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\title{Math 413 [513] (Phillips) Solutions to Homework 9}
\date{28 Nov.\ 2018}
\begin{document}
% \setcounter{section}{-1}
\maketitle
Generally, a ``solution'' is something that would be acceptable
if turned in in the form presented here, although the solutions
given are usually close to minimal in this respect.
A ``solution (sketch)'' is too sketchy to be considered a
complete solution if turned in;
varying amounts of detail would need to be filled in.
\vspace{3ex}
\noindent
{\textbf{Problem 4.10:}}
Use the fact that infinite subsets of compact sets have limit
points to give an alternate proof that if $X$ and $Z$ are \ms{s}
with $X$ \cpt, and $f \colon X \to Z$ is \ct, then $f$ is \uct.
\vspace{1ex}
\begin{proof}[Solution]
Assume that $f$ is not \uct.
Choose $\ep > 0$ for which the definition of uniform continuity fails.
Then for every $n \in \N$ there are $x_n, \, y_n \in X$ such that
$d (x_n, y_n) < \frac{1}{n}$
and $d \bigl( f (x_n), \, f (y_n) \bigr) \geq \ep$.
Since $X$ is compact,
by Theorem 3.6(a) of Rudin
the sequence $(x_n)_{n \in \N}$
has a convergent subsequence, say $(x_{k (n)})_{n \in \N}$.
Let $x = \limi{n} x_{k (n)}$.
Since
$d \bigl( x_{k (n)}, \, y_{k (n)} \bigr)
< \frac{1}{k (n)} \leq \frac{1}{n}$,
we also have $\limi{n} y_{k (n)} = x$.
We claim that $f$ is discontinuous at~$x$.
This will complete the proof.
To prove the claim,
suppose that $f$ is \ct{} at $x$.
Then
\[
\limi{n} f (x_{k (n)} ) = \limi{n} f (y_{k (n)} ) = f (x).
\]
Choose $N \in \N$ such that $n \geq N$ implies
\[
d \bigl( f (x_{k (n)} ), \, f (x) \bigr) < \frac{\ep}{3}
\A
d \bigl( f (y_{k (n)} ), \, f (x) \bigr) < \frac{\ep}{3}.
\]
Then
\[
d \bigl( f (x_{k (N)} ), \, f (y_{k (N)} \bigr)
\leq d \bigl( f (x_{k (n)} ), \, f (x) \bigr)
+ d \bigl( f (x), \, f (y_{k (n)} ) \bigr)
< \frac{\ep}{3} + \frac{\ep}{3}
= \frac{2 \ep}{3},
\]
but by construction
$d \bigl( f (x_{k (N)} ), \, f (y_{k (N)} \bigr) \geq \ep$.
This contradiction proves the claim.
\end{proof}
\vspace{1ex}
\noindent
{\emph{Remark:}}
It is a serious mistake to simply claim that
the sequence $(x_n)_{n \in \N}$
has a convergent subsequence $(x_{k (n)})_{n \in \N}$
and the sequence $(y_n)_{n \in \N}$
has a convergent subsequence $(y_{k (n)})_{n \in \N}$.
If one chooses convergent subsequences
of $(x_n)_{n \in \N}$ and $(y_n)_{n \in \N}$,
they must be called, say,
$(x_{k (n)})_{n \in \N}$ and $(y_{l (n)})_{n \in \N}$ for
{\emph{different}} functions $k, \, l \colon \N \to \N$.
It is nevertheless possible to carry out a proof by passing to
convergent subsequences of $(x_n)_{n \in \N}$ and $(y_n)_{n \in \N}$.
The following solution shows how it can be done.
This solution is not recommended here, but in other situations it may
be the only way to proceed.
\vspace{1ex}
\noindent
\begin{proof}[Alternate solution]
Assume that $f$ is not \uct.
Choose $\ep > 0$ for which the definition of uniform continuity fails.
Then for every $n \in \N$ there are $x_n, \, y_n \in X$ such that
$d (x_n, y_n) < \frac{1}{n}$
and $d \bigl( f (x_n), \, f (y_n) \bigr) \geq \ep$.
Since $X$ is compact,
by Theorem 3.6(a) of Rudin
the sequence $(x_n)_{n \in \N}$
has a convergent subsequence, say $(x_{k (n)})_{n \in \N}$.
