城南讲马堂

Analysis on Manifolds 5. The Derivative

2021年1月24日christangdt留下评论

这一章开始进入多元微分板块，本节是导数（多元导数）概念，通过一元的导数等价条件（牛顿近似）引入多元导数定义，并介绍了方向导数和偏导数（单位向量的方向导数）。本节的多元导数是以矩阵形式定义的（可看成线性变换），导数定义可以满足：可导必可微、复合函数可导、所有方向导数存在等三个条件。这一节开始的论述就比较详细，几个例子较好地体现了多元微分和一元微分的差异之处。
Theorem 5.1说明可微推出所有方向导数存在，且给出方向导数计算方法。
Theorem 5.2说明可微推出连续。
Theorem 5.3说明值域为R的多维函数的导数计算方法，即如果函数可微，其导数是一个行向量，行向量的子项是偏导数。
Theorem 5.4将上一定理的结论推广到了一般的多维空间。

Exercises

Exercise 1. Let $A\subset \mathbf{R}^m$ ; let $f:A\to\mathbf{R}^n$ . Show that if $f'(\mathbf{a};\mathbf{u})$ exists, then $f'(\mathbf{a};c\mathbf{u})$ exists and equals $cf'(\mathbf{a};\mathbf{u})$ .

Solution: If $f'(\mathbf{a},\mathbf{u})$ exists, then we have

$\displaystyle{f'(\mathbf{a},\mathbf{u})=\lim_{t\to 0}\frac{f(\mathbf{a}+t\mathbf{u})-f(\mathbf{a})}{t}}$

thus to calculate $f'(\mathbf{a};c\mathbf{u})$ , notice if $t\to 0$ , then $h:=ct\to 0$ , thus

$\displaystyle{\begin{aligned}f'(\mathbf{a},c\mathbf{u})&=\lim_{t\to 0}\frac{f(\mathbf{a}+tc\mathbf{u})-f(\mathbf{a})}{t}=c\lim_{t\to 0}\frac{f(\mathbf{a}+tc\mathbf{u})-f(\mathbf{a})}{tc}\\&=c\lim_{h\to 0}\frac{f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})}{h}=c'f(\mathbf{a};\mathbf{u})\end{aligned}}$

Exercise 2. Let $f:\mathbf{R}^2\to\mathbf{R}$ be defined by setting $f(\mathbf{0})=0$ and

$\displaystyle{f(x,y)=xy/(x^2+y^2)\quad\text{ if }\quad (x,y)\neq 0.}$

( a ) For which vectors $\mathbf{u}\neq\mathbf{0}$ does $f'(\mathbf{0};\mathbf{u})$ exist? Evaluate it when it exists.
( b ) Do $D_1f$ and $D_2f$ exist at $\mathbf{0}$ ?
( c ) Is $f$ differentiable at $\mathbf{0}$ ?
( d ) Is $f$ continuous at $\mathbf{0}$ ?

Solution:
( a ) Let $\mathbf{u}=[h,k]^T$ , then

$\displaystyle{\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{u})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{t^2hk}{t^2h^2+t^2k^2}=\lim_{t\to 0}\frac{hk}{t(h^2+k^2)}}$

this limit exists if $h=0,k\neq 0$ or $h\neq 0,k=0$ , so $f'(\mathbf{0};\mathbf{u})$ exists when $\mathbf{u}$ is along the x-asis or the y-axis, and when it exists, $f'(\mathbf{0};\mathbf{u})=0$ .
( b ) We have

$\displaystyle{D_1f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e}_1)-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=0}$
$\displaystyle{D_2f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e}_2)-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=0}$

( c ) Since not all directional derivative exists, $f$ is not differentiable at $\mathbf{0}$ , otherwise contradicting Theorem 5.1.
( d ) $f$ is not continuous at $\mathbf{0}$ , let $\mathbf{h}=[a,a]^T$ , then when $a\to 0, \mathbf{h}\to 0$ , but $f(a,a)=1/2\neq f(\mathbf{0})$ .

