\documentclass[%
 version=last,%
 a5paper,%
 12pt,%
 headings=small,%
% twoside,%
% open=any,%
 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,mathrsfs}
\allowdisplaybreaks

\theoremstyle{definition}
\newtheorem{problem}{Problem}
\newtheorem*{solution}{\c C\"oz\"um}

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

\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}


\DeclareMathVersion{false}
\SetSymbolFont{operators}{false}{OT1}{cmbr}{m}{n}
\SetSymbolFont{letters}{false}{OML}{cmbrm}{m}{it}
\SetSymbolFont{symbols}{false}{OMS}{cmbrs}{m}{n}
%\SetSymbolFont{largesymbols}{false}{OMX}{cmex}{m}{n}
\usepackage{relsize}
\newenvironment{falseproof}{%
\begin{quote}\mathversion{false}\relscale{0.9}}%
{\end{quote}}

\renewenvironment{quote}{%
\relscale{.90}
\begin{list}{}{%
\setlength{\leftmargin}{0.05\textwidth}
%\setlength{\topsep}{0pt}
\setlength{\topsep}{0.25\baselineskip}
\setlength{\rightmargin}{\leftmargin}
\setlength{\parsep}{\parskip}
}
\item\relax}
{\end{list}}

\renewenvironment{quotation}{%
\begin{list}{}{%
\relscale{.90}
\setlength{\leftmargin}{0.05\textwidth}
%\setlength{\topsep}{0\baselineskip}
\setlength{\topsep}{0.25\baselineskip}
\setlength{\rightmargin}{\leftmargin}
\setlength{\listparindent}{1em}
\setlength{\itemindent}{\listparindent}
\setlength{\parsep}{\parskip}
}
\item\relax}
{\end{list}}


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

\title{Aksiyomatik\\ K\"umeler Kuram\i}
\subtitle{MAT \newstylenums{340} Finali \c C\"oz\"umleri}
\date{16 Ocak 2020}
\author{David Pierce}
\maketitle
\thispagestyle{empty}

\begin{problem}
  Verilen ordinal i\c slemler s\"urekli midir?  K\i saca a\c c\i klay\i n.
  \begin{enumerate}[(a)]
  \item
    $\xi\mapsto\upomega^{\upomega+\xi}$
    
  \item
    $\xi\mapsto(\upomega+\xi)^{\upomega}$
    
  \end{enumerate}
\end{problem}

\begin{solution}
  \begin{enumerate}[(a)]
  \item
    S\"ureklidir \c c\"unk\"u
    \begin{itemize}
    \item
      verilen i\c slem,
      $(\eta\mapsto\upomega^{\eta})\circ(\xi\mapsto\upomega+\xi)$
      bile\c skesidir;
    \item
      bile\c senlerden her biri s\"ureklidir
      (\c c\"unk\"u normaldir).
    \end{itemize}
  \item
    S\"urekli de\u gildir \c c\"unk\"u artand\i r ama (\"orne\u gin)
    \begin{equation*}
      \sup_{\xi<\upomega^{\upomega}}(\upomega+\xi)^{\upomega}
      =\sup_{x<\upomega}(\upomega^x)^{\upomega}
      =\upomega^{\upomega}
      <\upomega^{\upomega^2}=(\upomega+\upomega^{\upomega})^{\upomega}.
    \end{equation*}
  \end{enumerate}
\end{solution}

\begin{problem}
  $k$ ve $n$ do\u gal say\i s\i\ i\c cin $(\upomega+n)^k$
  kuvvetinin Cantor normal bi\c cimini bulun.
\end{problem}

\begin{solution}
$(\upomega+n)^0=1$,
  $(\upomega+n)^1=\upomega+n$,
  \begin{equation*}
    (\upomega+n)^2
    =(\upomega+n)\cdot\upomega+(\upomega+n)\cdot n
    =\upomega^2+\upomega\cdot n+n,
  \end{equation*}
  \begin{align*}
    (\upomega+n)^3
    &=(\upomega^2+\upomega\cdot n+n)\cdot\upomega
    +(\upomega^2+\upomega\cdot n+n)\cdot n\\
    &=\upomega^3+\upomega^2\cdot n+\upomega\cdot n+n,
  \end{align*}
  ve genelde
\fbox{$(\upomega+n)^k
    =\upomega^k+\upomega^{k-1}\cdot n+\dots+\upomega\cdot n+n.$}
\end{solution}




\begin{problem}
  A\c sa\u g\i daki kan\i t\i n t\"um yanl\i\c s ad\i mlar\i n\i\ g\"osterin.
  
  \.Ilk olarak
  \begin{align}
    0\cdot(\beta+\gamma)
    &=0\\
    &=0+0\\
    &=0\cdot\beta+0\cdot\gamma.
  \end{align}
    Ayr\i ca e\u ger
    \begin{equation}
      \alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma
    \end{equation}
    ise,
  o zaman
\begin{align}
\alpha'\cdot(\beta+\gamma)
&=\alpha\cdot(\beta+\gamma)+(\beta+\gamma)\\
&=(\alpha\cdot\beta+\alpha\cdot\gamma)+(\beta+\gamma)\\
&=(\alpha\cdot\beta+\beta)+(\alpha\cdot\gamma+\gamma)\\
&=\alpha'\cdot\beta+\alpha'\cdot\gamma.
\end{align}
Son olarak
$\alpha$ limit oldu\u gunda
\begin{equation}
  \Forall{\xi}\bigl(\xi<\alpha\lto
  \xi\cdot(\beta+\gamma)=\xi\cdot\beta+\xi\cdot\gamma\bigr)
\end{equation}
ise, o zaman
\begin{align}
	\alpha\cdot(\beta+\gamma)
	&=\sup_{\xi<\alpha}\bigl(\xi\cdot(\beta+\gamma)\bigr)\\
	&=\sup_{\xi<\alpha}\bigl(\xi\cdot\beta+\xi\cdot\gamma)\\
	&=\sup_{\xi<\alpha}(\xi\cdot\beta)+\sup_{\xi<\alpha}(\xi\cdot\gamma)\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma.
\end{align}
B\"oylece her $\alpha$, $\beta$, ve $\gamma$ i\c cin
  \begin{equation}
     \alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma.
  \end{equation}
\end{problem}

