# Topological Spaces: Compactness

Heine-Borel Theorem

Any cover of a closed interval $$[a, b]$$ by a sytem of open intervals (or, more generally, open sets) has a finite subcover.

Definition A topological space $$T$$ is said to be compact if every open cover of $$T$$ has a finite subcover. A compact Hausdorff space is called a compactum.

Definition A system of subsets $$\{A_\alpha\}$$ of a set $$T$$ is said to be centered if every finite intersection $$\cap_{k=1}^n A_k$$ is nonempty (also called the finite intersection property).

Theorem 1

A topological space $$T$$ is compact if and only if every centered system of closed subsets of $$T$$ has a nonempty intersection.

Proof [Credit to Dan Ma for this proof; I was unable to fully understand the proof given in the book.]

Let $$T$$ be compact and $$\{F_\alpha\}$$ any centered system of closed subsets of $$T$$. For contradiction, assume $$\cap_\alpha F_\alpha = \emptyset$$, then $$\{G_\alpha\} = \{T - F_\alpha\}$$ is an open cover of $$T$$. By the compactness of $$T$$, $$G_\alpha$$ has a finite subcover $$\{T - F_{\alpha_1}, \ldots, T - F_{\alpha_n}\}$$ implying that $$\cap_{k = 1}^n F_{\alpha_k} = \emptyset$$. Contradiction. Therefore $$\cap_\alpha F_\alpha \not= \emptyset$$.

Conversely, let $$\{G_\alpha\}$$ be an open cover of $$T$$ with no finite subcover, then $$\{F_\alpha\} = \{T - G_\alpha\}$$ is a centered system of closed subsets of $$T$$ with $$\cap_\alpha F_\alpha = \emptyset$$.

Theorem 2

Every closed subset $$F$$ of a compact topological space $$T$$ is itself compact.

Proof

Let $$\{F_\alpha\}$$ be any centered system of closed subset of the subspace $$F \subseteq T$$. Then every $$F_\alpha$$ is closed in $$T$$ as well, therefore $$\cap_\alpha F_\alpha \not= \emptyset$$. Then by Theorem 1, $$F$$ is compact.

Corollary

Every closed subset of a compactum is itself a compactum.

Proof

By Theorem 2 and the fact that every subset of a Hausdorff space is itself a Hausdorff space.

Theorem 3

Let $$K$$ be a compactum and $$T$$ any Hausdorff space containing $$K$$. Then $$K$$ is closed in $$T$$.

Proof

Suppose $$y \notin K$$, so that $$y \in T- K$$. Then, for any point $$x \in K$$, there is a neighbourhood $$U_x$$ of $$x$$ and $$V_x$$ of $$y$$ such that $$U_x \cap V_x = \emptyset$$. The neighbourhoods $$\{U_x\}$$ for $$x \in K$$ form an open cover of $$K$$. Hence, by the compactness of $$K$$, $$\{U_x\}$$ has a finite subcover consisting of sets $$U_{x_1}, \ldots U_{x_n}$$. Let $$V = V_{x_1} \cap \ldots \cap V_{x_n}$$. Then $$V$$ is a neighbourhood of the point $$y$$ which does not intersect the set $$U_{x_1} \cup \ldots \cup U_{x_n} \supseteq K$$, and hence $$y \notin [K]$$. It follows that $$K$$ is closed (in $$T$$).

Remark It is a consequence of Theorem 2 and Theorem 3 that compactness is an ‘intrinsic property’, in the sense that a compactum remains a compactum after being ‘embedded’ in any larger Hausdorff space.

Theorem 4

Every compactum $$K$$ is a normal space.

Proof

Let $$X$$ and $$Y$$ be any two disjoint closed subsets of $$K$$. Repeating the argument given in the proof of Theorem 3, we easily see that, given any point $$y \in Y$$, there exists a neighbourhood $$U_y$$ containing $$y$$ and an open set $$O_y \supseteq X$$ such that $$U_y \cap O_y = \emptyset$$. Since $$Y$$ is compact, by Theorem 2, the cover $$\{U_y\} (y \in Y)$$ of the set $$Y$$ has a finite subcover $$U_{y_1}, \ldots, U_{y_n}$$. The open sets $$O^{(1)} = O_{y_1} \cap \ldots \cap O_{y_n}$$, $$O^{(2)} = U_{y_1} \cap \ldots \cap U_{y_n}$$ then satisfy the normality conditions $$O^{(1)} \supseteq X$$, $$O^{(2)} \supseteq Y$$, $$O^{(1)} \cap O^{(2)} = \emptyset$$.

