Exercise 4.4.1
(a) It is already shown in Example 4.3.5.
(b) Let $x_n=n+\dfrac{1}{n}$ and $y_n=n$, then we have
$$
|x_n-y_n|\to 0.
$$On the other hand, we have
\begin{align*}
|f(x_n)-f(y_n)|=&\ \left|\Big(n+\dfrac{1}{n}\Big)^3-n^3\right|\\
= &\ \left|n^3+3n+\dfrac{3}{n}+\frac{1}{n^3}-n^3\right|\\
\geqslant &\ 3n \geqslant 1.
\end{align*}Hence we can choose $\epsilon_0=1$ in Theorem 4.4.5. It follows from Theorem 4.4.5 that $f$ is not uniformly continuous on $\mathbf R$.
Lemma: If a function $f$ is uniformly continuous on a set $A$, then it is also uniformly continuous on any subset $B$ of $A$.
Proof: Since $f$ is uniformly continuous on $A$, for any $\epsilon >0$, there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$.
Now for any $x,y\in B$, $|x-y|<\delta$, as $B$ is a subset of $A$, we also have $x,y\in A$, $|x-y|<\delta$, therefore we have $|f(x)-f(y)|<\epsilon$.
(c) Let $A$ be a bounded subset of $\mathbf R$. Then there exists $M>0$ such that $A\subset [-M,M]$. The closed interval $[-M,M]$ is a compact set. As $f(x)$ is continuous on $[-M,M]$, by Theorem 4.4.7, it follows that $f(x)$ is uniformly continuous on $[-M,M]$. Now it follows from the lemma above that $f(x)$ is uniformly continuous on $A$.
Exercise 4.4.2
(a) No, let $x_n=\dfrac{1}{n}$ and $y_n=\dfrac{1}{2n}$, then\[|x_n-y_n|=\frac{1}{2n}\to 0,\quad |f(x_n)-f(y_n)|=n\geqslant 1.\]Hence by Theorem 4.4.5, $f(x)$ is not uniformly continuous on $(0,1)$.
(b) Yes. Note that $g(x)$ is continuous on $[0,1]$. Since $[0,1]$ is bounded and closed, $[0,1]$ is compact. By Theorem 4.4.7, $g(x)$ is uniformly continuous on $[0,1]$. Therefore, $g(x)$ is also uniformly continuous on $(0,1)$.
(c) Yes. We extend the function $h(x)$ by defining $h(0)=0$, then it is clear that $h(x)$ is continuous on $[0,1]$. Since $[0,1]$ is bounded and closed, $[0,1]$ is compact. By Theorem 4.4.7, $h(x)$ is uniformly continuous on $[0,1]$. Therefore, $h(x)$ is also uniformly continuous on $(0,1)$.
Exercise 4.4.3
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.2.
Exercise 4.4.4
(a) See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.4.
(b) True. We argue it by contradiction. Suppose $f(A)$ is not bounded. Then there exists a sequence $(x_n)$ such that $x_n\in A$ and $|f(x_{n+1})|-|f(x_n)|\geqslant 1$ for all $n\geqslant 1$ (this sequence can be obtained inductively). Since $A$ is bounded, by Theorem 2.5.5, then we have a convergent subsequence $(x_{n_k})$. Let $y_{n_k}=x_{n_{k+1}}$, then\[|x_{n_k}-y_{n_k}|\to 0,\] but\[|f(x_{n_k})-f(y_{n_{k}})|\geqslant |f(y_{n_{k}})|-|f(x_{n_k})|\geqslant 1.\]Hence by Theorem 4.4.5, $f$ is not uniformly continuous, which is a contradiction. Therefore, we are done.
(c) False. Define a function $f(x)$ as follows,\[f(x)=\begin{cases}0,&\text{ if }x\in \mathbf Q;\\ 1, &\text{ otherwise. }\end{cases}\]Then $f(K)$ is always a finite set and hence compact. But $f$ is not continuous on $\mathbf R$.
Exercise 4.4.5
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.7.
Exercise 4.4.6
(a) Let $f(x)=1/x$ for $x\in (0,1)$. Then $f$ is continuous on $(0,1)$. Let $x_n=1/n$, then $(x_n)$ is Cauchy. But $f(x_n)=n$, so the sequence $(f(x_n))$ is not Cauchy.
(b) Impossible. Since $f$ is uniformly continuous on $(0,1)$, for any $\epsilon>0$, there exists $\delta >0$ such that $|f(x)-f(y)|<\epsilon$ provided $|x-y|<\delta$. Since $(x_n)$ is Cauchy, there exists $N>0$ such that $|x_n-x_m|<\delta$ whenever $n,m>N$. Hence for $n,m>N$, we have\[|f(x_n)-f(x_m)|<\epsilon.\]This implies that $(f(x_n))$ is Cauchy.