Then $(y_{k (n)})_{n \in \N}$ is a sequence in a compact \ms,
and therefore,
again by Theorem 3.6(a) of Rudin, it has a convergent subsequence,
say $(y_{k ( r ( n ))})_{n \in \N}$.
Let $l = k \circ r$.
Then $(x_{l (n)})_{n \in \N}$ is a subsequence of the convergent
sequence $(x_{k (n)})_{n \in \N}$, and therefore converges.
Let $x = \limi{n} x_{l (n)}$ and $y = \limi{n} y_{l (n)}$.
We claim that $x = y$.
To prove the claim,
let $\ep > 0$; we show that $d (x, y) < \ep$.
Choose $N_1, \, N_2, \, N_3 \in \N$
so large that $n \geq N_1$ implies
$d (x_{l (n)}, \, x ) < \frac{\ep}{3}$,
so large that $n \geq N_2$ implies
$d (y_{l (n)}, \, y ) < \frac{\ep}{3}$,
and so large that $n \geq N_3$ implies $\frac{1}{n} < \frac{\ep}{3}$.
Set $n = \max (N_1, \, N_2, \, N_3)$.
Then
\[
d (x, y)
\leq d (x, \, x_{l (n)}) + d (x_{l (n)}, \, y_{l (n)} )
+ d (y_{l (n)}, \, y )
< \frac{\ep}{3} + \frac{1}{l (n)} + \frac{\ep}{3}
\leq \frac{\ep}{3} + \frac{1}{n} + \frac{\ep}{3}
< \ep.
\]
The claim is proved.
We now know that $\limi{n} y_{l (n)} = x = \limi{n} x_{l (n)}$.
We claim that $f$ is discontinuous at~$x$.
This will complete the proof.
To prove the claim,
suppose that $f$ is \ct{} at $x$.
Then
\[
\limi{n} f (x_{l (n)} ) = \limi{n} f (y_{l (n)} ) = f (x).
\]
Choose $N \in \N$ such that $n \geq N$ implies
\[
d \bigl( f (x_{l (n)} ), \, f (x) \bigr) < \frac{\ep}{3}
\A
d \bigl( f (y_{l (n)} ), \, f (x) \bigr) < \frac{\ep}{3}.
\]
Then
\[
d \bigl(f (x_{l (N)} ), \, f (y_{l (N)} \bigr)
\leq d \bigl( f (x_{l (n)} ), \, f (x) \bigr)
+ d \bigl( f (x), \, f (y_{l (n)} ) \bigr)
< \frac{\ep}{3} + \frac{\ep}{3}
= \frac{2 \ep}{3},
\]
but by construction
$d \bigl( f (x_{l (N)} ), \, f (y_{l (N)} \bigr) \geq \ep$.
This contradiction proves the claim.
\end{proof}
\vspace{2ex}
\noindent
{\textbf{Problem 4.11:}}
Let $X$ and $Y$ be \ms{s}.
%
\begin{enumerate}
%
\item\label{Item_8Y20_411_UCCauchy}
Let $f \colon X \to Y$ be \uct.
Prove that if $(x_n)_{n \in \N}$ is a Cauchy sequence in $X$, then
$(f (x_n))_{n \in \N}$ is a Cauchy sequence in~$Y$.
%
\item\label{Item_8Y20_411_UCExt}
Use Part~(\ref{Item_8Y20_411_UCCauchy})
to prove that if $Y$ is complete, $E \S X$ is
dense, and $f_0 \colon E \to Y$ is \uct,
then there is a unique \cfn{} $f \colon X \to Y$
such that $f |_E = f_0$.
%
\end{enumerate}
%
\begin{proof}[Solution to~(\ref{Item_8Y20_411_UCCauchy})]
We prove the first statement.
Let $(x_n)_{n \in \N}$ be a Cauchy sequence in~$X$.
Let $\ep > 0$.
Choose $\dt > 0$ such that if $s_1, \, s_2 \in X$ satisfy
$d (s_1, s_2) < \dt$, then $d (f (s_1), \, f (s_2) ) < \ep$.
Choose $N \in \N$ such that if $m, \, n \in \N$ satisfy
$m, \, n \geq N$, then $d (x_m, x_n ) < \dt$.
Then whenever $m, \, n \in \N$ satisfy
$m, \, n \geq N$, we have $d (x_m, x_n ) < \dt$,
so that $d (f (x_m), \, f (x_n) ) < \ep$.
This shows that $(f (x_n))_{n \in \N}$ is a Cauchy sequence.
