This page is dedicated to proving the equivalence of two notions of compactness, one being the classical open-cover formulation, and the other a “stable closure” condition (the property that “$X \to 1$ is a closed map” is stable under pullback).

One direction is a very classical and straightforward fact, proved in every textbook on general topology.

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 $\pi(C)$.

The converse statement requires more ingenuity to prove. A preliminary observation is that the proof above was a bit convoluted because it was phrased throughout in terms of complements of closed sets; this suggests it would be convenient to 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

$f^{-1}(B) \subseteq U \qquad iff \qquad B \subseteq \forall_f U$

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 $X \times Y$.

Next, a slight reformulation of the concept of compactness. Recall that 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.

This is a topology.

Clearly such $W$ are closed under arbitrary unions. If $W$ and $W'$ are open and $U \in \Sigma \cap W$ and $U' \in \Sigma \cap W'$, then any upper bound of $U$ and $U'$ in $\Sigma$ belongs to both $W$ and $W'$ since these are upward-closed.

Moreover, whenever $U$ belongs to $\Sigma$, the principal up-set $prin(U) = \{V \in \mathcal{O}(X): U \subseteq V$ 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 $U' \in \Sigma$ such that $x \in U'$ (because $\Sigma$ is a cover), and then for $U'' = U \cap U'$, the set $U'' \times prin(U'')$ is an open set which contains $(x, U)$, and $U \times prin(U) \subseteq E$ because for every $(y, V) \in U'' \times prin(U'')$, 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

- Martín Escardó, Intersections of compactly many open sets are open, 2009 (pdf)

See also this exchange at Math Overflow, where the question was raised as to whether the axiom of choice (or possibly a weaker choice principle like the ultrafilter theorem) is required to prove the equivalence of these two notions of compactness (examination of the proofs above show it is not).

Last revised on June 20, 2019 at 16:10:39. See the history of this page for a list of all contributions to it.