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 is regular and every open cover of has a refinement that can be decomposed into acountablecollection of locally finite collections of open sets. Then is paracompact.

Note that the ‘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 2Suppose is a regular space. Suppose every open cover of has a locally finite refining cover (which refinement is not required to be an open cover). Then is paracompact.

We start by proving this. We will first show that every open cover of has a locally finite refining *closed* cover. Let be an open cover of . By regularity, we can find a refinement such that each closure is contained in some .

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 of . The closures are locally finite (this is true for any locally finite cover) and refine the , hence refine the . 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 to a locally finite, closed cover. Now, the lemma will follow from the next result.

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

Let be an open cover of . We shall find an *open*, locally finite refinement. This will prove paracompactness.

First, choose a locally finite closed refinement of . There is a cover of consisting of open sets such that each intersects only finitely many of the . This is precisely the definition of local finiteness. By assumption, we can find a locally finite closed refinement of the .

Then define

This is an open set because the are locally finite. Moreover, , so the are an open cover of .

I claim that the are locally finite. Indeed, intersects if and only if . (This means, for instance, that isn’t excised in the definition of .) However, is contained in a , so it intersects only finitely many . Thus only finitely many intersect each , so the are locally finite (because the are and cover ).

However, the are not necessarily a refinement of . So, to each , choose some containing it (since the refine ), and intersect with this . This replacement shrinks the so the new collection is locally finite, but it is still a cover since the ‘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 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 and decompose a refinement of it, by the hypothesis, into a countable collection , where each is a locally finite collection (not necessarily a cover) of open sets. Write and

Then the form a cover of , and they are locally finite—indeed, each can only intersect the (finitely many) with . Consider the intersections

These form a locally finite (non-open) cover of . Call this cover . The union is a cover of that refines (because it refines a refinement of ). Moreover, it is locally finite on each open set because only the finitely many with factor in, and each of these is locally finite.

In particular, the is a locally finite (non-open) refinement of . The previous lemma now implies that is paracompact. This completes the proof of Michael’s theorem.

**3. Applications **

First of all, we can rederive the result of DieudonnĂ©, which I discussed earlier, that a locally compact, -compact space is paracompact. Indeed, note that a locally compact space is regular. Moreover, if for the compact, we can decompose each open cover of into a countable collection of *finite* open covers of . So these are thus locally finite, and we can apply Michael’s theorem to show that is paracompact.

Corollary 4A 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.

August 19, 2010 at 9:53 am

[…] of that can be decomposed into a countable collection of locally finite families. Thanks to Michael’s theorem, this will prove the result. First, suppose that we have a countable cover . Then the idea is to […]

August 21, 2010 at 8:44 pm

[…] meanwhile, Ankit Mathew is working on a sequence on paracompactness, starting here and continuing here and […]

January 15, 2011 at 11:34 am

Could you pls provide the proof of

1. Show that one point compactification of Sw is homeomorphic with Sw-

2. Sw is not paracompact if X has discrete topology then X is paracompact.

3. Show that if f:X->>Y is continous and X is compact then the image of f:X->>Y neednot be paracompact.

4. Give an example if X is paracompact, it doesnot follow, for every open covering A of X there exists locally finite sucollection A that covers X

January 15, 2011 at 1:33 pm

Dear Buddhi, I’m not sure that I can entirely parse your questions. What is Sw?

3 is definitely false (at least if $Y$ has Hausdorff): the continuous image of a compact space is compact, so paracompact.

4 is also false, since the definition of paracompactness is that such an example does not exist.

March 28, 2012 at 6:55 pm

I’d like to know whether the converse of Lemma 3 holds.