Exercise 3. Repeat Exercise 2 for the function $f$ defined by setting $f(\mathbf{0})=0$ and

$\displaystyle{f(x,y)=x^2y^2/(x^2y^2+(y-x)^2)\quad\text{ if }\quad (x,y)\neq 0.}$

Solution:
( a ) Let $\mathbf{u}=[h,k]^T$ , then

$\displaystyle{\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{u})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{t^4hk}{t(t^4h^2k^2+t^2(k-h)^2)}=\lim_{t\to 0}\frac{thk}{t^2h^2h^2+(k-h)^2}}$

this limit exists if $k\neq h$ , so $f'(\mathbf{0};\mathbf{u})$ exists when $\mathbf{u}$ is not along line $y=x$ , and when it exists, $f'(\mathbf{0};\mathbf{u})=1/(k-h)^2$ .
( b ) We have

Exercise 4. Repeat Exercise 2 for the function $f$ defined by setting $f(\mathbf{0})=0$ and

$\displaystyle{f(x,y)=x^3/(x^2+y^2)\quad\text{ if }\quad (x,y)\neq 0.}$

Solution:
( a ) Let $\mathbf{u}=[h,k]^T$ , then

$\displaystyle{\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{u})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{t^3h^3}{t(t^2h^2+t^2k^2)}=\lim_{t\to 0}\frac{h^3}{h^2+k^2}}$

this limit exists if $h=0,k\neq 0$ , so $f'(\mathbf{0};\mathbf{u})$ exists when $\mathbf{u}$ is along the y-axis, and when it exists, $f'(\mathbf{0};\mathbf{u})=0$ .
( b ) We have

$\displaystyle{D_1f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e}_1)-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=1}$
$\displaystyle{D_2f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e}_2)-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(0,t)}{t}=0}$

( c ) Since not all directional derivative exists, $f$ is not differentiable at $\mathbf{0}$ , otherwise contradicting Theorem 5.1.
( d ) $f$ is continuous at $\mathbf{0}$ , since for every open set $V$ which contains $f(\mathbf{0})=0$ , we can find a $\epsilon>0$ s.t. $(-\epsilon,\epsilon)\subset V$ , and for any $\mathbf{h}=[a,b]^T,|\mathbf{h}|<\epsilon$ , we have

$\displaystyle{|f(\mathbf{h})|=\left|\frac{a^3}{a^2+b^2}\right|\leq |a|<|\mathbf{h}|<\epsilon}$

Let $U=B(\mathbf{0},\epsilon)$ , then $U$ is open and $f(U)\subset V$ .

Exercise 5. Repeat Exercise 2 for the function

$\displaystyle{f(x,y)=|x|+|y|.}$

Solution:
( a ) Let $\mathbf{u}=[h,k]^T$ , then

$\displaystyle{\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{u})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{|th|+|tk|}{t}=\lim_{t\to 0}\frac{|t|}{t}(|h|+|k|)}$

this limit does not exist if $\mathbf{u}\neq 0$ , so $f'(\mathbf{0};\mathbf{u})$ does not exist.
( b ) We have

$\displaystyle{D_1f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e}_1)-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=\lim_{t\to 0}\frac{|t|}{t}}$
$\displaystyle{D_2f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e}_2)-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(0,t)}{t}=\lim_{t\to 0}\frac{|t|}{t}}$

thus $D_1f(\mathbf{0})$ and $D_2f(\mathbf{0})$ does not exist at $\mathbf{0}$ .
( c ) Since not all directional derivative exists, $f$ is not differentiable at $\mathbf{0}$ , otherwise contradicting Theorem 5.1.
( d ) $f$ is continuous at $\mathbf{0}$ , since for every open set $V$ which contains $f(\mathbf{0})=0$ , we can find a $\epsilon>0$ s.t. $(-\epsilon,\epsilon)\subset V$ , and for any $\mathbf{h}=[a,b]^T,|\mathbf{h}|<\epsilon/2$ , we have

$\displaystyle{0\leq f(\mathbf{h})=|a|+|b|<\epsilon\implies f(\mathbf{h})\in(-\epsilon,\epsilon)\implies f(\mathbf{h})\in V}$

Let $U=B(\mathbf{0},\epsilon/2)$ , then $U$ is open and $f(U)\subset V$ .

