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\def\classname{Math 6210 --- Fall 2012}

\hw{2}
\due{Choose 10 of the problems below to submit by Weds., Sep.\ 19.\marg{this line was omitted in the original post}}

\begin{exercise}
Let $X$ and $Y$ be metric spaces and $f : X \rightarrow Y$ a function.  We have two definitions of what it means for $f$ to be continuous:
\begin{enumerate}[label=(\alph{*})]
\item $f$ is continuous as a function between metric spaces if, for any $x \in X$ and any $\epsilon > 0$ there is a $\delta > 0$ such that $d(x, x') < \delta$ implies $d(f(x), f(x')) < \epsilon$;
\item $f$ is continuous if, when $X$ and $Y$ are given their metric topologies, $f^{-1}(U)$ is open for every open $U \subset Y$.
\end{enumerate}
Show that these definitions are equivalent.
\end{exercise}

\begin{exercise}
\cite[Exercise~1-4]{HY}.
\begin{enumerate}[label=(\alph{*})]
\item Show that the collection of all open half-spaces is a subbasis for the standard topology on $\bR^n$.  (A subset $S \subset \bR^n$ is called an \textbf{open half-space} if there is an affine linear function $f : \bR^n \rightarrow \bR$ such that $S = \set{x \in \bR^n \big| f(x) > 0}$.  For $f$ to be \textbf{affine linear} means that $f(x) = c + g(x)$ where $c \in \bR$ is a constant and $g : \bR^n \rightarrow \bR$ is a linear function.)
\item Show that you can still get a subbasis by taking the open half spaces defined over $\bQ$.  An open half space is defined over $\bQ$ if it is given by an inequality $f(x) > 0$ with $f(x)$ an affine linear function defined over $\bQ$; this means that $f(x) = c + g(x)$ where $c \in \bQ$ and $g : \bR^n \rightarrow \bR$ is a linear function whose matrix representation has only rational entries.
\end{enumerate}
\end{exercise}

\begin{exercise}
Let $X$ be a set.  Recall that a basis for a topology on $X$ is a collection $\sC$ of subsets of $X$ such that a subset of $X$ is open if and only if it is a union of elements of $\sC$.  This definition is not the same as \cite[\S13, Definition]{Munkres}.  Show that this definition is equivalent to the one given in Munkres.

In other words, you should demonstrate that the statements below concerning a collection $\sC$ of subsets of $X$ are equivalent:
\begin{enumerate}[label=(\alph{*})]
\item There is a topology $T$ on $X$ such that a subset $U \subset X$ is open in $T$ if and only if $U = \bigcup_i V_i$ for some collection of $V_i \in \sC$.
\item 
\begin{enumerate}[label=(\roman{*})]
\item For each $x \in X$ there is at least one $V \in \sC$ with $x \in V$,\marg{typo corrected:  some words were accidentally repeated}
\item if $V, W \in \sC$ and $x \in V \cap W$ then there is some $U \in \sC$ with $x \in U \subset V \cap W$.
\end{enumerate}
\end{enumerate}
\end{exercise}

\begin{exercise}
Suppose that $f : X \rightarrow Y$ is a surjection of topological spaces and $Y$ has the quotient topology.  Must $f$ be an open map?  (Recall that for $f$ to be \textbf{open} means that $f(U) \subset Y$ is open whenever $U \subset X$ is open.)  Prove it or give a counterexample.
\end{exercise}

\begin{exercise}
Let $f : X \rightarrow Y$ be a continuous map of topological spaces.  Check the following properties:
\begin{enumerate}[label=(\alph{*})]
\item Suppose that $Y$ has the topology induced from $X$ (the quotient topology).  Let $g : Y \rightarrow Z$ be a function.  Then $gf$ is continuous if and only if $g$ is.
\item Suppose that $X$ has the topology induced from $Y$ (the generalized subspace topology).  Let $h : W \rightarrow X$ be a function.  Then $fh$ is continuous if and only if $h$ is.
\end{enumerate}
\end{exercise}

\begin{exercise}
Consider the map $f : \bR \rightarrow S^1$ defined by
\begin{equation*}
f(t) = (\cos(2 \pi t), \sin(2 \pi t)) .
\end{equation*}
Let $\alpha$ be an irrational number and give $\alpha \bZ \subset \bR$ the subspace topology.  Let $X \subset S^1$ be the image of the map
\begin{equation*}
\alpha \bZ \subset \bR \rightarrow S^1 .
\end{equation*}
Give $X$ the subspace topology from $S^1$.  Show that $X$ \textbf{does not} have the quotient topology from the map
\begin{equation*}
p : \alpha \bZ \rightarrow X .
\end{equation*}
(Hint:  show that $p$ is injective.)
\end{exercise}

\begin{exercise}
Let $X = [0,1]$, let $Y = [0,1)$, and let $Z = (0,1)$.  Give all of these the subspace topology from $\bR$.  Show that no two of $X$, $Y$, and $Z$ are homeomorphic.
\end{exercise}

\begin{exercise}
Let $I$ be the interval $[0,1]$.  Let $R$ be the equivalence relation on $I$ wherein $x \equiv y$ if $x =y$ or $\abs{x - y} = 1$.\marg{typo corrected:  originally I forgot to include the possibility $x = y$; thanks Jim van Meter}  That is $0 \equiv 1$ is the only equivalence other than $x \equiv x$ for all $x \in I$.  Show that $I / R$, with its quotient topology, is homeomorphic to $S^1$.  You may use the fact that $S^1 \cong \bR /\bZ$ without proof.\marg{added this suggestion}

(Hint:  You may find it helpful to prove that if $U$ is an open subset of $I$ that contains both $0$ and $1$ then $\bigcup_{n \in \bZ} (n + U) \subset \bR$ is open.)
\end{exercise}

\begin{exercise}
A \textbf{directed set} is a pair $(I, \leq)$ where $I$ is a set and $\leq$ is a filtered pre-order on $I$.  Recall that a \textbf{pre-order} is a relation $\leq$ on $I$ that is \emph{reflexive} ($\alpha \leq \alpha$ for all $\alpha \in I$) and \emph{transitive} (if $\alpha \leq \beta \leq \gamma$ then $\alpha \leq \gamma$).  Note that we can have $\alpha \leq \beta$ and $\beta \leq \alpha$ without $\alpha = \beta$, so a pre-order is not quite the same thing as a partial order.

