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

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 Dieudonné, 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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