Exercise 6. Repeat Exercise 2 for the function

$\displaystyle{f(x,y)=|xy|^{1/2}.}$

Solution:
( a ) Let $\mathbf{u}=[h,k]^T$ , then

$\displaystyle{\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{u})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{|thtk|^{1/2}}{t}=\lim_{t\to 0}\frac{|t|}{t}|hk|}$

this limit exists if $h=0,k\neq 0$ or $h\neq 0,k=0$ , so $f'(\mathbf{0};\mathbf{u})$ exists when $\mathbf{u}$ is along the x-axis or the y-axis, and when it exists, $f'(\mathbf{0};\mathbf{u})=0$ .
( b ) We have

$\displaystyle{f(\mathbf{h})=|ab|^{1/2}<|\epsilon^2|^{1/2}=\epsilon\implies f(\mathbf{h})\in(-\epsilon,\epsilon)\implies f(\mathbf{h})\in V}$

Let $U=B(\mathbf{0},\epsilon/2)$ , then $U$ is open and $f(U)\subset V$ .

Exercise 7. Repeat Exercise 2 for the function $f$ defined by setting $f(\mathbf{0})=0$ and

$\displaystyle{f(x,y)=x|y|/(x^2+y^2)^{1/2}\quad\text{ if }\quad (x,y)\neq 0.}$

Solution:
( a ) Let $\mathbf{u}=[h,k]^T$ , then

$\displaystyle{\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{u})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{th|tk|}{t|t|(h^2+k^2)}=\frac{h|k|}{h^2+k^2}}$

this limit exists as long as $\mathbf{u}\neq 0$ , so $f'(\mathbf{0};\mathbf{u})$ exists and
$\displaystyle{f'(\mathbf{0};\mathbf{u})=\frac{h|k|}{h^2+k^2}}$
( b ) We have

( c ) We have, for $\mathbf{h}=(x,x)^T\to (0,0),x>0$ , that

$\displaystyle{\frac{f(\mathbf{0}+\mathbf{h})-f(\mathbf{0})-[0,0]\cdot\mathbf{h}}{|\mathbf{h}|}=\frac{x|x|}{|x|\sqrt{x^2+x^2}}=\frac{x}{\sqrt{2}|x|}=\frac{1}{\sqrt{2}}\nrightarrow 0}$

we see $f$ is not differentiable at $\mathbf{0}$ .
( d ) $f$ is continuous at $\mathbf{0}$ , since for every open set $V$ which contains $f(\mathbf{0})=0$ , we can find a $\epsilon>0$ s.t. $(-\epsilon,\epsilon)\subset V$ , and for any $\mathbf{h}=[a,b]^T,|\mathbf{h}|<\epsilon$ , we have

$\displaystyle{|f(\mathbf{h})|=\left|\frac{a|b|}{\sqrt{a^2+b^2}}\right|\leq |a|<\epsilon\implies f(\mathbf{h})\in(-\epsilon,\epsilon)\implies f(\mathbf{h})\in V}$

Let $U=B(\mathbf{0},\epsilon/2)$ , then $U$ is open and $f(U)\subset V$ .

Analysis on Manifolds 4. Compact Subspaces and Connected Subspaces of R^n

2021年1月3日christangdt留下评论

这一节回顾两个重要的概念：紧致集和联通集。由于本书主要讨论 $\mathbf R^n$ 上的情况，因此紧致和有界闭集是等价的，但本节练习中也提到了非 $\mathbf R^n$ 上的情况。
Theorem 4.1其实是很多书上紧致集的定义。
Theorem 4.2很简单， $\mathbf R$ 上闭区间是compact的。
Theorem 4.3说明在 $\mathbf R^n$ 上compact可以推出closed和bounded，推论是最大最小值定理。
Theorem 4.5是极值定理，说明compact区间上的连续函数的值域也是compact的（连续函数保留compact特性）
Theorem 4.6是 $\epsilon$ 邻域定理，如果 $U$ 是包含紧致集 $X$ 的一个开集，那么可以找到一个 $X$ 的 $\epsilon$ 邻域包含在 $U$ 中。
Theorem 4.7是一致连续定理，compact集上的连续函数是一致连续的。
Theorem 4.9说明在 $\mathbf R^n$ 上closed和bounded可以推出compact
Theorem 4.10说明 $\mathbf R$ 上闭区间是联通的。
Theorem 4.11是介值定理。

