Do compactness and sequential compactness coincide in countable spaces?

It's well-known that compactness (every open cover has a finite subcover) and sequential compactness (every sequence has a convergent subsequence) coincide in metric spaces, and that neither one implies the other in general topological spaces.

Given that both imply countable compactness (every countable open cover has a finite subcover), it is natural to ask if the notions coincide in countable spaces, i.e., can we prove the following?

Let $$X$$ be a countable topological space. Then $$X$$ is compact if and only if $$X$$ is sequentially compact.

I'm particularly interested in this because countable spaces are rather prevalent on pi-base, and it would be good to auto-populate the sequential compactness property for them.

One direction, already known to pi-base, follows quickly from the fact that if $$X$$ is sequentially compact, then $$X$$ is countably compact. Since $$X$$, as a countable space, is also Lindelöf, $$X$$ is compact.

Conversely, if $$X$$ fails to be sequentially compact, and $$X$$ is countable, then let $$X=\{x_n\mid n\in \mathbb N\}$$, and let $$(y_i)$$ be a sequence with no convergent subsequence.

We construct an open cover $$\mathcal U=\{U_n\mid n\in \mathbb N\}$$ as follows. For any set $$S$$ in which $$(y_i)$$ leaves "frequently" (i.e., infinitely often), denote by $$(y_i)^S$$ the subsequence taken from those terms lying outside of $$S$$. Note that $$(y_i)^S$$ can have no convergent subsequence either, and that $$(y_i)^{S\cup T}=[(y_i)^S]^T$$

Since $$(y_i)$$ has no convergent subsequence, there is some open neighborhood $$U_1$$ of $$x_1$$ so that $$(y_i)$$ frequently leaves $$U_1$$. Since $$(y_i)^{U_1}$$ has no convergent subsequence, there is an open neighborhood $$U_2$$ of $$x_2$$ which $$(y_i)^{U_1}$$ frequently leaves. Continuing in this fashion, having picked $$U_1,\dots, U_k$$ so that $$(y_i)$$ frequently leaves $$U_1\cup U_2\cup\dots\cup U_k$$, we choose some open neighborhood $$U_{k+1}$$ of $$x_{k+1}$$ so that $$(y_i)^{U_1\cup U_2\cup \dots\cup U_k}$$ frequently leaves $$U_{k+1}$$.

Then by construction, $$\mathcal U$$ is an open cover with no finite subcover, so $$X$$ is not compact.

Here is a rather different approach to proving that a countable compact space $$X$$ is sequentially compact which I find enlightening. Let $$(y_n)$$ be a sequence in $$X$$ and for each $$x\in X$$, let $$C_x$$ be the set of nonprincipal ultrafilters on $$\mathbb{N}$$ with respect to which $$(y_n)$$ converges to $$x$$. Each $$C_x$$ is closed in $$\beta\mathbb{N}\setminus\mathbb{N}$$ (it is just the set of ultrafilters that contain $$\{n:y_n\in U\}$$ for each neighborhood of $$U$$ of $$x$$), and they cover all of $$\beta\mathbb{N}\setminus\mathbb{N}$$ since $$X$$ is compact. Since $$X$$ is countable, the Baire category theorem now implies that some $$C_x$$ must have nonempty interior in $$\beta\mathbb{N}\setminus\mathbb{N}$$. That means there is some infinite $$A\subseteq\mathbb{N}$$ such that $$(y_n)$$ converges to $$x$$ with respect to every ultrafilter that contains $$A$$, which just means that the subsequence $$(y_n)_{n\in A}$$ converges to $$x$$.

(If you unravel this proof, it is actually basically the same as the one in M W's answer! The framework given by spaces of ultrafilters makes it more intuitive to me, though. Maybe I am weird in that respect.)

• Magnificent! Thank you very much!
– M W
Commented Jan 25 at 20:40