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% Edit This:
\def\classtitle{Real Analysis}
\def\classnumber{Math 70100}
\def\examtitle{Homework 5}
\def\duedate{October 11, 2019}

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\begin{document}
\noindent
{\large \bf \classnumber: \classtitle (Prof. Hooper) \hfill Due \duedate}\\
{\large \bf \examtitle}\\

\ifprintanswers 
\noindent
{\bf Solutions:}
\else
\noindent
%{\bf Instructions:}
%Think about how to write clear and concise answers. Write the solution in your own words.
\fi

\begin{questions}

\question (Folland \S 4.3 \# 31a) Let $\langle x_n \rangle_{n \in N}$ be a sequence and let $k \mapsto n_k$ be a map from $\N$ to itself. Prove that $\langle x_{n_k} \rangle_{k \in \N}$ is a subnet if and only if $n_k \to \infty$ as $k \to \infty$. (On the other hand, $\langle x_{n_k} \rangle_{k \in \N}$ is a subsequence if and only if $n_k$ is strictly increasing in $k$. This is basically the definition of subsequence.)

\begin{sol}
\end{sol}

\question (Folland \S 4.3 \# 34) If $X$ has the weak topology generated by a family $\sF$ of functions, then a net $\langle x_\alpha \rangle$
converges to $x \in X$ iff $\langle f(x_\alpha)\rangle$ converges to $f (x)$ for all $f \in  \sF$. (In particular, if $X=\prod_{i \in I} X_i$ then $x_\alpha \to x$ if and only if $\pi_i(x_\alpha) \to x_i$ for all $i \in I$.)

\begin{sol}
\end{sol}

\question The Hilbert cube is the countable product $H=[0,1]^\N$
of all functions $\N \to [0,1]$ endowed with the product topology.
Give a direct proof that the Hilbert cube is sequentially compact.
That is, given a sequence $\{\alpha^n \in H\}_{n \in \N}$,
find a convergent subsequence. 
({\em Hint:} You may want to use a version of the Cantor diagonal argument.)
\begin{sol}
\end{sol}

\question {\em (Modified from Pugh Chapter 2 \#79)} A space $X$ is locally path-connected if given any $x \in X$
and any open set $U \subset X$ containing $x$, there is an open set $V \subset U$ containing $x$ which is path-connected.

Let $X$ be a topological space which is non-empty, compact, locally path-connected and connected.
Prove that $X$ is path-connected.

\begin{sol}
\end{sol}

\question {\em (Spring 2007 Qual)} Let $X$ be a compact metric space and let ${\mathcal U}$ be an open cover of $X$. Prove that there is an $\epsilon>0$
so that for every $x \in X$ there is a $U \in {\mathcal U}$ containing the open ball of radius $\epsilon$ about $x$. (Such an $\epsilon>0$ is called a {\em Lebesgue number} for the cover.)

\begin{sol}
\end{sol}

\question {\em (Spring 2015 Qual)} Suppose $X$ is a compact metric space and $f:X \to X$ is an isometry in the sense that 
$$d\big(f(x_1),f(x_2)\big)=d(x_1, x_2) \quad \text{for all $x_1, x_2 \in X$}.$$
Prove that $f(X)=X$. ({\em Hint:} Give a dynamical proof, i.e., consider repeatedly applying $f$.)

\begin{sol}
\end{sol}

\question {\em (Folland \S 4.4 \#43)}
Let $A = \{0,1\}^{[0,1)}$. (Recall that $\{0,1\}^{[0,1)}$ denotes the set of functions from $[0,1)$ into $\{0,1\}$, and comes with the product topology.)
Define the sequence $a_n \in A$ such that 
$$x = \sum_{n=1}^\infty \frac{a_n(x)}{2^n}$$
and such that if $x$ is a dyadic rational, $a_n(x)$ is eventually zero. Prove that the sequence $\{a_n\}$ has no convergent subsequence. Thus $A$ is not sequentially compact. (However, it is compact by Tychonoff's theorem.)

\begin{sol}
\end{sol}

\end{questions}
\end{document}
