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 “ 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 is compact, then for any space the projection is a closed map.
Let be a closed subset, and suppose that does not belong to . We want to find an open neighborhood of that does not intersect , or so that does not intersect . Consider the collection of all open for which there exists an open containing , such that does not intersect . Since , for any we have , and since is closed in the product topology, there exist containing and containing such that does not intersect . Therefore, covers , so it has a finite subcover . For each of the finitely many there is a corresponding such that does not intersect , and the intersection of the is a neighborhood of which does not intersect .
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 is closed iff is open in for every open in . Here is defined by the adjunction condition
for every .
The traditional formulation is that is closed in whenever is closed in , which is the same as that is closed in whenever is open in , i.e., is open in whenever is open in .
In the case of a projection map , this says
is open in whenever is open in .
Next, a slight reformulation of the concept of compactness. Recall that a collection of subsets of is directed if every finite subcollection has an upper bound. Then, a space is compact if every directed open cover of contains .
If is a closed map for every space , then is compact.
Let be a directed open cover of . Define a space as follows: the points of are open sets of (so the underlying set of is the topology ), and the open sets of are upward-closed subsets of such that is nonempty whenever is nonempty.
This is a topology.
Clearly such are closed under arbitrary unions. If and are open and and , then any upper bound of and in belongs to both and since these are upward-closed.
Moreover, whenever belongs to , the principal up-set is open in .
Now consider the set . Claim: this is open in . Proof: for every , there exists such that (because is a cover), and then for , the set is an open set which contains , and because for every , we have .
By the open-set reformulation of the closed map condition, the set
is open in , so this set is upward-closed and intersects , so that for some . But then is all of ! So for any directed open cover ; therefore is compact.
The proof of the theorem above was extracted from
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).