Compact Spaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We now define the notion of compactness.

Compact Sets

Definition 1 A topological space \(X\) is compact if for every open covering \((U_i)_{i\in I}\) of \(X\), there exists a finite subset \(J\subseteq I\) such that \((U_j)_{j\in J}\) is still an open covering of \(X\).

By definition, a finite union of compact spaces is again compact. Thus compactness is related to an appropriate kind of finiteness.

Consider a subspace \(Y\) of a topological space \(X\). Then \(Y\) is itself a topological space, so we may ask whether it is compact. The following proposition shows that to verify the compactness of \(Y\), it suffices to consider coverings by open sets in \(X\).

Proposition 2 Let \(Y\) be a subspace of a topological space \(X\). Then the following are equivalent: \(Y\) is compact, and for every family \((U_i)_{i\in I}\) of open sets in \(X\) satisfying \(Y\subseteq\bigcup U_i\), there exists a finite subset \(J\subseteq I\) such that the union of \((U_j)_{j\in J}\) still contains \(Y\).

Proof

First, assume that \(Y\) is compact, and let \((U_i)_{i\in I}\) be a family of open sets in \(X\) satisfying \(Y\subseteq\bigcup U_i\). Then the sets \(Y\cap U_i\) are open in \(Y\), so \((U_i\cap Y)_{i\in I}\) is an open covering of \(Y\). Hence we can choose a finite subset \(J\) such that \((U_j\cap Y)_{j\in J}\) is still an open covering of \(Y\). It is then obvious that the union of the \((U_j)\) still contains \(Y\).

Conversely, assume that the stated condition holds, and let \((V_i)_{i\in I}\) be an arbitrary open covering of \(Y\). By the definition of the topology on \(Y\), there exist open sets \((U_i)\) in \(X\) such that \(V_i=U_i\cap Y\), and \(\bigcup U_i\) contains \(Y\). Thus there exists a finite subset \(J\) such that the union of \((U_j)_{j\in J}\) contains \(Y\). Then \((V_j)_{j\in J}\) is the desired finite subcover of \((V_i)_{i\in I}\).

By the proposition above, to prove that \(Y\) is compact it suffices to cover \(Y\) by open sets in the ambient space \(X\) and then show that these satisfy the condition of Definition 1. Therefore, by a slight abuse of terminology, we call open sets \(U_i\) in \(X\) satisfying \(Y\subseteq \bigcup U_i\) an open cover of \(Y\), and we shall make this distinction explicit only when there is a risk of confusion.

Lemma 3 A closed subset of a compact space is compact.

Proof

Let \(X\) be a compact space, let \(Y\) be a closed subset of \(X\), and let \((U_i)\) be an open covering of \(Y\). Then \(X\setminus Y\) is open, and adding this set to the covering \((U_i)\) yields an open covering of \(X\). Since \(X\) is compact, this new covering has a finite subcover; removing \(X\setminus Y\) from it again gives a covering of \(Y\), which is a finite subcover of the original \((U_i)\).

Compact Hausdorff Spaces

For a Hausdorff space, defined in the previous post, imposing the compact condition yields stronger separation properties. To see this, we first establish the following lemma.

Lemma 4 Let \(X\) be a Hausdorff space, let \(x\) be a point of \(X\), and let \(Y\) be a compact subspace of \(X\) not containing \(x\). Then the two sets \(\{x\}\) and \(Y\) can be separated by neighborhoods.

Proof

Since \(X\) is a Hausdorff space, for each \(y\in Y\) there exist an open neighborhood \(U_{xy}\) of \(x\) and an open neighborhood \(V_y\) of \(y\) such that \(U_{xy}\cap V_y=\emptyset\). Now by Lemma 3 there exists a finite subcover \(V_{y_1},\ldots,V_{y_n}\) of \((V_y)_{y\in Y}\) such that

\[Y\subseteq V_{y_1}\cup\cdots\cup V_{y_n}\]

still holds. Then

\[U_{xy_1}\cap \cdots\cap U_{xy_n}\]

is an open neighborhood of \(\{x\}\) disjoint from \(\bigcup_{i=1}^n V_{y_i}\).