The pre-order is said to be \textbf{filtered} if, for every pair $\alpha, \beta \in I$, there is some $\gamma \in I$ with $\alpha \leq \gamma$ and $\beta \leq \gamma$.
\begin{enumerate}[label=(\alph{*})]
\item \cite[ch.\ 3, Supplementary Exercise \#1(c)]{Munkres} Let $X$ be a topological space containing a point $x$.  Define $I$ to be the set of open subsets of $X$ that contain $x$.  Give $I$ the relation $U \leq V$ if $U \supset V$.  Show that $(I, \leq)$ is a net.
\end{enumerate}
Let $X$ be a topological space.  A \textbf{net} in $X$ is a function $I \rightarrow X : \alpha \mapsto x_\alpha$ where $I$ is a directed set.  We say that $\set{x_\alpha}$ converges to $x \in X$ if, for any open set $U \subset X$ containing $x$, there is an index $\alpha \in I$ such that for all $\beta \geq \alpha$ we have $x_\beta \in U$.
\begin{enumerate}[resume, label=(\alph{*})]
\item \cite[ch.\ 3, Supplementary Exercise \#6]{Munkres} Let $X$ be a topoological space.  Show that $Z \subset X$ is closed if and only if whenever $(x_\alpha)_{\alpha \in I}$ is a net in $Z$ having a limit $x \in X$, the limit is actually contained in $X$.
\item \cite[ch.\ 3, Supplementary Exercise \#7]{Munkres} Let $X$ and $Y$ be topological spaces and $f : X \rightarrow Y$ a function.  Show that $f$ is continuous if and only if $f$ takes convergent nets to convergent nets (i.e., if $(x_\alpha)_{\alpha \in I}$ is convergent then $(f(x_\alpha))_{\alpha \in I}$ is also convergent).
\item \cite[ch.\ 3, Supplementary Exercise \#10]{Munkres}.  Let $X$ be a topological space.  Show that $X$ is compact if and only if every net in $X$ has a convergent subnet.  (See the problem in \cite{Munkres} for a hint.)
\end{enumerate}
\end{exercise}

\begin{exercise}
In the last problem set, we saw that there is a $p$-adic metric on $\bQ$ defined by $d(x,y) = \abs{x-y}_p$.  We also saw that every metric space has a \emph{completion}.  The completion of $\bQ$ in the $p$-adic metric is called $\bQ_p$ and the completion of $\bZ$ in the $p$-adic metric is called $\bZ_p$ and is called the ring of \textbf{$p$-adic integers}.
\begin{enumerate}[label=(\alph{*})]
\item Suppose that for all $k \geq 0$ we have an integer $a_k$ with $0 \leq a_k < p$.  Show that the sequence
\begin{equation} \label{eqn:p-adic}
\sum_{k = 0}^\infty a_k p^k = a_0 + a_1 p + a_2 p^2 + \cdots
\end{equation}
form a Cauchy sequence in $\bZ$ with respect to the $p$-adic metric.  The sum therefore has a limit in $\bZ_p$.
\item Show that every element of $\bZ_p$ can be represented as a sum of the form~\eqref{eqn:p-adic}.  
\item   Show that if $x$ and $y$ are two points in $\bQ_p$ then there are disjoint open sets $U$ and $V$ in $\bQ_p$ such that $x \in U$ and $y \in V$ and $U \cup V = \bQ_p$.  (Hint:  show that $B(x,\epsilon)$ is open and closed for most values of $\epsilon$.)

Conclude that $\bQ_p$ is \textbf{totally disconnected}:  for any $x \in \bQ_p$, the only connected subset of $\bQ_p$ containing $x$ is $\set{x}$.\marg{corrected the definition of totally disconnected}
\item Show that the topology on $\bQ_p$ is \emph{not discrete} by showing that the set $\set{0}$ is not open.
\item Prove that $\bZ_2$ is homeomorphic to the middle third Cantor set.
\end{enumerate}
\end{exercise}



\begin{exercise}
Let $X$ be the topological space
\begin{equation*}
X = \Bigl( \set{0} \times [-1,1] \Bigr) \cup \Bigl( [-1,1] \times \set{0} \Bigr) \subset \bR^2,
\end{equation*}
with the subspace topology.  Let $Y$ be the topological space
\begin{equation*}
Y = \Bigl( \set{0} \times [0,1] \Bigr) \cup \Bigl( [-1,1] \times \set{0} \Bigr)  \subset \bR^2,
\end{equation*}
also with the subspace topology.  Show that $X$ and $Y$ are not homeomorphic.
\end{exercise}

\begin{exercise}
Let $I = [0,1]$ with the subspace topology from $\bR$.  Show that any continuous\marg{typo corrected:  I forgot the word ``continuous''!  Thanks Megan Ly and the person who asked about this in class (I apologize for forgetting who it was)} function $f : I \rightarrow I$ must have a fixed point.  (A \textbf{fixed point} is a point $x \in I$ such that $f(x) = x.$)

(Hint:  Consider $\set{(x,f(x)) \big| x \in I} \subset I \times I$.  This is called the \textbf{graph} of $f$.)
\end{exercise}

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