N]$,
      sonsuzk\"u\c c\"uk ger\c cel\"ust\"u bir say\i d\i r.
\end{example}

\begin{theorem}
  S\i f\i r olmayan bir $[\seq a]$ ger\c cel\"ust\"u say\i s\i\ i\c cin
  \begin{equation*}
    \text{$[\seq a]$ sonsuzk\"u\c c\"ukt\"ur}
    \liff
    \text{$[\seq a]\inv$ sonsuzdur.}
  \end{equation*}
\end{theorem}

\section{Sonsuzyak\i nl\i k}

Tan\i ma g\"ore
\begin{equation*}
  [\seq a]\approx[\seq b]\liff
  [\seq a]-[\seq b]\text{ sonsuzk\"u\c c\"ukt\"ur}
\end{equation*}
olsun.

\begin{theorem}\mbox{}\label{thm:0}
  \begin{enumerate}
  \item 
  $\stR$ cisminde $\approx$ ba\u g\i nt\i s\i\
    bir denklik ba\u g\i nt\i s\i d\i r.
  \item
    $[\seq a]\approx0\liff\text{$[\seq a]$ sonsuzk\"u\c c\"ukt\"ur}$.
\item
  $[\seq a]\approx[\seq b]\liff[\seq a]-[\seq b]\approx0$.
  \end{enumerate}
\end{theorem}

E\u ger $[\seq a]\approx[\seq b]$ ise,
o zaman $[\seq a]$ ve $[\seq b]$
birbirine \textbf{sonsuzyak\i nd\i r}
(\eng{infinitely close}).

\begin{example}\label{ex:=approx}
  E\c sit ger\c cel\"ust\"u say\i lar
  birbirine sonsuzyak\i nd\i r.
\end{example}

\begin{theorem}\label{thm:olamaz}
    \.Iki farkl\i\ ger\c cel say\i\ sonsuzyak\i n olamaz.
\end{theorem}

\begin{theorem}\label{thm:finite}
  Ger\c cel bir say\i ya sonsuzyak\i n olan
  her ger\c cel\"ust\"u say\i\ sonludur.
\end{theorem}

\c Simdi \Teoremin{thm:finite} tersini kan\i tlayaca\u g\i z.

\begin{theorem}\label{thm:stp}
  Her sonlu ger\c cel\"ust\"u say\i,
  ger\c cel bir say\i ya sonsuzyak\i nd\i r.
\end{theorem}

\begin{proof}
  Sonlu bir $[\seq a]$ verilsin, ve $[\seq a]\geq0$ olsun.
  (Di\u ger durumda $-[\seq a]$'ya bakar\i z.)
  \begin{equation*}
    \{x\in\R\colon x\leq[\seq a]\}=A
  \end{equation*}
  olsun.
  O zaman
  \begin{equation*}
    \Forall x\Forall y(x<y\land y\in A\lto x\in A).
  \end{equation*}
  Ayr\i ca
  $0\in A$,
  ama (sonlulu\u gun tan\i m\i ndan)
  $A$'n\i n \"usts\i n\i r\i\ vard\i r,
  dolay\i s\i yla
  $\R$'nin taml\i\u g\i ndan
  $A$'n\i n supremumu vard\i r.
  O supremum $b$ olsun.
  O zaman her $n$ sayma say\i s\i\ i\c cin
  \begin{align*}
    b-\frac1n&\in A,&
    b+\frac1n&\notin A,
  \end{align*}
  dolay\i s\i yla
    \begin{equation*}
      b-\frac1n\leq[\seq a]<b+\frac1n.
    \end{equation*}
    Sonu\c c olarak
  $b\approx[\seq a]$.
\end{proof}

Kan\i tta bulunan $b$,
$[\seq a]$'n\i n \textbf{standart par\c cas\i d\i r}
(\eng{standard part}).

\begin{theorem}\label{thm:inf-close}
  E\u ger
  $[\seq a]\approx[\seq c]$, $[\seq b]\approx[\seq d]$,
  ve $[\seq e]$ sonlu ise,
  o zaman
  \begin{align*}
    [\seq a]+[\seq b]&\approx[\seq c]+[\seq d],
    &[\seq a][\seq e]&\approx[\seq c][\seq e].
  \end{align*}
\end{theorem}

\begin{theorem}\label{thm:inv}
  E\u ger
  $[\seq a]\approx[\seq c]$
  ve $[\seq a]$ sonsuzk\"u\c c\"uk de\u gilse, o zaman
  \begin{equation*}
    [\seq a]\inv\approx[\seq c]\inv.
  \end{equation*}
\end{theorem}

\chapter{S\"ureklilik}\label{ch:continuity}

\section{K\"umeler ve \c ce\c sitleri}

$\stR$, $\R$'nin \emph{b\"uy\"utmesidir.}
Daha genelde,
e\u ger $A\included\R$ ise,
o zaman $\stR$'nin
(\sayfada{eqn:*R}ki) tan\i m \onunesitliginde{eqn:*R}
$\R$'nin yerine koyarak
$A$'n\i n $\stA$ \textbf{b\"uy\"utlemesini} elde ederiz.
B\"oylece
\begin{equation*}
\{[\seq x]\colon\seq x\in A^{\N}\}=  \stA.
\end{equation*}

\begin{theorem}\label{thm:u-}
  E\u ger $A\included\R$ ve $B\included\R$ ise,
  o zaman
  \begin{align*}
%    \stA\cap\stB&=\yildiz(A\cap B),&
    \stA\cup\stB&=\yildiz(A\cup B),&
    \stA\setminus\stB&=\yildiz(A\setminus B).
  \end{align*}
\end{theorem}

Elemanlar\i\ ger\c cel say\i\ olan bir k\"umenin b\"uy\"utlemesinin
her sonsuzyak\i n olan
ger\c cel\"ust\"u say\i y\i\ i\c cerdi\u gi
ger\c cel bir say\i,
k\"umenin \textbf{i\c c noktas\i d\i r.}
B\"oylece bir $a$ ger\c cel say\i s\i n\i n
$\R$'nin bir $A$ altk\"umesinin i\c c noktas\i\ olmak i\c cin
\begin{equation*}
  \Forall{[\seq x]}\bigl([\seq x]\approx a\lto[\seq x]\in\stA\bigr),
\end{equation*}
gerek ve yeter bir ko\c suldur.

\begin{theorem}\label{thm:int-suf}
  E\u ger pozitif bir $\delta$ i\c cin
  \begin{equation*}
    (a-\delta,a+\delta)\included A
  \end{equation*}
  ise, o zaman $a$, $A$'n\i n bir i\c c noktas\i d\i r.
\end{theorem}

\begin{proof}
  E\u ger
  $[\seq x]\approx a$ ise, o zaman $[\seq x]\in\yildiz(a-\delta,a+\delta)$,
  dolay\i s\i yla $[\seq x]\in\stA$.
\end{proof}

\Teoremin{thm:int-suf} tersi,
\sayfada{thm:int-nec}ki Teorem \numaradir{thm:int-nec}.



