Todd Trimble
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.

Compactness and stable closure

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


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).

(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: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.

Now, let us reformulate the concept of compactness slightly. 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.

Claim: this is a topology. Proof: 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 and UUU' \subseteq U, 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

Compactness and filters

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 XX be compact and let FF be a filter on (the underlying set of) XX, and suppose FF has no cluster point. Consider the collection of all open sets which have empty intersection with some element of FF. Because each point xx is not a cluster point, there exists a neighborhood of xx belonging to this collection, and therefore the collection is a cover. Let U 1,,U nU_1, \ldots, U_n be a finite subcover, and let C 1,,C nC_1, \ldots, C_n be elements of the filter such that U iC i=U_i \cap C_i = \emptyset. Then C 1C nC_1 \cap \ldots \cap C_n intersects each of the U iU_i in the empty set and is therefore empty since the U iU_i cover, contradiction.

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

Proof of Theorem 1 via filters

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

Now let DD be the closure in X×YX \times Y of the diagonal subset

Δ={(x,x):xX}X×X dX×Y\Delta = \{(x, x): x \in X\} \hookrightarrow X \times X_d \hookrightarrow X \times Y

Then, by hypothesis, π(D)\pi(D) is closed in YY, but it also clearly contains X dX_d so it is dense. Hence π(D)=Y\pi(D) = Y, and therefore there exists xXx \in X such that (x,ω)D(x, \omega) \in D. This means that every neighborhood U×OU \times O of (x,ω)(x, \omega) intersects the diagonal Δ\Delta, but this intersection can be identified with UOU \cap O. Hence UOU \cap O is nonempty for every neighborhood UU of xx and element OO of the filter FF, which is to say that xx is a cluster point of FF. Hence every filter FF has a cluster point xx, and therefore XX is compact.

Third proof

The following very direct proof was transcribed from email from Tom Leinster.

…OK, here’s what I found. Let XX be a space such that the projection X×YYX \times Y \to Y is closed for all YY. Let (V i)(V_i) be a family of closed sets in XX with the finite intersection property. Define YY to be the disjoint union of XX with a single point, \infty. Let BP(Y)\mathbf{B} \subseteq P(Y) be the collection of all subsets of X together with all sets of the form V iV_i \cup {\infty}. Take the topology on YY generated by B\mathbf{B}. Apparently the closedness of the projection implies that the family (V i)(V_i) has nonempty intersection.

Here’s what my notes say as to why. Take the diagonal D={(x,x):xX}X×YD = \{(x, x): x \in X\} \subseteq X \times Y. Consider Cl(D)Cl(D), the closure of DD in X×YX \times Y. The projection p(Cl(D))Yp(Cl(D)) \subseteq Y is closed and contains XX; but {}\{\infty\} is not open in YY by the finite intersection property, so p(Cl(D))=Yp(Cl(D)) = Y. Hence (x,)Cl(D)(x, \infty) \in Cl(D) for some xXx \in X. Thus, for each open neighbourhood UU of xx in XX, and each ii, the set U×(V i)U \times (V_i \cup {\infty}) meets DD, i.e. UU meets V iV_i. It follows that xx is in V iV_i for each ii (otherwise we could take U=XV iU = X \setminus V_i). Hence the intersection of all the V iV_i‘s contains xx, and is in particular nonempty.

Tychonoff’s theorem

In this section we transcribe a proof of Tychonoff’s theorem (due to Clementino and Tholen, who work in rather greater generality), which exploits the closed-projection formulation of compactness. Obviously now we assume the axiom of choice.


Let (X i) i:I(X_i)_{i: I} be a family of spaces. Then for a point xx and subset AA of i:IX i\prod_{i: I} X_i, we have xCl(A)x \in Cl(A) if, for every finite FIF \subseteq I, we have π F(x)Cl(π F(A))\pi_F(x) \in Cl(\pi_F(A)) under the projection operator π F: i:IX i i:FX i\pi_F: \prod_{i: I} X_i \to \prod_{i: F} X_i.


Suppose xCl(A)x \notin Cl(A), so that there is an open neighborhood UU of xx such that UA=U \cap A = \emptyset. Sets of the form π i 1(U i)\pi_i^{-1}(U_i) for open U iX iU_i \subseteq X_i for a subbasis of the product topology, so there is a basis element π F 1(V)\pi_F^{-1}(V) (for some V=U 1××U nV = U_1 \times \ldots \times U_n) that is a neighborhood of xx and contained in UU, so we have Aπ F 1(V)=A \cap \pi_F^{-1}(V) = \emptyset or A¬π F 1(V)=π F 1(¬V)A \subseteq \neg \pi_F^{-1}(V) = \pi_F^{-1}(\neg V) or what is the same, π F(A)¬V\pi_F(A) \subseteq \neg V or π F(A)V=\pi_F(A) \cap V = \emptyset. Since VV is a neighborhood of π F(x)\pi_F(x) not intersecting AA, we conclude π F(x)Cl(π F(A))\pi_F(x) \notin Cl(\pi_F(A)).

