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\begin{center}
{\bf Solutions to Homework 6}
\end{center}
\vspace*{.3in}

\noindent{\bf Supplementary problem 1.}
Prove directly from Definition 4.5 that the function 
$f:[0,\infty)\to[0,\infty)$ given by $f(x) = \sqrt{x}$
is continuous at every point in its domain.
\vspace*{.1in}

\noindent{\bf Solution.}
First consider the point $0$.  If $\epsilon > 0$ is given, then
take $\delta = \epsilon^2$.  Then $|x| < \delta$ (and $x\in [0,\infty)$)
implies that 
$$
|\sqrt{x} - \sqrt{0}| = \sqrt{x} < \sqrt{\delta} = \epsilon.
$$
So $f$ is continuous at the point $0$.

Now let $a \in (0,\infty)$ be some other point.  If $\epsilon > 0$ is
given, set $\delta = \sqrt{a}\epsilon$.   Then for any $x\in [0,\infty)$
such that $|x - a|<\delta$, we have
$$
|\sqrt{x} - \sqrt{a}| = \left|\frac{x - a}{\sqrt{x} + \sqrt{a}}\right|
	\leq \frac{|x-a|}{\sqrt{a}} < \frac{\delta}{\sqrt{a}} = \epsilon.
$$
So $f$ is continuous at $a$.  We conclude that $f$ is continuous at every
point in $[0,\infty)$.
\qed

\vspace*{.2in}
\noindent{\bf Supplementary problem 2.}
Prove each item in Theorem 4.4 directly from Definition 4.1---i.e. do
not use the corresponding facts about limits of sequences.
\vspace*{.1in}

\noindent{\bf Part a.} Let $\epsilon > 0$ be given.  By the hypotheses
on $f$ and $g$, we can choose $\delta_1,\delta_2 > 0$
so that for each $x\in E$,
\begin{itemize}
\item $0< d(x,p) < \delta_1$ implies that $|f(x) - A|<\epsilon/2$;
\item $0< d(x,p) < \delta_2$ implies that $|g(x) - B|<\epsilon/2$.
\end{itemize}
Therefore let us set $\delta = \min\{\delta_1,\delta_2\}$.  If $x\in E$
satisfies $0 < d(x,p) < \delta$, then the triangle inequality gives
$$
|(f+g)(x) - (A + B)| \leq |f(x) - A| + |g(x) - B|
	< \epsilon/2 + \epsilon/2 = \epsilon.
$$
Hence $\lim_{x\to p} (f+g)(x) = A+B$.
\vspace*{.1in}

\noindent{\bf Part b.} Let $\epsilon > 0$ be given.
By hypothesis on $f$ and $g$, we can choose $\delta_1,\delta_2 > 0$
so that for each $x\in E$,
\begin{itemize}
\item $0< d(x,p) < \delta_1$ implies that $|f(x) - A|<\epsilon/4|B|$;
\item $0< d(x,p) < \delta_2$ implies that $|g(x) - B|<
	\min\{\epsilon/2|A|,|B|\}$ (in particular, $|g(x)| < 2|B|$).
\end{itemize}
Therefore let us set $\delta = \min\{\delta_1,\delta_2\}$.  If $x\in E$
satisfies $0 < d(x,p) < \delta$, then
\begin{eqnarray*}
|(fg)(x) - AB| & \leq & |f(x)g(x) - Ag(x)| + |Ag(x) - AB|
	< 2|B||f(x) - A| + |A||g(x) - B| \\
	& < & 2|B|\frac{\epsilon}{4|B|} + |A|\frac{\epsilon}{2|A|} = \epsilon.
\end{eqnarray*}
Hence $\lim_{x\to p} (fg)(x) = AB$.
\newpage

\noindent{\bf Part c.} Let $\epsilon > 0$ be given.
By hypothesis on $f$ and $g$, we can choose $\delta_1,\delta_2 > 0$
so that for each $x\in E$,
\begin{itemize}
\item $0< d(x,p) < \delta_1$ implies that $|f(x) - A|<|B|\epsilon/4$;
\item $0< d(x,p) < \delta_2$ implies that $|g(x) - B|< 
	\min\{\epsilon,|B|^2\epsilon/|A|\}$ (in particular, $|g(x)| > |B|/2$).
\end{itemize}
Therefore let us set $\delta = \min\{\delta_1,\delta_2\}$.  If $x\in E$
satisfies $0 < d(x,p) < \delta$, then
\begin{eqnarray*}
|(f/g)(x) - A/B| & = & \left|\frac{f(x)B - Ag(x)}{g(x)B}\right| 
	\leq \frac{2}{|B|^2}|f(x)B - Ag(x)| \\
	& \leq & \frac{2}{|B|^2}(|B||f(x) - A| + |A||g(x) - B|) 
	< \frac{2}{|B|^2} \left(|B|\frac{|B|\epsilon}{4} 
	+ |A|\frac{|B|^2\epsilon}{4|A|}\right)
	 = \epsilon.
\end{eqnarray*}
Hence $\lim_{x\to p} (f/g)(x) = A/B$.
\qed

