[Munkres Topology] 27. Compact Subspaces of the Real Line
This article is one of the posts in the Textbook Commentary Project.
Theorem 27.1. Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.
Proof. Step 1. Given $a<b$, let $\mathcal{A}$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace topology (which is the same as the order topology). We wish to prove the existence of a finite subcollevtion of $\mathcal{A}$ covering $[a,b]$. First we prove the following: If $x$ is a point of $[a,b]$ different from $b$, then there is a point $y>x$ of $[a,b]$ such that the interval $[x,y]$ can be covered by at most two elements of $\mathcal{A}$.
If $x$ has an immediate successor in $X$(there is no element on $(x,y)$), let $y$ be this immediate successor. Then $[x,y]$ consists of the two points $x$ and $y$, so that it can be covered by at most two elements of $\mathcal{A}$. If $x$ has no immediate successor in $X$, choose an element $A$ of $\mathcal{A}$ contining $x$. Because $x\ne b$ and $A$ is open, $A$ contains an interval of the form $[x,c)$, for some $c$ in $[a,b]$. Choose a point $y$ in $(x,c)$; then the interval $[x,y]$ is covered by the single element $A$ of $\mathcal{A}$.
Step 2. Let $C$ be the set of all points $y>a$ of $[a,b]$ such that the interval $[a,y]$ can be covered by finitely many elements of $\mathcal{A}$. Applying Step 1 to the case $x=a$, we see that there exists at least one such $y$, so $C$ is not empty. Let $c$ be the least upper bound of the set $C$; then $a<c\le b$.
Step 3. We show that $c$ belongs to $C$; that is, we show that the interval $[a,c]$ can be covered by finitely many elements of $\mathcal{A}$. Choose an element $A$ of $\mathcal{A}$ containing $c$; since $A$ is open, it contains an interval of the form $(d,c]$ for some $d$ in $[a,b]$. If $c$ is not int $C$, there must be a point $z$ of $C$ lying in the interval $(d,c)$, because otherwise $d$ would be a smaller upper bound on $C$ than $c$. See Figure 27.1. Since $z$ is in $C$, the interval $[a,z]$ can be covered by finitely many, say $n$, elements of $\mathcal{A}$. Now $[z,c]$ lies in the single element $A$ of $\mathcal{A}$, hence $[a,c]=[a,z]\cup [z,c]$ can be covered by $n+1$ elements of $\mathcal{A}$. Thus $c$ is in $C$, contrary to assumption.
Step 4. Finally, we show that $c=b$, and our theorem is proved. Suppose that $c<b$. Applying Step 1 to the case $x=c$, we conclude that there exists a point $y>c$ of $[a,b]$ such that the interval $[c,y]$ can be covered by finitely many elements of $\mathcal{A}$. See Figure 27.2. We proved in Step 3 that $c$ is in $C$, so $[a,c]$ can be covered by finitely many elements of $\mathcal{A}$. Therefore, the interval $$[a,y]=[a,c]\cup [c,y]$$ can also be covered by finitely many elements of $\mathcal{A}$. This means that $y$ is in $C$, contradicting the fact that $c$ is an upper bound on $C$.
(Isn't this is to complicated?)
Corollary 27.2. Every closed interval in $\mathbb{R}$ is compact.
Theorem 27.3. A subspace $A$ of $\mathbb{R}^n$ is compact if and only if it is closed and is bounded in the euclidean metric $d$ or the square metric $\rho$.
Proof. It will suffice to consider only the metric $\rho$; the inequalities $$\rho(x,y)\le d(x,y)\le \sqrt{n} \rho (x,y)$$ imply that $A$ is bounded under $d$ is and only if it is bounded under $\rho$.
[$\Rightarrow$] Suppose that $A$ is compact. Then, by Theorem 26.3, it is closed. Consider the collection of open sets $$\{B_\rho (\mathbf{0},m)|m\in \mathbb{Z}_+\},$$ whose union is all of $\mathbb{R}^n$. Some finite subcollection covers $A$. It follows that $A\subset B_\rho(\mathbf{0},M)$ for some $M$ (exists because finite! as always.). Therefore, for any two points $x$ and $y$ of $A$, we have $\rho(x,y)\le 2M$. Thus $A$ is bounded under $\rho$.
