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\begin{document}
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%\frontmatter
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\mbox{}
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\title{Set theory exercises}
\author{David Pierce}
\date{Spring semester, 2010/11}

\maketitle

\section{}

The first set of 21 problems was available by March 29, 2011; the
first exam was on April 8.

\begin{xca}
  Without using the Completeness Theorem, show that it is a logical
  theorem that sets exist.
\end{xca}

\begin{xca}
  Ordered pairs are defined by
    $(a,b)=\{\{a\},\{b\}\}$,
and this ensures
\begin{equation*}
  (a,b)=(c,d)\liff a=c\land b=d.
\end{equation*}
If $\class C$ and $\class D$ are possibly-proper classes, find a way
to define a class $(\class C,\class D)$ so that
\begin{equation*}
  (\class C,\class D)=
  (\class E,\class F)\liff
\class C=\class E\land\class D=\class F.
\end{equation*}
\end{xca}

\begin{xca}
\textbf{Triples} can be defined by
\begin{equation*}
  \{a,b,c\}=\{a,b\}\cup\{c\}.
\end{equation*}
Given a class $\class C$, show that there is a bijection between
$(\class C\times\class C)\times\class C$ and the class defined by
\begin{equation*}
  \Exists y\Exists z\Exists w(y\in\class C\land z\in\class C\land
  w\in\class C\land x=\{\{y\},\{y,z\},\{y,z,w\}\}).
\end{equation*}
\end{xca}

\begin{xca}
We define
  $a'=a\cup\{a\}$.
Prove that
\begin{equation*}
a'=\bigcap\{x\colon a\in x\land a\included x\}.
\end{equation*}
\end{xca}

\begin{xca}
We have defined an \textbf{ordinal} as a transitive set that is
well-ordered by membership.
Show that, if $a=\{a\}$, then $a$ is not an ordinal.
\end{xca}

\begin{xca}
  The class of all ordinals is $\on$.  Let $\vnn$ be defined as the
  class of ordinals that neither are limits nor contain limits.  Using
  this only, show
  \begin{equation*}
    \vnn=\bigcup\vnn.
  \end{equation*}
\end{xca}

\begin{xca}
Prove that an ordinal $\alpha$ is $0$ or a limit if and only if
$\alpha=\bigcup\alpha$. 
\end{xca}

\begin{xca}
  Using that $\vnn$ satisfies the Peano Axioms (Theorem 29 of the
  notes), prove the various unproved lemmas and theorems about
  addition, multiplication, and exponentiation on $\vnn$ in \S3.4,
  `Arithmetic', of the notes.
\end{xca}

\begin{xca}
We know by the Recursion Theorem that there is a unique homomorphism
from $(\vnn,0,{}')$ to $(\universe,0,x\mapsto\{x\})$.  Let $\class Z$
be the image of $\vnn$ under this homomorphism.  Find a way to define
$\class Z$, and to prove that $(\class Z,0,x\mapsto\{x\})$ satisfies
the Peano Axioms, \emph{without} any reference to $\vnn$.  (Here
$\class Z$ stands for Zermelo, because \emph{his} class of natural
numbers was this.) 
\end{xca}

\begin{xca}
Prove that if $\vnn$ is a proper class, it is $\on$.
\end{xca}

\begin{xca}
By the Replacement Axiom, every function whose domain is a set is a
set, since if $\class F$ is a function with domain $a$, then $\class
F$ is the range of the function $x\mapsto(x,\class F(x))$. 
Hence there is a class of all functions from $a$ to a given class
$\class D$.  We can denote this class of functions by
\begin{equation*}
  \mapset a{\class D}.
\end{equation*}
Given also a good order $(\class C,<)$, suppose $\class E$ is the
class of all functions whose domains are sections of $\class C$ and
whose ranges are included in $\class D$. 
That is,
\begin{equation*}
\class E=\{x\colon\Exists y(y\in\class C\land
x\in\mapset{\pred[]y}{\class D})\}. 
\end{equation*}
Say $\class F$ is a function from $\class E$ to $\class D$.  Show that
there is a unique function $\class G$ from $\class C$ to $\class D$
such that, for all $a$ in $\class C$, 
\begin{equation*}
\class G(a)=\class F(\class G\restriction{\pred[]a}).
\end{equation*}
\end{xca}

