10.6 Compact spaces

中文

A collection of sets $\mathcal{U}=\{U_i|i\in I\}$ is said to be a covering of a subset $A$ of a topological space $X$ if every point $x\in A$ belongs to some member of the collection. If every member is an open set it is called an open covering. A subset of the covering, ${\mathcal{U}}^{\prime }\subseteq \mathcal{U}$ , which covers $A$ is referred to as a subcovering. If ${\mathcal{U}}^{\prime }$ consists of finitely many sets $\left\{U_1,U_2,\dots ,U_n\right\}$ it is called a finite subcovering.

A topological space $(X,\mathcal{O})$ is said to be compact if every open covering of $X$ contains a finite subcovering. The motivation for this definition lies in the following theorem, the proof of which can be found in standard books on analysis [10–12].

Theorem 10.8 · Heine–Borel theorem

(Heine–Borel) A subset $A$ of ${\mathbb{R}}^n$ is closed and bounded (included in a central ball, $A\subset B_a(0)$ for some $a>0)$ if and only if every open covering of $A$ has a finite subcovering.

Theorem 10.9 · Closed subspaces of compact spaces

Every closed subspace $A$ of a compact space $X$ is compact in the relative topology.

Proof

Let $\mathcal{U}$ be any covering of $A$ by sets that are open in the relative topology. Each member of this covering must be of the form $U\cap A,$ , where $U$ is open in $Y$. The sets $\{U\}$ together with the open set $X-A$ form an open covering of $X$ that, by compactness of $X$, must have a finite subcovering $\{U_1,U_2,\dots ,U_n,X-A\}$ . The sets $\{U_1\cap A,\dots ,U_n\cap A\}$ are thus a finite subfamily of the original open covering of $A$. Hence $A$ is compact in the relative topology.

Theorem 10.10 · Continuous images of compact sets

If $f:X\to Y$ is a continuous map from a compact topological space $X$ into a topological space $Y$, then the image set $f(X)\subseteq Y$ is compact in the relative topology.

Proof

Let $\mathcal{U}$ be any covering of $f(X)$ consisting entirely of open sets in the relative topology. Each member of this covering is of the form $U\cap f(X)$ , where $U$ is open in $Y$. Since $f$ is continuous, the sets $f^{-1}(U)$ form an open covering of $X$. By compactness of $X$, a finite subfamily $\{f^{-1}(U_i)|i=1,\dots ,n\}$ serves to cover $X$, and the corresponding sets $U_i\cap f(X)$ evidently form a finite subcovering of $f(X)$

Compactness is therefore a topological property, invariant under homeomorphisms.

Example 10.16

If $E$ is an equivalence relation on a compact topological space $X$, the map $i_E:X\to X/E$ is continuous in the topology on $X/E$ obtained by identification from $X$. By Theorem 10.10 the topological space $X/E$ is compact. For example, the torus $T^2$ formed by identifying opposite sides of the closed and compact unit square in ${\mathbb{R}}^2$ is a compact space.

Theorem 10.11 · Compactness of finite products

The topological product $X\times Y$ is compact if and only ifboth $X$ and $Y$ are compact.

Proof

If $X\times Y$ is compact then $X$ and $Y$ are compact by Theorem 10.10 since both the projection maps ${\mathrm{pr}}_1:X\times Y\to X$ and ${\mathrm{pr}}_2:X\times Y\to Y$ are continuous in the product topology.

Conversely, suppose $X$ and $Y$ are compact. Let $\mathcal{W}=\{W_i|i\in I\}$ be an open covering of $X\times Y$ . Since each set $W_i$ is a union of such sets of the form $U\times V$ where $U$ and $V$ are open sets of $X$ and $Y$ respectively, the family of all such sets $U_j\times V_j(j\in J)$ that are subsets of $W_i$ for some $i\in I$ is an open cover of $X\times Y$ . Given any point $y\in Y$ , the set of all $U_j$ such that $y\in V_j$ is an open cover of $X$, and since $X$ is compact there exists a finite subcover $\{U_{j_1},U_{j_2},\dots ,U_{j_n}\}$ . The set $A_y=V_{j_1}\cap V_{j_2}\cap \cdot \cdot \cdot \cap V_{j_n}$ is an open set in $Y$ by condition (Top2), and $y\in A_y$ since $y\in V_{j_k}$ for each $k=1,\dots ,n$ . Thus the family of sets $\{A_y|y\in Y\}$ forms an open cover of $Y$. As $Y$ is compact, there is a finite subcovering $A_{y_1},A_{y_2},\dots ,A_{y_m}$ . The totality of all the sets $U_{j_k}\times V_{j_k}$ associated with these sets $A_{y_a}$ forms a finite open covering of $X\times Y$ . For each such set select a corresponding member $W_i$ of the original covering of which it is a subset. The result is a finite subcovering of $X\times Y,$ proving that $X\times Y$ is compact.

