Compactness

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.

Now we examine the remaining results related to compactness and variations of this definition.

Tychonoff’s Theorem

An arbitrary product of compact spaces is again compact. If the product is finite, this result can be shown in a more intuitive manner, but if the product is infinite, the following lemma is needed. This generalizes §Compactness and Convergence of Filters, ⁋Proposition 5 to the language of filters.

Lemma 1 A topological space \(X\) is compact if and only if every ultrafilter converges.

Proof

First, assume that \(X\) is compact, and let an arbitrary ultrafilter \(\mathcal{F}\) be given. Suppose, for the sake of contradiction, that \(\mathcal{F}\) has no limit point. That is, for each \(x\in X\), there exists an open neighborhood \(U_x\) such that \(U_x\not\in \mathcal{F}\). Then by the compactness of \(X\), there exists a finite subcover \(U_{x_1},\ldots, U_{x_n}\) of \(X\).

On the other hand, by [Set Theory] §Filters and Ideals, Galois Correspondence, ⁋Proposition 5, \(\mathcal{F}\) is prime. That is, for any subset \(A\subseteq X\), exactly one of \(A\in \mathcal{F}\) or \(X\setminus A\in \mathcal{F}\) holds. Then for any \(A\in \mathcal{F}\),

\[A=A\cap X=(A\cap U_{x_1})\cup \cdots\cup (A\cap U_{x_n})\in \mathcal{F}\]

and since \(U_{x_i}\not\in \mathcal{F}\) by assumption, each \(A\cap U_{x_i}\) also does not belong to \(\mathcal{F}\); since \(\mathcal{F}\) is maximal, \(X\setminus (A\cap U_{x_i})\in \mathcal{F}\) must hold. Then the finite intersection

\[X\setminus A=(X\setminus (A\cap U_{x_1}))\cap\cdots\cap (X\setminus (A\cap U_{x_n}))\]

must also belong to \(\mathcal{F}\), which contradicts the maximality of \(\mathcal{F}\).

Conversely, suppose that every ultrafilter \(\mathcal{F}\) has a limit point, and let \(\mathcal{A}\) be a family of closed subsets of \(X\) satisfying the finite intersection property. Then we can consider an ultrafilter \(\mathcal{F}\) containing the filter generated by \(\mathcal{A}\); by assumption \(\mathcal{F}\) has a limit point \(x\). That is, \(\mathcal{N}(x)\subseteq \mathcal{F}\), and thus for every \(F\in \mathcal{F}\), there exists a neighborhood \(U\) of \(x\) such that \(U\cap F\neq\emptyset\). In particular, for any \(A\in \mathcal{A}\), there exists a neighborhood \(U\) of \(x\) such that \(A\cap U\neq\emptyset\), and therefore \(x\in \cl(A)=A\) always holds. From this we see that \(x\in\bigcap_{A\in \mathcal{A}}A\), and thus we obtain the desired result by §Compact Spaces, ⁋Proposition 11.

Then the following holds.

Theorem 2 (Tychonoff) The product \(X=\prod_{i\in I} X_i\) of compact spaces \((X_i)_{i\in I}\) is compact. Conversely, if the product space \(X\) is compact, then each \(X_i\) is compact.

Proof

If \(X\) is compact, then the fact that each \(X_i\) is compact follows immediately from the continuity of \(\pr_i\) and §Compact Spaces, ⁋Proposition 8.

For the converse, given any ultrafilter \(\mathcal{F}\) on \(X\), one checks that \(\pr_i(\mathcal{F})\) defines an ultrafilter base on \(X_i\); from the assumption that \(X_i\) is compact and Lemma 1, we obtain a limit point \(x_i\) of this ultrafilter. Since one can show that \(x=(x_i)_{i\in I}\) is a limit point of \(\mathcal{F}\), the proof is completed by Lemma 1.

Locally Compact Spaces

Compactness of a space \(X\) is such a strong property that many spaces we encounter are not compact. For example, even the real numbers \(\mathbb{R}\) are not compact, as one sees by considering the open cover consisting of intervals \((n,n+2)\). Therefore, it is necessary to weaken this condition somewhat.

