Characterizations of compactness

This page is mostly concerned with conditions necessary and sufficient for compactness, where the proof of equivalence can be carried out in Zermelo set theory.

As a warm-up, let us prove a simple and very classical fact:

If $X$ is compact, then for any space $Y$ the projection $\pi :X\times Y\to Y$ is a closed map.

Let $C\subseteq X\times Y$ be a closed subset, and suppose that $y$ does not belong to $\pi (C)$. We want to find an open neighborhood of $y$ that does not intersect $\pi (C)$, or so that $X\times V$ does not intersect $C$. Consider the collection $\mathcal{C}$ of all open $U\subseteq X$ for which there exists an open $V\subseteq Y$ containing $y$, such that $U\times V$ does not intersect $C$. Since $y\notin \pi (C)$, for any $x\in X$ we have $(x,y)\notin C$, and since $C$ is closed in the product topology, there exist $V$ containing $y$ and $U$ containing $x$ such that $U\times V$ does not intersect $C$. Therefore, $\mathcal{C}$ covers $X$, so it has a finite subcover ${U}_{i}$. For each of the finitely many $i$ there is a corresponding ${V}_{i}$ such that ${U}_{i}\times {V}_{i}$ does not intersect $C$, and the intersection of the ${V}_{i}$ is a neighborhood of $y$ which does not intersect $\pi (C)$.

(Thanks to Mike Shulman for suggesting a proof which avoids the axiom of choice; we have adapted his proof here.)

In any event, let’s consider how to prove the converse of this statement. As a preliminary, observe that the proof above was a bit convoluted because it was phrased throughout in terms of complements of closed sets; this suggests that we reformulate the condition of being a closed map directly in terms of *open* sets:

A map $f:X\to Y$ is closed iff ${\forall}_{f}U$ is open in $Y$ for every open $U$ in $X$. Here ${\forall}_{f}U$ is defined by the adjunction condition

$${f}^{-1}(B)\subseteq U\phantom{\rule{2em}{0ex}}\mathrm{iff}\phantom{\rule{2em}{0ex}}B\subseteq {\forall}_{f}U$$

for every $B\subseteq Y$.

The traditional formulation is that ${\exists}_{f}C$ is closed in $Y$ whenever $C$ is closed in $X$, which is the same as that ${\exists}_{f}\neg U=\neg {\forall}_{f}U$ is closed in $Y$ whenever $U$ is open in $X$, i.e., ${\forall}_{f}U$ is open in $Y$ whenever $U$ is open in $X$.

In the case of a projection map $f=\pi :X\times Y\to Y$, this says

$$\{y\in Y:X\times \{y\}\subseteq U\}$$

is open in $Y$ whenever $U$ is open in $X\times Y$.

Now, let us reformulate the concept of compactness slightly. A collection of subsets of $X$ is *directed* if every finite subcollection has an upper bound. Then, a space $X$ is compact if every directed open cover $\Sigma $ of $X$ contains $X$.

If $\pi :X\times Y\to Y$ is a closed map for every space $Y$, then $X$ is compact.

Let $\Sigma $ be a directed open cover of $X$. Define a space $Y$ as follows: the points of $Y$ are open sets of $X$ (so the underlying set of $Y$ is the topology $\mathcal{O}(X)$), and the open sets of $Y$ are upward-closed subsets $W$ of $\mathcal{O}(X)$ such that $\Sigma \cap W$ is nonempty whenever $W$ is nonempty.

Claim: this is a topology. Proof: Clearly such $W$ are closed under arbitrary unions. If $W$ and $W\prime $ are open and $U\in \Sigma \cap W$ and $U\prime \in \Sigma \cap W\prime $, then any upper bound of $U$ and $U\prime $ in $\Sigma $ belongs to both $W$ and $W\prime $ since these are upward-closed.

Moreover, whenever $U$ belongs to $\Sigma $ and $U\prime \subseteq U$, the principal up-set $\mathrm{prin}(U\prime )=\{V\in \mathcal{O}(X):U\subseteq V$ is open in $Y$.

Now consider the set $E=\{(x,U)\in X\times Y:x\in U\}$. Claim: this is open in $X\times Y$. Proof: for every $(x,U)\in E$, there exists $U\prime \in \Sigma $ such that $x\in U\prime $ (because $\Sigma $ is a cover), and then for $U\u2033=U\cap U\prime $, the set $U\u2033\times \mathrm{prin}(U\u2033)$ is an open set which contains $(x,U)$, and $U\u2033\times \mathrm{prin}(U\u2033)\subseteq E$ because for every $(y,V)\in U\u2033\times \mathrm{prin}(U\u2033)$, we have $y\in V$.