Exercises

Exercise 1. Let $\mathbf R_{+}$ denote the set of positive real numbers.
( a ) Show that the continuous function $f:\mathbf{R_+}\to\mathbf{R}$ given by $f(x)=1/(1+x)$ is bounded but has neither a maximum value nor a minimum value.
( b ) Show that the continuous function $g:\mathbf{R_+}\to\mathbf{R}$ given by $g(x)=\sin{(1/x)}$ is bounded but does not satisfy the condition of uniform continuity on $\mathbf{R_+}$ .
Solution:
(a) $x>0$ means $1+x>1$ and thus $f(x)<1$ , also $f(x)>0$ is obvious, thus $00$ , choose $k\in\mathbf{N}^+$ such that $4k\delta>1$ , if we let $x_1=1/2k\pi,x_2=1/(2k\pi+\pi/2)$ , then

$\displaystyle{|x_1-x_2|=\left|\frac{1}{2k\pi}-\frac{1}{2k\pi+\pi/2}\right|=\frac{\pi/2}{2k\pi(2k\pi+\pi/2)}=\frac{1}{4k(2k\pi+\pi/2)} <\frac{1}{4k}<\delta}$

but $|g(x_1)-g(x_2)|=|\sin(2k\pi)-\sin(2k\pi+\pi/2)|=1$ .

Exercise 2. Let $X$ denote the subset $(-1,1)\times 0$ of $\mathbf{R^2}$ , and let $U$ be the open ball $B(\mathbf{0};1)$ in $\mathbf{R^2}$ , which contains $X$ . Show there is no $\epsilon>0$ such that the $\epsilon$ -neighborhood of $X$ in $\mathbf{R^n}$ is contained in $U$ .
Solution: Given any $\epsilon>0$ , the point $\mathbf{a}=(1-\epsilon/2,0)\in X$ , and $B(\mathbf{a},\epsilon)$ is a $\epsilon$ -neighborhood of $X$ , but for the point $\mathbf{b}=(1+\epsilon/4,0)$ , we have

$\displaystyle{|\mathbf{b}-\mathbf{a}|=3\epsilon/4<\epsilon \implies b\in B(\mathbf{a},\epsilon),\quad |\mathbf{b}-\mathbf{0}|=1+\epsilon/4>1 \implies \mathbf{b}\notin B(\mathbf{0},1)=U}$

thus $B(\mathbf{a},\epsilon)\nsubseteq U$ .

Exercise 3. Let $\mathbf{R}^{\infty}$ be the set of all “infinite-tuples” $\mathbf{x}=(x_1,x_2,\dots)$ of real numbers that end in an infinite string of 0’s. Define an inner product on $\mathbf{R}^{\infty}$ by the rule $\langle x,y\rangle=\sum{x_iy_i}$ . (This is a funite sum, since all but finitely many terms vanish.) Let $|\mathbf{x}-\mathbf{y}|$ be the corresponding metric on $\mathbf{R}^{\infty}$ . Define

$\displaystyle{\mathbf{e}_i=(0,\dots,0,1,0,\dots,0,\dots)}$

where 1 appears in the $i^{th}$ place. Then the $\mathbf{e}_i$ form a basis for $\mathbf{R}^{\infty}$ . Let $X$ be the set of all the points $\mathbf{e}_i$ . Show that $X$ is closed, bounded and non-compact.
Solution: $X$ contains only isolated points, thus is closed. Since $|\mathbf{e}_i|=\sqrt{\langle \mathbf{e}_i,\mathbf{e}_i\rangle}=1,\forall i$ , $X$ is bounded.
To see $X$ is not compact, notice that $|\mathbf{e}_i-\mathbf{e}_j|=\sqrt{2},\forall i\neq j$ , then $B(\mathbf{e}_i,1)\cap X=\mathbf{e}_i,\forall i$ , the open sets ${B(\mathbf{e}_i,1):i\in \mathbf{N}^+ }$ covers $X$ , but no finite subcover of $X$ exists.

