\documentclass[11pt]{article}
\usepackage{amsmath,amsfonts,graphicx}

\title{Solutions to Homework 3}
\author{}
\date{}

\newcommand \Q {\mathbb Q}
\newcommand \R {\mathbb R}

\begin{document}
\maketitle

\begin{enumerate}


\item[2.6.] Without using Proposition 2.12,

\begin{enumerate}

\item Prove that if $U, V \subset X$ are open, then the intersection $U \cap V$ is again open.

{\bf Solution:} Let $p \in U \cap V$.  Then there is an $r_1 > 0$ such that $B_{r_1}(p) \subset U$, and an $r_2 > 0$ such that $B_{r_2}(p) \subset V$.  Let $r = \min\{r_1,r_2\}$; then $B_r(p) \subset B_{r_1}(p) \subset U$, and $B_r(p) \subset B_{r_2}(p) \subset V$, so $B_r(p) \subset U \cap V$.

\item Give an example of countably many open sets $U_1, U_2, U_3, \dotsc \subset X$ such that their intersection $U_1 \cap U_2 \cap U_3 \cap \dotsb$ is not open.

{\bf Solution:} Let $X = \R$ with the usual metric, and let $U_n$ be the open interval $(-\tfrac{1}{n}, \tfrac{1}{n})$.  Then $U_1 \cap U_2 \cap \dotsb = \{ 0 \}$, which is not open.

\item Let $I$ be a set, and suppose that for each $i \in I$ we have an open set $U_i \subset X$.  Prove that the union $\bigcup_{i \in I} U_i$ is again open.

{\bf Solution:} Let $p \in \bigcup_{i \in I} U_i$.  Then there is some $i \in I$ with $p \in U_i$, and because $U_i$ is open, there is an $r > 0$ such that $B_r(p) \subset U_i$.  Thus $B_r(p) \subset \bigcup_{i \in I} U_i$.

\end{enumerate}
\pagebreak


\item[3.1.] Prove that the sequence of piecewise-linear functions $f_2, f_3, f_4, \dotsc \in C([0,1])$ introduced at the beginning of the section is Cauchy in the $L^1$ metric.

{\bf Solution:} Recall that $f_m$ was defined as going piecewise linearly from $f_n(0) = 0$ to $f_n(\frac12 - \frac1n) = 0$ to $f_n(\frac12 + \frac1n) = 1$ to $f_n(1) = 1$, so the graphs of $f_m$ and $f_n$ look like this:
\begin{center}
\includegraphics[width=1in]{hw3_sol_3.1.png}
\end{center}

The integral of $|f_m - f_n|$ is the area between the two graphs, that is, the area of two triangles whose height is $\frac12$ and whose base is $|\frac1m - \frac1n|$, so
\[ d_1(f_m,f_n) = \tfrac12 \left|\tfrac1m - \tfrac1n \right|. \]
So we must prove that for every $\epsilon > 0$ there is an integer $N$ such that for all $m,n \ge N$ we have
\[ \tfrac12 \left|\tfrac1m - \tfrac1n \right| < \epsilon. \]

If we're feeling sneaky, we could point out that this statement is equivalent to saying the sequence of real numbers $\frac{1}{2n}$ is Cauchy, and we know that $\frac{1}{2n} \to 0$, and that a convergent sequence is Cauchy, so we're done.

Or we could just prove the statement by hand.  Let $\epsilon > 0$ be given, choose an $N$ such that $\tfrac1{2N} < \epsilon$, and let $m,n \ge N$.  If $n \ge m$ then $\tfrac1m \ge \tfrac1n$, so
\[ \tfrac12 \left|\tfrac1m - \tfrac1n \right| \le \tfrac12 \cdot \tfrac1m \le \tfrac12 \cdot \tfrac 1N < \epsilon, \]
and if $m \ge n$ then it's similar.
\pagebreak


\item[3.4.] Let $(X,d)$ be a metric space, let $f\colon X \to X$, suppose there is a ``Lipschitz constant'' $r \in [0,1)$ such that for all $p,q \in X$ we have
\[ d(f(p), f(q)) \le r\cdot d(p,q). \]
Prove that $f$ is continuous.

{\bf Solution:} Let $p \in X$ and $\epsilon > 0$ be given, and set $\delta = \epsilon$.  If $d(p,q) < \delta$, then
\[ d(f(p), f(q)) \le r\cdot d(p,q) < d(p,q) < \delta = \epsilon, \]
where in the second step we used the fact that $r < 1$.
\bigskip


\item[3.3.] (Optional.)  Let $p_1, p_2, p_3, \dotsc$ be a Cauchy sequence in a metric space $(X,d)$.  Prove that the sequence is bounded, meaning that there is a point $q \in X$ and a radius $R > 0$ such that $p_n \in B_R(q)$ for all $n$.  In fact, for any $q \in X$ you can find such a radius $R$, and in particular for $X = \R$ you can take $q=0$.

{\bf Solution:} I'll prove the stronger statement, ``for any $q$ there is a radius $R$\dots'' rather than ``there is a $q$ and a radius $R$\dots''

Let $q \in X$ be given.  Apply the definition of Cauchy with $\epsilon = 1$ to get an $N$ such that for all $m,n \ge N$ we have $d(p_m, p_n) < 1$, and in particular $d(p_N, p_n) < 1$.  Let $R$ be the maximum of $d(q, p_1) + 1$, $d(q, p_2)+1$, and so on up through $d(q, p_N) + 1$.

I claim that $p_n \in B_R(q)$ for all $n$, that is, $d(q,p_n) < R$ for all $n$.  If $n \le N$ then this is clear by construction.  If $n \ge N$ then $d(p_N, p_n) < 1$, so
\[ d(q, p_n) \le d(q, p_N) + d(p_N, p_n) < d(q, p_N) + 1 \le R. \]


\end{enumerate}

\end{document}