Elemanlar\i\ ger\c cel say\i\ olan,
her eleman\i\ bir i\c c noktas\i\ olan bir k\"ume
\textbf{a\c c\i kt\i r.}
B\"oylece bir $O$ k\"umesinin a\c c\i k olmas\i\ i\c cin
\begin{equation*}
  \Forall x\Forall{[\seq y]}\bigl(x\in O\land x\approx[\seq y]
  \lto[\seq y]\in\stO\bigr),
\end{equation*}
gerek ve yeter bir ko\c suldur.

\begin{theorem}\mbox{}
  \begin{enumerate}
  \item 
    Her a\c c\i k aral\i k, a\c c\i k bir k\"umedir.
  \item
    A\c c\i k k\"umelerin birle\c simi de a\c c\i kt\i r.
  \item
    \.Iki a\c c\i k k\"umelerin kesi\c simi de a\c c\i kt\i r.
  \end{enumerate}
\end{theorem}

A\c c\i k bir k\"umenin t\"umleyeni
\textbf{kapal\i d\i r.}
\Teorem{thm:u-} sayesinde
bir $F$ k\"umesinin kapal\i\ olmas\i\ i\c cin
\begin{equation*}
  \Forall{[\seq x]}\Forall y\bigl([\seq x]\in\stF\land[\seq x]\approx y
  \lto y\in F\bigr),
\end{equation*}
gerek ve yeter bir ko\c suldur.

\begin{theorem}\mbox{}
  \begin{enumerate}
  \item 
    Her kapal\i\ aral\i k, kapal\i\ bir k\"umedir.
  \item
    Kapal\i\ k\"umelerin kesi\c simi de kapal\i d\i r.
  \item
    \.Iki kapal\i\ k\"umelerin birle\c simi de kapal\i d\i r.
  \end{enumerate}
\end{theorem}

\begin{theorem}\label{thm:bdd}\sloppy
  $\R$'nin bir $A$ altk\"umesinin s\i n\i rl\i\ olmas\i\ i\c cin
  $\stA$'n\i n her eleman\i n\i n sonlu olmas\i,
  gerek ve yeter bir ko\c suldur.
\end{theorem}

\begin{theorem}\label{thm:compact}
  $\R$'nin bir $K$ altk\"umesinin s\i n\i rl\i\ ve kapal\i\ olmas\i\ i\c cin
  \begin{equation*}
    \Forall{[\seq x]}\Exists y\bigl([\seq x]\in\stK\lto
    [\seq x]\approx y\land y\in K\bigr),
  \end{equation*}
  gerek ve yeter bir ko\c suldur.
\end{theorem}

\begin{theorem}\label{thm:cbd}
  $\R$'nin kapal\i\ ve s\i n\i rl\i\ olan bir altk\"umesinin
  en b\"uy\"uk ve en k\"u\c c\"uk eleman\i\ vard\i r.
\end{theorem}

\begin{proof}
  $K$,
  $\R$'nin kapal\i\ ve s\i n\i rl\i\ olan bir altk\"umesi olsun.
  \begin{itemize}
  \item 
  $K$ s\i n\i rl\i\ oldu\u gundan $b$ supremumu vard\i r.
  Her $n$ sayma say\i s\i\ i\c cin,
  $K$'nin bir $a_n$ eleman\i\ i\c cin,
  \begin{equation*}
    b-\frac1n<a_n\leq b.
  \end{equation*}
  O halde
    $[\seq a]\in\stK$ ve $[\seq a]\approx b$.
  \item
    $K$ kapal\i\ oldu\u gundan $b\in K$.
    \qedhere
  \end{itemize}
\end{proof}

\section{Do\u gal\"ust\"u say\i lar}

$\N$, $\R$'nin altk\"umesi oldu\u gundan
$\N$'nin
\begin{equation*}
  \stN
\end{equation*}
b\"uy\"utlemesi vard\i r.
O zaman $\N$ gibi $\stN$, t\"umevar\i ml\i d\i r.

\begin{example}
    $[k\colon k\in\N]\in\stN\setminus\N$.
\end{example}

Bununla birlikte $\stN$'de t\"umevar\i m bir \c sekilde kullan\i labilir:

\begin{theorem}\label{thm:ind}
  E\u ger $\N$'nin bir $A$ altk\"umesinin $\stA$ b\"uy\"utlemesi
  t\"umevar\i ml\i\ ise,
  o zaman $\stA=\stN$,
  dolay\i s\i yla $A=\N$.
\end{theorem}

\begin{proof}
  $\stA$ t\"umevar\i ml\i\ oldu\u gundan
  \begin{enumerate}[1)]
  \item 
    $1\in\stA$, dolay\i s\i yla $1\in A$;
  \item
    e\u ger $n\in A$ ise,
  o zaman $n\in\stA$, dolay\i s\i yla $n+1\in\stA$,
  dolay\i s\i yla $n+1\in A$.
  \end{enumerate}
  T\"umevar\i mdan $A=\N$,
  dolay\i s\i yla $\stA=\stN$.
\end{proof}

\begin{example}
  $\N$, $\stN$'nin t\"umevar\i ml\i\ bir altk\"umesidir,
  dolay\i s\i yla $\N$'nin her $A$ altk\"umesi i\c cin
  $\stA\neq\N$.
\end{example}

\begin{theorem}
  $\stN=\{1\}\cup\{[\seq x]+1\colon[\seq x]\in\stN\}$.
\end{theorem}

\begin{proof}
  Ya \Teoremi{thm:ind} kullanabiliriz,
  ya da a\c sa\u g\i daki hesaplamalar\i:
  \begin{align*}
  &\phantom{{}={}}\{1\}\cup\{[\seq x]+1\colon[\seq x]\in\stN\}&&\\
          &=\{1\}\cup\yildiz\{x+1\colon x\in\N\}&&\text{[Teorem \ref{thm:f}]}\\
      &=\yildiz(\{1\}\cup\{x+1\colon x\in\N\})&&\text{[Teorem \ref{thm:u-}]}\\
      &=\stN&&\text{[Teorem \ref{thm:N1n+1}]}.\qedhere
    \end{align*}
\end{proof}

\textbf{Do\u gal say\i lar} k\"umesi $\upomega$
(\emph{omega}) ile yazal\i m.
O zaman
\begin{equation*}
  \upomega=\{0,1,2,\dots\}=\{0\}\cup\N
\end{equation*}
ve ayr\i ca
\begin{equation*}
  \sto=\{0\}\cup\stN.
\end{equation*}
Burada $\sto$'n\i n elemanlar\i na
\textbf{do\u gal\"ust\"u say\i} densin.
Her do\u gal say\i,
do\u gal\"ust\"u olarak d\"u\c s\"un\"ulebilir.