Exercise 4.
( a ) Show that open balls and open cubes in $\mathbf{R}^n$ are convex.
( b ) Show that (open and closed) rectangles in $\mathbf{R}^n$ are convex.
Solution:
( a ) Let $B(\mathbf{y},{\epsilon})$ be an open ball in $\mathbf{R}^n$ , $C(\mathbf{y},{\epsilon})$ be an open cube in $\mathbf{R}^n$ . First for any $\mathbf{a},\mathbf{b}\in B(\mathbf{y},{\epsilon})$ , there is the relation $|\mathbf{a}-\mathbf{y}|<{\epsilon},|\mathbf{b}-\mathbf{y}|<{\epsilon}$ , thus if $\mathbf{x}=\mathbf{a}+t(\mathbf{b}-\mathbf{a}),0\leq t\leq 1$ , then

$\displaystyle{\begin{aligned}|\mathbf{x}-\mathbf{y}|&=|\mathbf{a}+t(\mathbf{b}-\mathbf{a})-\mathbf{y}|=|(1-t)(\mathbf{a}-\mathbf{y})+t(\mathbf{b}-\mathbf{y})|\\&\leq (1-t)|\mathbf{a}-\mathbf{y}|+t|\mathbf{b}-\mathbf{y}|<{\epsilon}\end{aligned}}$

which means $\mathbf{x}\in B(\mathbf{y},{\epsilon})$ , thus $B(\mathbf{y},{\epsilon})$ is convex.
Next for any $a,b\in C(\mathbf{y},{\epsilon})$ , there is the relation $|\mathbf{a}-\mathbf{y}|<{\epsilon},|\mathbf{b}-\mathbf{y}|<{\epsilon}$ , thus if $\mathbf{x}=\mathbf{a}+t(\mathbf{b}-\mathbf{a}),0\leq t\leq 1$ , then

which means $\mathbf{x}\in C(\mathbf{y},{\epsilon})$ , thus $C(\mathbf{y},{\epsilon})$ is convex.
( b ) If $Q$ is a rectangle in $\mathbf{R}^n$ , we can write $Q=[c_1,d_1 ]\times{\cdots}\times[c_n,d_n]$ , for any $a,b\in Q$ , we write $a=(a_1,{\dots},a_n ),b=(b_1,{\dots},b_n)$ , then there is the relation

$\displaystyle{c_i\leq a_i,b_i\leq d_i,\quad 1\leq i\leq n}$

thus if $\mathbf{x}=\mathbf{a}+t(\mathbf{b}-\mathbf{a}),0\leq t\leq 1$ , then $x_i=a_i+t(b_i-a_i )=(1-t) a_i+tb_i\in [c_i,d_i ]$ , so $\mathbf{x}\in Q$ .
If $Q$ is an open rectangle in $\mathbf{R}^n$ , the proof is very similar.

Analysis on Manifolds 3. Review of Topology in Rn

2021年1月2日christangdt留下评论

这一节回顾了在 $R^n$ 中的基本拓扑知识，包括metric space，极限和连续，内点外点和边界点。这部分内容在大部分实分析中都有比较详尽的阐述。
Theorem 3.1是比较经典的开闭集交并结论：开集的有限交和无限并仍是开集，闭集的无限交合有限并仍是闭集。Theorem 3.2也比较经典，子空间的开/闭集是子空间和母空间开/闭集的交集。Theorem 3.3是闭包的概念以及闭集的判别条件。Theorem 3.4说明开闭集与选取的metric无关。
Theorem 3.5说明连续函数的限制仍是连续的，复合连续函数仍是连续函数。Theorem 3.6说明多维连续函数的每个分量函数必须是连续的，反之亦然，四则运算不影响连续性，投影函数连续。
Theorem 3.7说明连续性和极限的关系（在非isolated point上满足当 $x\to x_0$ 时, $f(x)\to f(x_0)$ ）。Theorem 3.8将上一定理的结果拓展到n维。