\begin{xca}
Prove that every initial segment of a well-ordered class is either the
class itself or a section of it.  
\end{xca}

\begin{xca}
Prove that there is at most one embedding of one well-ordered class in
another such that the range of the embedding is an initial segment. 
\end{xca}

\begin{xca}
Prove that the union of a set of transitive sets is transitive, and
the union of a set of ordinals is either an ordinal or $\on$ itself.
\end{xca}

\begin{xca}\label{p:sub}
Prove that, if $b$ is a set of ordinals, then $\bigcup\set{\vscr
  x}{x\in b}$ is the least strict upper bound of $b$. 
\end{xca}

\begin{xca}
Prove from the definition that $\on$ contains $0$ and is closed under
$x\mapsto x'$. 
\end{xca}

\begin{xca}
If $c\included\on$, prove that $\bigcup\{x'\colon x\in c\}$ is the
least of the upper bounds of $c$ that are in \emph{in} $c$.  [This
  doesn't make sense; `in \emph{in}' should have been `\emph{not in}'.
  This would have made the problem a repetition of 
  \ref{p:sub}: so the problem should have been deleted.]
\end{xca}

\begin{xca}
Prove that $a\times b$ is a set.
\end{xca}

\begin{xca}\label{ex:lex-tot}
Prove that the lexicographic ordering is indeed a linear ordering.
\end{xca}

\begin{xca}
Use transfinite induction to prove that, if $\alpha\leq\beta$, then
$\alpha+x=\beta$ has a solution. 
\end{xca}

\begin{xca}
Find an ordinal $\alpha$ such that $\vnn+\alpha=\alpha$.
\end{xca}

\section{}

The next exam, on May 6, concerned the following two problems:

\begin{xca}
  Find the Cantor normal forms of sums, products, and powers of
  ordinals given in Cantor normal form.
\end{xca}

\begin{xca}
  Using transfinite induction, prove all of our theorems about ordinal
  arithmetic from the \emph{recursive} definitions of addition,
  multiplication, and exponentiation of ordinals.
\end{xca}

\section{}

The third exam, May 25, concerns cardinality and some basics of the
`well-founded universe' $\wf$ as discussed in class.  Such things are treated
by the following problems. First, the Axiom of Choice is not
  assumed:

\begin{xca}
  Show that Zorn's Lemma (every ordered set whose every linearly
  ordered subset has an upper bound has a maximal element) implies the
  Axiom of Choice (every set has a choice-function).
  \emph{Suggestion:}  Given a set $a$, find an appropriate 
  ordered set $(b,\pincluded)$ of functions such that a maximal
  element of $b$ will be a choice-function for $a$.  \emph{Note:}  The set of choice-functions for subsets of $a$ is \emph{not} an appropriate ordered set $b$.  (Earlier editions of the course notes suggested that it was; this was a mistake.) 
\end{xca}

Now the Axiom of Choice is assumed, so that cardinal exponentiation is
defined. 

\begin{xca}
  Show that, for all cardinals $\kappa$, $\lambda$, $\mu$, and $\nu$,
  \begin{gather*}
\begin{aligned}
\kappa^0&=1,&
0^\lambda&=\begin{cases}
  1,&\text{ if }\lambda=0,\\
  0,&\text{ if }\lambda>0,
\end{cases}\\
\kappa^1&=\kappa,&
1^\lambda&=1,\\
\kappa^{\lambda+\mu}&=\kappa^\lambda\cdot\kappa^\mu,&
\kappa^{\lambda\cdot\mu}&=(\kappa^\lambda)^\mu,
\end{aligned}
\\
\kappa\leq\mu\land\lambda\leq\nu\lto\kappa^\lambda\leq\mu^\nu.
  \end{gather*}
\end{xca}

\begin{xca}
  We have various operations on sets, such as $\setminus$, $\cap$, $\cup$,
  $\bigcap$, $\bigcup$, $\mathscr P$, $\times$, and
  $(x,y)\mapsto\mapset xy$.  If various compositions of these
  operations are applied to sets of known cardinality, what are the
  possible cardinalities of the results?  (These cardinalities may
  have to be given in the form $\beth_{\alpha}$ or $2^{\aleph_{\alpha}}$.)
\end{xca}