(c) Impossible. Since $(x_n)$ is Cauchy, $(x_n)$ is bounded by Lemma 2.6.3. Let $|x_n|\leqslant M$ where $M>0$ for all $n\geqslant 1$. Then we can consider the restriction of $f$ on the interval $[0,M]$. Because $[0,M]$ is compact, by Theorem 4.4.7, $f$ is uniformly continuous on $[0,M]$. Using the same argument as in (b), you can show that $(f(x_n))$ is Cauchy.
Exercise 4.4.7
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.8 (b).
Exercise 4.4.8
(a) Impossible. Because $f$ is continuous and $[0,1]$ is compact, by Theorem 4.4.1, the range is compact as well. But $(0,1)$ is not compact.
(b) Let $f(x)=\left|\sin\dfrac{1}{x}\right|$, see Example 4.2.6.
(c) Let\[f(x)=\frac{|\sin\frac{1}{x}|+x}{1+2x}.\]
Exercise 4.4.9
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.9.
Exercise 4.4.10
The functions $f(x)+g(x)$ and $f(g(x))$ are necessarily uniformly continuous.
$f(x)+g(x)$ is uniformly continuous. This follows from the triangle inequality. Let $\epsilon>0$. Because $f(x)$ and $g(x)$ are uniformly continuous on $A$, there exist $\delta_1>0$ and $\delta_2>0$ such that $$|f(x_1)-f(y_1)|<\epsilon/2,\quad |g(x_2)-g(y_2)|<\epsilon/2$$for all $|x_1-y_1|<\delta_1$ and $|x_2-y_2|<\delta_2$ on $A$. Let $\delta=\min\{\delta_1,\delta_2\}$. Then for all $x,y\in A$ and $|x-y|<\delta$, we have\begin{align*}|(f(x)+g(x))-(f(y)+g(y))|&\leqslant |f(x)-f(y)|+|g(x)-g(y)|\\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\end{align*}Thus $f(x)+g(x)$ is uniformly continuous.
$f(x)g(x)$ may bot be uniformly continuous. Let $f(x)=g(x)=x$, then $f(x)g(x)=x^2$ is not uniformly continuous. Because we can choose $x_n=n$ and $y_n=n+1/n$, then\[|x_n-y_n|\to 0,\quad |x_n^2-y_n^2|=|2+\frac{1}{n^2}|\geqslant 2.\]By Theorem 4.4.5, $x^2$ is not uniformly continuous.
$f(x)/g(x)$ may bot be uniformly continuous. Let $f(x)=1$ and $g(x)=x$ defined on $(0,1)$, then $f(x)/g(x)=1/x$ is not uniformly continuous, see Exercise 4.4.2.
$f(g(x))$ is uniformly continuous. Let $\epsilon>0$. Since $f(x)$ is uniformly continuous, there exists $\delta_1>0$ such that for all $x,y\in A$ and $|x-y|<\delta_1$ we have $$|f(x)-f(y)|<\epsilon.$$ Since $g(x)$ is uniformly continuous, there exists $\delta>0$ such that for all $x,y\in A$ and $|x-y|<\delta$ we have $$|g(x)-g(y)|<\delta.$$ Therefore, if $x,y\in A$ and $|x-y|<\delta$, we have $|g(x)-g(y)|<\delta_1$ and therefore $$|f(g(x))-f(g(y))|<\epsilon.$$Thus $f(g(x))$ is uniformly continuous.
Exercise 4.4.11
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.11.
Exercise 4.4.12
(a) (b) (c) may be false. Consider the example $f(x)=1$ for all $x\in\mathbf R$. Let $B=K=A$, then $f^{-1}(B)=f^{-1}(K)=f^{-1}(A)=\mathbf R$ which is clear not finite, not compact, not bounded.
(d) is true. To show $f^{-1}(F)$ is closed, it suffices to show that $f^{-1}(F)^c$ is open. We show that $f^{-1}(F)^c=f^{-1}(F^c)$. Then because $F^c$ is open, by definition of Exercise 4.4.11, $f^{-1}(F)^c=f^{-1}(F^c)$ is also open.
Now we show $f^{-1}(F)^c=f^{-1}(F^c)$. First we show $f^{-1}(F)^c\subset f^{-1}(F^c)$. Let $a\in f^{-1}(F)^c$, then $a\not\in f^{-1}(F)$, i.e. $f(a)\not\in F$. Therefore $f(a)\in F^c$, hence $a\in f^{-1}(F^c)$.
Then we show $f^{-1}(F^c)\subset f^{-1}(F)^c$. Let $a\in f^{-1}(F^c)$, then $f(a)\in F^c$. Thus $f(a)\not\in F$ and $a\not\in f^{-1}(F)$. Therefore $a\in f^{-1}(F)^c$, this implies $f^{-1}(F^c)\subset f^{-1}(F)^c$.
Now we have $f^{-1}(F)^c=f^{-1}(F^c)$.
Exercise 4.4.13
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.13.
