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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.

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2 Answers 2

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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.

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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.)

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

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