Baby Rudin Problems
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 19 March 2014
Chapter 4

Suppose $f$ is a real function defined on ${R}^{1}$ which satisfies
$$\underset{h\to 0}{\text{lim}}[f(x+h)f(xh)]=0$$
for every $x\in {R}^{1}\text{.}$ Does this imply that $f$ is continuous?

If $f$ is a continuous mapping of a metric space $X$ into a metric space
$Y,$ prove that
$$f\left(\stackrel{\u203e}{E}\right)\subset \stackrel{\u203e}{f\left(E\right)}$$
for every set $E\subset X\text{.}$
$\text{(}\stackrel{\u203e}{E}$ denotes the closure of
$E\text{.)}$ Show, by an example, that
$f\left(\stackrel{\u203e}{E}\right)$ can be a proper subset of
$\stackrel{\u203e}{f\left(E\right)}\text{.}$

Let $f$ be a continuous real function on a metric space $X\text{.}$ Let
$Z\left(f\right)$ (the zero set of
$f\text{)}$ be the set of all $p\in X$ at which
$f\left(p\right)=0\text{.}$ Prove that
$Z\left(f\right)$ is closed.

Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space
$Y,$ and let $E$ be a dense subset of $X\text{.}$ Prove
that $f\left(E\right)$ is dense in
$f\left(X\right)\text{.}$ If
$g\left(p\right)=f\left(p\right)$
for all $p\in E,$ prove that
$g\left(p\right)=f\left(p\right)$
for all $p\in X\text{.}$ (In other words, a continuous mapping is determined by its values
on a dense subset of its domain.)

If $f$ is a real continuous function defined on a closed set $E\subset {R}^{1},$
prove that there exist continuous real functions $g$ on ${R}^{1}$ such that
$g\left(x\right)=f\left(x\right)$ for all
$x\in E\text{.}$ (Such functions $g$ are called continuous extensions of
$f$ from $E$ to ${R}^{1}\text{.)}$ Show that the result becomes false
if the word "closed" is omitted. Extend the result to vectorvalued functions. Hint: Let the graph of $g$ be a straight line on each of the
segments which constitute the complement of $E$ (compare Exercise 29, Chap. 2). The result remains true if
${R}^{1}$ is replaced by any metric space, but the proof is not so simple.

If $f$ is defined on $E,$ the graph of $f$ is the set of points
$(x,f\left(x\right)),$ for
$x\in E\text{.}$ In particular, if $E$ is a set of real numbers, and
$f$ is realvalued, the graph of $f$ is a subset of the plane.
Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.

If $E\subset X$ and if $f$ is a function defined on $X,$
the restriction of $f$ to $E$ is the function $g$ whose domain of definition is
$E,$ such that $g\left(p\right)=f\left(p\right)$
for $p\in E\text{.}$ Define $f$ and $g$ on
${R}^{2}$ by: $f(0,0)=g(0,0)=0,$
$f(x,y)=x{y}^{2}/({x}^{2}+{y}^{4}),$
$g(x,y)=x{y}^{2}/({x}^{2}+{y}^{6})$ if
$(x,y)\ne (0,0)\text{.}$
Prove that $f$ is bounded on ${R}^{2},$ that $g$ is
unbounded in every neighborhood of $(0,0),$ and that
$f$ is not continuous at $(0,0)\text{;}$
nevertheless, the restrictions of both $f$ and $g$ to every straight line in ${R}^{2}$
are continuous!

Let $f$ be a real uniformly continuous function on the bounded set $E$ in
${R}^{1}\text{.}$ Prove that $f$ is bounded on $E\text{.}$
Show that the conclusion is false if boundedness of $E$ is omitted from the hypothesis.

Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every
$\epsilon >0$ there exists a $\delta >0$ such that
$\text{diam}\hspace{0.17em}f\left(E\right)<\epsilon $
for all $E\subset X$ with $\text{diam}\hspace{0.17em}E<\delta \text{.}$

Complete the details of the following alternative proof of Theorem 4.19: If $f$ is not uniformly continuous, then for some
$\epsilon >0$ there are sequences $\left\{{p}_{n}\right\},\left\{{q}_{n}\right\}$
in $X$ such that ${d}_{X}({p}_{n},{q}_{n})\to 0$
but ${d}_{Y}(f\left({p}_{n}\right),f\left({q}_{n}\right))>\epsilon \text{.}$
Use Theorem 2.37 to obtain a contradiction.

Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$
and prove that $\left\{f\left({x}_{n}\right)\right\}$ is a Cauchy sequence in
$Y$ for every Cauchy sequence $\left\{{x}_{n}\right\}$ in $X\text{.}$
Use this result to give an alternative proof of the theorem stated in Exercise 13.

