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## Solution to Principles of Mathematical Analysis Chapter 4 Part B

### Chapter 4 Continuity

#### Exercise 10

(By analambanomenos) By Theorem 2.37, $\{p_n\}$ has a subsequence which converges to $p\in X$. Replace $\{p_n\}$ with this subsequence and replace $\{q_n\}$ with the corresponding subsequence. Similarly, $\{q_n\}$ has a subsequence which converges to $q\in X$. Again replace $\{q_n\}$ with this subsequence and replace $\{p_n\}$ with the corresponding subsequence. Since $d_X(p_n,q_n)$ converges to 0, we must have $p=q$. Hence by continuity, $f(p_n)$ and $f(q_n)$ must both converge to $f(p)=f(q)$, contradicting the assumption that $d_Y\bigl(f(p_n),f(q_n)\bigr)>\varepsilon$.

#### Exercise 11

(By analambanomenos) Let $\varepsilon>0$ and let $\delta>0$ such that $d_Y\bigl(f(x),f(y)\bigr)<\varepsilon$ if $d_X(x,y)<\delta$. Let $\{x_n\}$ be a Cauchy sequence in $X$. Then there is an integer $N$ such that $d_X(x_n,x_m)<\delta$ if $n\ge N$ and $m\ge N$. Hence $$d_Y\bigl(f(x_n),f(x_m)\bigr)<\varepsilon$$ if $n\ge N$ and $m\ge N$, so $\{f(x_n)\}$ is a Cauchy sequence in $Y$.

Let $E$ be a dense subset of a metric space $X$, and let $f$ be a uniformly continuous mapping of $X$ into a complete metric space $Y$. We can extend $f$ to a function on all of $X$ as follows. Let $x\in X$ and $\{p_n\}$ be a sequence in $E$ which converges to $x$. Then $\{f(p_n)\}$ is a Cauchy sequence in $Y$ which converges to $y\in Y$ since $Y$ is complete. This sets up a well-defined function $g$ from $X$ to $Y$ since if $\{p_n’\}$ is another sequence in $X$ converging to $x$, then $\{f(p_n)\}$ converges to $y’\in Y$. Since $f$ is uniformly continuous on $D$, for $\varepsilon>0$ there is $\delta>0$ such that if $p,p’\in E$ with $d_X(p,p’)<\delta$ then $d_Y\bigl(f(p),f(p’)\bigr)<\epsilon/3$. There is an integer $N$ such that $d_X(p_n,x)<\delta/2$ and $d_X(p_n’,x)<\delta/2$ for all $n\ge N$. Then for such $n$ we have
$$d_X(p_n,p_n’)\le d_X(p_n,x)+d_X(x,p_n’)< \delta.$$ Also, there is an integer $M\ge N$ such that for all $n>M$ we have $d_Y\bigl(y,f(p_n)\bigr)<\varepsilon/3$ and $d_Y\bigl(y’,f(p_n’)\bigr)<\varepsilon/3$. Hence, for large enough $n$ we have $$d_Y(y,y’)\le d_Y\bigl(y,f(p_n)\bigr)+d_Y\bigl(f(p_n),f(p_n’)\bigr)+ d_Y\bigl(f(p_n’),y’\bigr)<\varepsilon.$$

Considering the constant sequence $\{x\}$ for $x\in E$, we see that $g$ extends $f$ to all of $X$. To see that $g$ is uniformly continuous, let $\varepsilon>0$. Since $f$ is uniformly continuous on $E$, there is $\delta>0$ such that if $p,p’\in E$ with $d_X(p,p’)<\delta$ then $d_Y\bigl(f(p),f(p’)\bigr)<\epsilon/3$. Let $x,x’\in X$ such that $d_X(x,x’)<\delta/3$, and let $\{p_n\}$ and $\{p_n’\}$ be sequences in $E$ which converge to $x$ and $x’$, respectively. Let $n$ be a large enough integer so that $d_X(p_n,x)<\delta/3$ and $d_X(p_n’,x’)<\delta/3$. Then $$d_X(p_n,p_n’)\le d(p_n,x)+d(x,x’)+d(x’,p_n’)<\delta$$ so that $d_Y\bigl(f(p_n),f(p_n’)\bigr)<\epsilon/3$. Also, if $n$ is large enough we have $d_Y\bigl(g(x),f(p_n)\bigr)<\varepsilon/3$ and $d_Y\bigl(g(x’),f(p_n’)\bigr)<\varepsilon/3$. Hence, for large enough $n$ we have
$$d_Y\bigl(g(x),g(x’)\bigr)\le d_Y\bigl(g(x),f(p_n’)\bigr)+ d_Y\bigl(f(p_n),f(p_n’)\bigr)+ d_Y\bigl(f(p_n’),g(x’)\bigr)<\varepsilon.$$ Hence $g$ is uniformly continuous on $X$.