\end{proof}
For~(\ref{Item_8Y20_411_UCExt}),
the neatest arrangement I can think of is to prove the following
lemmas first.
\begin{lem}\label{L_8Y20_LimExist}
Let $X$ and $Y$ be \ms{s}, with $Y$ complete,
let $E \S X$, and let $f \colon E \to Y$ be \uct.
Let $(x_n)_{n \in \N}$ be a sequence in $E$
which converges to some point in~$X$.
Then $\limi{n} f (x_n)$ exists in~$Y$.
\end{lem}
\begin{proof}
We know that convergent sequences are Cauchy.
This is therefore immediate from the first part of the problem and the
definition of completeness.
\end{proof}
\begin{lem}\label{L_8Y20_LimUniq}
Let $X$ and $Y$ be \ms{s}, with $Y$ complete,
let $E \S X$, and let $f \colon E \to Y$ be \uct.
Then for every $\ep > 0$ there is $\dt > 0$ such that whenever
$w, x \in X$ satisfy $d (w, x) < \dt$,
and whenever $(r_n)_{n \in \N}$ and $(s_n)_{n \in \N}$
are sequences in $E$
such that $r_n \to w$ and $s_n \to x$, then
\[
d \left( \limi{n} f (r_n), \, \limi{n} f (s_n) \right) < \ep.
\]
\end{lem}
The limits exist by Lemma~\ref{L_8Y20_LimExist}.
\begin{proof}[Proof of Lemma~\ref{L_8Y20_LimUniq}]
Let $\ep > 0$.
Choose $\rh > 0$ such that whenever
$w, \, x \in E$ satisfy $d (w, x) < \rh$,
then $d (f (w), \, f (x) ) < \frac{\ep}{2}$.
Set $\dt = \frac{\rh}{2} > 0$.
Let $w, \, x \in X$ satisfy $d (w, x) < \dt$,
and let $(r_n)_{n \in \N}$ and $(s_n)_{n \in \N}$ be sequences in $E$
such that $r_n \to w$ and $s_n \to x$.
Let $y = \limi{n} f (r_n)$ and $z = \limi{n} f (s_n)$.
(These exist by Lemma~\ref{L_8Y20_LimExist}.)
Choose $N$ so large that for all $n \in \N$ with $n \geq N$,
the following four conditions are all satisfied:
\begin{itemize}
\item
$d (r_n, w) < \frac{\rh}{4}$.
\item
$d (s_n, x) < \frac{\rh}{4}$.
\item
$d (f (r_n), \, y) < \frac{\ep}{4}$.
\item
$d (f (s_n), \, z) < \frac{\ep}{4}$.
\end{itemize}
We then have
\[
d (r_N, s_N) \leq d (r_N, w) + d (w, x) + d (x, s_N)
< \frac{\rh}{4} + \frac{\rh}{2}
+ \frac{\rh}{4}
= \rh.
\]
Therefore $d (f (r_N), \, f (s_N) ) < \frac{\ep}{2}$.
So
\[
d (y, z)
\leq d (y, \, f (r_N))
+ d (f (r_N), \, f (s_N) ) + d (f (s_N), \, z)
< \frac{\ep}{4} + \frac{\ep}{2}
+ \frac{\ep}{4}
= \ep,
\]
as desired.
\end{proof}
\begin{thm}\label{T_8Y20_Extend}
Let $X$ and $Y$ be \ms{s}, with $Y$ complete,
let $E \S X$, and let $f \colon E \to Y$ be \uct.
Then there is a unique \cfn{} $f \colon X \to Y$ such that $f |_E = f_0$.
\end{thm}
\begin{proof}
If $f$ exists, then it is unique by Problem~4.4, which was in the
previous assignment.
So we prove existence.
For $x \in X$, we want to define $f (x)$ by choosing
a sequence $(r_n)_{n \in \N}$ in $E$ with $\limi{n} r_n = x$ and then
setting $f (x) = \limi{n} f_0 (r_n)_{n \in \N}$.
We know that such a sequence exists because $E$ is dense in~$X$.
We know that $\limi{n} f_0 (r_n)$ exists, by Lemma~\ref{L_8Y20_LimExist}.
However, we must show that $\limi{n} f_0 (r_n)$ only depends on
$x$, not on the sequence $(r_n)_{n \in \N}$.