\vspace*{.2in}
\noindent{\bf Solution to \#2 on Page 98.}
By Theorem 3.2d in Rudin, we can choose a sequence $\{x_n\}\subset E$
converging to $x$ (if $x\in E$, then we can take $x_n = x$ for all $n\in\N$).
Then by continuity of $f$ and Theorem 4.2, we see that 
$$
f(x) = f(\lim_{n\to\infty} x_n) = \lim_{n\to\infty} f(x_n).
$$
Since $f(x_n)\in f(E)$ by definition, we conclude that $f(x)\in \overline{E}$.
And since $x\in\overline{E}$ was arbitrary, $f(\overline{E})\subset
\overline{f(E)}$.

To see that equality need not hold, consider $f:\R\to\R$ given by $f(x) = e^x$.
Then $f(\overline{R}) = f(\R) = (0,\infty) \neq \overline{(0,\infty)} 
= [0,\infty)$.
\qed

\vspace*{.2in}
\noindent{\bf Solution to \#6 on Page 98.}
Let $E\subset\R$ be compact, $f:E\to\R$ be a function on $E$, and 
$G = \{(x,f(x))\in\R^2:x\in E\}$ be the graph of $f$.

Suppose first that $f$ is continuous. Then by Theorem 4.10a, so is the 
function $F:E\to\R^2$ given by $F(x) = (x,f(x))$.  As $E$ is compact, we
conclude that $F(E) = G$ is compact, as well.

Now suppose instead that $G$ is compact.  Again by Theorem 4.10a, it is
enough to show that the map $F$ defined in the previous paragraph is 
continuous.  To this end, let $K\subset\R^2$ be closed, and consider 
$F^{-1}(K) = F^{-1}(K\cap G) = \pi(K\cap G)$ where $\pi:\R^2 \to \R$
is the continuous map $(x,y)\to x$.  Now $K\cap G$ is a closed subset
of a compact set and therefore compact.  Hence $\pi(K\cap G)$ is compact
and therefore closed.  That is, the preimage of a closed set under $F$
is closed, and $F$ is therefore continuous.  We conclude that $f$ is
continuous.
\qed

\vspace*{.2in}
\noindent{\bf Solution to \#10 on Page 99.}
Suppose by way of contradiction that $f$ is not uniformly continuous on $X$.
That is, there exists $\epsilon>0$ such that for \emph{any} $\delta>0$,
we can find points $p,q\in X$ such that $d(p,q) < \delta$ while 
$d(f(p),f(q)) \geq \epsilon$.  Let us fix this $\epsilon$, and choose points
$p = p_n$, $q = q_n$ as above for each $\delta$ of the form $1/n$, $n\in\N$.
In other words, for each natural number $n$, we have
$$
0 < d(p_n,q_n) < 1/n,\qquad d(f(p_n),f(q_n)) \geq \epsilon.
$$
In particular $\lim_{n\to\infty} d(p_n,q_n) = 0$.

Now since $X$ is compact, we can apply Theorem 3.6a from Rudin to obtain a 
subsequence $\{p_{n_k}\}\subset \{p_n\}$ that converges to some point
$p\in X$.  On the other hand $\lim_{k\to\infty} d(p_{n_k},q_{n_k}) \to 0$,
so it follows that $\lim q_{n_k} = p$, too.  Therefore by continuity
$$
\lim_{k\to\infty} f(p_{n_k}) = f(p) = \lim_{k\to\infty} f(q_{n_k}).
$$
In particular,
$$
\lim_{k\to\infty} d(f(p_{n_k}),f(q_{n_k})) = 0
$$
This contradicts the fact that $d(f(p_{n_k}),f(q_{n_k})) \geq \epsilon$ for
all $k\in\N$.  Therefore, it must be the case that we started on the wrong
hypothetical foot and that $f$ must be uniformly continuous on $X$ after all.
\qed

\vspace*{.2in}
\noindent{\bf Solution to \#14 on Page 100.}
Let $I = [a,b]$.  If $f(a) = a$ or $f(b) = b$, we are already done, so we 
can assume (because $f(I)\subset I$) that $f(a) > a$ and $f(b) < b$.  
Consider the function $g(x) = f(x) - x$, which is also continuous.
Then by our assumptions, $g(a) > 0$ and $g(b) < 0$.  So by the 
intermediate value theorem, there exists $x\in (a,b)$ such that 
$g(x) = 0$---i.e. $f(x) = x$.
\qed
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