\begin{xca}
  We can define a function $\class F$ from $\on\times\cn$ to $\cn$ by
  \begin{align*}
    \class F(0,\kappa)&=1,&
\class F(\alpha',\kappa)&=\kappa^{\class F(\alpha,\kappa)},&
\class F(\beta,\kappa)&=\sup_{\alpha<\beta}\class F(\alpha,\kappa),
  \end{align*}
where $\beta$ is a limit.  Now define $\class G$ on the class of
infinite cardinals by
\begin{equation*}
  \class G(\kappa)=\class F(\alpha,\kappa),
\end{equation*}
where $\kappa=\aleph_{\alpha}$.  Show that
\begin{equation*} 
\class G(\aleph_{\alpha})=
\begin{cases}
1,&\text{ if }\alpha=0,\\
\beth_{\alpha-1}(\aleph_{\alpha}),&\text{ if }0<\alpha<\vnn,\\
\beth_{\alpha}(\aleph_{\alpha}),&\text{ if }\vnn\leq\alpha,
\end{cases}
\end{equation*}
for all $\alpha$, where
\begin{align*}
\beth_0(\kappa)&=\kappa,&
\beth_{\alpha+1}(\kappa)&=2^{\beth_{\alpha}(\kappa)},&
\beth_{\beta}(\kappa)&=\sup_{\alpha<\beta}\beth_{\alpha}(\kappa)
\end{align*}
where $\beta$ is a limit.  
\end{xca}

In the last problem, note that $\beth_{\alpha}(\aleph_0)=\beth_{\alpha}$.
Recall that the function $\rwf$ is defined on $\on$ by
  \begin{align*}
    \rwf(0)&=0,& 
\rwf(\alpha')&=\pow{\rwf(\alpha)},&
\rwf(\beta)&=\bigcup_{\alpha<\beta}\rwf(\alpha).   
  \end{align*}
 
\begin{xca}
Show that
\begin{align*}
  \alpha&<\beta\lto\rwf(\alpha)\included\rwf(\beta),&
  \alpha&<\beta\lto\rwf(\alpha)\in\rwf(\beta).
\end{align*}
\end{xca}
 
  \begin{xca}
  Show that 
  \begin{equation*}
  \card{\rwf(\vnn+\alpha)}=\beth_{\alpha}.
  \end{equation*}
  \end{xca}
  
\begin{xca}
If $\alpha\leq\beta$, show
\begin{align*}
\aleph_{\alpha}{}^{\aleph_{\beta}}&=2^{\aleph_{\beta}},&
\beth_{\alpha}{}^{\beth_{\beta}}&=\beth_{\beta+1}.
\end{align*}
\end{xca}

\begin{xca}
If $\alpha>\beta$, show
\begin{align*}
\aleph_{\alpha}&\leq\aleph_{\alpha}{}^{\aleph_{\beta}}\leq2^{\aleph_{\alpha}},&
\beth_{\alpha}&\leq\beth_{\alpha}{}^{\beth_{\beta}}\leq\beth_{\alpha+1}.
\end{align*}
\end{xca}

\begin{xca}
Supposing
\begin{align*}
2&\leq\kappa,& 1&\leq\lambda,& \aleph_0\leq\kappa+\lambda,
\end{align*}
show that
\begin{equation*}
\max(\kappa,2^{\lambda})\leq\kappa^{\lambda}\leq\max(2^{\kappa},2^{\lambda}).
\end{equation*}
\end{xca}

\begin{xca}
If $k\in\vnn$, show that
\begin{equation*}
\beth_{\alpha+k}{}^{\beth_{\alpha}}=\beth_{\alpha+\max(1,k)}.
\end{equation*}
\end{xca}

\begin{xca}
Compute
\begin{equation*}
\aleph_{\vnn^2+\vnn}{}^{\aleph_{\vnn^3}}.
\end{equation*}
\end{xca}

\begin{xca}
Find
\begin{equation*}
\sup\{\beth_1,\beth_1{}^{\beth_1},\beth_1{}^{\beth_1{}^{\beth_1}},\dots\}.
\end{equation*}
\end{xca}

\end{document}