To prove this,
let $(r_n)_{n \in \N}$ and $(s_n)_{n \in \N}$
be sequences in $E$ with
\[
\limi{n} r_n = \limi{n} s_n = x.
\]
We claim that $\limi{n} f_0 (r_n) = \limi{n} f_0 (s_n)$.
To prove the claim, let $\ep > 0$; we show that
\[
d \left( \limi{n} f_0 (r_n), \, \limi{n} f_0 (s_n) \right) < \ep.
\]
Choose $\dt > 0$ according to Lemma~\ref{L_8Y20_LimUniq}.
We certainly have $d (x, x) < \dt$.
Therefore the conclusion of Lemma~\ref{L_8Y20_LimUniq} gives
\[
d \left( \limi{n} f_0 (r_n), \, \limi{n} f_0 (s_n) \right) < \ep,
\]
proving the claim.
We now get a well defined function $f \colon X \to Y$ by,
for $x \in X$,
choosing any sequence $(r_n)_{n \in \N}$ in $E$ with $\limi{n} r_n = x$
and defining $f (x) = \limi{n} f_0 (r_n)$.
By considering the constant sequence $x_n = x$ for all $n$,
we see immediately that $f (x) = f_0 (x)$ for $x \in E$.
We claim that $f$ is \ct, in fact \uct.
Let $\ep > 0$.
Choose $\dt > 0$ according to Lemma~\ref{L_8Y20_LimUniq}.
For $x_1, \, x_2 \in X$ with $d (x_1, x_2) < \dt$,
choose (by density of $E$, as above)
sequences $(r_n)_{n \in \N}$ and $(s_n)_{n \in \N}$ in $E$
such that $r_n \to x_1$ and $s_n \to x_2$.
Then
\[
d \left( \limi{n} f_0 (r_n), \, \limi{n} f_0 (s_n) \right) < \ep.
\]
By construction, we have $f (x_1) = \limi{n} f_0 (r_n)$
and $f (x_2) = \limi{n} f_0 (s_n)$.
Therefore we have shown that $d (f (x_1), \, f (x_2)) < \ep$,
as desired.
The claim is proved.
\end{proof}
The point of stating Lemma~\ref{L_8Y20_LimUniq} separately
is that the proof that
$f$ is well defined, and the proof that $f$ is \ct,
use essentially
the same argument.
By putting that argument in a lemma, we avoid repeating it.
\vspace{2ex}
\noindent
{\textbf{Problem 4.12:}}
State precisely and prove the following:
``A \ucfn{} of a \ucfn{} is \uct.''
\bigskip
Here is the precise statement.
\begin{prp}\label{P_8Y20_CompUC}
Let $X$, $Y$, and $Z$ be \ms{s}.
Let $f \colon X \to Y$ and $g \colon Y \to Z$ be \ucfn s.
Then $g \circ f$ is \uct.
\end{prp}
\begin{proof}[Solution]
Let $\ep > 0$.
Choose $\rh > 0$ such that if $y_1, \, y_2 \in Y$ satisfy
$d (y_1, y_2) < \rh$, then $d (g (y_1), \, g (y_2) ) < \ep$.
Choose $\dt > 0$ such that if $x_1, \, x_2 \in X$ satisfy
$d (x_1, x_2) < \dt$, then $d (f (x_1), \, f (x_2) ) < \rh$.
Then whenever $x_1, \, x_2 \in X$ satisfy
$d (x_1, x_2) < \dt$, we have $d (f (x_1), \, f (x_2) ) < \rh$,
so that $d (g (f (x_1)), \, g ( f (x_2) ) ) < \ep$.
\end{proof}
\vspace{2ex}
\noindent
{\textbf{Problem 4.14:}}
Let $f \colon [0, 1] \to [0, 1]$ be \ct.
Prove that there is $x \in [0, 1]$ such that $f (x) = x$.
\vspace{1ex}
\begin{proof}[Solution]
Define $g \colon [0, 1] \to \R$ by $g (x) = x - f (x)$.
Then $g$ is \ct.
Since $f (0) \in [0, 1]$, we have $g (0) = - f (0) \leq 0$,
while since $g (1) \in [0, 1]$, we have $g (1) = 1 - f (1) \geq 0$.
If $g (0) = 0$ then $x = 0$ satisfies the conclusion, while if
$g (1) = 0$ then $x = 1$ satisfies the conclusion.
Otherwise, $g (0) < 0$ and $g (1) > 0$, so Theorem~4.23 of Rudin
provides $x \in (0, 1)$ such that $g (x) = 0$.
