Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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

2 Answers 2

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.

share|improve this answer

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?

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.