\section{Fonksiyonlar ve limitler}

Bir k\"umeninki gibi bir fonksiyonun b\"uy\"utlemesi vard\i r.

\begin{theorem}\label{thm:f}
  E\u ger $A\included\R$ ve $f\colon A\to\R$ ise,
  o zaman iyitan\i mlanm\i\c s bir $\stf$ i\c cin
  \begin{equation*}
    \stf\colon\stA\to\stR,
  \end{equation*}
  ve $\stA$'n\i n her $[\seq x]$ eleman\i\ i\c cin
  \begin{equation}\label{eqn:*f}
    \stf[\seq x]=[f(x_k)\colon k\in\N].
  \end{equation}
  Ayr\i ca
  \begin{equation}\label{eqn:*f[A]}
    \yildiz\bigl(f[A]\bigr)=
    \stf[\stA].
  \end{equation}
\end{theorem}

\begin{proof}
  Verilen $f$ fonksiyonu,
  elemanlar\i\ $\bigl(a,f(a)\bigr)$ s\i ral\i\ ikisi olan bir k\"umedir.
  O halde $f^{\N}$'nin elemanlar\i,
  $\seq a\in A^{\N}$ olmak \"uzere
  \begin{equation*}
    \Bigl(\bigl(a_k,f(a_k)\bigr)\colon k\in\N\Bigr)
  \end{equation*}
  dizileridir.
  Tan\i m \ref{eqn:D} gibi bir tan\i ma g\"ore
  \begin{multline*}
    \Bigl(\bigl(a_k,f(a_k)\bigr)\colon k\in\N\Bigr)
    \denk
    \Bigl(\bigl(b_k,f(b_k)\bigr)\colon k\in\N\Bigr)
    \liff{}\\
    \Bigl\{k\in\N\colon\bigl(a_k,f(a_k)\bigr)
    =\bigl(b_k,f(b_k)\bigr)\Bigr\}\text{ b\"uy\"ukt\"ur}
  \end{multline*}
  olsun.
  O halde
  \begin{multline*}
    \Bigl\{k\in\N\colon\bigl(a_k,f(a_k)\bigr)
    =\bigl(b_k,f(b_k)\bigr)\Bigr\}={}\\
    \{k\in\N\colon a_k=b_k\}\cap
    \bigl\{k\in\N\colon f(a_k)=f(b_k)\bigr\}
  \end{multline*}
  oldu\u gundan
  \begin{multline*}
    \Bigl(\bigl(a_k,f(a_k)\bigr)\colon k\in\N\Bigr)
    \denk
    \Bigl(\bigl(b_k,f(b_k)\bigr)\colon k\in\N\Bigr)
    \liff{}\\
         [\seq a]=[\seq b]\land
         \bigl[f(a_k)\colon k\in\N\bigr]=
         \bigl[f(b_k)\colon k\in\N\bigr].
  \end{multline*}
  B\"oylece $\stf$,
  \esitlikte{eqn:*f}ki gibi olarak anla\c s\i labilir,
  ve bundan \eqref{eqn:*f[A]} \c c\i kar.
\end{proof}

\c Simdi
\begin{align*}
  A&\included\R,&
  a&\in A,&
  f&\colon A\setminus\{a\}\to\R,&
  L&\in\R
\end{align*}
olsun.  E\u ger
\begin{equation*}%\label{eqn:ns-2}
  \Forall{[\seq x]}
  \bigl([\seq x]\in\stA\setminus\{a\}\land
         [\seq x]\approx a
         \lto
    \stf[\seq x]\approx L\bigr)
\end{equation*}
ise,
o zaman $L$,
$f$'nin $a$'daki \textbf{limitidir.}
%(Gerektirme \eqref{eqn:ns-2}, \eqref{eqn:ns} ile ayn\i d\i r.)
$L$ limitini
\begin{equation*}
  \lim_{x\to a}f(x)\quad\text{ veya }\quad\lim_af
\end{equation*}
ile
yazar\i z.
Limit tan\i m\i m\i z\i n
``standart'' tan\i ma denk oldu\u gunu,
\Sayfada{ch:ed}
Teorem \numarada{thm:ed} kan\i tlayaca\u g\i z.

\begin{theorem}\label{thm:+.}
  $\lim_af=L$ ve $\lim_ag=M$ ise
  \begin{equation*}
    \lim_a(f+g)=L+M\land\lim_a(fg)=LM.
  \end{equation*}
\end{theorem}

\begin{proof}
  E\u ger $[\seq x]\approx a$ ve $[\seq x]\neq a$ ise,
  o zaman varsay\i ma g\"ore
  \begin{align*}
    \stf[\seq x]&\approx L,&\stg[\seq x]&\approx M.
  \end{align*}
  \"Ozellikle \Teorem{thm:finite} sayesinde
  $\stf[\seq x]$ sonludur.  \c Simdi
  \Teorem{thm:inf-close} sayesinde
  \begin{align*}
    \stf[\seq x]+\stg[\seq x]&\approx L+M,&
    \stf[\seq x]\stg[\seq x]&\approx\stf[\seq x]M\approx LM.\qedhere
  \end{align*}
\end{proof}

\begin{theorem}\label{thm:L-inv}
  $\lim_af=L$ ve $L\neq0$ ise
  \begin{equation*}
    \lim_a\frac1f=\frac1L.
  \end{equation*}  
\end{theorem}

\begin{proof}
  \Teorem{thm:inv}.
\end{proof}

\section{S\"ureklilik}

\begin{itemize}
\item 
Tan\i m k\"umesi
bir $a$ noktas\i n\i\ i\c ceren bir $f$ fonksiyonu i\c cin,
e\u ger
\begin{equation*}
  \lim_af=f(a)
\end{equation*}
ise, o zaman $f$, \textbf{$a$'da s\"ureklidir.}
B\"oylece tan\i m k\"umesi $A$ olan
ve $a$'y\i\ i\c ceren bir $f$'nin $a$ s\"urekli olmas\i\ i\c cin
\begin{equation*}%\label{eqn:ns-2}
  \Forall{[\seq x]}
  \bigl([\seq x]\in\stA\land
         [\seq x]\approx a
         \lto
    \stf[\seq x]\approx f(a)\bigr),
\end{equation*}
gerek ve yeter bir ko\c suldur.
\item
Tan\i m k\"umesi
bir $A$ k\"umesini kapsayan,
$A$'n\i n her noktas\i nda s\"urekli olan bir fonksiyon,
\textbf{$A$'da s\"ureklidir.}
\item
Kendinin tan\i m k\"umesinde s\"urekli olan bir fonksiyon,
\textbf{s\"ureklidir.}
\end{itemize}