In particular, the following holds.

Corollary 5 A compact subset of a Hausdorff space is closed.

Proof

From the proof of Lemma 4, setting

\[U_x=U_{xy_1}\cap \cdots\cap U_{xy_n}\]

we have \(X\setminus Y=\bigcup_{x\not\in Y} U_x\).

As mentioned above, a compact Hausdorff space satisfies the following additional separation axiom. (§Hausdorff Spaces, ⁋Definition 3)

Lemma 6 A compact Hausdorff space is regular.

Proof

Fix a compact Hausdorff space \(X\), and let \(x\in X\) be a point and \(Y\) a closed subset of \(X\) not containing \(x\). Then \(Y\) is compact by Lemma 3, and the desired result follows immediately from Lemma 4.

Moreover, applying this once more yields the following proposition.

Proposition 7 A compact Hausdorff space is normal.

Proof

Let \(A\) and \(B\) be any two disjoint closed subsets of a compact Hausdorff space. Then for each \(a\in A\), by Lemma 6 there exist an open neighborhood \(U_a\) of \(a\) and an open neighborhood \(V_a\) of \(B\) such that \(U_a\cap V_a=\emptyset\). Now, in the same way as in Lemma 4, \((U_a)_{a\in A}\) is an open covering of \(A\), so again by Lemma 3 we can take a finite subcover \(U_{a_1},\ldots, U_{a_n}\) of \((U_a)\). Then the two open sets

\[U_{a_1}\cup\cdots \cup U_{a_n},\qquad V_{a_1}\cap\cdots\cap V_{a_n}\]

are open neighborhoods separating the two closed subsets \(A\) and \(B\).

Compact Spaces and Continuous Functions

Compactness also behaves well with respect to continuous functions.

Proposition 8 For a continuous function \(f:X \rightarrow Y\) and any compact subspace \(A\) of \(X\), the image \(f(A)\) is also compact.

Proof

For any open covering \((U_i)\) of \(f(A)\), the family \((f^{-1}(U_i))\) covers \(A\), and since \(A\) is compact, a finite subcover exists. The corresponding \(U_i\) then form a finite open subcover of \(f(A)\).

Meanwhile, we have seen that a bijective continuous function \(f:X \rightarrow Y\) need not be a homeomorphism; however, if \(X\) is compact and \(Y\) is Hausdorff, then this is indeed the case.

Proposition 9 If \(X\) is compact and \(Y\) is Hausdorff, then any bijective continuous function \(f:X \rightarrow Y\) is a homeomorphism.

Proof

To show this, we must prove that \(f^{-1}\) is continuous. We use the third condition of §Continuous Functions, ⁋Theorem 4: namely, we show that \(f\) is a closed map. Given a closed subset \(A\) of \(X\), this follows by applying Lemma 3, Proposition 8, and Corollary 5 in order.

Finite Intersection Property

Definition 10 A family \(\mathcal{A}\) of subsets of a set \(X\) has the finite intersection property if for any finitely many elements \(A_1,\ldots, A_n\) of \(\mathcal{A}\),

\[A_1\cap\cdots\cap A_n\]

is nonempty.

Then in particular \(\emptyset\not\in \mathcal{A}\). Moreover, given a family \(\mathcal{A}\) satisfying this condition, we may add all finite intersections of elements of \(\mathcal{A}\) to obtain a filter base \(\mathcal{B}\) on \(X\). (§Other Definitions of Topological Spaces, ⁋Definition 5) For this reason \(\mathcal{A}\) is sometimes called a subbase of \(\uparrow \mathcal{B}\).

The following proposition gives an alternative characterization of compactness.

Proposition 11 A topological space \(X\) is compact if and only if for every family \(\mathcal{A}\) of closed sets satisfying the finite intersection property, \(\bigcap \mathcal{A}\neq\emptyset\).

Proof

It suffices to take complements.