\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amsfonts,latexsym,amssymb,xcolor}
\usepackage{fancyhdr}
\pagestyle{fancy}
\setlength{\headheight}{50pt}

\begin{document}
\lhead{Set Theory \\ Assignment 10} 
\rhead{Solution Key \\ December 11}

\noindent
\begin{enumerate}
\item[1.]
Using Zorn's Lemma, show that every connected graph has a spanning tree. 
\end{enumerate}

\bigskip
\color{blue}
Let $G = \langle V; E\rangle$ be a graph with vertex-set $V$
and edge-set $E$. Let's agree to call
a subgraph $H=\langle V_H; E_H\rangle$
with $V_H=V$ and $E_H\subseteq E$
a `subforest' if it is acyclic, and 
a `spanning tree' if it is acyclic and connected.
That is, a subforest or spanning tree will have the
same vertices as $G$ and will have a subset of the
edges of $G$, both are required to be acyclic,
but a spanning tree is also required to be connected.

Define a poset $\langle P; < \rangle$
whose elements are edge-sets $E_H$ where
$H$ is a subforest of $G$,
and the set $P$ is ordered by $E_H < E_K$
iff $E_H\subsetneq E_K$ (the proper inclusion order).
\medskip

\noindent
    {\bf Claim.} $\langle P; < \rangle$ is inductively ordered.

    \medskip

\noindent    
    {\bf Proof.} We argue that if $C\subseteq P$ is a chain,
    then $\bigcup C\in P$. Since each element
    of $C$ is the edge-set of a subforest, 
    $\bigcup C$ is a set of edges. To see that
    $\bigcup C\in P$, we must argue that
    $\bigcup C$ is acyclic. If not, then there
    is a cycle $v_1, v_2,\ldots, v_n, v_{n+1}=v_1$
    with $\{v_i,v_{i+1}\}\in \bigcup C$ for all $i$.
    (I will write undirected edges as doubletons.)
    For each $i=1,\ldots,n$, there exists $E_i\in C$
    with $(v_i,v_{i+1})\in E_i$.  Since $C$ is a chain
    and is ordered by inclusion, there is an
    $i_0\in \{1, 2, \ldots, n\}$ such that
    $E_{i_0}$ contains $E_i$ for all $i=1,\ldots,n$,
    and that means $E_{i_0}$ contains the cycle
    $v_1, v_2,\ldots, v_n, v_{n+1}=v_1$.
    This is impossible, since $E_{i_0}\in C$
    and all sets in $C$ are edge-sets of subforests.
    The conclusion is that $\bigcup C$ is acyclic,
    hence $\bigcup C\in P$, hence
    $\bigcup C$ is an upper bound to $C$ in $P$. $\Box$

    \medskip

    By the Claim, together with Zorn's Lemma, there must exist a
    maximal element $E_H$ of $P$.
    The subgraph $H = \langle V; E_H\rangle$
    is a subforest of $G$ whose edge-set is maximal
    in the ordering of $P$. This maximality implies
    that adding any single new edge to $E_H$
    will result in a cycle. We need
    to argue that $H = \langle V; E_H\rangle$ is a spanning
    tree for $G$. For this, we need to
    show that $H$ is connected.

    To complete the proof, assume that $H$ is not connected.
    This means that there exist $u, w\in V$  which are
    not connected by any path in $H$.
    Since $G$ is connected, the vertices
    $u$ and $w$ are connected by a path in $G$,
    say a path of length $m$:
    $u = u_1,\ldots, u_m = w$. Among all such choices
    of $u$ and $w$ satisfying these conditions, we may
    assume that our choices have been made to minimize $m$.
    Necessarily $m=1$, as I now explain.
    If $u = u_1,\ldots, u_m = w$ with $m>1$
    and $m$ chosen minimally, then $u_2$ is connected
    to both $u$ and $w$ in $G$, but cannot be connected
    to both $u$ and $w$ in $H$. This means that there are $G$-paths
    $u--u_2$ and $u_2--w$ that are both of length $<m$
    and they are not both $H$-paths, so one
    of these paths contradicts the minimality of $m$.

    So far we have learned that $H$ has vertices $u$ and $w$
    that are adjacent (= connected by a path of length $m=1$)
    in $G$ but not connected by any path in $H$.
    In particular, this means that $\{u,w\}\notin E_H$.
 Let $E_{H'} = E_H\cup \{\{u,w\}\}$. The maximality 
 of $E_H$ implies that $H':=\langle V; E_{H'}\rangle$
 has a cycle. Since $H=\langle V; E_{H}\rangle$
 did not have a cycle, the cycle in $H'$
 must contain the new edge $\{u,w\}$.
 Assume that the cycle is $u, w = w_1, \ldots, w_k = u$,
 where we may assume that the vertices are distinct
 except for the first and last.
 But now $w = w_1, \ldots, w_k = u$
 is a path in $H$ that connects $w$ to $u$, and we
 chose these two elements so that there was no such path.
 This contradicts our assumption that $H$
 is not connected. The proof is complete.    
\color{black}
\bigskip

  
\begin{enumerate}
\item[2.] 
(Exercise 9.1.10.)
  Prove that $\kappa\cdot\kappa \cdots$ ($\lambda$ times) = $\kappa^{\lambda}$.
\end{enumerate}

\bigskip

\color{blue}
The lefthand side of
\[
\kappa\cdot\kappa \cdots \stackrel{?}{=} \kappa^{\lambda}
\]
is the unique cardinal equipotent with the set $A = \prod_{\alpha<\lambda} \kappa$,
while the righthand side is the unique cardinal that is
equipotent with the set
$B = \{f\colon \lambda\to \kappa\;|\;\textrm{$f$ is a function}\}$.
To prove the cardinals are equal, it suffices to produce
a bijection $F\colon B\to A$. One such bijection is
\[
B\to A\colon f\mapsto (f(0),f(1),f(2),\ldots).
\]
If you prefer to give a bijection in the other
direction, i.e. $G\colon A\to B$, then you might choose
\[
A\mapsto B\colon t\mapsto \widehat{t}, \quad\textrm{where}\quad \widehat{t}(\alpha) = t_{\alpha}.
\]
\color{black}
\bigskip

\begin{enumerate}
\item[3.]
  Show that if $\alpha$ is any ordinal, then there
  is an ordinal $\beta$ of countable cofinality satisfying $\beta>\alpha$.
\end{enumerate}
  
\bigskip

\color{blue}
If $\alpha$ is an ordinal and $\beta:=\alpha+\omega$,
then $\beta>\alpha$ and $\beta$ has countable
cofinality (since $S=\{\alpha+n\;|\;n\in \omega\}$ is a countable
cofinal subset of $\beta$).
\color{black}

\end{document}
