A topological space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. But I'm interested in a stronger condition. Suppose that every countable cover of $X$ by Borel sets has a finite subcover. Then what properties must X have?

And what are examples of spaces that do and don't have this property?

up vote 9 down vote accepted

This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.

Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel sets. Let us call a set $A\subseteq X$ large if $A$ is Borel and has infinitely many Borel subsets. Note that if $A$ is large and $A$ is the disjoint union of two Borel sets $B$ and $C$, then at least one of $B$ and $C$ is large, since every Borel subset of $A$ is the union of a Borel subset of $B$ and a Borel subset of $C$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).

We can now use this to construct a countable Borel cover of $X$ with no finite subcover. Starting with $A_0=X$, we can pick a large proper subset $A_1\subset A_0$, and then a large proper subset $A_2\subset A_1$, and so on. The sets $A_0\setminus A_1, A_1\setminus A_2,\dots$ and $\bigcap_n A_n$ are then all Borel and cover $X$, and have no finite subcover (they are disjoint and all except possibly $\bigcap_n A_n$ are nonempty).

(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.