Definition 3 A topological space \(X\) is locally compact at a point \(x\in X\) if there exists a compact neighborhood \(A\) of \(x\) in \(X\). If \(X\) is locally compact at every point, it is called a locally compact space.

Then by an argument similar to §Compact Spaces, ⁋Lemma 6, one can show that any locally compact Hausdorff space is regular. The following theorem gives important information about locally compact Hausdorff spaces.

Theorem 4 (Alexandroff) Let \(X\) be a locally compact Hausdorff space. Then there exist a compact Hausdorff space \(X'\), a point \(\ast_{X'}\in X'\), and a homeomorphism \(f: X \rightarrow X'\setminus\{\ast_{X'}\}\). Moreover, if another datum \(g: X \rightarrow X''\setminus\{\ast_{X''}\}\) satisfying this condition is given, there exists a unique homeomorphism \(h: X' \rightarrow X''\) such that \(g=h\circ f\).

To make the proof more intuitive, let us state the claim in a slightly less rigorous way: given a locally compact Hausdorff space, one can add a single point to make it a compact Hausdorff space, and this way of adding a point is unique. The new space obtained in this way is called the one-point compactification of \(X\).

Proof of Theorem 4

First, we show uniqueness. Naturally, one defines \(h\) by sending \(f(x)\) to \(g(x)\) on \(X'\setminus\{\ast_{X'}\}\) and \(\ast_{X'}\) to \(\ast_{X''}\). Then \(h\) is continuous. To prove this, let \(V\) be an arbitrary open subset of \(X''\). If \(V\) does not contain \(\ast_{X''}\), then by definition

\[h^{-1}(V)=f(g^{-1}(V))\]

and since \(f\) is a homeomorphism, \(h^{-1}(V)\) is an open set in \(X'\setminus\{\ast_{X'}\}\). On the other hand, since \(X'\setminus\{\ast_{X'}\}\) is an open set because \(X'\) is Hausdorff, by §Subspaces, ⁋Lemma 2, \(h^{-1}(V)\) is an open set in \(X'\).

If \(V\) contains \(\ast_{X''}\), then \(X''\setminus V\) is a closed subset of \(X''\), hence compact, and thus also a compact subset of \(f(X)\). Since \(g\) is a homeomorphism, \(g^{-1}(X''\setminus V)\) is a compact subspace of \(X\), and therefore the set

\[h^{-1}(X''\setminus V)=f(g^{-1}(X''\setminus V))\]

is a compact subspace of \(X'\), hence closed. (§Compact Spaces, ⁋Corollary 5) Therefore, by the third condition of §Continuous Functions, ⁋Theorem 4, we know that \(h\) is continuous.

From the above proof, if a topological space \(X'=X\cup \{\ast_{X'}\}\) satisfying the given condition exists, then the open sets of \(X'\) must be of the following two types:

\[f(U)\quad\text{for $U$ open in $X$},\qquad X'\setminus f(C)\quad\text{for $C$ compact in $X$}\]

Then one can easily verify that this indeed satisfies the conditions of §Open Sets, ⁋Definition 1. It remains only to show that the topology given in this way is compact Hausdorff.

First, we show that \(X'\) is compact. Given an arbitrary open covering \((U_i)_{i\in I}\) of \(X'\), this family must contain an open set \(U_j\) containing \(\ast_{X'}\); by the topology defined above, this can be written as \(U_j=X'\setminus C\) for some compact subset \(C\) of \(X\). On the other hand, consider the collection of open sets \((f^{-1}(U_i))_{i\neq j}\) of \(X\). Since \((U_i)_{i\neq j}\) must cover \(f(C)\), the family \((f^{-1}(U_i))_{i\neq j}\) is an open covering of \(C\), and from the assumption that \(C\) is compact, we can choose a finite subcover and thereby obtain a finite subcover of the open covering \((U_i)_{i\in I}\) of \(X'\).

Now, to show that \(X'\) is Hausdorff, let \(x,y\in X'\) be any two points. If both belong to \(f(X)\), there is nothing to prove, so we may assume \(y=\ast_{X'}\). Then there exists a compact neighborhood \(C\) of \(x\) in \(X\) and an open neighborhood \(U\) of \(x\) contained in \(C\), so \(f(U)\) and \(X'\setminus f(C)\) are open sets separating \(x\) and \(y\).