\begin{example}
  Her sabit fonksiyon s\"ureklidir;
  \"ozde\c slik fonksiyonu s\"ureklidir.
  Sonu\c c olarak
  \begin{itemize}
  \item 
  \Teorem{thm:+.} sayesinde
  her polinom fonksiyonu s\"ureklidir;
\item
  \Teorem{thm:L-inv} sayesinde
  her kesirli fonksiyonu s\"ureklidir.
  \end{itemize}
\end{example}

\begin{example}
  E\u ger
  kapal\i\ olan,
  s\i n\i rl\i\ olmayan
  $[0,\infty)$ aral\i\u g\i nda $f(x)=x^2$ ise,
    $f$'nin s\"urekli oldu\u gu halde
    \begin{gather*}
    [k+k\inv\colon k\in\N]\approx[k\colon k\in\N],\\
    \stf[k+k\inv\colon k\in\N]=\stf[k\colon k\in\N]+2.
    \end{gather*}
\end{example}

\begin{example}
  E\u ger s\i n\i rl\i\ olan,
  kapal\i\ olmayan $(0,1]$ aral\i\u g\i nda $f(x)=x\inv$ ise,
    $f$'nin s\"urekli oldu\u gu halde
    \begin{gather*}
      [n\inv\colon n\in\N]\approx
      [(n+1)\inv\colon n\in\N],\\
      \stf[n\inv\colon n\in\N]+1
      =\stf[(n+1)\inv\colon n\in\N].      
    \end{gather*}
\end{example}

\begin{theorem}\label{thm:cont}
  $f$'nin
  kapal\i\ ve s\i n\i rl\i\ bir $A$ k\"umesinde s\"urekli olmas\i\ i\c cin
  $\stA$'n\i n t\"um sonlu $[\seq x]$ ve $[\seq y]$
  eleman\i\ i\c cin
  \begin{multline*}
    \Forall{\bigl([\seq x],[\seq y]\bigr)}
    \bigl([\seq x]\in\stA\land[\seq y]\in\stA\land{}\\
         [\seq x]\approx[\seq y]\lto\stf[\seq x]\approx\stf[\seq y]\bigr),
  \end{multline*}
   gerek ve yeter bir ko\c suldur.
\end{theorem}

\begin{theorem}[Arade\u ger]\label{thm:ivt}
  E\u ger
  $f$, $[a,b]$'de s\"urekli ve
  \begin{equation*}
    \bigl(f(a)-c\bigr)\bigl(c-f(b)\bigr)>0
%    f(a)<c<f(b)\quad\text{ veya }\quad f(a)>c>f(b)
  \end{equation*}
  ise,
  o zaman bir $d$ i\c cin
  \begin{equation}\label{eqn:f(d)=c}
    a<d<b\land f(d)=c.
  \end{equation}
\end{theorem}

\begin{proof}
$f(a)<c<f(b)$ varsay\i labilir.
\.Istedi\u gimiz $d$ ger\c cel say\i s\i\
bir $[d_n\colon n\in\N]$ ger\c cel\"ust\"u say\i s\i nin
standart par\c cas\i\ olacak.

Her $n$ sayma say\i s\i\ i\c cin
  \begin{equation*}%\label{eqn:delta-def}
    \delta_n=\frac{b-a}{2^{n-1}}
  \end{equation*}
  olsun.
  O zaman
  \begin{align*}
    \delta_1&=b-a,&\delta_{m+1}=\frac{\delta_m}2.
  \end{align*}
  \Teoremi{thm:rec} kullanarak
  \begin{enumerate}[1)]
  \item 
    $a=  d_1$,
    \item
    $\left.
    \begin{array}{ll}
      \text{$f(d_m+\delta_{m+1})\leq c$ ise}&d_m+\delta_{m+1}\\
      \text{$c<f(d_m+\delta_{m+1})$ ise}&d_m
    \end{array}
    \right\}=d_{m+1}$
  \end{enumerate}
olsun.
T\"umevar\i mdan her $n$ sayma say\i s\i\ i\c cin
  \begin{align*}%\label{eqn:delta}
    a&\leq d_n<d_n+\delta_n\leq b,&
    f(d_n)\leq c<f(d_n+\delta_n).
  \end{align*}
  Sonu\c c olarak
  \begin{equation*}
\stf[d_n\colon n\in\N]\leq c<\stf[d_n+\delta_n\colon n\in\N],    
  \end{equation*}
  ama $[\delta_n\colon n\in\N]$ sonsuzk\"u\c c\"ukt\"ur.
  \Teorem{thm:cont} sayesinde
  \begin{equation*}
    \stf[d_n\colon n\in\N]\approx\stf[d_n+\delta_n\colon n\in\N].
  \end{equation*}
  Bundan dolay\i
  \begin{equation*}
    \stf[d_n\colon n\in\N]\approx c.
  \end{equation*}
  \c Simdi $d$,
  $[d_n\colon n\in\N]$ ger\c cel\"ust\"u say\i s\i nin
  standart par\c cas\i\ olsun.
  O zaman
  \begin{equation*}
    f(d)\approx\stf[d_n\colon n\in\N],
  \end{equation*}
dolay\i s\i yla \eqref{eqn:f(d)=c} sa\u glan\i r.
\end{proof}

\begin{example}
  $[1,\infty)$ aral\i\u g\i\ kapal\i d\i r ama s\i n\i rl\i\ de\u gildir;
    $(0,1]$ aral\i\u g\i\ kapal\i\ de\u gildir ama s\i n\i rl\i d\i r.
  S\"urekli $x\mapsto x\inv$ fonksiyonu alt\i nda
  verilen aral\i klar\i n her biri,
  di\u gerin imgesidir.
\end{example}

\begin{theorem}\label{thm:icc}
  S\"urekli bir fonksiyon alt\i nda
  kapal\i\ ve s\i n\i rl\i\ olan bir k\"umenin imgesi de
  kapal\i\ ve s\i n\i rl\i d\i r.
\end{theorem}

\begin{proof}
  Kapal\i\ ve s\i n\i rl\i\ olan $K$ k\"umesinde
  $f$ s\"urekli olsun, ve
  \begin{equation*}
    [\seq a]\in\yildiz(f[K])
  \end{equation*}
  olsun.
    \Teorem{thm:f} sayesinde
  \begin{equation*}
    \yildiz(f[K])=\stf[\stK],
  \end{equation*}
  dolay\i s\i yla bir $[\seq b]$ i\c cin
  \begin{align*}
    [\seq b]&\in\stK,&\stf[\seq b]&=[\seq a].
  \end{align*}
  Bu durumda \Teorem{thm:compact} sayesinde
  bir $b$ i\c cin
  \begin{align*}
    b&\in K,&
    [\seq b]&\approx b.
  \end{align*}
  $K$'de $f$ s\"urekli oldu\u gundan
  \begin{equation*}
    [\seq a]\approx f(b).
  \end{equation*}
  Tekrar \Teorem{thm:compact} sayesinde
  $\yildiz(f[K])$ kapal\i\ ve s\i n\i rl\i d\i r.
  \begin{comment}
    
