TITLE: Noncompact sequentially compact space
QUESTION [5 upvotes]: Have you an example of a noncompact sequentially compact space, without using ordinal?

REPLY [5 votes]: $\newcommand{\supp}{\operatorname{supp}}$Let $A$ be any uncountable index set, for each $\alpha\in A$ let $D_\alpha=\{0,1\}$ with the discrete topology, and let $$X=\left\{x\in\prod_{\alpha\in A}D_\alpha:|\supp(x)|\le\omega\right\}\;,$$ where $\supp(x)=\{\alpha\in A:x(\alpha)=1\}$, the support of $x$. (This is the $\Sigma$-product of the $D_\alpha$’s.)
For $\alpha\in A$ let $B_\alpha=\{x\in X:\alpha\notin\supp(x)\}$; clearly $B_\alpha$ is open in $X$. Since $A$ is uncountable, but each $x\in X$ has countable support, $\{B_\alpha:\alpha\in A\}$ is an open cover of $X$, but it obviously has no countable subcover. Thus, $X$ is not Lindelöf.
Now let $\langle x_n:n\in\omega\rangle$ be any sequence in $X$. Let $S=\bigcup_{n\in\omega}\supp(x_n)$; $S$ is countable, so $K\triangleq\prod_{\alpha\in S}D_\alpha$ is a Cantor set (or a finite discrete space). Let $\pi:X\to K$ be the obvious projection map. The sequence $\langle \pi(x_n):n\in\omega\rangle$ in the compact metrizable space $K$ has a subsequence $\langle \pi(x_{n(k)}):k\in\omega\rangle$ converging to some $p\in K$. Let $x$ be the unique point of $X$ that agrees with $p$ on $S$ and is $0$ on $A\setminus S$; clearly $\langle x_{n(k)}:k\in\omega\rangle$ converges to $x$ in $X$. Thus, $X$ is sequentially compact.