This $x$ satisfies $f (x) = x$.
\end{proof}
\vspace{1ex}
Something much more general is true, namely the Brouwer Fixed Point
Theorem.
\begin{thm}\label{T_8Y20_brouwer}
Let $n \geq 1$, and let $B = \{ x \in \R^n \colon \| x \| \leq 1 \}$.
Let $f \colon B \to B$ be \ct.
Then there is $x \in B$ such that $f (x) = x$.
\end{thm}
The proof requires higher orders of connectedness, and is best
done with algebraic topology.
\vspace{2ex}
\noindent
{\textbf{Problem 4.15:}}
Prove that every \ct{} open map $f \colon \R \to \R$ is monotone.
\vspace{1ex}
Two solutions are presented,.
the second of which is sketchy.
The second is what I expect people to have done.
The first is essentially a careful rearrangement of the ideas of the
second, done so as to minimize the number of cases.
(You will see when reading the second solution why this is desirable.)
\begin{proof}[Solution]
It is convenient to break the solution into several lemmas.
\begin{lem}\label{L_8Z13_HW9_415_1}
Let $f \colon \R \to \R$ be \ct{} and open.
Let $a, \, b \in \R$ satisfy $a < b$.
Then $f (a) \neq f (b)$.
\end{lem}
\begin{proof}[Proof]
Since $f$ is \ct{} and $[a, b]$ is \cpt,
there are $c_1, c_2 \in [a, b]$
such that $m_1 = f (c_1)$ is the minimum value of $f$ on $[a, b]$
and $m_2 = f (c_2)$ is the maximum value of $f$ on $[a, b]$.
We claim that $c_1 \in \{ a, b \}$.
Suppose not.
Then $m_1 \in f \bigl( (a, b) \bigr)$,
but $(- \infty, m_1) \cap f \bigl( [a, b] \bigr) = \E$.
Therefore $m_1$ is not an interior point of $f \bigl( (a, b) \bigr)$.
This contradicts the assumption that $f$ is an open map,
and proves the claim.
Applying this argument to $- f$,
we see that $c_2 \in \{ a, b \}$.
Now, if $f (a) = f (b)$,
then $m_1 = m_2$.
So $f \bigl( (a, b) \bigr) = \{ m_1 \}$,
which is not an open set.
This contradiction proves the lemma.
\end{proof}
\begin{lem}\label{L_8Z13_HW9_415_2}
Let $f \colon \R \to \R$ be \ct{} and open.
Let $a, \, b \in \R$ satisfy $a < b$.
If $f (a) < f (b)$, then whenever $x \in \R$ satisfies $x < a$,
we have $f (x) < f (a)$.
\end{lem}
\begin{proof}
We can't have $f (x) = f (a)$, by Lemma~\ref{L_8Z13_HW9_415_1}.
If $f (x) = f (b)$, we again have a contradiction
by Lemma~\ref{L_8Z13_HW9_415_1}.
If $f (x) > f (b)$, then the Intermediate Value Theorem
provides $z \in (x, a)$ such that $f (z) = f (b)$.
Since $z < b$, this contradicts Lemma~\ref{L_8Z13_HW9_415_1}.
If $f (a) < f (x) < f (b)$, then the Intermediate Value Theorem
provides $z \in (a, b)$ such that $f (z) = f (x)$.
Since $x < z$, this again contradicts Lemma~\ref{L_8Z13_HW9_415_1}.
The only remaining possibility is $f (x) < f (a)$.
\end{proof}
\begin{lem}\label{L_8Z13_HW9_415_3}
Let $f \colon \R \to \R$ be \ct{} and open.
Let $a, \, b \in \R$ satisfy $a < b$.
If $f (a) < f (b)$, then whenever $x \in \R$ satisfies $b < x$,
we have $f (b) < f (x)$.
\end{lem}
\begin{proof}
Apply Lemma~\ref{L_8Z13_HW9_415_2} to the function $x \mapsto - f (- x)$.
\end{proof}
\begin{lem}\label{L_8Z13_HW9_415_4}
Let $f \colon \R \to \R$ be \ct{} and open.
Let $a, \, b \in \R$ satisfy $a < b$.
If $f (a) < f (b)$, then whenever $x \in \R$ satisfies $a < x < b$,
we have $f (a) < f (x) < f (b)$.