Theorem 5

Let $$X$$ be a compact space and $$f$$ a continuous mapping of $$X$$ onto a topological space $$Y$$. Then $$Y = f(X)$$ is itself compact.

Proof

Let $$\{V_\alpha\}$$ be any open cover of $$Y$$, and let $$U_\alpha = f^{-1}(V_\alpha)$$. Then the sets $$U_\alpha$$ are open (being preimages of open sets under a continuous mapping) and cover the space $$X$$. Since $$X$$ is compact, $$\{U_\alpha\}$$ has a finite subcover $$U_{x_1}, \ldots, U_{x_n}$$. Then the sets $$V_{x_1}, \ldots V_{x_n}$$, where $$V_k = f(U_k)$$, cover $$Y$$. It follows that $$Y$$ is compact.

Theorem 6

A one-to-one continuous mapping of a compactum $$X$$ onto a compactum $$Y$$ is necessarily a homeomorphism.

Proof

We must show that $$f^{-1}$$ is continuous too. Let $$F$$ be a closed set in $$X$$ and $$P = f(F)$$ be its image in $$Y$$. Then $$P$$ is a compactum, by Theorem 5. Hence, by Theorem 3, $$P$$ is closed in $$Y$$. Therefore the preimage under $$f^{-1}$$ of any closed set $$F \subseteq X$$ is closed. It follows from Theorem 10’ (from Topological Spaces: Basic Concepts) that $$f^{-1}$$ is continuous.

Theorem 7

If $$T$$ is a compact space, then any infinite subset of $$T$$ has at least one limit point.

Proof

Suppose $$T$$ contains an infinite set $$X$$ with no limit point. Then $$T$$ contains a countable set $$X = \{x_1, \ldots, x_n\}$$ with no limit point. But then the sets $$F_n = \{x_n, x_{n+1}, \ldots\} (n = 1, 2, \ldots)$$ form a centered system of closed sets in $$T$$ with an empty intersection ,i.e., $$T$$ is not compact.

Definition A topological space $$T$$ is said to be countably compact if every infinite subset of $$T$$ has at least one limit point (in $$T$$).

Remark Theorem 7 says that every compact set is countably compact. The converse, however, is not true (see Problem 1).

Theorem 8

Each of the following two conditions is necessary and sufficient for a topological space $$T$$ to be countably compact:

1. every countable open cover of $$T$$ has a finite subcover;

2. every countable centered system of closed subsets of $$T$$ has a non-empty intersection.

Proof

The equivalence of conditions 1 and 2 is an immediate consequence of the duality principle. Moreover, if $$T$$ is not countably compact, then, repeating the argument given in proving Theorem 7, we find that there is a countable centered system of closed subsets of $$T$$ with an empty intersection. This proves the sufficiency of condition 2. Thus we need only prove the necessity of condition 2.

Let $$T$$ be countably compact, and let $$\{F_n\}$$ be a countable centered system of closed sets in $$T$$. Then, as we now show, $$\cap_n F_n \not= \emptyset$$. Let $$\Phi_n = \cap_{k=1}^n F_k$$. Then none of the $$\Phi_n$$ is empty, since $$\{F_k\}$$ is centered. Moreover, $$\Phi_1 \supseteq \ldots \supseteq \Phi_n \supseteq \ldots$$ and $$\cap_n \Phi_n = \cap_n F_n$$. There are now two possibilities:

1. $$\Phi_{n_0} = \Phi_{n_0 + 1} = \ldots$$ starting from some index $$n_0$$, in which case it is obvious that $$\cap_n \Phi_n = \Phi_{n_0} \not= \emptyset$$ .

