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The proof that I have seen for the result "countably compact paracompact implies compact" involves metacompactness, which follows from paracompactness.

I wonder if it can be proved without going through metacompactness.

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up vote 7 down vote accepted

Perhaps the easiest argument uses the following lemmata.

Lemma 1. Let $\mathscr{A}$ be a locally finite family of sets in a space $X$; then $\mathscr{A}$ is closure-preserving, i.e., $\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$ is closed.

Proof. Let $F=\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$, and suppose that $x\in X\setminus F$. $\mathscr{A}$ is locally finite, so $x$ has an open nbhd $G$ such that $\mathscr{A}(x)=\{A\in\mathscr{A}:G\cap A\ne\varnothing\}$ is finite. Let $$V=G\setminus\bigcup\{\operatorname{cl}A:A\in\mathscr{A}(x)\}\;;$$ then $V$ is an open nbhd of $x$ disjoint from $F$, and $F$ is closed. $\dashv$

Lemma 2. If $X$ is countably compact, every locally finite family of non-empty subsets of $X$ is finite.

Proof. Suppose that there is an infinite locally finite family $\mathscr{A}$ of non-empty subsets of $X$. Any subcollection of $\mathscr{A}$ is still locally finite, so we may assume that $\mathscr{A}=\{A_n:n\in\omega\}$ is countably infinite. For $n\in\omega$ let $$F_n=\bigcup_{k\ge n}\operatorname{cl}A_k\;;$$ by Lemma 1 each $F_n$ is closed. Clearly each $F_n$ is non-empty, $F_0\supseteq F_1\supseteq F_2\supseteq\ldots\,$, and $\bigcap_{n\in\omega}F_n=\varnothing$. But then $\{X\setminus F_n:n\in\omega\}$ is an open cover of $X$ with no finite subcover, and $X$ is not countably compact. $\dashv$

Now the desired result is trivial: if $X$ is paracompact, every open cover of $X$ has a locally finite open refinement, and if $X$ is also countably compact, that locally finite open refinement must in fact be finite.

I should point out that the usual proof doesn’t really use metacompactness: it just uses the fact that a point-finite open cover has an irreducible subcover and the obvious fact that a locally finite open cover is point-finite. One could easily combine all of this into a direct proof.

Let $X$ be paracompact, $T_1$, and countably compact, and let $\mathscr{U}$ be an open cover of $X$. Let $\mathscr{V}$ be a locally finite open refinement of $\mathscr{U}$, and index $\mathscr{V}=\{V_\xi:\xi<\kappa\}$ for some cardinal $\kappa$. Let $\mathscr{W}_0=\varnothing$. Suppose that we’ve already chosen $W_\xi\in\mathscr{V}$ for $\xi<\eta$. Let $\mathscr{W}_\eta=\{W_\xi:\xi<\eta\}$. If $\mathscr{W}_\eta$ covers $X$, stop. Otherwise, $A_\eta=\{\xi<\kappa:V_\xi\nsubseteq\bigcup\mathscr{W}_\eta\}\ne\varnothing$, and we set $W_\eta=V_{\min A_\eta}$ and continue. Clearly this construction must stop at some $\mathscr{W}_\eta$.

By construction, for each $\xi<\eta$ there is a point $x_\xi\in W_\xi\setminus\bigcup\mathscr{W}_\xi$; let $D=\{x_\xi:\xi<\eta\}$. $\mathscr{W}_\eta$ is locally finite, so each $x\in X$ has an open nbhd $G_x$ such that $G_x\cap D$ is finite. $X$ is $T_1$, so if $x\notin D$, then $G_x\setminus D$ is an open nbhd of $x$ disjoint from $D$, and if $x\in D$, then $\{x\}\cup(G_x\setminus D)$ is an open nbhd of $x$ disjoint from $D\setminus\{x\}$. Thus, $D$ is closed and discrete in $X$, and since $X$ is countably compact, $D$ must be finite. But then $\mathscr{W}_\eta$ is a finite open refinement of $\mathscr{U}$, which therefore has a finite subcover, and $X$ is compact. $\dashv$

There’s no mention here of metacompactness, and no explicit mention of irreducible covers, but it’s really the same proof: I’ve just rolled all of the pieces into a single argument.

Yet, another argument uses the (non-trivial) result that if $X$ is a paracompact Hausdorff space, then every open cover of $X$ has a $\sigma$-discrete open refinement:

Let $\mathscr{U}$ be an open cover of $X$, and let $\mathscr{V}=\bigcup_{n\in\omega}\mathscr{V}_n$ be an open refinement of $\mathscr{U}$ such that each $\mathscr{V}_n$ is discrete. Since $X$ is countably compact, each $\mathscr{V}_n$ is finite, and $\mathscr{V}$ is therefore a countable open cover of $X$. But then $\mathscr{V}$ has a finite subcover, which is a finite open refinement of $\mathscr{U}$, so $\mathscr{U}$ has a finite subcover, and $X$ is compact. $\dashv$

That looks nice and short, but the characterization of paracompactness that it uses is harder than the first the argument that I gave.

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I was always amused by "lemmata", as in Hebrew it means "down" or "below". It always seemed very fitting because the lemmata are below the theorem, in some sense! – Asaf Karagila Jul 15 '12 at 22:05

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