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.

How to prove that every pseudocompact, paralindelöf Tychonoff space is compact?

  • pseudocompact = every continuous real-valued function is bounded.
  • paralindelöf = every open cover has a locally countable open refinement.

Thanks for any help:)

share|improve this question
1  
I'd try playing around here austinmohr.com/home/?page_id=146 –  Byron Schmuland Jun 27 '12 at 1:31
    
I can't open it. –  Paul Jun 27 '12 at 1:36
    
I have no problem with the site. Unfortunately, paralindelof is not an option, however they suggest the cocountable topology (en.wikipedia.org/wiki/Cocountable_topology) as a non-compact, pseudo compact, Lindelof space. Does that help? –  Byron Schmuland Jun 27 '12 at 1:42
    
It's not Tychonoff:) –  Paul Jun 27 '12 at 1:57
    
Sorry I only read the title not the body of your problem. –  Byron Schmuland Jun 27 '12 at 1:59

1 Answer 1

up vote 6 down vote accepted

Let $X$ be a para-Lindelöf, pseudocompact Tikhonov space. To show that $X$ is compact, it suffices to show that $X$ is Lindelöf, since a Lindelöf Tikhonov space is normal, a normal pseudocompact space is countable compact, and a countably compact Lindelöf space is compact.

Let $\mathscr{U}$ be an open cover of $X$, and let $\mathscr{R}$ be a locally countable open refinement of $\mathscr{U}$; I’ll show that $\mathscr{R}$ is countable, from which it follows immediately that $\mathscr{U}$ has a countable subcover.

Let $\mathscr{V}$ be an open cover of $X$ such that each $V\in\mathscr{V}$ meets only countably many members of $\mathscr{R}$, and let $\mathscr{W}$ be a locally countable open refinement of $\mathscr{V}$. Note that since $\mathscr{W}$ refines $\mathscr{V}$, each $W\in\mathscr{W}$ meets only countably many members of $\mathscr{R}$.

Suppose that $\langle G_n:n\in\omega\rangle$ is an infinite sequence of non-empty open sets such that $$G_n\subseteq X\setminus\bigcup_{k<n}\operatorname{st}(G_k,\mathscr{W})\tag{1}$$ for each $n\in\omega$. (Here $\operatorname{st}(V_k,\mathscr{W})=\bigcup\{W\in\mathscr{W}:W\cap V_k\ne\varnothing\}$.) For $n\in\omega$ let $$H_n=\bigcup_{k\ge n}G_k\;,$$ and let $\mathscr{H}=\{H_n:n\in\omega\}$. $\mathscr{H}$ is an open filterbase in a pseudocompact Tikhonov space, so it clusters at some $x\in X$. Pick any $W\in\mathscr{W}$ such that $x\in W$; $x\in\bigcap_{n\in\omega}\operatorname{cl}H_n$, so $W\cap H_n\ne\varnothing$ for each $n\in\omega$. This implies that there are $m,n\in\omega$ such that $m<n$ and $W\cap G_m\ne\varnothing\ne W\cap G_n$, contradicting $(1)$.

It follows that any attempt to construct such a sequence recursively must halt after only finitely many $G_n$ have been chosen. Thus, there is a finite sequence $\langle G_0,\dots,G_m\rangle$ of open sets such that $(1)$ holds for $n<m$, and $$\operatorname{cl}\bigcup_{k\le m}\operatorname{st}(G_k,\mathscr{W})=X\;.$$ Moreover, since $\mathscr{W}$ is locally countable, we may at each stage of the construction choose $G_n$ so that it meets only countably many members of $\mathscr{W}$. Then $$\mathscr{W}_0\triangleq\{W\in\mathscr{W}:W\cap\bigcup_{k\le m}G_k\}$$ is countable, and $\operatorname{cl}\bigcup\mathscr{W}_0=X$, so $R\cap\bigcup\mathscr{W}_0\ne\varnothing$ for every non-empty $R\in\mathscr{R}$. On the other hand, $\mathscr{W}_0$ is countable, and every member of $\mathscr{W}_0$ meets only countably many members of $\mathscr{R}$, so $\mathscr{R}$ must be countable.

This is a result of Dennis Burke and Sheldon Davis. Note that para-Lindelöf cannot be weakened to meta-Lindelöf: a bit earlier I constructed an example (assuming CH) of a meta-Lindelöf, pseudocompact Tikhonov space that is not compact.

Dennis K. Burke and Sheldon Davis, Pseudocompact paralindelöf spaces are compact, Abstracts Amer. Math. Soc. 3 (1982), 213.

Brian M. Scott, Pseudocompact, metacompact spaces are compact, Top. Procs. 4 (1979), 577-587. link

share|improve this answer
    
Congratulations on your gold specialist badge in general topology. –  Martin Sleziak Jun 30 '12 at 6:35
    
Another realted paper: W.Stephen Watson: A pseudocompact meta-lindeöf space which is not compact, Topology and its Applications 20 (1985) 237-243; DOI: 10.1016/0166-8641(85)90091-4. We settle the remaining question of whether pseudocompact is compatible with meta-Lindelöf by constructing, in ZFC, a pseudocompact meta-Lindelöf space which is not compact. –  Martin Sleziak Jun 30 '12 at 7:12

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.