I need some help to clarify something.
I understand that if $X$ is a Hausdorff space but not metric, compact and sequentially compact are not equivalent. This means that there can be sequences without convergent subsequences. Is this correct?
Then I read here that
If $T$ is a compact space, then any infinite subset of $T$ has at least one limit point.
Is this true in any compact space of any separation class? Does not this imply sequential compactness? If we view any sequence as the set formed by the union of all its terms, then if it is infinite it has a limit point. Doesn't this imply that there is a subsequence converging to this point? At least in a Hausdorff space?
Best Answer
A compact Hausdorff space need not be sequentially compact, there is an example here (near the end of the page), and a sequentially compact space need not be compact (e.g. the first uncountable ordinal $\omega_1$ in the order topology, which is first countable, hereditarily normal, sequentially compact, countably compact, but not compact). Of course a compact space is trivially countably compact: if all open covers have finite subcovers, all countable ones certainly do.
A sequence $(x_n)$ (say of all distinct terms) in a countably compact space $X$ can be seen as an infinite set, and so it has an $\omega$-accumulation point (every neighbourhood of $p$ contains infinitely many terms of the sequence). See my answer here.
But this does not guarantee that there is a subsequence that converges to $p$ (see the proof in the link I gave), because there can be a lot of neighbourhoods of $p$, and we need every neighbourhood of $p$ to contain a tail of the subsequence.
If however $p$ has a countable local base $O_n$ (like in metric spaces, where we can use the balls around of $p$ of radius $\frac{1}{n}$), then we can find such a subsequence: pick $n_1$ with $x_{n_1} \in O_1$, $n_2 > n_1$ with $x_{n_2} \in O_1 \cap O_2$, $n_3 > n_2$ such that $x_{n_3} \in O_1 \cap O_2 \cap O_3$ and so on. This uses that all these (finite!) intersections are neighbourhoods of $p$ and thus contain infinitely many terms of the sequence. And the fact that these $O_n$ form a local base, ensures that the subsequence $x_{n_k}$ converges to $p$ as $k$ tends to infinity: if $O$ is open and contains $p$, for some $m$, $O_n \subset O$, and so all $x_{n_k}$ with $k \ge m$ are by construction in $O_m$ and hence in $O$.
So in first countable (countably) compact spaces we do have sequential compactness, and many spaces are indeed first countable (like metrisable spaces), so most people's intuition is more tuned to that case. But in general we do not have that compact spaces are sequentially compact.