\begin{solution}
  (5), (7), (8), (10), (12), ve (13) yanl\i\c st\i r.
\end{solution}



\begin{problem}
  A\c sa\u g\i daki e\c sitli\u gi kan\i tlay\i n.
  \begin{equation*}
     \alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma
  \end{equation*}  
\end{problem}

\begin{solution}
  \.Ilk olarak
  \begin{align*}
    \alpha\cdot(\beta+0)
    &=\alpha\cdot\beta&&\text{[tan\i m]}\\
    &=\alpha\cdot\beta+0&&\text{[tan\i m]}\\
    &=\alpha\cdot\beta+\alpha\cdot0.&&\text{[tan\i m]}
  \end{align*}
  Ayr\i ca
 $\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$
  ise, o zaman
  \begin{align*}
    \alpha\cdot(\beta+\gamma')
    &=\alpha\cdot(\beta+\gamma)'&&\text{[tan\i m]}\\
    &=\alpha\cdot(\beta+\gamma)+\alpha&&\text{[tan\i m]}\\
    &=(\alpha\cdot\beta+\alpha\cdot\gamma)+\alpha&&\text{[varsay\i m]}\\
    &=\alpha\cdot\beta+(\alpha\cdot\gamma+\alpha)&&\text{[birle\c smelilik]}\\
    &=\alpha\cdot\beta+\alpha\cdot\gamma'.&&\text{[tan\i m]}
  \end{align*}
  Son olarak
$\gamma$ limit oldu\u gunda
\begin{equation*}
  \Forall{\xi}\bigl(\xi<\gamma\lto
  \alpha\cdot(\beta+\xi)=\alpha\cdot\beta+\alpha\cdot\xi\bigr)
\end{equation*}
ise, o zaman
\begin{align*}
  \alpha\cdot(\beta+\gamma)
  &=\alpha\cdot\sup_{\xi<\gamma}(\beta+\xi)&&\text{[tan\i m]}\\
  &=\sup_{\xi<\gamma}\bigl(\alpha\cdot(\beta+\xi)\bigr)&&\text{[normallik]}\\
  &=\sup_{\xi<\gamma}(\alpha\cdot\beta+\alpha\cdot\xi)&&\text{[varsay\i m]}\\
  &=\alpha\cdot\beta+\sup_{\xi<\gamma}(\alpha\cdot\xi)&&\text{[normallik]}\\
  &=\alpha\cdot\beta+\alpha\cdot\gamma.&&\text{[tan\i m]}
\end{align*}
\end{solution}



\begin{problem}
  \c C\"oz\"un.
  \begin{enumerate}[(a)]
  \item
    $(\aleph_{\upomega}\oplus\aleph_{\xi})\otimes\aleph_{\upomega}
    =\aleph_{\upomega^2}$
    
  \item
    $\xi+\upomega^{\upomega}+\eta=\upomega+\upomega^{\upomega}+\upomega$
    
  \end{enumerate}
\end{problem}

\begin{solution}
  \begin{enumerate}[(a)]
  \item
  $\xi=\upomega^2$.
  \item
    $\xi<\upomega^{\upomega}$ ve $\eta=\upomega$.
  \end{enumerate}
\end{solution}

\begin{problem}
  A\c sa\u g\i daki k\"umelerin kardinallerini,
  $\aleph_{\alpha}$ veya $\beth_{\alpha}$
  bi\c ciminde yaz\i n.
  \begin{enumerate}[(a)]
  \item
    $\mathscr P\bigl(\mathscr P(\mathbb R)\bigr)$
    
  \item
    $\sup\left\{\beth_5,(\beth_5)^{\beth_5},(\beth_5)^{(\beth_5)^{\beth_5}},
    (\beth_5)^{(\beth_5)^{(\beth_5)^{\beth_5}}},\dots\right\}$
    
  \end{enumerate}
\end{problem}

\begin{solution}
  \begin{enumerate}[(a)]
  \item
    $\mathscr P\bigl(\mathscr P(\mathbb R)\bigr)
    \approx\mathscr P\Bigl(\mathscr P\bigl(\mathscr P(\upomega)\bigr)\Bigr)
    \approx 2^{2^{2^{\aleph_0}}}
    =\beth_3$.
  \item
    Sonsuz $\kappa$ i\c cin $\kappa^{\kappa}=2^{\kappa}$,
    dolay\i s\i yla
    verilen k\"ume
    \begin{equation*}
      \{\beth_5,\beth_6,\beth_7,\beth_8,\dots\}
    \end{equation*}
    olur
    ve supremumu $\beth_{\upomega}$ olur.
  \end{enumerate}
\end{solution}

\end{document}