A uniformly continuous function of a uniformly continuous function is uniformly continuous.
State this more precisely and prove it.

Let $E$ be a dense subset of a metric space $X,$ and let $f$
be a uniformly continuous real function defined on $E\text{.}$ Prove that $f$
has a continuous extension from $E$ to $X$ (see Exercise 5 for terminology). (Uniqueness follows from Exercise 4.)
Hint: For each $p\in X$ and each positive integer $n,$ let
${V}_{n}\left(p\right)$ be the set of all
$q\in E$ with $d(p,q)<1/n\text{.}$
Use Exercise 9 to show that the intersection of the closures of the sets $f\left({V}_{1}\left(p\right)\right),f\left({V}_{2}\left(p\right)\right),\dots ,$
consists of a single point, say $g\left(p\right),$ of
${R}^{1}\text{.}$ Prove that the function $g$ so defined on $X$
is the desired extension of $f\text{.}$
Could the range space ${R}^{1}$ be replaced by ${R}^{k}\text{?}$
By any compact metric space? By any complete metric space? By any metric space?

Let $I=[0,1]$ be the closed unit interval. Suppose
$f$ is a continuous mapping of $I$ into $I\text{.}$ Prove that
$f\left(x\right)=x$ for at least one
$x\in I\text{.}$

Call a mapping of $X$ into $Y$ open if $f\left(V\right)$
is an open set in $Y$ whenever $V$ is an open set in $X\text{.}$
Prove that every continuous open mapping of ${R}^{1}$ into ${R}^{1}$ is monotonic.

Let $\left[x\right]$ denote the largest integer contained in $x,$ that is,
$\left[x\right]$ is the integer such that
$x1<\left[x\right]\le x\text{;}$ and let
$\left(x\right)=x\left[x\right]$ denote the fractional part of
$x\text{.}$ What discontinuities do the functions $\left[x\right]$ and
$\left(x\right)$ have?

Let $f$ be a real function defined on $(a,b)\text{.}$
Prove that the set of points at which $f$ has a simple discontinuity is at most countable. Hint: Let $E$
be the set on which $f(x)<f(x+)\text{.}$
With each point $x$ of $E,$ associate a triple
$(p,q,r)$ of rational numbers such that

$f(x)<p<f(x+),$

$a<q<t<x$ implies $f\left(t\right)<p,$

$x<t<r<b$ implies
$f\left(t\right)>p\text{.}$
The set of all such triples is countable. Show that each triple is associated with at most one point of $E\text{.}$
Deal similarly with the other possible types of simple discontinuities.

Every rational $x$ can be written in the form $x=m/n,$
where $n>0,$ and $m$ and $n$ are integers without
any common divisors. When $x=0,$ we take $n=1\text{.}$
Consider the function $f$ defined on ${R}^{1}$ by
$$f\left(x\right)=\{\begin{array}{cc}0& \left(x\hspace{0.17em}\text{irrational}\right),\\ \frac{1}{n}& (x=\frac{m}{n})\text{.}\end{array}$$
Prove that $f$ is continuous at every irrational point, and that $f$ has a simple discontinuity at every rational point.

Suppose $f$ is a real function with domain ${R}^{1}$ which has the intermediate value property: If
$f\left(a\right)<c<f\left(b\right),$
then $f\left(x\right)=c$ for some $x$ between
$a$ and $b\text{.}$
Suppose also, for every rational $r,$ that the set of all $x$ with
$f\left(x\right)=r$ is closed.
Prove that $f$ is continuous.
Hint: If ${x}_{n}\to {x}_{0}$ but
$f\left({x}_{n}\right)>r>f\left({x}_{0}\right)$
for some $r$ and all $n,$ then
$f\left({t}_{n}\right)=r$ for some
${t}_{n}$ between ${x}_{0}$ and
${x}_{n}\text{;}$ thus ${t}_{n}\to {x}_{0}\text{.}$
Find a contradiction. (N. J. Fine, Amer. Math. Monthly, vol. 73, 1966, p. 782.)

If $E$ is a nonempty subset of a metric space $X,$ define the distance from
$x\in X$ to $E$ by
$${\rho}_{E}\left(x\right)=\underset{x\in E}{\text{inf}}\hspace{0.17em}d(x,z)\text{.}$$

Prove that ${\rho}_{E}\left(x\right)=0$ if and only if
$x\in \stackrel{\u203e}{E}\text{.}$

Prove that ${\rho}_{E}$ is a uniformly continuous function on $X,$
by showing that
$${\rho}_{E}\left(x\right){\rho}_{E}\left(y\right)\le d(x,y)$$
for all $x\in X,$ $y\in X\text{.}$
Hint: ${\rho}_{E}\left(x\right)\le d(x,z)\le d(x,y)+d(y,z),$ so that
$${\rho}_{E}\left(x\right)\le d(x,y)+{\rho}_{E}\left(y\right)\text{.}$$

Suppose $K$ and $F$ are disjoint sets in a metric space $X,$
$K$ is compact, $F$ is closed. Prove that there exists $\delta >0$
such that $d(p,q)>\delta $ if
$p\in K,$ $q\in F\text{.}$
Hint: ${\rho}_{F}$ is a continuous positive function on $K\text{.}$
Show that the conclusion may fail for two disjoint closed sets if neither is compact.

