# Every pseudocompact, paralindelöf, Tychonoff space is compact?

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:)

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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

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

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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