\documentclass[%
 version=last,%
 a5paper,%
 12pt,%
 headings=small,%
 twoside,%
 parskip=half,%  this option takes 2.5% more space than parskip
% draft=true,%
%DIV=12,%
headinclude=false,%
 pagesize]%
{scrartcl}
%\documentclass[a4paper,12pt]{article}
\usepackage[turkish]{babel}
\usepackage{hfoldsty}
\usepackage[neverdecrease]{paralist}
\usepackage{amsmath,amsthm,amssymb,upgreek,bm}

\theoremstyle{definition}
\newtheorem{problem}{Problem}

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

\usepackage{verbatim}
%\let\solution=\comment
%\let\endsolution=\endcomment

\newcommand{\card}[1]{\operatorname{kard}(#1)}
\newcommand{\R}{\mathbb R}
\usepackage{mathrsfs}
\newcommand{\pow}[1]{\mathscr P(#1)}
\newcommand{\Forall}[1]{\forall{#1}\;}
\newcommand{\Exists}[1]{\exists{#1}\;}
\newcommand{\included}{\subseteq}
\renewcommand{\phi}{\varphi}
\newcommand{\lto}{\Rightarrow}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}
\renewcommand{\setminus}{\smallsetminus}

%\renewcommand{\familydefault}{cmfib}
\pagestyle{empty}
\begin{document}

\title{Aksiyomatik K\"umeler Kuram\i\ (MAT 340)}
\date{29 May\i s 2014}
\author{David Pierce}
%\abstract{Final s\i nav\i}
\maketitle
\thispagestyle{empty}

\begin{problem}
$\bm A=\{x\colon\phi(x)\}$ ve $\bm R=\{(x,y)\colon\theta(x,y)\}$ olsun.
  \begin{enumerate}[a)]
\item
$\bigcap\bm A=\{x\colon\psi(x)\}$ ko\c sulunu sa\u glayan 
bir $\psi$ form\"ul\"un\"u bulun.
\item
$\bigcap\bigcap\bm A=\{x\colon\rho(x)\}$ ko\c sulunu sa\u glayan 
bir $\rho$ form\"ul\"un\"u bulun.
\item
Hangi $\sigma$ c\"umlesi i\c cin 
$\sigma$ do\u grudur ancak ve ancak 
$\bm A$ s\i n\i f\i\ $\bm R$ ba\u g\i nt\i s\i\ taraf\i ndan 
[do\u grusal] s\i ralan\i r?
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[a)]
  \item 
$\Forall y(y\in\bm A\lto x\in y)$.
\item
$\Forall y(\Forall z(z\in\bm A\lto y\in z)\lto x\in y)$.
\item
$\Forall x\Forall y\Forall z\bigl(x\in\bm A\land y\in\bm A\land z\in\bm A\lto
(\lnot{}x\mathrel Rx)\land(x\mathrel Ry\land y\mathrel Rz\lto x\mathrel Rz)\land(x\mathrel Ry\lor y\mathrel Rx\lor x=y)\bigr)$
  \end{enumerate}
\end{solution}

\begin{problem}
A\c sa\u g\i daki her iddia i\c cin 
ya bir kan\i t ya da bir kar\c s\i t \"ornek verin.
Kan\i tlar\i n\i zda bildi\u gimiz herhangi bir teoremi kullanabilirsiniz.
  \begin{enumerate}[a)]
  \item 
Her $n$ do\u gal say\i s\i\ i\c cin $(n+1)\cdot\alpha=n\cdot\alpha+\alpha$.
\item
Her $n$ do\u gal say\i s\i\ i\c cin $\alpha\cdot(n+1)=\alpha+\alpha\cdot n$.
\item
$(\alpha\cdot\beta)^{\gamma}=\alpha^{\gamma}\cdot\beta^{\gamma}$.
\item
$\left(\alpha^{\beta}\right)^{\gamma+\delta}
=\alpha^{\beta\cdot\gamma}\cdot\alpha^{\beta\cdot\delta}$.
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[a)]
  \item 
%Her $n$ do\u gal say\i s\i\ i\c cin $(n+1)\cdot\alpha=n\cdot\alpha+\alpha$.
$n=1$ ve $\alpha=\upomega$ bir kar\c s\i t \"ornektir.
\item
%Her $n$ do\u gal say\i s\i\ i\c cin $\alpha\cdot(n+1)=\alpha+\alpha\cdot n$.
$n=0$ ise do\u grudur.  $n=m$ durumunda do\u gru olsun.
O zaman
\begin{align*}
\alpha\cdot\bigl((m+1)+1\bigr)
&=\alpha\cdot(m+1)+\alpha\\
&=(\alpha+\alpha\cdot m)+\alpha\\
&=\alpha+(\alpha\cdot m+\alpha)\\
&=\alpha+\alpha\cdot(m+1).
\end{align*}
\item
%$(\alpha\cdot\beta)^{\gamma}=\alpha^{\gamma}\cdot\beta^{\gamma}$.
$(\upomega\cdot2)^2=\upomega\cdot2\cdot\upomega\cdot2=\upomega^2\cdot2
<\upomega^2\cdot2^2$.
\item
$\left(\alpha^{\beta}\right)^{\gamma+\delta}
=\alpha^{\beta\cdot(\gamma+\delta)}
=\alpha^{\beta\cdot\gamma}\cdot\alpha^{\beta\cdot\delta}$.
  \end{enumerate}  
\end{solution}