    O halde
    \begin{itemize}
    \item
      $K$ s\i n\i rl\i\ oldu\u gundan,
      \Teorem{thm:bdd} sayesinde
      $[\seq b]$ sonludur;
    \item
      \Teorem{thm:stp} sayesinde bir $b$ i\c cin
      $[\seq b]\approx b$;
    \item
      $K$ kapal\i\ oldu\u gundan $b\in K$,
      dolay\i s\i yla
      $f(b)\in f[K]$;
    \item
      $f$'nin s\"urekli oldu\u gundan
      $f(b)\approx[\seq a]$;
    \item
      \Teorem{thm:finite} sayesinde $[\seq a]$ sonludur;
    \item
      \Teorem{thm:olamaz} sayesinde $[\seq a]\approx c$ ise,
      o zaman $c=f(b)$, dolay\i s\i yla $c\in f[K]$.
    \end{itemize}
    \"Oyleyse $f[K]$ kapal\i d\i r,
    ve tekrar \Teorem{thm:bdd} sayesinde $f[K]$ s\i n\i rl\i d\i r.
  \end{comment}
\end{proof}

\begin{theorem}[Maksimum-minimum]\label{thm:max-min}
  Kapal\i\ ve s\i n\i rl\i\ bir k\"umede
  s\"urekli bir fonksiyon
  en b\"uy\"uk bir de\u geri ve en k\"u\c c\"uk bir de\u geri al\i r.
\end{theorem}

\begin{proof}
  Teoremler \ref{thm:cbd} ve \ref{thm:icc}.
\end{proof}

\chapter{Hesaplama}

\section{T\"urevler}

E\u ger
\begin{equation*}
  \lim_{x\to a}\frac{f(x)-f(a)}{x-a}
\end{equation*}
varsa,
\begin{equation*}
  f'(a)
\end{equation*}
olsun.
Bu \c sekilde tan\i mlanan $f'$ fonksiyonu, $f$'nin \textbf{t\"urevidir.}
E\u ger $[\seq h]$ sonsuzk\"u\c c\"uk ama $0$ de\u gilse,
o zaman $f'(a)$,
\begin{equation*}
  \frac{\stf(a+[\seq h])-f(a)}{[\seq h]}
\end{equation*}
b\"ol\"um\"un\"un standart par\c cas\i d\i r.
Bu standart par\c ca
\begin{equation*}
  \frac{\dee f(x)}{\dee x}(a)
\end{equation*}
olarak da yaz\i labilir.

\begin{theorem}\label{thm:x'}
  $\dee1/\dee x=0$ ve
  $\dee x/\dee x=1$.
\end{theorem}

\begin{theorem}\label{thm:+'}
  $(f+g)'=f'+g'$.
\end{theorem}

\begin{theorem}\label{thm:.'}
  $(fg)'=f'g+fg'$.
\end{theorem}

\begin{proof}
  \Teoremi{thm:+.} kullanarak
  \begin{equation*}
    \frac{f(a)g(a)-f(x)g(x)}{a-x}
    =f(a)\frac{g(a)-g(x)}{a-x}+\frac{f(a)-f(x)}{a-x}g(x)
  \end{equation*}
  e\c sitli\u ginden istedi\u gimiz sonucu elde ederiz.
\end{proof}

\begin{example}
  Teoremler \ref{thm:x'}, \ref{thm:+'}, ve \ref{thm:.'} sayesinde
  \begin{equation*}
    \frac{\dee(a_0+a_1x+\dots+a_nx^n)}{\dee x}(t)
    =a_1+a_2t+\dots+na_nt^{n-1}.
  \end{equation*}
\end{example}

\begin{theorem}
  E\u ger $f$'nin t\"urevi varsa,
  o zaman $f$ s\"ureklidir.
\end{theorem}

\begin{theorem}[Rolle]\label{thm:Rolle}
  E\u ger
  \begin{itemize}
  \item
    $[a,b]$ aral\i\u g\i nda $f$ s\"urekli,
  \item
    $(a,b)$ aral\i\u g\i nda $f$'nin t\"urevlenebilir,
  \item
    $f(a)=f(b)$
  \end{itemize}
  ise,
o zaman $(a,b)$ aral\i\u g\i n\i n bir $c$ eleman\i\ i\c cin
  \begin{equation*}
    0=f'(c).
  \end{equation*}  
\end{theorem}

\begin{proof}
  \Teoreme{thm:max-min} g\"ore $[a,b]$'de $f$,
  en k\"u\c c\"uk de\u gerini bir $c$ noktas\i nda ve
  en b\"uy\"uk de\u gerini bir $d$ noktas\i nda al\i r.
  Bu noktalar\i n biri ne $a$ ne $b$'dir; $c$ olsun.
  O halde $x\in[a,c)$ ise
  \begin{equation*}
    \frac{f(x)-f(c)}{x-c}\leq0,
  \end{equation*}
  dolay\i s\i yla
    $f'(c)\leq0$.
  Ayn\i\ zamanda $x\in(c,b]$ ise
  \begin{equation*}
    \frac{f(x)-f(c)}{x-c}\geq0,
  \end{equation*}
  dolay\i s\i yla $f'(c)\geq0$.
\end{proof}

\begin{theorem}[Ortade\u ger]\label{thm:MVT}
  E\u ger bir $[a,b]$ aral\i\u g\i nda $f$'nin t\"urevi varsa,
  o zaman aral\i\u g\i n bir $c$ eleman\i\ i\c cin
  \begin{equation*}
    \frac{f(b)-f(a)}{b-a}=f'(c).
  \end{equation*}
\end{theorem}

\begin{proof}
Tan\i mlayan denklemi
  \begin{equation*}
    g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)
  \end{equation*}
  olan $g$ fonksiyonu ile
    \Teoremi{thm:Rolle} kullan\i n.
\end{proof}

\begin{theorem}\label{thm:f'=g'}
  Bir aral\i kta e\u ger $f'=g'$ ve $f(a)=g(a)$ ise,
  o zaman $f=g$.
\end{theorem}

\begin{proof}
  E\u ger $f(b)\neq g(b)$ ise, o zaman
  \Teorem{thm:MVT} sayesinde
  $[a,b]$ veya $[b,a]$ aral\i\u g\i n\i n bir noktas\i nda
  $f-g$ fonksiyonunun $f'-g'$ t\"urevi s\i f\i r de\u gildir.
\end{proof}

