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 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 .
(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 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 .
Now, let us reformulate the concept of compactness slightly. 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.
Claim: this is a topology. Proof: 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 and , 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
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 be compact and let be a filter on (the underlying set of) , and suppose has no cluster point. Consider the collection of all open sets which have empty intersection with some element of . Because each point is not a cluster point, there exists a neighborhood of belonging to this collection, and therefore the collection is a cover. Let be a finite subcover, and let be elements of the filter such that . Then intersects each of the in the empty set and is therefore empty since the cover, contradiction.
In the other direction, suppose every filter has a cluster point, and let be a directed open cover of . We show that belongs to . If not, then the dual collection of closed complements generates a filter, and if is a cluster point of the filter, then every neighborhood of intersects every closed set of the collection. Then belongs to the closure of the intersection , which is already closed (being an intersection of closed sets), so is nonempty, which is to say that 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 is a closed map for every space , and suppose is a filter on . Construct a space as follows: the underlying set is the disjoint union of and an ideal point (which we thinking of as a formal point that accumulates to), and the topology is defined by saying that every set not containing is open, and sets of the form are open if belongs to . Clearly the complement of is a discrete space , and every neighborhood of intersects so is dense in .
Now let be the closure in of the diagonal subset
Then, by hypothesis, is closed in , but it also clearly contains so it is dense. Hence , and therefore there exists such that . This means that every neighborhood of intersects the diagonal , but this intersection can be identified with . Hence is nonempty for every neighborhood of and element of the filter , which is to say that is a cluster point of . Hence every filter has a cluster point , and therefore is compact.
The following very direct proof was transcribed from email from Tom Leinster.
…OK, here’s what I found. Let be a space such that the projection is closed for all . Let be a family of closed sets in with the finite intersection property. Define to be the disjoint union of with a single point, . Let be the collection of all subsets of X together with all sets of the form . Take the topology on generated by . Apparently the closedness of the projection implies that the family has nonempty intersection.
Here’s what my notes say as to why. Take the diagonal . Consider , the closure of in . The projection is closed and contains ; but is not open in by the finite intersection property, so . Hence for some . Thus, for each open neighbourhood of in , and each , the set meets , i.e. meets . It follows that is in for each (otherwise we could take ). Hence the intersection of all the ‘s contains , and is in particular nonempty.