# 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:

every countable open cover of \(T\) has a finite subcover;

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:

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

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

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.

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.]

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.]