Contents

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:

Proposition

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

Proof

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 $\pi(C)$.

(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:

Proposition

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 $B \subseteq Y$.

Proof

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 $X \times Y$.

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

Theorem

If $\pi: X \times Y \to Y$ is a closed map for every space $Y$, then $X$ is compact.

Proof

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.

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

Reference

The proof of the theorem above was extracted from

• Martin Escardo, Intersections of compactly many open sets are open, 2009 (pdf)

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 $X$ be compact and let $F$ be a filter on (the underlying set of) $X$, and suppose $F$ has no cluster point. Consider the collection of all open sets which have empty intersection with some element of $F$. Because each point $x$ is not a cluster point, there exists a neighborhood of $x$ belonging to this collection, and therefore the collection is a cover. Let $U_1, \ldots, U_n$ be a finite subcover, and let $C_1, \ldots, C_n$ be elements of the filter such that $U_i \cap C_i = \emptyset$. Then $C_1 \cap \ldots \cap C_n$ intersects each of the $U_i$ in the empty set and is therefore empty since the $U_i$ cover, contradiction.

In the other direction, suppose every filter $F$ has a cluster point, and let $\Sigma$ be a directed open cover of $X$. We show that $X$ belongs to $\Sigma$. If not, then the dual collection $\Sigma^\perp$ of closed complements generates a filter, and if $x$ is a cluster point of the filter, then every neighborhood of $x$ intersects every closed set of the collection. Then $x$ belongs to the closure of the intersection $\bigcap \Sigma^\perp$, which is already closed (being an intersection of closed sets), so $\bigcap \Sigma^\perp$ is nonempty, which is to say that $\Sigma$ 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 $\pi: X \times Y \to Y$ is a closed map for every space $Y$, and suppose $F$ is a filter on $X$. Construct a space $Y$ as follows: the underlying set is the disjoint union of $X$ and an ideal point $\omega$ (which we thinking of as a formal point that $F$ accumulates to), and the topology is defined by saying that every set not containing $\omega$ is open, and sets of the form $O \cup \{\omega\}$ are open if $O$ belongs to $F$. Clearly the complement of $\omega$ is a discrete space $X_d$, and every neighborhood of $\omega$ intersects $X_d$ so $X_d$ is dense in $Y$.

Now let $D$ be the closure in $X \times Y$ of the diagonal subset

$\Delta = \{(x, x): x \in X\} \hookrightarrow X \times X_d \hookrightarrow X \times Y$

Then, by hypothesis, $\pi(D)$ is closed in $Y$, but it also clearly contains $X_d$ so it is dense. Hence $\pi(D) = Y$, and therefore there exists $x \in X$ such that $(x, \omega) \in D$. This means that every neighborhood $U \times O$ of $(x, \omega)$ intersects the diagonal $\Delta$, but this intersection can be identified with $U \cap O$. Hence $U \cap O$ is nonempty for every neighborhood $U$ of $x$ and element $O$ of the filter $F$, which is to say that $x$ is a cluster point of $F$. Hence every filter $F$ has a cluster point $x$, and therefore $X$ is compact.

Third proof

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

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

Here’s what my notes say as to why. Take the diagonal $D = \{(x, x): x \in X\} \subseteq X \times Y$. Consider $Cl(D)$, the closure of $D$ in $X \times Y$. The projection $p(Cl(D)) \subseteq Y$ is closed and contains $X$; but $\{\infty\}$ is not open in $Y$ by the finite intersection property, so $p(Cl(D)) = Y$. Hence $(x, \infty) \in Cl(D)$ for some $x \in X$. Thus, for each open neighbourhood $U$ of $x$ in $X$, and each $i$, the set $U \times (V_i \cup {\infty})$ meets $D$, i.e. $U$ meets $V_i$. It follows that $x$ is in $V_i$ for each $i$ (otherwise we could take $U = X \setminus V_i$). Hence the intersection of all the $V_i$‘s contains $x$, 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.

Lemma

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

Proof

Suppose $x \notin Cl(A)$, so that there is an open neighborhood $U$ of $x$ such that $U \cap A = \emptyset$. Sets of the form $\pi_i^{-1}(U_i)$ for open $U_i \subseteq X_i$ for a subbasis of the product topology, so there is a basis element $\pi_F^{-1}(V)$ (for some $V = U_1 \times \ldots \times U_n$) that is a neighborhood of $x$ and contained in $U$, so we have $A \cap \pi_F^{-1}(V) = \emptyset$ or $A \subseteq \neg \pi_F^{-1}(V) = \pi_F^{-1}(\neg V)$ or what is the same, $\pi_F(A) \subseteq \neg V$ or $\pi_F(A) \cap V = \emptyset$. Since $V$ is a neighborhood of $\pi_F(x)$ not intersecting $A$, we conclude $\pi_F(x) \notin Cl(\pi_F(A))$.

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

$\pi^\kappa: Y \times \prod_{\alpha \lt \kappa} X_\alpha \to Y$

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

It will be convenient to introduce some notation. For $\gamma \leq \kappa$, let $X^\gamma$ denote the product $Y \times \prod_{\alpha \lt \gamma} X_\alpha$ (so $X^0 = Y$ in this notation), and for $\beta \leq \gamma$ let $\pi_\beta^\gamma: X^\gamma \to X^\beta$ be the obvious projection map. Let $K \subseteq X^\kappa$ be closed, and put $K_\beta \coloneqq Cl(\pi_{\beta}^\kappa(K))$. In particular $K_\kappa = K$ since $K$ is closed, and we are done if we show $\pi_0^\kappa(K) = K_0$.

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

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

$\pi_\beta^\kappa: X^\beta \times X_\beta \to X^\beta$

is a closed map since $X_\beta$ is compact. Thus $\pi_\beta^\kappa(K) = Cl(\pi_\beta^\kappa(K)) = K_\beta$ since $K$ is closed, so there exists $x_\kappa \in K$ with $\pi_\beta^\kappa(x_\kappa) = x_\beta$, and then

$\pi_0^\kappa(x_\kappa) = \pi_0^\beta \pi_\beta^\kappa (x_\kappa) = \pi_0^\beta(x_\beta) = x_0$

as desired.

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

$\pi_F(x_\kappa) = \pi_F^\beta \pi_\beta^\kappa(x_\kappa) = \pi_F^\beta(x_\beta) \in \pi_F^\beta(K_\beta) = \pi_F^\beta(Cl(\pi_\beta^\kappa(K))) \subseteq Cl(\pi_F^\beta \pi_\beta^\kappa(K)) = Cl(\pi_F(K))$

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

Reference

• N. Bourbaki, General Topology, Part I, Hermann, 1966. See especially 10.2, lemma 1, page 101.
Revised on May 7, 2017 at 10:31:35 by Todd Trimble