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Have you an example of a noncompact sequentially compact space, without using ordinal?

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Try searching this web site. – David Mitra Oct 5 '12 at 14:15
@DavidMitra: All the spaces it returns are fundamentally based on the uncountable ordinal $\omega_1$. – Nate Eldredge Oct 5 '12 at 14:22
up vote 4 down vote accepted

$\newcommand{\supp}{\operatorname{supp}}$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.

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Here is an idea which I have not completed: Take a non locally compact space, $X$, e.g. the rationals $\mathbb Q$, and take its one-point sequential compactification described in

R. Brown, Sequentially proper maps and a sequential compactification, J. London Math Soc. (2) 7 (1973) 515-522.

and written here as $X^+$, which is $X$ with an extra point $\omega$. For the topology on $X^+$, let $S(X)$ be the set of sequences in $X$ with no convergent subsequence. The intuitive idea is that if $s \in S(X)$ is a sequence in $X$ with no convergent subsequence, then $s$ should converge to the extra point $\omega$ of $X^+$.

For the topology on $X^+$: if $U$ is open in $X$ then $U$ is also open in $X^+$, and $U^+$, the union of $U$ with $\omega$, is open in $X^+$ if and only if every element $S(X)$ is eventually in $U$. (Try this out with $X$ the space of positive integers.)

I am not so clear, though, if this gives a non-compact space when applied to $\mathbb Q$. Any comments?

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What makes you think it wouldn't be first countable? – JSchlather Oct 5 '12 at 16:12
So? Actually, I have not studied the relation between this sequential compactification and the Alexandoff compactification. – Ronnie Brown Oct 5 '12 at 17:55
Well, second countable sequentially compact spaces are compact. So it seems like a good first check would be to see whether or not the new point $\omega$ has a countable neighborhood basis. – JSchlather Oct 5 '12 at 19:54
So it was probably not a good idea to start with a countable space such as $\mathbb Q$! – Ronnie Brown Oct 5 '12 at 21:14
Example 3.3 of the paper points out that $\Bbb Q^+$, being sequentially compact and countable, must be compact. – Brian M. Scott Oct 5 '12 at 21:34

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