\begin{problem}
  Cantor normal bi\c cimlerini bulun.
%(Her \"us te Cantor normal bi\c ciminde olsun.)
  \begin{enumerate}[a)]
  \item 
$1+\upomega+\upomega^{\upomega}+\upomega^{\upomega^{\upomega}}$.
\item
$(\upomega+1)^{\upomega+1}$.
\item
$(\upomega+1)^{(\upomega+1)^{\upomega+1}}$.
\item
$(\upomega^4+\upomega^3\cdot2+\upomega^2\cdot3+\upomega+4)\cdot5$.
\item
$(\upomega^4+\upomega^3\cdot2+\upomega^2\cdot3+\upomega+4)\cdot\upomega^5$.
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
    \begin{enumerate}[a)]
  \item 
$%1+\upomega+\upomega^{\upomega}+\upomega^{\upomega^{\upomega}}=
\upomega^{\upomega^{\upomega}}$.
\item
$%(\upomega+1)^{\upomega+1}=\upomega^{\upomega}\cdot(\upomega+1)=
\upomega^{\upomega+1}+\upomega^{\upomega}$.
\item
$%(\upomega+1)^{(\upomega+1)^{\upomega+1}}=
(\upomega+1)^{\upomega^{\upomega+1}+\upomega^{\upomega}}
=\upomega^{\upomega^{\upomega+1}+\upomega^{\upomega}}$.
\item
$%(\upomega^4+\upomega^3\cdot2+\upomega^2\cdot3+\upomega+4)\cdot5=
\upomega^4\cdot5+\upomega^3\cdot2+\upomega^2\cdot3+\upomega+4$.
\item
$%(\upomega^4+\upomega^3\cdot2+\upomega^2\cdot3+\upomega+4)\cdot\upomega^5=
\upomega^9$.
  \end{enumerate}
\end{solution}

\begin{problem}
  Do\u gru mu, yanl\i\c s m\i, ZFC aksiyomlar\i ndan bilinemez mi?
(Yanl\i\c s cevaplar puan kaybeder.)
  \begin{enumerate}[a)]
  \item
$\aleph_1\oplus\aleph_2=\aleph_3$.
\item
$\aleph_{\alpha}\otimes(\aleph_{\beta}\oplus\aleph_{\gamma})
=\aleph_{\max\{\alpha,\beta,\gamma\}}$.
\item
$\card{\R}=\aleph_1$.
\item
$\card{\{x\in\pow{\R}\colon\card x<\aleph_0\}}=2^{\aleph_0}$.
\item
$\left(\beth_5\right)^{\beth_4}=\beth_5$.
\item
$\left(\beth_4\right)^{\beth_5}=\beth_5$.
  \end{enumerate}
\end{problem}

\begin{solution}\mbox{}
    \begin{enumerate}[a)]
  \item
%$\aleph_1\oplus\aleph_2=\aleph_3$ 
Yanl\i\c st\i r.
\item
%$\aleph_{\alpha}\otimes(\aleph_{\beta}\oplus\aleph_{\gamma})=\aleph_{\max\{\alpha,\beta,\gamma\}}$ 
Do\u grudur.
\item
%$\card{\R}=\aleph_1$ 
Bilinemez.
\item
%$\card{\{x\in\pow{\R}\colon\card x<\aleph_0\}}=2^{\aleph_0}$ 
Do\u grudur.
\item
%$\left(\beth_5\right)^{\beth_4}=\beth_5$ 
Do\u grudur.
\item
%$\left(\beth_4\right)^{\beth_5}=\beth_5$ 
Yanl\i\c st\i r.
  \end{enumerate}
\end{solution}

\end{document}