Let $A$ and $B$ be disjoint nonempty closed sets in a metric space $X,$ and define
$$f\left(p\right)=\frac{{\rho}_{A}\left(p\right)}{{\rho}_{A}\left(p\right)+{\rho}_{B}\left(p\right)}\phantom{\rule{1em}{0ex}}(p\in X)\text{.}$$
Show that $f$ is a continuous function on $X$ whose range lies in $[0,1],$
that $f\left(p\right)=0$ precisely on $A$ and
$f\left(p\right)=1$ precisely on $B\text{.}$
This establishes a converse of Exercise 3: Every closed set $A\subset X$ is
$Z\left(f\right)$ for some continuous real $f$ on $X\text{.}$
Setting
$$V={f}^{1}\left([0,\frac{1}{2}]\right),\phantom{\rule{2em}{0ex}}W={f}^{1}\left((\frac{1}{2},1]\right),$$
show that $V$ and $W$ are open and disjoint, and that
$A\subset V,$ $B\subset W\text{.}$
(Thus pairs of disjoint closed sets in a metric space can be covered by pairs of disjoint open sets. This property of metric spaces is called normality.)

A realvalued function $f$ defined in $(a,b)$ is said to be
convex if
$$f(\lambda x+(1\lambda )y)\le \lambda f\left(x\right)+(1\lambda )f\left(y\right)$$
whenever $a<x<b,$
$a<y<b,$ $0<\lambda <1\text{.}$
Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. (For example, if
$f$ is convex, so is ${e}^{f}\text{.)}$
If $f$ is convex in $(a,b)$ and if
$a<s<t<u<b,$ show that
$$\frac{f\left(t\right)f\left(s\right)}{ts}\le \frac{f\left(u\right)f\left(s\right)}{us}\le \frac{f\left(u\right)f\left(t\right)}{ut}\text{.}$$

Assume that $f$ is a continuous real function defined in $(a,b)$ such that
$$f\left(\frac{x+y}{2}\right)\le \frac{f\left(x\right)+f\left(y\right)}{2}$$
for all $x,y\in (a,b)\text{.}$
Prove that $f$ is convex.

If $A\subset {R}^{k}$ and $B\subset {R}^{k},$
define $A+B$ to be the set of all sums $\text{x}+\text{y}$
with $\text{x}\in A,$ $\text{y}\in B\text{.}$

If $K$ is compact and $C$ is closed in ${R}^{k},$
prove that $K+C$ is closed.
Hint: Take $\text{z}\in K+C,$ put
$F=\text{z}C,$ the set of all
$\text{z}\text{y}$ with $\text{y}\in C\text{.}$
Then $K$ and $F$ are disjoint. Choose $\delta $ as in Exercise 21. Show that the open ball with center
$\text{z}$ and radius $\delta $ does not intersect $K+C\text{.}$

Let $\alpha $ be an irrational real number. Let ${C}_{1}$ be the set of all integers, let
${C}_{2}$ be the set of all $n\alpha $ with
$n\in {C}_{1}\text{.}$ Show that ${C}_{1}$
and ${C}_{2}$ are closed subsets of ${R}^{1}$ whose sum
${C}_{1}+{C}_{2}$ is not closed, by showing that
${C}_{1}+{C}_{2}$ is a countable dense subset of ${R}^{1}\text{.}$

Suppose $X,Y,Z$ are metric spaces, and $Y$ is compact. Let
$f$ map $X$ into $Y,$ let $g$ be a continuous onetoone
mapping of $Y$ into $Z,$ and put $h\left(x\right)=g\left(f\left(x\right)\right)$
for $x\in X\text{.}$
Prove that $f$ is uniformly continuous if $h$ is uniformly continuous.
Hint: ${g}^{1}$ has compact domain
$g\left(Y\right),$ and
$f\left(x\right)={g}^{1}\left(h\left(x\right)\right)\text{.}$
Prove also that $f$ is continuous if $h$ is continuous.
Show (by modifying Example 4.21, or by finding a different example) that the compactness of $Y$ cannot be omitted from the hypotheses, even when
$X$ and $Z$ are compact.
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