#### Exercise 12

(By analambanomenos) Let $\varepsilon>0$. Then there is $\eta>0$ such that $d_Z\bigl(g(x),g(y)\bigr)<\varepsilon$ if $d_Y(x,y)<\eta$. There is also $\delta>0$ such that $d_Y\bigl(f(x),f(y)\bigr)<\eta$ if $d_X(p,q)<\delta$. Hence if $d_X(p,q)<\delta$, then $d_Z\bigl(g(f(p)),g(f(q))\bigr)<\varepsilon$, so $g\circ f$ is uniformly continuous.

#### Exercise 13

(By analambanomenos) Following the hint, for $p\in X$ let $V_n(p)$ be the set of $q\in E$ such that $d(p,q)<1/n$. By Exercise 9, there is an integer $N$ such that $\mathop{\rm diam} f\bigl(V_N(p)\bigr)<1$ so the closure $F_n$ of $f\bigl(V_n(p)\bigr)$ for $n\ge N$ is a compact subset of $\mathbf R$. Since $F_N\supset F_{N+1}\supset\cdots$, the intersection $\cap_N^\infty F_n$ is nonempty by Theorem 2.36. This intersection consists of a single point. For suppose $x,y\in\cap F_n$ and $|x-y|=\varepsilon>0$. Let $n\ge N$ be large enough so that if $q_1,q_2\in V_n(p)$ then $\big|f(q_1)-f(q_2)\big|<\varepsilon/3$. Since both $x$ and $y$ are in $F_n$, they have open neighborhoods of radius $\varepsilon/3$ which intersect $f\bigl(V_n(p)\bigr)$, let $q_1,q_2\in V_n(p)$ such that $f(q_1)$ and $f(q_2)$ are in these neighborhoods of $x$ and $y$, respectively. Then $$\varepsilon=|x-y|\le\big|x-f(q_1)\big|+\big|f(q_1)-f(q_2)\big|+\big|f(q_2)-y\big|<\varepsilon,$$ which is a contradiction.

Let $g(p)$ be this single point in $\cap F_n$. If $p\in E$, then $f(p)\in F_n$ for all $n$, so $g(p)=f(p)$, hence $g$ is an extension of $f$ to all of $X$. To show that $g$ is continuous, let $\varepsilon>0$. Since $f$ is uniformly continuous on $E$, there is $\delta>0$ so that $\big|f(q_1)-f(q_2)\big|<\varepsilon/3$ if $q_1,q_2\in E$ and $d(q_1,q_2)<\delta$. Let $p_1,p_2\in X$ such that $d(p_1,p_2)<\delta/3$, and let $n$ be a large enough integer so that $1/n<\delta/3$ and large enough so that, if $q_1\in V_n(p_1)$ and $q_2\in V_n(p_2)$, then $\big|g(p_1)-f(q_1)\big|<\varepsilon/3$ and $\big|g(p_2)-f(q_2)\big|<\varepsilon/3$. Then
$$d(q_1,q_2)\le d(q_1,p_1)+d(p_1,p_2)+d(p_2,q_2)<\delta$$ so that $$\big|g(p_1)-g(p_2)\big|\le\big|g(p_1)-f(q_1)\big|+\big|f(q_1)-f(q_2)\big|+\big|f(q_2)-g(p_2)\big|< \varepsilon.$$ Hence $g$ is uniformly continuous on $X$.

The only topological property of $\mathbf R$ that was used was that closed and bounded sets are compact, so the assertion can probably also be extended to $\mathbf R^k$. Also, the $F_n$ will also be compact if the target space is compact, so the assertion is probably also true for that case. For a complete metric space, the assertion was proved in the answer to Exercise 11. The assertion is probably not true for a general metric space, but I can’t come up with an example.

#### Exercise 14

(By Matt frito Lundy) Because both $h(x) = x$ and $f$ are continuous mappings on $I$, $g(x) = f(x) – x$ is also continuous on $I$. We are searching for an $x$ such that $g(x) = 0$. $g(0) \in [0,1]$ and $g(1) \in [-1,0]$, and if either $g(0) = 0$ or $g(1) = 0$, we are done, so assume that both $g(0) \neq$ and $g(1) \neq 0$. Then $g(0) \in (0,1]$ and $g(1) \in [-1,0)$, and because $g$ is continuous on $I$, with $I$ connected, the intermediate value theorem applies, and there exists an $x \in (0,1)$ such that $f(x) = 0$.