Exercises

Throughout, let $X$ be a metric space with metric $d$ .
Exercise 1. Show that $U(x_0;\epsilon)$ is an open set.
Solution: For $\forall x\in U(x_0;\epsilon )$ , we have $d(x,x_0 )<\epsilon$ , thus $\delta :=\epsilon -d(x,x_0 )>0$ , consider the set $U(x,\delta)$ , for any $y\in U(x,\delta )$ , we can have $d(y,x_0 )\leq d(y,x)+d(x,x_0 )<\delta +d(x,x_0 )=\epsilon$ , so $y\in U(x_0;\epsilon)$ and $U(x,\delta )\subset U(x_0;\epsilon )$ , which means $U(x_0;\epsilon )$ is an open set.

Exercise 2. Let $Y\subset X$ . Give an example where $A$ is open in $Y$ but not open in $X$ . Give an example where $A$ is closed in $Y$ but not closed in $X$ .
Solution: Let $X=(-1,2),Y=A=[0,1]$ , then $A$ is open in $Y$ , but not open in $X$ .
Let $X=[0,1],Y=A=[0,1)$ , then $A$ is closed in $Y$ , but not closed in $X$ .

Exercise 3. Let $A\subset X$ . Show that if $C$ is a closed set of $X$ and $C$ contains $A$ , then $C$ contains $\overline{A}$ .
Solution: Let $x$ be any limit point of $A$ , then every $\epsilon$ -neighborhood of $x$ intersects $A$ with a point $y\neq x$ , thus every $\epsilon$ -neighborhood of $x$ intersects $C$ with a point $y\neq x$ , so $x$ is a limit point of $C$ . Since $C$ is closed, $x\in C$ , and $\overline{A}\subset C$ .

Exercise 4. ( a ) Show that if $Q$ is a rectangle, then $Q$ equals the closure of $\text{Int} Q$ .
( b ) If $D$ is a closed set, what is the relation in general between the set $D$ and the closure of $\text{Int }D$ ?
( c ) If $U$ is an open set, what is the relation in general between the set $U$ and the interior of $\overline{U}$ ?
Solution:
( a ) Since $\text{Int }Q\subseteq Q$ and $Q$ is closed, by Exercise 3 we know $\overline{\text{Int }Q}\subseteq Q$ , now for any $\mathbf{q}\in Q$ , if $\mathbf{q}\notin \text{Int }Q$ , then $\mathbf{q}\in\text{Bd }Q$ , if we denote $Q=[a_1,b_1 ]\times \dots \times [a_n,b_n ]$ and $\mathbf{q}=(q_1,\dots ,q_n)$ , then at least one $q_i$ equals $a_i$ or $b_i$ , for $\forall \epsilon >0$ , we choose x_i as follows:

$\displaystyle{x_i=\begin{cases}a_i+\epsilon /n,\quad q_i=a_i\\b_i-\epsilon /n,\quad q_i=b_i\\q_i,\quad a_i<q_i<b_i \end{cases},\quad i=1,\dots ,n}$