Topological Manifolds

Now let us weaken the finiteness demanded by compactness somewhat. For example, one might consider the situation where every open covering of a topological space \(X\) has a locally finite open subcover. (§Interior, Closure, and Boundary, ⁋Definition 3) However, by a slight play on words, one can show that this condition is in fact equivalent to the compactness of \(X\), so this alone does not yield a new definition. Now, for two open covers \((U_i)_{i\in I}\) and \((V_j)_{j\in J}\) of a topological space, if for every \(V_j\) there exists a \(U_i\) that entirely contains it, we call \((V_j)_{j\in J}\) an (open) refinement of \((U_i)_{i\in I}\).

Definition 5 A topological space \(X\) is paracompact if every open covering of \(X\) has a locally finite open refinement.

If \(X\) is compact, then a finite subcover of any open covering is immediately a locally finite open refinement, so every compact space is paracompact.

Almost all spaces we deal with are Hausdorff. Although we will not prove it separately, one of the important properties characterizing a paracompact Hausdorff space is the existence of a partition of unity. (Theorem 7)

Definition 6 Let a topological space \(X\) be given. Then a set of continuous functions

\[\Phi=\{\phi:X \rightarrow [0,1]\mid \text{$\phi$ continuous}\}\]

is called a partition of unity if the following two conditions hold:

1. For any \(x\in X\), there exists an open neighborhood \(U\) of \(x\) such that only finitely many \(\phi\in \Phi\) satisfy \(\phi\vert_U\neq 0\).
2. For any \(x\in X\), \(\sum_{\phi\in \Phi} \phi(x)=1\).

In particular, for an open covering \((U_i)_{i\in I}\) of \(X\), a partition of unity \(\Phi=(\phi_i)_{i\in I}\) satisfying \(\supp \phi_i\subseteq U_i\) is called a partition of unity subordinate to \((U_i)\).

Thus the following holds.

Theorem 7 A topological space \(X\) is a paracompact Hausdorff space if and only if for every open cover \((U_i)\) of \(X\), there exists a partition of unity subordinate to \((U_i)\).

The proof of this can be found on Wikipedia. Instead of copying it verbatim, let us briefly examine how it is used. To do so, we first define a topological manifold. The model for a topological manifold is the familiar space, namely Euclidean space \(\mathbb{R}^m\).

Definition 8 A topological space \(M\) is locally Euclidean of dimension \(m\) if for every \(x\in M\), there exists an open neighborhood \(U\) of \(x\) such that \(U\) is homeomorphic to an open subset of \(\mathbb{R}^m\).

This condition is the essence of a topological manifold, but it is weaker than one might expect, so we add some further conditions.

Definition 9 A space that is second countable, Hausdorff, and locally Euclidean of dimension \(m\) is called a topological manifold of dimension \(m\).

Then there is another theorem we mention without proof.

Theorem 10 For any Hausdorff locally Euclidean topological space \(M\), \(M\) is second-countable if and only if \(M\) is paracompact and has countably many connected components. (§Connected Spaces, ⁋Definition 7)

We have not yet defined what a connected component is, but in any case, the above theorem logically implies that every topological manifold \(M\) is paracompact Hausdorff, and therefore for every open cover \((U_i)\) of a topological manifold \(M\), there exists a partition of unity subordinate to \((U_i)\). In particular, using the locally Euclidean condition, we can regard the open cover \((U_i)\) as consisting of pieces of Euclidean space; a continuous function to \(\mathbb{R}\) defined on such a piece is an object we have known since learning calculus. Then by the formula

\[f=\sum \phi_i f_i\]

we can define a continuous function \(f\) on all of \(M\), and conversely, if a continuous function \(f\) on all of \(M\) is given, each \(\phi_i f\) is a continuous function on \(U_i\). In most cases, there is no particular problem in thinking of a topological manifold as having only one connected component, so paracompactness can in fact be considered an essential condition when dealing with manifolds.