By the open-set reformulation of the closed map condition, the set

$$\{V\in Y:X\times \{V\}\subseteq E\}$$

is open in $Y$, so this set is upward-closed and intersects $\Sigma $, so that $X\times \{V\}\subseteq E$ for some $V\in \Sigma $. But then $V$ is all of $X$! So $X\in \Sigma $ for any directed open cover $\Sigma $; therefore $X$ is compact.

The proof of the theorem above was extracted from

- Martin Escardo, Intersections of compactly many open sets are open, 2009 (pdf)

A second way of characterizing compactness is

A space is compact if every filter has a convergent refinement.

Put slightly differently, every filter has a cluster point: a point such that every neighborhood intersects every element of the filter in a nonempty set. Then the desired refinement is the join of the neighborhood filter and the given filter within the poset of filters, viz., the filter generated by binary intersections of elements coming from each of the filters.

In one direction, let $X$ be compact and let $F$ be a filter on (the underlying set of) $X$, and suppose $F$ has no cluster point. Consider the collection of all open sets which have empty intersection with some element of $F$. Because each point $x$ is not a cluster point, there exists a neighborhood of $x$ belonging to this collection, and therefore the collection is a cover. Let ${U}_{1},\dots ,{U}_{n}$ be a finite subcover, and let ${C}_{1},\dots ,{C}_{n}$ be elements of the filter such that ${U}_{i}\cap {C}_{i}=\varnothing $. Then ${C}_{1}\cap \dots \cap {C}_{n}$ intersects each of the ${U}_{i}$ in the empty set and is therefore empty since the ${U}_{i}$ cover, contradiction.

In the other direction, suppose every filter $F$ has a cluster point, and let $\Sigma $ be a directed open cover of $X$. We show that $X$ belongs to $\Sigma $. If not, then the dual collection ${\Sigma}^{\perp}$ of closed complements generates a filter, and if $x$ is a cluster point of the filter, then every neighborhood of $x$ intersects every closed set of the collection. Then $x$ belongs to the closure of the intersection $\bigcap {\Sigma}^{\perp}$, which is already closed (being an intersection of closed sets), so $\bigcap {\Sigma}^{\perp}$ is nonempty, which is to say that $\Sigma $ is not a cover, contradiction.

The following proof of theorem 1 appears in Bourbaki’s book on topology. It devolves on the equivalent characterization of compactness as saying that every filter has a cluster point.

Suppose $\pi :X\times Y\to Y$ is a closed map for every space $Y$, and suppose $F$ is a filter on $X$. Construct a space $Y$ as follows: the underlying set is the disjoint union of $X$ and an ideal point $\omega $ (which we thinking of as a formal point that $F$ accumulates to), and the topology is defined by saying that every set not containing $\omega $ is open, and sets of the form $O\cup \{\omega \}$ are open if $O$ belongs to $F$. Clearly the complement of $\omega $ is a discrete space ${X}_{d}$, and every neighborhood of $\omega $ intersects ${X}_{d}$ so ${X}_{d}$ is dense in $Y$.

Now let $D$ be the closure in $X\times Y$ of the diagonal subset

$$\Delta =\{(x,x):x\in X\}\hookrightarrow X\times {X}_{d}\hookrightarrow X\times Y$$

Then, by hypothesis, $\pi (D)$ is closed in $Y$, but it also clearly contains ${X}_{d}$ so it is dense. Hence $\pi (D)=Y$, and therefore there exists $x\in X$ such that $(x,\omega )\in D$. This means that every neighborhood $U\times O$ of $(x,\omega )$ intersects the diagonal $\Delta $, but this intersection can be identified with $U\cap O$. Hence $U\cap O$ is nonempty for every neighborhood $U$ of $x$ and element $O$ of the filter $F$, which is to say that $x$ is a cluster point of $F$. Hence every filter $F$ has a cluster point $x$, and therefore $X$ is compact.

- N. Bourbaki,
*General Topology*, Part I, Hermann, 1966. See especially 10.2, lemma 1, page 101.

Revised on October 17, 2010 21:38:36
by
Todd Trimble