#### Exercise 15

(By analambanomenos) By Theorem 4.16, there is an $x$ and $y$ in any closed interval $[a,b]$ such that $f(x)$ is the minimum value of $f$ on $[a,b]$ and $f(y)$ is the maximum value. If $a<x<y<b$, then $f\bigl((a,b)\bigr)$ is the closed interval $\bigl[f(x),f(y)\bigr]$, contradicting the openness of $f$. Similarly, if $a=x<y<b$ or $a<x<y=b$ the image of $(a,b)$ is a half-closed interval. Hence $f$ must attain its maximum or minimum values at the endpoints of any closed interval. Suppose $f(a)$ is the minimum value and $f(b)$ is the maximum value of $f$ on $[a,b]$, and let $a<y\le b$. Then $f(y)$ would be the maximum value of $f$ on $[a,y]$, so $f(x)\le f(y)$ for $a\le x<y$, that is, $f$ is monotonically increasing on $[a,b]$.

Hence $f$ is monotonic on any closed interval, and if $f$ is monotonically increasing on one closed interval, then it must also be monotonically increasing on any larger interval. Hence $f$ is monotonic on all of $\mathbf R$.

#### Exercise 16

(By analambanomenos) Let $f(x)=[x]$ and $g(x)=(x)$. Then $f$ and $g$, being constant or linear functions on $(n,n+1)$ for each integer $n$, are continuous except at the integers. If $n$ is an integer, then $f(n-)=n-1$ and $f(n+)=f(n)=n$, and $g(n-)=1$ and $g(n+)=g(n)=0$. These are all simple discontinuities.

#### Exercise 17

(By analambanomenos) Following the hint, let $E_1$ be the set of points in $(a,b)$ on which $f(x-)<f(x+)$. With each $x\in E_1$ associate a triple $(p,q,r)$ of rational numbers such that $f(x-)<p<f(x+)$, $f(y)<p$ for $a<q<y<x$, and $f(y)>p$ for $x<y<r<b$. Each triple is associated with at most one point of $E_1$. Otherwise there would be $x_1$ and $x_2$ in $E_1$ such that $f(y)<p$ for $q<y<x_1$ and $q<y<x_2$, and $f(y)>p$ for $x_1<y<r$ and $x_2<y<r$, which is a set of contradictory conditions on $f(y)$ for $y$ between $x_1$ and $x_2$. Hence $E_1$ can be mapped in a one-to-one fashion to a subset of the set of all rational triples, which is countable, and so $E_1$ is at most countable. Similarly, the set $E_2$ of points in $(a,b)$ on which $f(x-)>f(x+)$ is also at most countable.

Let $F_1$ be the set of points in $(a,b)$ on which $f(x-)=f(x+)<f(x)$. With each $x\in F_1$ associate a triple $(p,q,r)$ of rational numbers such that $f(y)<p<f(x)$ for $a<q<y<x$ and $x<y<r<b$. Each triple is associated with at most one point of $F_1$. Otherwise there would be $x_1$ and $x_2$ in $F_1$ such that $f(y)<p<f(x_1)$ for $q<y<x_1$ and $x_1<y<r$, and $f(y)<p<f(x_2)$ for $q<y<x_2$ and $x_2<y<r$, which is a set of contradictory conditions on $f(x_1)$ and $f(x_2)$. Hence, as with $E_1$, $F_1$ is at most countable. Similarly, the set $F_2$ of points in $(a,b)$ on which $f(x-)=f(x+)>f(x)$ is also at most countable.

Since the set of points in $(a,b)$ at which $f$ has a simple discontinuity is $E_1\cup E_2 \cup F_1\cup F_2$, the set of such points is at most countable.

#### Exercise 18

(By analambanomenos) Let $p$ be any real number, let $\varepsilon>0$, and let $N$ be an integer large enough so that $1/N<\varepsilon$. Since in any interval around $p$ there are only a finite number of rational numbers $m/n$ such that $n<N$, there is a $\delta>0$ small enough so that if $0<|x-p|<\delta$ then $\big|f(x)|<1/N<\varepsilon$. Hence $\lim_{x\rightarrow p}f(x)=0$ for all real $p$. Since $f(p)=0$ for irrational $p$, $f$ is continuous at such $p$. And since $f(p)>0$ for rational $p$, $f$ has a simple discontinuity at such $p$.

Baby Rudin 数学分析原理完整第四章习题解答