\end{lem}
\begin{proof}
Lemma~\ref{L_8Z13_HW9_415_1}
implies that $f (x)$ is equal to neither $f (a)$ nor $f (b)$.
If $f (x) < f (a)$,
we apply Lemma~\ref{L_8Z13_HW9_415_3} to $- f$, with $b$ and $x$
interchanged, to get $f (b) < f (x)$.
This implies $f (b) < f (a)$, which contradicts the hypotheses.
If $f (x) > f (b)$,
we apply Lemma~\ref{L_8Z13_HW9_415_2} to $- f$, with $a$ and $x$
interchanged, to get $f (a) > f (x)$.
This again contradicts the hypotheses.
\end{proof}
\begin{cor}\label{C_8Z13_HW9_415_4}
Let $f \colon \R \to \R$ be \ct{} and open.
Let $a, \, b \in \R$ satisfy $a < b$, and suppose that $f (a) < f (b)$.
Let $x \in \R$.
Then:
\begin{enumerate}
\item\label{C_8Z13_HW9_415_4_P1}
If $x \leq a$ then $f (x) \leq f (a)$.
\item\label{C_8Z13_HW9_415_4_P2}
If $x \leq b$ then $f (x) \leq f (b)$.
\item\label{C_8Z13_HW9_415_4_P3}
If $x \geq a$ then $f (x) \geq f (a)$.
\item\label{C_8Z13_HW9_415_4_P4}
If $x \geq b$ then $f (x) \geq f (b)$.
\end{enumerate}
\end{cor}
\begin{proof}
In any of the four cases,
if we have equality ($x = a$ or $x = b$), the conclusion is obvious.
With strict inequality,
Part~(\ref{C_8Z13_HW9_415_4_P1})
follows from Lemma~\ref{L_8Z13_HW9_415_2}, and
Part~(\ref{C_8Z13_HW9_415_4_P4})
follows from Lemma~\ref{L_8Z13_HW9_415_3}.
Part~(\ref{C_8Z13_HW9_415_4_P2})
follows from Lemma~\ref{L_8Z13_HW9_415_4} if $x > a$,
from Lemma~\ref{L_8Z13_HW9_415_2} if $x < a$,
and is trivial if $x = a$.
Part~(\ref{C_8Z13_HW9_415_4_P3})
follows from Lemma~\ref{L_8Z13_HW9_415_4} if $x < b$,
from Lemma~\ref{L_8Z13_HW9_415_3} if $x > b$,
and is trivial if $x = b$.
\end{proof}
I won't actually use Part~(\ref{C_8Z13_HW9_415_4_P2});
it is included for symmetry.
\vspace{1ex}
Now we prove the result.
Choose arbitrary $c, \, d \in \R$ with $c < d$.
We have $f (c) \neq f (d)$ by Lemma~\ref{L_8Z13_HW9_415_1}.
Suppose first that $f (c) < f (d)$.
Let $r, \, s \in \R$ satisfy $r \leq s$.
We prove that $f (r) \leq f (s)$, and there are several cases.
I will try to arrange this to keep the number of cases as small
as possible.
Case 1: $r \leq c \leq s$.
Then $f (r) \leq f (c) \leq f (s)$
by Parts (\ref{C_8Z13_HW9_415_4_P1}) and~(\ref{C_8Z13_HW9_415_4_P3})
of Corollary~\ref{C_8Z13_HW9_415_5},
taking $a = c$ and $b = d$.
Case 2: $r \leq s \leq a$.
Then $f (s) \leq f (a)$
by Part~(\ref{C_8Z13_HW9_415_4_P1}) of Corollary~\ref{C_8Z13_HW9_415_5},
taking $a = c$ and $b = d$.
Further, $f (r) \leq f (s)$
by Part~(\ref{C_8Z13_HW9_415_4_P1}) of Corollary~\ref{C_8Z13_HW9_415_5},
taking $a = s$ and $b = d$.
Case 3: $a \leq r \leq s$.
Then $f (a) \leq f (r)$
by Part~(\ref{C_8Z13_HW9_415_4_P3}) of Corollary~\ref{C_8Z13_HW9_415_5},
taking $a = c$ and $b = d$.
Further, $f (r) \leq f (s)$
by Part~(\ref{C_8Z13_HW9_415_4_P4}) of Corollary~\ref{C_8Z13_HW9_415_5},
taking $a = c$ and $b = r$.
