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.
Best Answer
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.