Somewhat surprisingly, this statement extends to arbitrary infinite products (Tychonoff’s theorem). The interested reader is referred to [8] or [2] for a proof of this more difficult result.

Theorem 10.12 · Accumulation points of infinite subsets of compact spaces

Every infinite subset of a compact topological space has an accumulation point.

Proof

Suppose $X$ is a compact topological space and $A\subset X$ has no accumulation point. The aim is to show that $A$ is a finite set. Since every point in $x\in A-X$ has an open neighbourhood $U_x$ such that $U_x\cap A=\varnothing$ it follows that $A\subseteq X$ is closed since its complement $\begin{array}{r}A-X={\bigcup }_{x\in X-A}{U}_x\end{array}$ is open. Hence, by Theorem 10.9, $A$ is compact. Since each point $a\in A$ is not an accumulation point, there exists an open neighbourhood $U_a$ of $a$ such that $U_a\cap A=\{a\}$ . Hence each singleton $\{a\}$ is an open set in the relative topology induced on $A$, and the relative topology on $A$ is therefore the discrete topology. The singleton sets $\{a\mid a\in A\}$ therefore form an open covering of $A$, and since $A$ is compact there must be a finite subcovering $\{a_1\},\{a_2\},\dots ,\{a_n\}$ . Thus $A=\{a_1,a_2,\dots ,a_n\}$ is a finite set.

Theorem 10.13 · Compact subsets of Hausdorff spaces are closed

Every compact subspace of a Hausdorff space is closed

Proof

Let $X$ be a Hausdorff space and $A$ a compact subspace in the relative topology. If $a\in A$ and $x\in X-A$ then there exist disjoint open sets $U_a$ and $V_a$ such that $a\in U_a$ and $x\in V_a$ . The family of open sets $U_a\cap A$ is an open covering of $A$ in the relative topology. Since $A$ is compact there is a finite subcovering $\{U_{a_1}\cap A,\dots ,U_{a_n}\cap A\}$ . The intersection of the corresponding neighbourhoods $W=V_{a_1}\cap \cdots \cap V_{a_n}$ is an open set that contains $x$. As all its points lie outside every $U_{a_i}\cap A$ we have $W\cap A=\varnothing .$ . Thus every point $x\in X-A$ has an open neighbourhood with no points in $A$. Hence $A$ includes all its accumulation points and must be a closed set.

In a metric space $(M,d)$ we will say a subset $A$ is bounded if $\sup \{d(x,y)\mid x,y\in A\}<\infty$.

Theorem 10.14 · Compact subsets of metric spaces are closed and bounded

Every compact subspace of a metric space is closed and bounded.

Proof

Let $A$ be a compact subspace of a metric space $(M,d)$ . Since $M$ is a Hausdorf space by Theorem 10.4 it follows by the previous theorem that $A$ is closed. Let $\mathcal{U}=\{B_1(a)\cap A\mid a\in A\}$ be the open covering of $A$ consisting of intersections of $A$ with unit open balls centred on points of $A$. Since $A$ is compact, a finite number of these open balls $\{B_1(a_1),B_1(a_2),\dots ,B_1(a_n)\}$ can be selected to cover $A$. Let the greatest distance between any pair of these points be $D=\max d(a_i,a_j)$ . For any pair of points $a$, $b\in A,$ if $a\in B_1(a_k)$ and $b\in B_1(a_l)$ then by the triangle inequality

$$d(a,b)\le d(a,a_i)+d(a_i,a_j)+\cdots +d(a_j,b)\le D+2.$$

Thus $A$ is a bounded set.

Problems

Problem 10.16

Show that every compact Hausdorff space is normal (see Problem 10.11).

Problem 10.17

Show that every one-to-one continuous map $f:X\to Y$ from a compact space $X$ onto a Hausdorff space $Y$ is a homeomorphism.