then there is at least one $x_i\neq q_i$ , so $\mathbf{x}=(x_1,\dots ,x_n )\neq \mathbf{q}$ , and $|\mathbf{x}-\mathbf{q}|\leq \epsilon <\epsilon$ , as $\mathbf{x}\in \text{Int }Q$ , we see $\mathbf{q}\in \overline{\text{Int }Q}$ and $Q\subseteq \overline{\text{Int }Q}$ .
( b ) We obviously have $\text{Int }D\subseteq D$ , thus $\overline{\text{Int }D}\subseteq D$ . We may not have the relation in the other direction, for if we define $D={(0,0),(1,0)}\subseteq \mathbf{R}^2$ , then $\text{Int }D=\emptyset$ .
( c ) In general, if $x\in U$ , then $x$ belongs to an open set that is contained in $U$ , thus in $\overline{U}$ , so $x\in \overline{\text{Int }U}$ , which means $U\subseteq \overline{\text{Int }U}$ . We may not have the relation in the other direction, for if we define $U=A\cup B$ , where $A=(0,1)\times (0,1),B=(-1,0)\times (0,1)$ be two open rectangles in $\mathbf{R}^2$ , then $\overline{U}=[-1,1]\times [0,1]$ and $\overline{\text{Int }U}=(-1,1)\times (0,1)$ .

Exercise 5. Let $f:X\to Y$ . Show that $f$ is continous if and only if for each $x\in X$ there is a neighborhood $U$ of $x$ such that $f|U$ is continuous.
Solution: If $f$ is continuous, then $f$ is continuous at each point $x$ of $X$ . If $x\in X$ , then for any neighborhood $U$ of $x$ , $f|U$ is continuous since $U\subseteq X$ .
Conversely, if $\forall x\in X$ , there exists a neighborhood $U$ of $x$ such that $f|U$ is continuous, then $x\in U$ means $f$ is continuous at $x$ , thus $f$ is continuous on $X$ .

Exercise 6. Let $X=A\cup B$ , where $A$ and $B$ are subspaces of $X$ . Let $f:X\to Y$ ; suppose that the restricted function

$\displaystyle{f|A:A\to Y\quad\text{and}\quad f|B:B\to Y}$

are continuous. Show that if both $A$ and $B$ are closed in $X$ , then $f$ is continuous.
Solution: Since $X=A\cup B=(X\backslash A)\cup (A\cap B)\cup (X\backslash B)$ , and the three sets are disjoint. Let $\forall x\in X$ .
If $x\in X\backslash A$ , then notice that $A$ is closed means $X\backslash A$ is open, and $X\backslash A\subset B$ , since $f|B$ is continuous, by Exercise 3.5 there is a neighborhood $U$ of $x$ which is open in $B$ such that $f|U$ is continuous, notice that we can assign an open set $U'$ in $X$ s.t. $U=U'\cap B$ , and $x\in U'$ , so $U_x:=U'\cap (X\backslash A)$ is still a neighborhood of $x$ and belongs to $B$ , we can conclude $f|U_x$ is continuous.
If $x\in X\backslash B$ , we can similarly find an open set $U_x$ in $X$ s.t. $f|U_x$ is continuous.
If $x\in A\cap B$ , then there is $U_1$ open in $A$ , s.t. $x\in U_1,f|U_1$ is continuous, and $U_2$ open in $B$ , s.t. $x\in U_2,f|U_2$ is continuous. Then we can find $U_1',U_2'$ open in $X$ , s.t. $U_1=U_1'\cap A,U_2=U_2'\cap B$ , and $x\in U_1',x\in U_2'$ , let $U_x:=U_1'\cap U_2'$ , then $U_x$ is open in $X$ and contains $x$ , thus a neighborhood of $x$ in $X$ . To prove $f|U_x$ is continuous, notice $U_x=U_x\cap X=U_x\cap (A\cup B)=(U_x\cap A)\cup (U_x\cap B)$ , and $U_x\cap A=U_1'\cap U_2'\cap A=U_1\cap U_2'\subseteq U_1$ , $U_x\cap B=U_1'\cap U_2'\cap B=U_1'\cap U_2\subseteq U_2$ , so $f$ is continuous on $U_x\cap A$ and $U_x\cap B$ , thus on $U_x$ .
We proved for each $x\in X$ there is a a neighborhood $U$ of $x$ s.t. $f|U$ is continuous, by Exercise 3.5 again, $f$ is continuous.

Exercise 7. Finding the limit of a composite function $g\circ f$ is easy if both $f$ and $g$ are continuous; see Theorem 3.5. Otherwise it can be a bit tricky:
Let $f:X\to Y$ and $g:Y\to Z$ . Let $x_0$ be a limit point of $X$ and let $y_0$ be a limit point of $Y$ . See Figure 3.1.

