compactness and stable closure

Compactness and stable closure

Compactness and stable closure


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 “X1X \to 1 is a closed map” is stable under pullback).

Compactness implies stable closure

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


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


Let CX×YC \subseteq X \times Y be a closed subset, and suppose that yy does not belong to π(C)\pi(C). We want to find an open neighborhood of yy that does not intersect π(C)\pi(C), or so that X×VX \times V does not intersect CC. Consider the collection 𝒞\mathcal{C} of all open UXU \subseteq X for which there exists an open VYV \subseteq Y containing yy, such that U×VU \times V does not intersect CC. Since yπ(C)y \notin \pi(C), for any xXx \in X we have (x,y)C(x,y) \notin C, and since CC is closed in the product topology, there exist VV containing yy and UU containing xx such that U×VU \times V does not intersect CC. Therefore, 𝒞\mathcal{C} covers XX, so it has a finite subcover U iU_i. For each of the finitely many ii there is a corresponding V iV_i such that U i×V iU_i \times V_i does not intersect CC, and the intersection of the V iV_i is a neighborhood of yy which does not intersect π(C)\pi(C).

Stable closure implies compactness

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:XYf: X \to Y is closed iff fU\forall_f U is open in YY for every open UU in XX. Here fU\forall_f U is defined by the adjunction condition

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

for every BYB \subseteq Y.


The traditional formulation is that fC\exists_f C is closed in YY whenever CC is closed in XX, which is the same as that f¬U=¬ fU\exists_f \neg U = \neg \forall_f U is closed in YY whenever UU is open in XX, i.e., fU\forall_f U is open in YY whenever UU is open in XX.

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

{yY:X×{y}U}\{y \in Y: X \times \{y\} \subseteq U\}

is open in YY whenever UU is open in X×YX \times Y.

Next, a slight reformulation of the concept of compactness. Recall that a collection of subsets of XX is directed if every finite subcollection has an upper bound. Then, a space XX is compact if every directed open cover Σ\Sigma of XX contains XX.


If π:X×YY\pi: X \times Y \to Y is a closed map for every space YY, then XX is compact.


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


This is a topology.

Proof of the claim

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

Moreover, whenever UU belongs to Σ\Sigma, the principal up-set prin(U)={V𝒪(X):UVprin(U) = \{V \in \mathcal{O}(X): U \subseteq V is open in YY.

Now consider the set E={(x,U)X×Y:xU}E = \{(x, U) \in X \times Y: x \in U\}. Claim: this is open in X×YX \times Y. Proof: for every (x,U)E(x, U) \in E, there exists UΣU' \in \Sigma such that xUx \in U' (because Σ\Sigma is a cover), and then for U=UUU'' = U \cap U', the set U×prin(U)U'' \times prin(U'') is an open set which contains (x,U)(x, U), and U×prin(U)EU \times prin(U) \subseteq E because for every (y,V)U×prin(U)(y, V) \in U'' \times prin(U''), we have yVy \in V.

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

{VY:X×{V}E}\{V \in Y: X \times \{V\} \subseteq E\}

is open in YY, so this set is upward-closed and intersects Σ\Sigma, so that X×{V}EX \times \{V\} \subseteq E for some VΣV \in \Sigma. But then VV is all of XX! So XΣX \in \Sigma for any directed open cover Σ\Sigma; therefore XX 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).

Last revised on June 20, 2019 at 12:10:39. See the history of this page for a list of all contributions to it.