Countable compactness implies sequential compactness in sequential Hausdorff spaces or Fréchet-Urysohn spaces

Question

I am trying to understand the relationship between sequential compactness and countable compactness, but am somewhat confused and not finding answers to my questions online. Would you help me clarify?

Let $X$ be a topological space. We know sequential compactness always implies countable compactness, since otherwise if $\{U_n\}$ was a countable open cover with no finite subcover, define $(x_n)$ such that $x_n \notin \cup_{k=1}^n U_n$. Since $(x_n)$ has a convergent subsequence it follows some $U_n$ must contain infinitely many terms. But this is a contradiction.

In the reverse direction, Wikipedia says if the space is sequential and Hausdorff, the two types of compactness are equivalent. But I couldn't prove this or find a proof.

I was able to prove if $X$ is Fréchet-Urysohn and countably compact then it is sequentially compact. Here is my proof.

Suppose $\exists (x_n)$ with no convergent subsequence. Then $\forall m \in \mathbb{N}, \text{Cl} \{x_n\}_{n \geq m} =\text{SeqCl} \{x_n\}_{n \geq m} = \{x_n\}_{n \geq m}$. So $\{x_n\}_{n \geq m}$ is closed. Now note $U_m = X\backslash \{x_n\}_{n=m}^\infty$ is a countable open cover of $X$ such that $U_1 \subset U_2 \subset \cdots $. By countable compactness $\exists U_N$ which covers $X$. A contradiction.

This made me wonder if sequential Hausdorfness implies a space is Fréchet-Urysohn, and Wikipedia's statement would follow, but I couldn't find a proof of this.

Answer 1

Yes, it is well-known that

Let $X$ be a sequential space. Then $X$ is countably compact iff $X$ is sequentially compact.

To be clear about definitions: $X$ is countably compact iff every countable open cover of $X$ has a finite subbcover. I'll use the convenient equivalence that $X$ is countably compact iff every infinite subset $A$ of $X$ has an $\omega$-accumulation point $p \in X$, i.e. every neighbourhood $U$ of $p$ has $U \cap A$ infinite. I wrote the proof of that equivalence (also free of separation axiom assumptions) here, in case this was unknown to you.

And $X$ is sequentially compact iff every sequence $(x_n)_n$ in $X$ has a convergent subsequence.

$X$ is sequential if for all subsets $A$ of $X$: $A$ sequentially closed iff $A$ is closed.

Proof (following the standard reference Engelking, General Topology 2nd ed. Theorem 3.10.31 (due to Franklin (1965), generalising this from metric spaces where it had been shown by Hausdorff in 1914). I'll be extra pedantic in my version of its proof to make sure I make no hidden separation axiom assumptions (Engelking includes Hausdorff in his definition of countable compactness and sequential compactness, so there are subtle instances in his proofs where he might use them, which I want to avoid for maximal generality)

If $X$ is sequentially compact, let $A$ be an infinite subset of $X$. Let $(a_n)_n$ be sequence of points from $A$ so that $a_n \neq a_m$ whenever $n \neq m$ (i.e. find an injection $\Bbb N \to A$). By sequential compactness, there is some $p \in X$ and some subsequence $(a_{n_k})_k$ of $(a_n)_n$ such that $a_{n_k} \to p$ as $k \to \infty$. Then if $U$ is any neighbourhood of $p$, there exists some $N$ so that for all $k \ge N$ we have $a_{n_k} \in U$. It follows that $\{a_{n_k} \mid k \ge N\} \subseteq U \cap A$ and so $U \cap A$ is infinite (by the injectivity). Hence $p$ is an $\omega$-limit point of $A$ and by the mentioned equivalence, $X$ is countably compact. This shows one implication.

Let $X$ be countably compact. Let $(x_n)_n$ be a sequence in $X$. We want to show it has a convergent subsequence.

We can assume WLOG that $n \neq m \to x_n \neq x_m$ (if you believe this read on, if you want my argument for it, reveal spoiler)

For $x \in X$ we can define $N_x=\{n \in \Bbb N\mid x_n = x\}$. If some $N_x$ is infinite, then $N_x$ defines a subsequence of $(x_n)_n$ that is constant with value $x$ and in any space this converges to $x$, and we'd be done. So assume all $N_x$ are empty (probably most of them are) or finite; picking one index from each non-empty one (yes I believe in AC) we get a subsequence as claimed, and as a subsequence of a subsequence is still a subsequence, we only have to consider that situation. We resume the proof under this injectivity assumption which is a handy technicality.

Define $A = \{x_n\mid n \in \Bbb N\}$ which is an infinite set. As $X$ is countably compact, $A$ has an $\omega$-accumulation point $p$. It's clear that the $A\setminus \{p\}$ is not closed, as $p$ is in its closure but not in the set. Because $X$ is sequential (finally we use it!), $A\setminus \{p\}$ is not sequentially closed, i.e. there is a sequence $(y_n)_n$ in $A \setminus \{p\}$ and some point $q \notin (A\setminus \{p\})$ so that $y_n \to q$.

By re-ordering this sequence we find a subsequence of $(x_n)_n$ converging to $q$. (details follow as before, skip if you see it already).

First note as all $y_n$ come from $A\setminus \{p\}$ and the sequence is injective, for every $n$ there is a unique $r(n) \in \Bbb N$ so that $y_n = x_{r(n)}$. Another folklore fact: if $h: \Bbb N \to \Bbb N$ is a bijection then $(y_{h(n)})_n \to q (n \to \infty)$ as well (any re-ordering of a convergent sequence has the same limit): let $U$ be any neighbourhood of $q$. There is an $N_1 \in \Bbb N$ so that $n > N_1$ implies $y_n \in U$. Then $h^{-1}[\{1,\ldots, N_1\}] \subseteq \{1,\ldots, N\}$ for some $N \in \Bbb N$, as $h^{-1}$ preserve finiteness. Then $h(n) > N$ implies $n > N_1$ so $q \in U$, and as $U$ was arbitrary, $y_{h(n)} \to q$ as $n \to \infty$. Then enumerate $r[\Bbb N]$ as $n_1< n_2< n_3 <n_4< \ldots$ and note that $x_{n_k} = (y_{r^{-1}(k)})$ converges to $q$ (as $k \to \infty$) as a re-ordering of $(y_n)_n$. So we have the desired convergent subsequence.

QED.

Answer 2

Your proof idea works with minor modification even if $X$ is just sequential and $T_1$, rather than Frechet-Urysohn. Indeed, if $(x_n)$ has no convergent subsequence, this means that no sequence in the set $\{x_n\}$ that takes infinitely many values can have a limit, since such a sequence would then have a subsequence which is also a subsequence of $(x_n)$. If $X$ is $T_1$ (so a convergent sequence that takes only finitely many values can only have one of those values as its limit), this implies that for any $m$, the set $\{x_n\}_{n\geq m}$ is sequentially closed and thus closed since $X$ is sequential, and then the rest of your proof goes through.

(With a bit more work, the $T_1$ assumption can in fact be dropped: instead of considering the sets $\{x_n\}_{n\geq m}$, consider $\bigcup_{n\geq m}\overline{\{x_n\}}$, and again the assumption that $(x_n)$ has no convergent subsequence can be used to show these sets are all sequentially closed, and also that their intersection as $m\to\infty$ is empty.)