Sequentially compact space

Question

Is every sequentially compact space metrisable? If not, then, can you give me an example of a sequentially compact space that is not compact.

Answer 1

Two classical examples: $X = [0,\omega_1)$ in the order topology, where $\omega_1$ is the first uncountable ordinal.

This is sequentially compact, because if $(\alpha_n)_n$ is a sequence in $X$, the set $\{ \alpha_n: n \in \mathbb{N} \}$ is bounded above by some ordinal $\beta < \omega_1$ (the union of the $\alpha_n$ is a countable ordinal, etc.). Then $[0,\beta]$ is compact metrisable and contains all $\alpha_n$ so also a convergent subsequence with its limit. A similar reasoning will show that $X$ is countably compact as well. But the cover $\{[0,\gamma): \gamma < \omega_1\}$ does not have a finite (or even countable) subcover.

Another: let $Y = \{ f \in [0,1]^{\mathbb{R}}: |\{x: f(x) \neq 0 \}| \le \aleph_0\}$, as a subspace of the product space $[0,1]^{\mathbb{R}}$. So these are all functions from $\mathbb{R}$ to $[0,1]$ that are $0$ except at an "exception set" that is at most countable.

Given some sequence $(f_n)$, we consider all countably many exception sets for all the $f_n$, and call their (countable!) union $E$. Then the sequence essentially "lives" in $\{0\}^{\mathbb{R} \setminus E} \times [0,1]^E$, which is compact metrisable again (as $E$ is countable) and we apply the same argument as before. But $Y$ is dense in $[0,1]^{\mathbb{R}}$ (which is compact) so cannot itself be compact (it would be closed and not dense, then).

Answer 2

HINT: In metrizable spaces sequential compactness is equivalent to compactness.