Does sequential compactness imply countable compactness?

Question

Let $X$ be a topological space which is sequentially compact. Does this imply that $X$ is countably compact?

Thank you!

Answer 1

The answer is yes. I'll put it in a context, to explain Stefan's comment.

First I'll define some terms, to avoid confusion: a point $x$ is an accumulation point (also called a cluster point) of the sequence $(x_n)_n$ in $X$ iff every (open) neighbourhood $O$ of $x$ contains infinitely many terms of the sequence (e.g. the set $\{n \in \omega: x_n \in O \} $ is infinite).

A point $x$ is an $\omega$-accumulation point of the set $A \subset X$ iff for every (open) neighbourhood $O$ of $x$ the set $O \cap A$ is infinite (note that this implies that $A$ is already infinite).

The following conditions on a topological space $X$ are equivalent:

1. Every countable open cover of $X$ has a finite subcover.
2. Every infinite subset $A$ of $X$ has an $\omega$-accumulation point.
3. Every sequence in $A$ has an accumulation point.

Note that 1. is the definition of $X$ being countably compact.

Suppose 1. holds and $A$ is an infinite subset of $X$ without an $\omega$-accumulation point. We can assume (w.l.o.g.) that $A$ is countable (or just use any countable subset of it). By assumption, every $x$ in $X$ has an open neighbourhood $O_x$ such that $O_x \cap A$ is finite (possibly empty), as every $x$ is not an $\omega$-accumulation point. For every finite subset $F$ of $A$ define $O_F = \cup \{O_x: O_x \cap A = F \}$. Then every $O_x$ is a subset of one of the $O_F$, so these subsets cover $X$, and clearly all $O_F$ intersect $A$ in a finite subset (namely $F$), so finitely many of them cannot even cover $A$, let alone $X$. So this contradicts 1., and so $A$ does have an $\omega$-accumulation point.

Suppose 2. holds, and let $(x_n)_n$ be a sequence in $X$. If there is any $k$ such that $\{n \in \omega: x_n = x_k \}$ is infinite (so an infinitely recurring point), then $x_k$ is clearly an accumulation point of the sequence. So assume all $x_k$ occur at most finitely many times. In particular the set $A = \{x_n: n \in \omega\}$ is infinite and so has an $\omega$-accumulation point $x$. This $x$ is then an accumulation point of the sequence, as is easily checked.

Suppose 3. holds, and suppose $O_n, n \in \omega$ is a countable open cover without a finite subcover. Then we pick points $x_n \in X$ that are not in $\bigcup_{i=1}^n O_i$. Now if $x$ were an accumulation point of $(x_n)_n$, $x$ is in some $O_k$, but then this is a neighbourhood of $x$ that misses all $x_n$ with $n > k$, so $x$ cannot be an accumulation point, contradiction. So such a cover cannot exist.

Ok, this being shown, let's assume $X$ is sequentially compact. Let $(x_n)$ be a sequence in $X$. Then by assumption, there are $n_1 < n_2 < n_3 < \dots$ in $\omega$ such that $(x_{n_k})_k$ converges to some point $x \in X$. Let $O$ be any neighbourhood of $x$, then there exists a $K$ such that $O$ contains all $x_{n_k}$ where $k > K$. In particular, the set $\{n \in \omega: x_n \in O \}$ contains all such $x_{n_k}$ and is infinite, so $x$ is an accumulation point of $(x_n)$. So $X$ is countably compact by the above.

Note that there is also a notion of weakly countably compact, or limit point compact, which states that every infinite subset of $X$ has a limit point $x$, i.e. a point $x$ such that every (open) neighbourhood of $x$ intersects $A \setminus \{x\}$.

Clearly countably compact implies limit point compact (as an $\omega$-accumulation point of $A$ is surely a limit point of $A$), and the reverse holds for $T_1$ spaces (using that finite subsets of $T_1$ are closed). A space like $\mathbb{R} \times \{0,1\}$, the last space in the indiscrete topology (the so-called double pointed reals) is a limit point compact space that is not countably compact.

And "large" compact spaces like $[0,1]^{[0,1]}$ and the Čech-Stone compactification of the integers ($\beta\omega$) are compact (so countably compact, trivially) spaces that are not sequentially compact (and are also Hausdorff, hence normal).

As is well-known, for metrisable spaces $X$, all these notions coincide (with compactness).