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\newtheorem*{solution}{\c C\"oz\"um}

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\begin{document}

\title{Aksiyomatik K\"umeler Kuram\i\ (MAT 340)}
\date{25 May\i s 2016\\
\c C\"OZ\"UMLER}
\author{David Pierce}
\maketitle
\thispagestyle{empty}

\begin{problem}
  Cantor normal bi\c cimlerini bulun.
  \begin{enumerate}[(a)]
  \item 
$\upomega^2+\upomega\cdot2+3+\upomega\cdot2+\upomega^2$
\item
$(\upomega\cdot3+1)\cdot(\upomega+2)$
\item
$(\upomega^4+\upomega^3\cdot2+\upomega^2\cdot3+\upomega+4)
\cdot\upomega^{\upomega}$
\item
$(\upomega+1)^3$
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[(a)]
    \item
$\upomega^2\cdot2$
\item
$\upomega^2+\upomega\cdot6+1$
\item
$\upomega^{\upomega}$
\item
$\upomega^3+\upomega^2+\upomega+1$
  \end{enumerate}
\end{solution}

\newpage
\begin{problem}
A\c sa\u g\i daki kan\i t yanl\i\c st\i r.
%\mbox{}\hrulefill\mbox{}
\begin{align}
\alpha+0
&=\alpha\\
&=0+\alpha.
\end{align}
%\mbox{}\hrulefill\mbox{}
Baz\i\ $\gamma$ i\c cin
\begin{equation}
\alpha+\gamma=\gamma+\alpha
\end{equation}
olsun.
O zaman
\begin{align}
	\alpha+\gamma'
	&=(\alpha+\gamma)'\\
	&=(\gamma+\alpha)'\\\label{wrong}
	&=\gamma'+\alpha.
\end{align}
Son olarak $\gamma$ limit olsun, 
ve $\beta<\gamma$ durumunda
\begin{equation}
\alpha+\beta=\beta+\alpha
\end{equation} 
olsun.
O zaman
\begin{align}
	\alpha+\gamma
&=\sup_{\xi<\gamma}(\alpha+\xi)\\
&=\sup_{\xi<\gamma}(\xi+\alpha)\\\label{wrong-again}
&=\gamma+\alpha.
\end{align}
Sonu\c c olarak her $\beta$ i\c cin
$\alpha+\beta=\beta+\alpha$.
  \begin{enumerate}[(a)]
  \item 
Kan\i t nerede yanl\i\c st\i r?
Anlat\i n.
\item
Kan\i t d\"uzeltilebilir mi?  Anlat\i n.
  \end{enumerate}
\end{problem}

\begin{solution}
  \begin{enumerate}[(a)]
\item
\eqref{wrong} ve \eqref{wrong-again} e\c sitlikleri yanl\i\c st\i r:
\begin{gather*}
(1+\upomega)'=\upomega'>\upomega=1'+\upomega;\\
\sup_{\xi<\upomega}(\xi+\upomega)
=\sup_{\xi<\upomega}(\upomega)
=\upomega<\upomega+\upomega.
\end{gather*}
\item
Kan\i t d\"uzeltilemez \c c\"unk\"u 
$\alpha+\beta=\beta+\alpha$ iddias\i\ yanl\i\c st\i r.
  \end{enumerate}
\end{solution}
\newpage

\begin{problem}\mbox{}
  \begin{enumerate}[(a)]
  \item 
S\i n\i f olmayan bir k\"ume yazabilir misiniz?
\item
K\"ume olmayan bir s\i n\i f yazabilir misiniz?
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[(a)]
    \item
Hay\i r, her $a$ k\"umesi, $\{x\colon x\in a\}$ s\i n\i f\i na e\c sittir.
\item
Evet, ne $\{x\colon x\notin x\}$ s\i n\i f\i\
ne $\mathbf{ON}$ s\i n\i f\i,
bir k\"umedir.
  \end{enumerate}
\end{solution}