2. There are infinitely many distinct sets $$\Phi_n$$. In this case, there is clearly no loss in generality in assuming that all the $$\Phi_n$$ are distinct. Let $$x_n \in \Phi_n - \Phi_{n+1}$$. Thn the sequence $$\{x_n\}$$ consists of infinitely many distinct points of $$T$$, and hence, by the countable compactness of $$T$$, must have at least one limit point, say $$x_0$$. But then $$x_0$$ must be a limit point of $$\Phi_n$$, since $$\Phi_n$$ contains all the points $$x_n, x_{n+1}, \ldots$$. Moreover $$x_0 \in \Phi_n$$, since $$\Phi_n$$ is closed. It follows that $$x_0 \in \cap_n \Phi_n$$ i.e. $$\cap_n \Phi_n \not= \emptyset$$.

Remark Thus compact topological spaces are those in which an arbitrary open cover has a finite subcover, while countably compoact spaces are those in which every countable open cover has a finite subcover.

Theorem 9

The concepts of compactness and countable compactness coincide for a topological space $$T$$ with a countable base.

Proof

By Theorem 6 (from Topological Spaces: Basic Concepts), every open cover $$\mathscr{O}$$ of $$T$$ has a countable subcover. Hence, if $$T$$ is countably compact, $$\mathscr{O}$$ has a finite subcover, by Theorem 8.

## Overview

 Compact Definition Finite subcover Compact Theorem 1 Nonempty intersection Compact Theorem 7 Limit point Countably compact Definition Limit point Countably compact Theorem 8 1) Finite subcover Countably compact Theorem 8 2) Nonempty intersection

## Problems

1. Let $$X$$ be the set of all ordinal numbers less than the first uncountable ordinal. Let $$(\alpha, \beta) \subseteq X$$ denote the set of all ordinal numbers $$\gamma$$ such that $$\alpha < \gamma < \beta$$, and let the open sets in $$X$$ be all the unions of the interval $$(\alpha, \beta)$$. Prove that the resulting topological space $$T$$ is countably compact but not compact.

$$T$$ is evidently countable, and every infinite subset has a limit point, as strictly increasing sequences of countable length converge to their supremum, hence it is countably compact. However, $$T$$ is not compact, as any open cover has no finite subcover. More details can be read here.

2. A topological space $$T$$ is said to be locally compact if every point $$x \in T$$ has at least one relatively compact neighbourhood. Show that a compact space is automatically locally compact, but not conversely. Prove that every closed subspace of a locally compact subspace is locally compact.

Let $$x$$ be an arbitrary point in a compact space $$T$$. Then for any open neighbourhood $$U$$ of $$x$$, the closure $$[U]$$ is a closed subset of $$T$$. By Theorem 2, $$[U]$$ is a compact space, so $$U$$ is a relatively compact neighbourhood. As $$x$$ is arbitrary, $$T$$ is locally compact.

$$\mathbb{R}$$ is locally compact but not compact.

Let $$F$$ be a closed subspace of locally compact $$K$$, and let $$x$$ be an arbitrary point of $$F$$. Then $$x \in T$$. Since $$T$$ is locally compact, there exists at least one neighbourhood $$U$$ of $$x$$ such that $$[U]$$ is compact with respect to $$T$$. Let $$[V] = [U] \cap F$$. Then $$[V] \subseteq [U]$$ and by Theorem 2, $$[V]$$ is compact with respect to $$T$$ (and $$F$$). Hence $$V$$ is a relatively compact neighbourhood of $$x$$ so $$F$$ is locally compact. [Thanks to CoveredInChocolate.]

3. A point $$x$$ is said to be a complete limit point of a subset $$A$$ of a topological space if, given any neighbourhood $$U$$ of $$x$$ the sets $$A$$ and $$A \cap U$$ have the same power (i.e. cardinal number). Prove that every infinite subset of a compact topological space has at least one complete limit point.

Comment. Conversely, it can be shown that if every infinite subset of a topological space $$T$$ has at least one complete limit point, then $$T$$ is compact.

Let $$T$$ be a compact space. Then any open covering $$\{U_\alpha\}$$ has a finite subcover, i.e $$T \subseteq \cup_{k=1}^N U_k$$. Let $$A \subseteq T$$ be an infinite set. Then there is at least one open set $$U_k$$ $$1 \leq k \leq N$$ that contains an infinite amount of points. Let $$x \in U_k$$, which is a neighbourhood containing an infinte amount of points, then $$A \cap U_k$$ has the same power, and $$x$$ is a complete limit point of $$A$$. [Thanks to CoveredInChocolate.]