\section{Integraller}

Tan\i m k\"umesi
bir $[a,b]$ kapal\i\ s\i n\i rl\i\ aral\i\u g\i n\i\ kapsayan
bir $f$ fonksiyonu i\c cin,
e\u ger bir $n$ sayma say\i s\i\ i\c cin
baz\i\ $x_i$ ger\c cel say\i lar\i
\begin{equation*}
  a=x_0<x_1<\dots<x_n=b
\end{equation*}
ko\c sulunu sa\u glarsa,
o zaman $\{x_0,x_1,\dots,x_n\}$ k\"umesi,
$[a,b]$'nin bir \textbf{par\c calamas\i d\i r.}
E\u ger ayr\i ca
baz\i\ $t_i$ i\c cin
\begin{equation}\label{eqn:xtx}
  x_{i-1}\leq t_i\leq x_i
\end{equation}
ise, o zaman
\begin{equation*}%\label{eqn:Rs}
  \sum_{i=1}^nf(t_i)(x_i-x_{i-1})
\end{equation*}
toplam\i,
$[a,b]$ aral\i\u g\i nda $f$'nin bir
\textbf{Riemann toplam\i d\i r.}

\c Simdi $i\mapsto x_i$ veya $x$,
tan\i m k\"umesi $\upomega$ olan,
artan bir fonksiyon olsun, ve bir $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
  a=x_0\leq x_1\leq\dots\leq x_n=x_{n+1}=\dots=b
\end{equation*}
olsun.  O zaman yukar\i daki gibi $\{x_i\colon i\in\upomega\}$,
$[a,b]$'nin bir par\c calamas\i s\i d\i r.
E\u ger ayr\i ca $i\mapsto t_i$ veya $t$,
tan\i m k\"umesi $\N$ olan bir fonksiyon ise,
ve her $i$ sayma say\i s\i\ i\c cin
e\c sitsizlik \eqref{eqn:xtx} do\u gru ise,
o zaman yukar\i daki Riemann toplam\i
\begin{equation*}
  \sum_{i=1}^{\infty}f(t_i)(x_i-x_{i-1})  
\end{equation*}
bi\c ciminde yaz\i labilir.

\c Simdi her $k$ do\u gal say\i s\i\ i\c cin
$x_k$ ve $t_k$, yukar\i daki gibi $x$ ve $t$ fonksiyonu olsun.
O zaman girdileri Riemann toplam\i\ olan bir
\begin{equation*}
  \left(\sum_{i=1}^{\infty}
  f(t_{k,i})(x_{k,i}-x_{k,i-1})\colon k\in\N\right)
\end{equation*}
dizisi vard\i r.
Ayr\i ca $\N^{\N}$'nin bir $\seq n$ eleman\i\ i\c cin,
her $k$ sayma say\i s\i\ i\c cin,
\begin{equation*}
  x_{k,n_k}=x_{k,n_k+1}=\dots=b.
\end{equation*}
Bu durumda yukar\i daki, girdileri Riemann toplam\i\ olan dizi
\begin{equation*}
  \left(\sum_{i=1}^{n_k}
  f(t_{k,i})(x_{k,i}-x_{k,i-1})\colon k\in\N\right)
\end{equation*}
olur.

\c Simdi
$\N^{\N}$'nin her $\seq i$ eleman\i\ i\c cin
\begin{align*}
  \seq t_{\seq i}&=(t_{k,i_k}\colon k\in\N),&
  \seq x_{\seq i}&=(x_{k,i_k}\colon k\in\N),
\end{align*}
ve
\begin{equation*}
  \seq x_{\seq i-1}=(x_{k,i_k-1}\colon k\in\N)
\end{equation*}
olsun.
Bu durumda
\begin{equation*}
  [\seq x_{\seq i-1}]\leq[\seq t_{\seq i}]\leq[\seq x_{\seq i}]
\end{equation*}
ve
\begin{equation*}
  [\seq i]=[\seq j]\lto[\seq t_{\seq i}]=[\seq t_{\seq j}]
  \land[\seq x_{\seq i}]=[\seq x_{\seq j}]
  \land[\seq x_{\seq i-1}]=[\seq x_{\seq j-1}].
\end{equation*}
E\u ger ayr\i ca
\begin{equation*}
  T=\bigl\{[\seq t_{\seq i}]\colon[\seq i]\in\stN\bigr\}
  =\bigl\{[\seq t_1],[\seq t_2],\dots,[\seq t_{\seq n}]\bigr\}
\end{equation*}
ise,
ve benzer \c sekilde
\begin{equation*}
  X=\bigl\{[\seq x_0],[\seq x_1],\dots,[\seq x_{\seq n}]\bigr\}
\end{equation*}
ise,
o zaman
\begin{equation*}
  \left[\sum_{i=1}^{n_k}f(t_{k,i})(x_{k,i}-x_{k,i-1})\colon
    k\in\N\right]
\end{equation*}
ger\c cel\"ust\"u say\i s\i
\begin{equation*}%\label{eqn:inf-Rs}
  \sum_{[\seq i]=1}^{[\seq n]}
  \stf[\seq t_{\seq i}]([\seq x_{\seq i}]-[\seq x_{\seq i-1}])
  \text{ veya }
  \RS TX
\end{equation*}
bi\c ciminde yaz\i labilir.
E\u ger her durumda
\begin{equation*}
  [\seq x_{\seq i}]
  \approx
  [\seq x_{\seq i-1}]
\end{equation*}
ise, o zaman
\begin{itemize}
\item
  $X$,
$[a,b]$'nin
\textbf{sonsuz bir par\c calamas\i d\i r;}
\item
  $\RS TX$,
$[a,b]$'de $f$'nin \textbf{sonsuz
  bir Riemann toplam\i d\i r.}
\end{itemize}
E\u ger
$[a,b]$'de $f$'nin her sonsuz Riemann toplam\i\
ayn\i\ $I$ ger\c cel say\i s\i na sonsuzyak\i n ise,
o zaman $I$,
$[a,b]$'de $f$'nin \textbf{integralidir,}
ve
\begin{equation*}
  I=\int_a^bf=\int_a^bf(x)\dee x
\end{equation*}
yaz\i l\i r.