\begin{problem}
Derste kulland\i\u g\i m\i z tan\i ma g\"ore
\begin{align*}
\beth_0&=\aleph_0=\upomega,&
\beth_{\alpha+1}&=\card\left(\pow{\beth_{\alpha}}\right)=2^{\beth_{\alpha}},
\end{align*}
ve limitlerde $\xi\mapsto\beth_{\xi}$ s\"ureklidir.
Bilinmeyen de\u gi\c skeni $\xi$ olan
a\c sa\u g\i daki denklemleri \c c\"oz\"un.
  \begin{enumerate}[(a)]
  \item 
$\left(\beth_{\alpha}\right)^{\beth_{\xi}}=\beth_{\alpha+2}$.
\item
$\left(\beth_{\alpha+1}\right)^{\beth_{\xi}}=\beth_{\alpha+1}$.
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[(a)]
    \item
\fbox{$\xi=\alpha+1$.}  Asl\i nda
$2<\beth_{\alpha}<\beth_{\alpha+2}=2^{\beth_{\alpha+1}}$ oldu\u gundan 
      \begin{equation*}
\beth_{\alpha+2}\leq\beth_{\alpha}{}^{\beth_{\alpha+1}}
\leq 2^{\beth_{\alpha+1}\otimes\beth_{\alpha+1}}=\beth_{\alpha+2},
      \end{equation*}
dolay\i s\i yla $\alpha+1$, denklemin bir \c c\"oz\"um\"ud\"ur.
Benzer \c sekilde $\beth_{\alpha}{}^{\beth_{\alpha}}=\beth_{\alpha+1}$
ve $\beth_{\alpha}{}^{\beth_{\alpha+2}}=\beth_{\alpha+3}$,
dolay\i s\i yla denklemin ba\c ska \c c\"oz\"um\"u yoktur.
\item
\fbox{$\xi\leq\alpha$} \c c\"unk\"u
\begin{equation*}
  \left(\beth_{\alpha+1}\right)^{\beth_{\xi}}
=2^{\beth_{\alpha}\otimes\beth_{\xi}}
=2^{\beth_{\max(\alpha,\xi)}}
=\beth_{\max(\alpha,\xi)+1}.
\end{equation*}
  \end{enumerate}
\end{solution}

\newpage
\begin{problem}
  Do\u gru mu, yanl\i\c s m\i?
(Yanl\i\c s cevaplar puan kaybeder.
$A$, $B$, ve $C$ her zaman k\"umedirler.)
  \begin{enumerate}[a)]
  \item
$\aleph_{11010}\oplus\aleph_{101}=\aleph_{11111}$.
\item
Baz\i\ $A$ ve $B$ i\c cin
\begin{align*}
A&\prec\pow A,&
B&\approx\pow B.
\end{align*}
\item
$A$, $B$, ve $C$ sonsuz ise
  \begin{equation*}
\card\bigl((A\sqcup B)\times C\bigr)=\max(\card A,\card B)\oplus\card C.
  \end{equation*}
\item
Herhangi
$A$ ve $B$ i\c cin,
bazen
\begin{equation*}
  \card(A\times B)=\max(\card A,\card B),
\end{equation*}
bazen
\begin{equation*}
\card(A\times B)=0.
\end{equation*}
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[a)]
  \item
	Yanl\i\c st\i r.
$\aleph_{11010}\oplus\aleph_{101}=\aleph_{11010}$.
\item
	Yanl\i\c st\i r.
Hi\c cbir $B$ i\c cin
$B\approx\pow B$ de\u gildir.
\item
Do\u grudur.  Herhangi $\kappa$ ve $\lambda$ kardinalleri i\c cin,
tan\i ma g\"ore
\begin{align*}
\kappa\oplus\lambda&=\card(\kappa\sqcup\lambda),\\
\kappa\otimes\lambda&=\card(\kappa\times\lambda),
\end{align*}
ama teoreme g\"ore $\kappa$ ve $\lambda$ sonsuz ise
\begin{equation*}
\kappa\oplus\lambda=\kappa\otimes\lambda=\max(\kappa,\lambda).
\end{equation*}
\item
Do\u grudur.  $A=\varnothing$ ise $\card(A\times B)=0$.
  \end{enumerate}
\end{solution}

\end{document}