[$\Leftarrow$] Conversely, suppose that $A$ is closed and bounded under $\rho$; suppose that $\rho(x,y)\le N$ for every pair $x,y$ of points of $A$. Choose a point $x_0$ of $A$, and let $\rho(x_0,\mathbf{0})=b$. The triangle inequality implies that $\rho(x,\mathbf{0})\le N+b$, then $A$ is a subset of the cube $[-P,P]^n$, which is compact. Being closed, $A$ is also compact.
Theorem 27.4 (Extreme value theorem). Let $f:X\rightarrow Y$ be continuous, where $Y$ is an ordered set in the order topology. If $X$ is compact, then there exist points $c$ and $d$ in $X$ such that $f(c)\le f(x)\le f(d)$ for every $x\in X$.
Proof. Since $f$ is continuous and $X$ is compact, the set $A=f(X)$ is compact. We show that $A$ has a largest element $M$ and a smallest element $m$. Then since $m$ and $M$ belong to $A$, we must have $m=f(c)$ and $M=f(d)$ for some points $c$ and $d$ of $X$.
If $A$ has no largest element, then the collection $$\{(-\infty,a) |a\in A\}$$ forms an open covering of $A$. Since $A$ is compact, some finite subcollection $\{(-\infty,a_1),\cdots,(-\infty,a_n)\}$$ covers $A$. If $a_i$ is the largest of the elements $a_1,\cdots,a_n$ (exists because finite!), then $a_i$ belongs to none of these sets, contrary to the fact that they cover $A$.
A similar argument shows that $A$ has a smallest element.
Definition. Let $(X,d)$ be a metric space; let $A$ be a nonempty subset of $X$. For each $x\in X$, we define the distance from $x$ to $A$ by the equation $$d(x,A)=\inf \{d(x,a)|a\in A\}.$$
It is easy to show that for fixed $A$, the function $d(x,A)$ is a continuous function of $x$: Given $x, y\in X$, one has the inequalities $$d(x,A)\le d(x,a)\le d(x,y)+d(y,a),$$ for each $a\in A$. It follows that $$d(x,A)-d(x,y)\le \inf d(y,a)=d(y,A),$$ so that $$d(x,A)-d(y,A)\le d(x,y).$$ The same inequality holds with $x$ and $y$ interchanged; continuity of the function $d(x,A)$ follows. (With $\epsilon-\delta$ statement)
Let the diameter of a bounded subset $A$ of a metric space $(X,d)$ is the number $$\sup \{ d(a_1,a_2)|a_1,a_2 \in A\}.$$
Lemma 27.5 (The Lebesgue number lemma). Let $\mathcal{A}$ be an open covering of the metric space $(X,d)$. If $X$ is compact, there is a Lebegue number $\delta >0$ such that for each subset of $X$ having diameter less than $\delta$, there exists an element of $\mathcal{A}$ containing it.
Proof. Let $\mathcal{A}$ be an open covering of $X$. If $X$ itself is an element of $\mathcal{A}$, then any positive number is a Lebesgue number for $\mathcal{A}$. So assume $X$ is not an element of $\mathcal{A}$.
Choose a finite subcollection $\{A_1,\cdots,A_n\}$ of $\mathcal{A}$ that covers $X$. For each $i$, set $C_i=X-A_i$, and define $f:X\rightarrow \mathcal{R}$ by letting $f(x)$ be the average of the numbers $d(x,C_i)$. That is, $$f(x)=\frac{1}{n}\sum_{i=1}^n d(x,C_i).$$ We show that $f(x)>0$ for all $x$. Given $x\in X$, choos $i$ so that $x\in A_i$. Then choose $\epsilon$ so the $\epsilon$-neighborhood of $x$ lies in $A_i$. Then $d(x,C_i)\ge \epsilon$, so that $f(x)\ge \epsilon/n$. (is $epsilon$ be minimum of whole $A_i$?)