Consider the following three conditions:
(i) $f(x)\to y_0$ as $x\to x_0$ .
(ii) $g(y)\to z_0$ as $y\to y_0$ .
(iii) $g(f(x))\to z_0$ as $x\to x_0$ .
( a ) Give an example where (i) and (ii) hold, but (iii) does not.
( b ) Show that if (i) and (ii) hold and if $g(y_0)=z_0$ , then (iii) holds.
Solution:
(a) let $f(x)=1$ on $X=[0,1]$ and $g(x)=\begin{cases}x,0\leq x<1\\0,x=1\end{cases}$ on $Y=[0,1]$ , then 1 is a limit point of $X$ and $Y$ . When $x\to 1,f(x)\to 1$ , and when $y\to 1,g(y)\to 1$ , but for $g(f(x))$ , when $x\to 1$ on $X$ , we have $g(f(x))=g(1)=0$ , so $g(f(x))\to 0$ .
(b) By Theorem 3.7, if $g(y_0 )=z_0$ and (ii) hold, then $g$ is continuous at $y_0$ , thus for an open set $V$ of $Z$ containing $z_0$ , we have some open set $W\subset Y$ s.t. $y_0\in W,g(W)\subset V$ , since (i) is true, for this $W$ we can find an open set $U$ of $X$ containing $x_0$ s.t. $f(x)\in W$ whenever $x\in U,x\neq x_0$ , then we have

$\displaystyle{g(f(x))\subset V,x\in U,x\neq x_0}$

this means (iii) holds.

Exercise 8. Let $f:\mathbf R\to\mathbf R$ be defined by setting $f(x)=\sin{x}$ if $x$ is rational, and $f(x)=0$ otherwise. At what points is $f$ continuous?
Solution: At points $x=k\pi,k\in \mathbf{Z}$ will $f$ be continuous.

Exercise 9. If we denote the general points of $\mathbf R^2$ by $(x,y)$ , determine $\text{Int }A,\text{Ext }A$ and $\text{Bd }A$ for the subset $A$ of $\mathbf R^2$ specified by each of the following conditions:
( a ) $x=0$ .
( b ) $0\leq x<1$ .
( c ) $0\leq x<1$ and $0\leq y<1$ .
( d ) $x$ is rational and $y>0$ .
( e ) $x$ and $y$ are rational.
( f ) $0<x^2+y^2<1$ .
( g ) $y<x^2$ .
( h ) $y\leq x^2$ .

Solution:
( a ) $\text{Int }A=\emptyset,\text{Ext }A=R^2-A,\text{Bd }A=A$
( b ) $\text{Int }A={(x,y):01},\text{Bd }A={(x,y):(x=0)\vee(x=1)}$
( c ) $\text{Int }A={(x,y):0<x<1,0<y<1}$ , $\text{Ext }A=\mathbf{R}^2-{(x,y):0\leq x\leq 1,0\leq y\leq 1}$ , $\text{Bd }A={(x,y):(x=0,0\leq y\leq 1)\vee(x=1,0\leq y\leq 1)\vee(y=0,0\leq x\leq 1)\vee(y=1,0\leq x\leq 1)}$
( d ) $\text{Int }A=\emptyset,\text{Ext }A={(x,y):y<0},\text{Bd }A={(x,y):y\geq 0}$
( e ) $\text{Int }A=\emptyset,\text{Ext }A=\emptyset,\text{Bd }A=\mathbf{R}^2$
( f ) $\text{Int }A=A,\text{Ext }A=\mathbf{R}^2-{(x,y):x^2+y^2\leq 1}$ , $\text{Bd }A={(x,y):x^2+y^2=1}\cup (0,0)$
( g ) $\text{Int }A=A,\text{Ext }A={(x,y):y>x^2 },\text{Bd }A={(x,y):y=x^2 }$
( h ) $\text{Int }A={(x,y):yx^2 },\text{Bd }A={(x,y):y=x^2 }$ .

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