\begin{theorem}
  E\u ger bir fonksiyon kapal\i\ s\i n\i rl\i\ bir aral\i kta s\"urekli ise,
  o zaman
  o aral\i kta fonksiyonun integrali vard\i r.
\end{theorem}

\begin{proof}
  Bir
  $\RS TX$ sonsuz Riemann toplam\i\ verilsin.
  \Teorem{thm:max-min} sayes\i nda
  her $[x_{k,i-1},x_{k,i}]$ aral\i\u ginda
  $f$'nin en k\"u\c c\"uk de\u geri
  bir $c_{k,i}$ noktas\i nda
  ve en b\"uyuk de\u geri bir $d_{k,i}$ noktas\i nda al\i n\i r.
  Bu durumda
  \begin{align*}
    \RS CX&=\RSmin X,&\RS DX&=\RSmax X
  \end{align*}
  olsun.
  O zaman
  \begin{equation*}
    \RSmin X\leq\RS TX\leq\RSmax X.
  \end{equation*}
    Ayr\i ca her $k$ i\c cin, bir $j_k$ i\c cin,
  \begin{equation*}
    \maks_{1\leq i\leq n_k}(f(d_{k,i})-f(c_{k,i}))
    =f(d_{k,j_k})-f(c_{k,j_k}),
  \end{equation*}
dolay\i s\i yla
  \begin{equation*}
    \RSmax X-\RSmin X\leq
    (\stf[\seq d_{\seq j}]-\stf[\seq c_{\seq j}])(b-a).
  \end{equation*}
  Ayr\i ca $[\seq d_{\seq j}]$ ve $[\seq c_{\seq j}]$
  sonsuzyak\i nd\i r,
  \c c\"unk\"u
  $[\seq x_{\seq j-1}]$ ve $[\seq x_{\seq j}]$'nin
  aras\i ndad\i r.
  Bundan dolay\i\
  \Teoremde{thm:cont}n
  \begin{equation*}
    \stf[\seq d_{\seq j}]\approx\stf[\seq c_{\seq j}],
  \end{equation*}
  ve sonu\c c olarak
  \begin{equation*}
    \RSmin X\approx\RS TX\approx\RSmax X.
  \end{equation*}
  E\u ger farkl\i\ bir $Y$ sonsuz par\c calamas\i\ verilirse,
  o zaman
  \begin{equation*}
    \RSmin X\leq\RSmin{X\cup Y}\leq\RSmax{X\cup Y}\leq\RSmax X,
  \end{equation*}
  dolay\i s\i yla
  \begin{equation*}
    \RSmin X\approx\RSmin{X\cup Y}\approx\RSmin Y.
  \end{equation*}
  Sonu\c c olarak $[a,b]$ aral\i\u g\i nda
  $f$'nin b\"ut\"un sonsuz Riemann toplam\i\
  sonsuzyak\i nd\i r.
\end{proof}

Tan\i ma g\"ore
\begin{equation*}
  \int_b^af=-\int_a^bf
\end{equation*}
olsun.

\begin{theorem}
  \begin{equation*}
    \int_a^bf=\int_a^cf+\int_c^bf.
  \end{equation*}
\end{theorem}

\begin{theorem}\label{thm:FTC}
  E\u ger $f$ s\"urekli ise, o zaman
  \begin{equation*}
    \frac{\dee}{\dee x}\int_a^xf=f.
  \end{equation*}
\end{theorem}

\begin{proof}
  $F(x)=\int_a^xf$ olsun, ve $[\seq h]$,
  s\i f\i r olmas\i n ama sonsuzk\"u\c c\"uk olsun.
  E\u ger $h_k\neq0$ ise,
  o zaman $f$ s\"urekli oldu\u gundan
  $[x,x+h_k]$ veya $[x+h_k,x]$ aral\i\u g\i nda,
  $f$ en k\"u\c c\"uk de\u gerini bir $c_k$ noktas\i nda,
  ve en b\"uy\"uk de\u gerini bir $d_k$ noktasinda, al\i r.
  O zaman
  \begin{equation*}
    f(c_k)\leq\frac{F(x+h_k)-F(x)}{h_k}\leq f(d_k),
  \end{equation*}
  dolay\i s\i yla
  \begin{equation*}
    \stf[\seq c]
    \leq\frac{\stF(x+[\seq h])-F(x)}{[\seq h]}\leq\stf[\seq d].
  \end{equation*}
  \.Istedi\u gimiz sonu\c c \c c\i kar
  \c c\"unk\"u $[\seq c]\approx x\approx[\seq d]$.
\end{proof}

\begin{theorem}
  E\u ger $F'=f$ ise, o zaman
  \begin{equation*}
    \int_a^bf=F(b)-F(a).
  \end{equation*}
\end{theorem}

\begin{proof}
  E\u ger $G(x)=\int_a^xf+F(a)$ ise,
  o zaman \Teorem{thm:FTC} sayesinde $G'=F'$.
  Ayr\i ca $G(a)=F(a)$,
  dolay\i s\i yla \Teorem{thm:f'=g'} g\"ore $G=F$.
\end{proof}

\chapter{Ek}\label{ch:app}

%\appendix

\section{A\c c\i k k\"umeler}

\Sayfada{thm:int-suf}ki \Teorem{thm:int-suf}
ve a\c sa\u g\i daki \Teorem{thm:int-nec} sayesinde
$\R$'nin a\c c\i k altk\"umeleri,
baz\i\ a\c c\i k aral\i klar\i n birle\c simidir.

\begin{theorem}\label{thm:int-nec}
  Bir $a$ ger\c cel say\i s\i n\i n,
  $\R$'nin bir $A$ altk\"umesinin
  bir i\c c noktas\i\ ise,
  o zaman
  pozitif bir $\delta$ ger\c cel say\i s\i\ i\c cin
  \begin{equation*}
    (a-\delta,a+\delta)\included A.
  \end{equation*}
\end{theorem}

\begin{proof}
  Her $k$ sayma say\i s\i\ i\c cin
  $(a-k\inv,a+k\inv)$ aral\i\u g\i,
  $A$'n\i n eleman\i\ olmayan
  bir $x_k$ ger\c cel say\i s\i n\i\ i\c cersin.
  O halde $[\seq x]\approx a$, ama $[\seq x]\notin A$.
  B\"oylece $a$, $A$'n\i n bir i\c c noktas\i\ de\u gildir.
\end{proof}


