A.H. Stone’s theorem says that a topological space is paracompact (and Hausdorff) if and only if it is fully normal. Today, I am going to give a brief introduction to the notions of paracompactness and full normality, prove some basic results, and then prove one direction of Stone’s theorem–that paracompact and Hausdorff fully normal. This direction is easier of the two. I don’t know if both directions are attributed to Stone, but in any case, he proves both directions in his 1948 article “Paracompactness and Product Spaces”. By the way, the Stone here is not the well-known Stone of the Stone-Weierstrass theorem or the Stone-Von Neumann theorem. According to Wikipedia, Arthur Harold“A.H.” Stone was a British mathematician and of no relation to the American mathematician Marshall Harvey “M.H.” Stone.

A topological space is paracompact if its Hausdorff (most authors don’t assume Hausdorff) and every open covering of has an open locally-finite refinement. A covering is locally finite if, for every point , there is a neighborhood which intersects finitely many sets in .

For any open covering , the star refinement of a set is defined to be the subcovering . If is a singleton , we denote the star refinement of the singleton by . An open cover is an open star refinement for an open cover if, the collection of stars forms an open refinement of . We say that a topological space is fully normal if every open cover has an open star refinement.

Full normality is a stronger property than normality, as the following lemma shows.

Lemma 1Every fully normal space is normal.

*Proof:* Let be disjoint closed sets in . Define open sets . Then form an open cover of By full normality, there exists an open star refinement . Define open sets and by

I claim that and are disjoint. Indeed,

Since is a star refinement, we have that or . But , which implies that and –this is a contradiction.

Every normal space is Hausdorff (take the disjoint closed sets to be singletons), so we obtain the following corollary. Every fully normal space is Hausdorff.

There are plenty of fully normal spaces because every metric space or metrizable space is fully normal. There are examples of fully normal spaces which are not metrizable, such as the Sorgenfrey Line.

Proposition 2Every metric space is fully normal.

*Proof:* Let be an open cover of . Given any , there is some index such that . Since is open, there exists some such that . Define an open cover . I claim that is a star refinement of . Indeed, fix and suppose that , where . Then by the triangle inequality,

Hence, . Noting that

completes the proof of the claim.

So far, we’e proved some basic results about full normality. We’ll now prove a few basic results about paracompactness, which we will use to prove the main theorem of this post, paracompactness full normality.

Given an infinite collection of sets , it is generally not true that ; the right-hand side of the equality need not be closed. For example, consider the open intervals , for . Then , whereas .

Lemma 3If is a locally finite collection, then .

*Proof:* If . then for some . For any open neighborhood , , which implies that . Note that this direction always holds, regardless of the assumptions we impose on .

For the reverse inclusion, let . We will show that belongs to the closure of finitely many . By local finiteness, there exists a neighborhood so that the set is finite. So for any open neighborhood , is an open neighborhood containing . Hence, the set is nonempty and finite. We conclude that

As is the case with compact spaces, closed subspaces of paracompact spaces are also paracompact.

Proposition 4Every closed subset of a paracompact space is paracompact.

*Proof:* Let be a paracompact space, and let be a closed subset. Let be an open cover of . Then is an open cover of . By paracompactness, there is a locally finite open refinement . Any set in satisfies , for some , or . We don’t care about the sets that satisfy the latter condition, so consider the collection . is an open refinement of which covers ; local finiteness follows from that of .

If we can show that paracompact spaces are normal, then they a fortiori regular. However, we’ll need use regularity as a stepping stone to proving normality.

Lemma 5Any paracompact space is regular.

*Proof:* Let be a paracompact space, let be closed, and let . For each , there is an open neighborhood such that . This construction is possible since is Hausdorff. The sets form an open cover of and therefore have a locally finite open refinement . Define an indexing set , and define an open set . I claim that . Indeed, by Lemma 3, , which contains in no set in the union. Hence, is an open neighborhood of tautologically disjoint from . Hence, and are disjoint open sets containing and , respectively.

We use the preceding lemma to show that a paracompact space is normal.

Lemma 6Every paracompact space is normal.

*Proof:* Let and be disjoint, closed subsets of . Lemma 3 tells us that every can be separated from by by disjoint open sets, which implies that there is an open neighborhood such that . Observe that forms an open cover of and therefore has a locally finite open refinement . If , then forms a locally finite open cover of . Since, for every , , for some , we have that . Define an open set . By Lemma 2, , which implies that ,

We now have all the tools ready to prove today’s main theorem. As mentioned in the introduction, I don’t know if Theorem 7 is attributed to A.H. Stone. A proof of the result is contained in his 1948 article, but given the difficulty of the proof relative to that of the converse, I would guess that it was proved earlier.

Theorem 7Every paracompact space is fully normal.

*Proof:* Let be a paracompact space, and let be an open cover of . Replacing with a locally finite refinement, if necessary, we may assume that is locally finite. We need to show that has an open star refinement.

By the shrinking lemma, there exists, for each , an open set such that and . Since is locally finite, each has an open neighborhood such that the set of indices is finite. Define subsets and . Note that is nonempty, since is an open cover. Also note that , but and are not necessarily disjoint, for an element could belong to .

The intuition for these definitions is that we seek an open cover such that the stars are contained in some . Observe that and are both finite intersections of open sets containing , hence are both open sets containing . We want to use local finiteness, so if we interesect the intersections with , we obtain an open neighborhood of .

The collection forms an open cover . Fix . Since is an open cover, , for some . If , for some , then the intersection is nonempty and contains , which implies that . Hence, .

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