% Let $r, \, s \in \R$ satisfy $r < s$.
% We prove that $f (r) < f (s)$, and there are several cases.
% First, the case $r < c < d < s$ is immediate from Lemmas~2 and~3;
% the case $r < c < s < d$ is immediate from Lemmas~2 and~4;
% and the case $c < r < d < s$ is immediate from Lemmas~3 and~4.
% Also, all the cases in which one or both of $r$ and $s$ is equal
% to one of $c$ and $d$ follow from one application of one of
% Lemmas~2, 3, or~4; details are omitted.
% Next, suppose $r < s < c$.
% Apply Lemma~\ref{L_8Z13_HW9_415_2} with $a = c$, $b = d$, and $x = s$
% to get $f (s) < f (c)$.
% Then apply Lemma~\ref{L_8Z13_HW9_415_2}
% with $a = s$, $b = d$, and $x = r$ to get $f (r) < f (s)$.
% If $r < c$ and $s = c$
% All cases with $r < c$ are now done.
% The remaining cases ($c < r < s < d$ and $d < r < s$)
% are similar to this one, and details are again omitted.
The case $f (c) > f (d)$ follows by applying the preceding argument
to $-f$.
\end{proof}
\begin{proof}[Alternate solution (brief sketch)]
% Suppose $f$ is not monotone; we prove that $f$ is not open.
% We first prove that there are $x, \, y \in \R$ with
% $x < y$ and $f (x) = f (y)$; this will yield a contradiction
% by Lemma~\ref{L_8Z13_HW9_415_1} of the other solution.
%
% Finding $x$ and $y$ is, as far as I can see, a messy case breakdown.
% To start:
% since $f$ isn't nondecreasing, there exist $a, \, b \in \R$
% such that $a < b$ and $f (a) > f (b)$;
%
Suppose $f$ is not monotone; we prove that $f$ is not open.
Since $f$ isn't nondecreasing, there exist $a, \, b \in \R$
such that $a < b$ and $f (a) > f (b)$;
and since $f$ isn't nonincreasing, there exist $c, \, d \in \R$
such that $c < d$ and $f (c) < f (d)$.
Now there are various cases depending on how $a$, $b$, $c$, and $d$
are arranged in $\R$, and depending on how $f (a)$ and $f (b)$ relate
to $f (c)$ and $f (d)$.
There are 13 possible ways for $a$, $b$, $c$, and $d$
to be arranged in $\R$, namely:
\begin{equation}\label{Case1}
a < b < c < d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case2}
a < b = c < d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case3}
a < c < b < d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case4}
a < c < b = d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case5}
a < c < d < b
\end{equation}
\begin{equation}\label{Eq_8Z13_Case6}
a = c < b < d
\end{equation}
\begin{equation}\label{2Eq}
a = c < b = d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case8}
a = c < d < b
\end{equation}
\begin{equation}\label{Eq_8Z13_Case9}
c < a < b < d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case10}
c < a < b = d
\end{equation}
\begin{equation}\label{Eq_8Z13_Case11}
c < a < d < b
\end{equation}
\begin{equation}\label{Eq_8Z13_Case12}
c < a = d < b
\end{equation}
\begin{equation}\label{Eq_8Z13_Case13}
c < d < a < b.
\end{equation}
Of these, the arrangement~(\ref{2Eq}) gives an immediate contradiction.
%
% Each of the remaining 12 possibilities yields $x$ and $y$ as above,
% using the Intermediate Value Theorem, but the cases in which no two
% of $a$, $b$, $c$, and $d$ are equal must each be split into two
% subcases, depending on the arrangement of the values of $f$, to
% get this.
%
For each of the others, we find $x < y < z$ such that
$f (y) < f (x), \, f (z)$
(so that $f$ is not open by Lemma~\ref{L_8Z13_HW9_415_2}
of the previous solution),
or such that $f (y) > f (x), \, f (z)$
(so that $f$ is not open by
Lemma~\ref{L_8Z13_HW9_415_3} of the previous solution).
Many cases break down into subcases depending on how the values of $f$
are arranged.
We illustrate by treating the arrangement~(\ref{Case1}).
Suppose $a < b < c < d$ and $f (b) < f (c)$.
Set
\[
x = a, \qquad y = b, \A z = d.
\]
Then $x < y < z$ and $f (y) < f (x), \, f (z)$,
so Lemma~\ref{L_8Z13_HW9_415_2} applies.