Assuming this lemma, suppose we have a family (X α) α<κ(X_\alpha)_{\alpha \lt \kappa} of compact spaces indexed by an ordinal κ\kappa, and let YY be any space. We must show that the projection

π κ:Y× α<κX αY\pi^\kappa: Y \times \prod_{\alpha \lt \kappa} X_\alpha \to Y

is closed. We do this by induction on κ\kappa. The case κ=0\kappa = 0 is trivial.

It will be convenient to introduce some notation. For γκ\gamma \leq \kappa, let X γX^\gamma denote the product Y× α<γX αY \times \prod_{\alpha \lt \gamma} X_\alpha (so X 0=YX^0 = Y in this notation), and for βγ\beta \leq \gamma let π β γ:X γX β\pi_\beta^\gamma: X^\gamma \to X^\beta be the obvious projection map. Let KX κK \subseteq X^\kappa be closed, and put K βCl(π β κ(K))K_\beta \coloneqq Cl(\pi_{\beta}^\kappa(K)). In particular K κ=KK_\kappa = K since KK is closed, and we are done if we show π 0 κ(K)=K 0\pi_0^\kappa(K) = K_0.

Assume as inductive hypothesis that starting with any x 0K 0x_0 \in K_0 there is x βK βx_\beta \in K_\beta such that whenever β<γ<κ\beta \lt \gamma \lt \kappa, the compatibility condition π β γ(x γ)=x β\pi_\beta^\gamma(x_\gamma) = x_\beta holds. In particular, π 0 β(x β)=x 0\pi_0^\beta(x_\beta) = x_0 for all β<κ\beta \lt \kappa, and we are now trying to extend this up to κ\kappa.

If κ=β+1\kappa = \beta + 1 is a successor cardinal, then the projection

π β κ:X β×X βX β\pi_\beta^\kappa: X^\beta \times X_\beta \to X^\beta

is a closed map since X βX_\beta is compact. Thus π β κ(K)=Cl(π β κ(K))=K β\pi_\beta^\kappa(K) = Cl(\pi_\beta^\kappa(K)) = K_\beta since KK is closed, so there exists x κKx_\kappa \in K with π β κ(x κ)=x β\pi_\beta^\kappa(x_\kappa) = x_\beta, and then

π 0 κ(x κ)=π 0 βπ β κ(x κ)=π 0 β(x β)=x 0\pi_0^\kappa(x_\kappa) = \pi_0^\beta \pi_\beta^\kappa (x_\kappa) = \pi_0^\beta(x_\beta) = x_0

as desired.

If κ\kappa is a limit ordinal, then we may regard X κX^\kappa as the inverse limit of spaces (X β) β<κ(X^\beta)_{\beta \lt \kappa} with the obvious transition maps π β γ\pi_\beta^\gamma between them. Hence the tuple (x β) β<κ(x_\beta)_{\beta \lt \kappa} defines an element x κx_\kappa of X κX^\kappa, and all that remains is to check that x κKx_\kappa \in K. But since KK is closed, the lemma indicates it is sufficient to check that for every finite set FF of ordinals below κ\kappa, that π F(x κ)Cl(π F(K))\pi_F(x_\kappa) \in Cl(\pi_F(K)) (as a subspace of αFX α\prod_{\alpha \in F} X_\alpha). But for every such FF there is some β<κ\beta \lt \kappa that dominates all the elements of FF. One then checks

π F(x κ)=π F βπ β κ(x κ)=π F β(x β)π F β(K β)=π F β(Cl(π β κ(K)))Cl(π F βπ β κ(K))=Cl(π F(K))\pi_F(x_\kappa) = \pi_F^\beta \pi_\beta^\kappa(x_\kappa) = \pi_F^\beta(x_\beta) \in \pi_F^\beta(K_\beta) = \pi_F^\beta(Cl(\pi_\beta^\kappa(K))) \subseteq Cl(\pi_F^\beta \pi_\beta^\kappa(K)) = Cl(\pi_F(K))

where the inclusion indicated as \subseteq just results from continuity of π F β\pi_F^\beta. This completes the proof.


Revised on May 7, 2017 10:31:35 by Todd Trimble