\section{Cebir}\label{ch:algebra}

Cebir a\c c\i s\i ndan
$\N$'nin altk\"umeleri,
bir $\pow{\N}$ halkas\i n\i\ olu\c sturur.
Halkan\i n iki $X$ ve $Y$ eleman\i n\i n
\begin{itemize}
\item
  $X+Y$
  toplam\i, $(X\setminus Y)\cup(Y\setminus X)$
  \emph{simetrik fark\i} olur;
\item
  $XY$
  \c carp\i m\i, $X\cap Y$ kesi\c simi olur.
\end{itemize}
$XX=X$ oldu\u gundan $\pow{\N}$
bir \textbf{Boole halkas\i d\i r.}
$\N$'nin k\"u\c c\"uk altk\"umeleri,
bu halkan\i n \emph{asal bir ideali} (\eng{prime ideal}) olu\c sturur.
Zorn Lemmas\i'ndan
her halkan\i n asal bir ideali var oldu\u gu
kan\i tlanabilir.
Asl\i nda
k\"umeler kuram\i n\i n Zermelo--Fraenkel Aksiyomlar\i\ alt\i nda
Zorn Lemmas\i\ ve Se\c cim Aksiyomu denktir
ama
sadece Asal \.Ideal Teoremi'nden
kan\i tlanamaz
\cite{MR0284328}.

$\{\seq x\in\R^{\N}\colon\seq x\denk 0\}$ k\"umesi,
$\R^{\N}$ halkas\i n\i n as\i l olmayan (\eng{non-principal})
asal bir $P$ idealidir, ve
bu \c sekilde
\begin{equation*}
  \stR=\R/P.
\end{equation*}
E\u ger tam tersine $P$,
$\R^{\N}$ halkas\i n\i n as\i l olmayan asal bir ideali ise,
o zaman $\N$'nin b\"uy\"uk altk\"umelerini,
$\seq a\in P$ olmak \"uzere
\begin{equation*}
  \{k\in\N\colon a_k=0\}
\end{equation*}
k\"umeleri
olarak tan\i mlayabiliriz.

$\stR$ cisminin sonlu elemanlar\i,
bir $\mathfrak O$ halkas\i n\i\ olu\c sturur.
Bu halka bir \textbf{de\u gerlendirme halkas\i d\i r}
(\eng{valuation ring}) \c c\"unk\"u
halkan\i n tek maksimal $\mathfrak M$ ideali vard\i r.
Asl\i nda
\begin{equation*}
  \mathfrak M=\{[\seq x]\colon[\seq x]\approx0\},
\end{equation*}
$\stR$'nin sonsuzk\"u\c c\"uk elemanlar\i n\i n olu\c sturdu\u gu
k\"umedir.
O zaman
\begin{equation*}
  \mathfrak O/\mathfrak M\cong\R.
\end{equation*}
Asl\i nda
tan\i m k\"umesi $\R$ olan
\begin{equation*}
  x\mapsto[x,x,x,\dots]+\mathfrak M
\end{equation*}
g\"ondermesi,
$\mathfrak O/\mathfrak M$ \"ust\"une bir $\phi$ izomorf\/izmas\i d\i r,
ve $[\seq a]$ sonlu ise $\phi^{-1}([\seq a]+\mathfrak M)$,
$[\seq a]$'n\i n standart par\c cas\i d\i r.

\section{Ger\c cel say\i lar\i n in\c sas\i}

\c Simdi $P$, $\Q^{\N}$ halkas\i n\i n as\i l olmayan asal bir ideali olsun.
O zaman $P$, maksimal bir idealdir.
Olsun
\begin{equation*}
  \stQ=\Q^{\N}/P.
\end{equation*}
Yukar\i daki gibi $\stQ$'n\"un sonlu elemanlar\i,
maksimal ideali $\mathfrak M$ olan bir $\mathfrak O$
de\u gerlendirme halkas\i n\i\ olu\c sturur.
Tekrar $\mathfrak O/\mathfrak M$ b\"ol\"um\"u,
tam do\u grusal s\i ralanm\i\c s bir cisim olur.
Bundan dolay\i\
$\mathfrak O/\mathfrak M$
olarak
$\R$'yi tan\i mlayabiliriz.

\section{Epsilon-delta tan\i m\i}\label{ch:ed}

\begin{theorem}\label{thm:ed}
  $f$'nin $a$'daki limitinin $L$ olmas\i\ i\c cin
  gerek ve yeter bir ko\c sul,
  $\R$'de her pozitif $\epsilon$ i\c cin,
  bir pozitif $\delta$ i\c cin,
  her $x$ i\c cin
  \begin{equation*}
    0<\abs{x-a}<\delta\lto\abs{f(x)-L}<\epsilon.
  \end{equation*}
\end{theorem}

\begin{proof}
  \.Ilk olarak
  verilen ko\c sulun do\u gru oldu\u gunu varsayal\i m.
  Bir $\seq x$ \emph{dizisi} i\c cin
  $[\seq x]\approx a$ ama $[\seq x]\neq a$ olsun.
  O halde $\stf[\seq x]\approx L$ g\"osterece\u giz.
  \c Simdi $\epsilon$, pozitif ger\c cel bir say\i\ olsun;
  $\{k\in\N\colon\abs{f(x_k)-L}<\epsilon\}$
  k\"umesinin b\"uy\"uk oldu\u gunu g\"ostermek yeter.
  Varsay\i ma g\"ore pozitif ger\c cel bir $\delta$ i\c cin
  o k\"ume $\{k\in\N\colon0<\abs{x_k-a}<\delta\}$ k\"umesini kapsar;
  ayr\i ca bu k\"ume b\"uy\"ukt\"ur.

  Tam tersine $\epsilon$,
  pozitif ger\c cel bir say\i\ olsun.
  M\"umk\"unse her pozitif ger\c cel $\delta$ i\c cin,
  ger\c cel bir $x$ i\c cin,
  \begin{equation*}
    0<\abs{x-a}<\delta\land\abs{f(x)-L}\geq\epsilon
  \end{equation*}
  olsun.
  O zaman bir $\seq x$ \emph{dizisi} i\c cin
  her $k$ sayma say\i s\i\ i\c cin
  \begin{equation*}
    0<\abs{x_k-a}<\frac1k\land\abs{f(x_k)-L}\geq\epsilon.
  \end{equation*}
  Bu durumda $[\seq x]\approx a$ ve $[\seq x]\neq a$,
  ama $\stf[\seq x]\not\approx L$.
\end{proof}

%\bibliographystyle{plain}
%\bibliography{../../references}
\def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'$}
  \def\rasp{\leavevmode\raise.45ex\hbox{$\rhook$}} \def\cprime{$'$}
  \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'$}
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\end{document}