Since $f$ is continuous, it has a minimum value $\delta$; we show that $\delta$ is our required Lebesgue number. Let $B$ be a subset of $X$ of diameter less than $\delta$. Choose a point $x_0$ of $B$; then $B$ lies in the $\delta$-neighborhood of $x_0$. Now $$\delta\le f(x_0)\le d(x_0,C_m),$$ where $d(x_0,C_m)$ is the largest of the numbers $d(x_0,C_i)$. Then the $\delta$-neighborhood of $x_0$ is contained in the element $A_m=X-C_m$ of the covering $\mathcal{A}$.
Definition. A function $f$ from the metric space $(X,d_X)$ to the metric space $(Y,d_Y)$ is said to be uniformly continuous if given $\epsilon>0$, there is a $\delta>0$ such that for every pair of points $x_0,x_1$ of $X$, $$d_X(x_0,x_1)<\delta \Rightarrow d_Y (f(x_0),f(x_1))<\epsilon.$$
Theorem 27.6 (Uniform continuity theorem). Let $f: X\rightarrow Y$ be a continuous map of the compact metric space $(X,d_X)$ to the metric space $(Y,d_Y)$. Then $f$ is uniformly continuous.
Proof. Given $\epsilon>0$, take the open covering of $Y$ by balls $B(y,\epsilon/2)$ of radius $\epsilon/2$. Let $\mathcal{A}$ be the open covering of $X$ by the inverse images of these balls under $f$. Choose $\delta$ to be a Lebesgue number for the covering $\mathcal{A}$. Then if $x_1$ and $x_2$ are two points of $X$ such that $d_X(x_1,x_2)<\delta$, the two-point set $\{x_1,x_2\}$ has diameter less than $\delta$, so that its image $\{f(x_1),f(x_2)\}$ lies in some ball $B(y,\epsilon/2)$. Then $d_Y(f(x_1),f(x_2))<\epsilon$, as desired.
Wow.. nice. perfect.
Definition. If $X$ is a space, a point $x$ of $X$ is said to be an isolated point of $X$ if the one-point set $\{x\}$ is open in $X$.
Theorem 27.7. Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.
Proof. Step 1. We show first that given any nonempty open set $U$ of $X$ and any point $x$ of $X$, there exists a nonempty open set $V$ contained in $U$ such that $x\notin \bar{V}$.
Choose a point of $X$ and it is possible if $x$ is not in $U$ because $x$ is not an isolated point of $X$ and it is possible if $x$ is not in $U$ simply because $U$ is nonempty. Now choose disjoint open sets $W_1$ and $W_2$ about $x$ and $y$, respectively. Then the set $V=W_2 \cap U$ is the desired open set; it is contained in $U$, it is nonempty because it contains $y$, and its closure does not contain $x$. See Figure 27.3.
Step 2. We show that given $f:\mathbb{Z}_+\rightarrow X$, the function $f$ is not surjective. It follows that $X$ is uncountable.
Let $x_n=f(n)$. Apply Step 1 to the nonempty open set $U=X$ to choose a nonempty open set $V_1 \subset X$ such that $\bar{V}_1$ does not contain $x_1$. In general, given $V_{n-1}$ open and nonempty, choose $V_n$ to be a nonempty open set such that $V_n\subset V_{n-1}$ and $\bar{V}_n$ does not contain $x_n$. Consider the nested sequence $$\bar{V}_1\supset \bar{V}_2\supset \cdots$$ of nonempty closed sets of $X$. Because $X$ is compact, there is a point $x\in \bigcap \bar{V}_n$, by Theorem 26.9. Now $x$ cannot equal $x_n$ for any $n$, since $x$ belongs to $\bar{V}_n$ and $x_n$ does not. ($x$ is out of countable set!)
Corollary 27.8. Every closed interval in $\mathbb{R}$ is uncountable.