Suppose, on the other hand, that $a < b < c < d$ and $f (b) \geq f (c)$.
Set
\[
x = a, \qquad y = c, \A z = d.
\]
Then again $x < y < z$ and $f (y) < f (x), \, f (z)$,
so Lemma~\ref{L_8Z13_HW9_415_2} applies.
\end{proof}
\vspace{2ex}
\noindent
{\textbf{Problem 4.16:}}
For $x \in \R$, define $[x]$ by the relations $[x] \in \Z$ and
$x - 1 < [x] \leq x$ (this is called the ``integer part of $x$'' or
the ``greatest integer function''), and define $(x) = x - [x]$
(this is called the ``fractional part of $x$'', but the notation $(x)$
is not standard).
What discontinuities do the functions $x \mapsto [x]$ and
$x \mapsto (x)$ have?
\begin{proof}[Solution (sketch)]
Both functions are continuous at all noninteger points,
since $x \in (n, \, n + 1)$ implies $[x] = n$ and $(x) = x - n;$
both expressions are \ct{} on the interval $(n, \, n + 1)$.
Both functions have jump discontinuities at all integers:
for $n \in \Z$, we have
\[
\lim_{x \to n^+} [x] = \lim_{x \to n^+} n = n = f (n)
{\qquad {\mbox{and}} \qquad}
\lim_{x \to n^-} [x] = \lim_{x \to n^-} (n - 1) = n - 1 \neq f (n),
\]
and also
\[
\lim_{x \to n^+} (x) = \lim_{x \to n^+} (x - n) = 0 = f (n)
% {\qquad {\mbox{and}} \qquad}
\]
and
\[
\lim_{x \to n^-} (x) = \lim_{x \to n^-} [x - (n - 1)] = 1 \neq f (n).
\]
This completes the sketch of the solution.
\end{proof}
A solution must include proofs that both functions
are continuous on $\R \setminus \Z$
and that both functions have jump discontinuities
at every point of~$\Z$.
Merely stating these facts gets little credit.
\vspace{2ex}
\noindent
{\textbf{Problem 4.18:}}
Define $f \colon \R \to \R$ by
\[
f (x) = \begin{cases}
0 & \hspace{1em} x \in \R \M \Q \\
\frac{1}{q} & \hspace{1em}
{\mbox{$x = \frac{p}{q}$ in lowest terms.}}
\end{cases}
\]
(By definition, we require $q > 0$.
If $x = 0$ we take $p = 0$ and $q = 1$.)
Prove that $f$ is \ct{} at each point $x \in \R \M \Q$, and that
$f$ has a simple
(removable) discontinuity at each point $x \in \Q$.
\begin{proof}[Solution]
We claim that $\lim_{x \to 0} f (x) = 0$ for all $x \in \R$.
To prove the claim,
let $x \in \R$, and let $\ep > 0$.
Choose $N \in \N$ such that $\frac{1}{N} < \ep$.
For $1 \leq n \leq N$, let
\[
S_n = \left\{ \frac{a}{n} \colon a \in \Z \,\, {\mbox{and}} \,\,
0 < \left| \frac{a}{n} - x \right| < 1 \right\}.
\]
Then $S_n$ is finite; in fact, ${\mathrm{card}} (S_n) \leq 2 n$.
Set
\[
S = \bigcup_{n = 1}^N S_n,
\]
which is a finite union of finite sets and hence finite.
Also, $x \not\in S$.
Set
\[
\dt = \min \left( 1, \, \, \min_{y \in S} | y - x | \right).
\]
Then $\dt > 0$ because $x \not\in S$ and $S$ is finite.
Let $y \in \R$ satisfy $0 < | y - x | < \dt$.
If $y \not\in \Q$, then $| f (y) - 0| = 0 < \ep$.
Otherwise, because $y \not\in S$, $| y - x | < 1$, and $y \neq x$,
it is not possible to write $y = \frac{p}{q}$ with $q \leq N$.
Thus, when we write $y = \frac{p}{q}$ in lowest terms, we have $q > N$,
so $f (y) = \frac{1}{q}$
satisfies $0 < f (y) < \frac{1}{N} < \ep$.
This shows that $| f (y) - 0| < \ep$ in this case also.
The claim immediately implies that
$f$ is \ct{} at all points $x$ for which $f (x) = 0$
and has a removable discontinuity at every $x$ for which $f (x) \neq 0$.
\end{proof}
\end{document}