Today I want to discuss an equivalent and seemingly weaker condition of paracompactness due to Ernest Michael (1953, Proc. of the AMS).

Theorem 1 (Michael) Suppose {X} is regular and every open cover {\mathfrak{A}} of {X} has a refinement that can be decomposed into a countable collection of locally finite {\mathfrak{A}_i} collections of open sets. Then {X} is paracompact.

Note that the {\mathfrak{A}_i}‘s are not required to be covers, only locally finite! This is a significant strengthening of the usual definition of paracompactness.

Following Michael’s original paper, I shall discuss the proof of this result. First, I shall give an auxiliary result, of independent result, that yields yet another variation on the theme of paracompactness: we don’t have to require the locally finite refinements to be open.

1. Reduction to arbitrary refinements

Lemma 2 Suppose {X} is a regular space. Suppose every open cover of {X} has a locally finite refining cover (which refinement is not required to be an open cover). Then {X} is paracompact.

We start by proving this. We will first show that every open cover of {X} has a locally finite refining closed cover. Let {\mathfrak{A} = \left\{U_{\alpha}\right\}} be an open cover of {X}. By regularity, we can find a refinement {\left\{V_{\beta}\right\}} such that each closure {\overline{V_{\beta}}} is contained in some {U_{\alpha}}.

Of course, this refinement by itself need not be locally finite or anything reasonable. However, we can apply the condition of the lemma to find a locally finite refining cover {\left\{G_{\gamma}\right\}} of {\left\{V_{\beta}\right\}}. The closures {\left\{\overline{G_{\gamma}}\right\} } are locally finite (this is true for any locally finite cover) and refine the {\overline{V_\beta}}, hence refine the {U_{\alpha}}. These form the cover in question.

So we have shown that under the hypotheses of the lemma, we can always refine each open cover of {X} to a locally finite, closed cover. Now, the lemma will follow from the next result.

Lemma 3 Suppose {X} is a Hausdorff space. Suppose every open cover of {X} has a closed, locally finite refinement. Then {X} is paracompact.

Let {\mathfrak{A} = \left\{U_{\alpha}\right\}} be an open cover of {X}. We shall find an open, locally finite refinement. This will prove paracompactness.

First, choose a locally finite closed refinement {\left\{F_{\beta}\right\}} of {\mathfrak{A}}. There is a cover {\left\{W_{\gamma}\right\}} of {X} consisting of open sets such that each {W_{\gamma}} intersects only finitely many of the {F_{\beta}}. This is precisely the definition of local finiteness. By assumption, we can find a locally finite closed refinement {\left\{G_{\kappa}\right\}} of the {\left\{W_{\gamma}\right\}}.

Then define

\displaystyle  V_{\beta} = X - \bigcup_{G_{\kappa} \cap F_{\beta} = \emptyset} G_{\kappa}.

This is an open set because the {G_{\kappa}} are locally finite. Moreover, {V_{\beta} \supset F_\beta}, so the {V_{\beta}} are an open cover of {X}.

I claim that the {V_{\beta}} are locally finite. Indeed, {V_\beta} intersects {G_\kappa} if and only if {G_\kappa \cap F_\beta \neq \emptyset}. (This means, for instance, that {G_\kappa} isn’t excised in the definition of {V_\beta}.) However, {G_\kappa} is contained in a {W_\gamma}, so it intersects only finitely many {F_\beta}. Thus only finitely many {V_\beta} intersect each {G_\kappa}, so the {V_\beta} are locally finite (because the {G_\kappa} are and cover {X}).

However, the {V_\beta} are not necessarily a refinement of {\mathfrak{A}}. So, to each {F_\beta}, choose some {U_\alpha} containing it (since the {F_\beta} refine {U_\alpha}), and intersect {V_\beta} with this {U_\alpha}. This replacement shrinks the {V_\beta} so the new collection is locally finite, but it is still a cover since the {F_\beta}‘s are still subsets. Thus we have found a locally finite refinement. In particular, the lemma is now proved.

2. Proof of Michael’s theorem

We shall now prove Michael’s theorem. Suppose that {X} is regular and every open cover has a refinement that can be decomposed into a countable collection of locally finite families. We will prove then that any open cover as a locally finite refinement by a not necessarily open cover, which will establish the result by the lemmas.

Pick an open cover {\mathfrak{A}} and decompose a refinement of it, by the hypothesis, into a countable collection {\bigcup \mathfrak{A}_i}, where each {\mathfrak{A}_i} is a locally finite collection (not necessarily a cover) of open sets. Write {U_i = \bigcup \mathfrak{A}_i} and

\displaystyle  F_i = U_i - \bigcup_{j<i} U_j.

Then the {F_i} form a cover of {X}, and they are locally finite—indeed, each {F_i} can only intersect the (finitely many) {U_j} with {j \leq i}. Consider the intersections

\displaystyle  F_i \cap U, \ U \in \mathfrak{A}_i.

These form a locally finite (non-open) cover of {F_i}. Call this cover {\mathfrak{B}_i}. The union {\bigcup \mathfrak{B}_i} is a cover of {X} that refines {\mathfrak{A}} (because it refines a refinement of {\mathfrak{A}}). Moreover, it is locally finite on each open set {U_i} because only the finitely many {\mathfrak{B}_j} with {j\leq i} factor in, and each of these is locally finite.

In particular, the {\bigcup \mathfrak{B}_i} is a locally finite (non-open) refinement of {\mathfrak{A}}. The previous lemma now implies that {X} is paracompact. This completes the proof of Michael’s theorem.

3. Applications

First of all, we can rederive the result of Dieudonn√©, which I discussed earlier, that a locally compact, {\sigma}-compact space is paracompact. Indeed, note that a locally compact space is regular. Moreover, if {X = \bigcup K_i} for the {K_i} compact, we can decompose each open cover {\mathfrak{A}} of {X} into a countable collection {\mathfrak{A}_i} of finite open covers of {K_i}. So these are thus locally finite, and we can apply Michael’s theorem to show that {X} is paracompact.

Corollary 4 A second-countable, regular space is paracompact.

Indeed, a second-countable space satisfies the condition that every open cover has a countable subcover, which obviously satisfies the conditions of Michael’s theorem.

Next time, I will discuss an application of Michael’s theorem and the associated lemmas to a result of Stone that a metric space is paracompact.

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