In the first installment, we proved that paracompact and Hausdorff implies fully normal, the “easy direction” of Stone’s theorem. We now prove the converse, which I consider to be the “hard direction” of Stone’s theorem. In the case reader has forgotten, I have provided the definitions of paracompact and fully normal below, along with the relevant auxiliary definitions.

A topological space is

paracompactif its Hausdorff and every open covering of has an open locally-finite refinement. A covering islocally finiteif, for every point , there is a neighborhood which intersects finitely many sets in .For any open cover , the

starof a set is defined to be the set –in words, is the union of all sets which intersect . If is a singleton , we denote the star of by . An open cover is anstar refinementfor an open cover if, the collection of stars forms an open refinement of . We say that a topological space isfully normalif every open cover has a star refinement.

*Proof of Hard Direction:* Let be a fully normal space, and let be an open covering of . We will construct a locally finite refinement of .

By full normality, there exists an open cover which star-refines . By induction, we may construct a sequence of open covers such that

For notational convenience, we denote the star of in by , and define

In the proof of Stone’s theorem, we will make frequent use of the following identities. Some of them are quite trivial; I encourage the reader to try to prove them on his or her own to become comfortable with the definitions of stars and star refinements.

*Proof:* 1.

Since is an open cover of , there is some such that .

2. We use (1) to obtain . Since , we conclude that , where is an element in the aforementioned set.

3. If , then , for , where . Noting that

,

let so that . By definition of star refinement, , for some . Since meets and is a subset of , we conclude that meets ; hence, .

4. and 5. are obvious.

6. Observe that

7. Observe that

A property of star refinements is that if two sets intersect, then their the union is contained in some member of the collection . To see this, observe that if , then both sets are contained in the star . Since star refines , we see that for some set .

Invoking the Well-Ordering Theorem, we may assume that the elements of are well-ordered.

For each , define a sequence of sets by

It is evident that , and is open for . Hence, is open, being the union of open sets. I claim that . Indeed, by Lemma 1 identity (2). If , then

by Lemma 1 identity (3) and the definition of . Induction completes the proof of the claim. Since , we also obtain for all ; whence, .

I claim that . It suffices to prove that . For , the star , for some , by the star-refinement construction. Identity (1) in Lemma 1 implies that . Furthermore, I claim that for , there exists such that . Indeed, by definition of , there exists such that . It follows from the definition of that .

We define a transfinite sequence of closed sets by

where we make use of the well ordering. The motivation for this definition is that we don’t want a set to intersect both and when . To see that we succeed, suppose that if and . Then , which is disjoint from .

I claim that . It suffices to prove the inclusion . Since the set is nonempty, our well ordering gives the existence of minimal element such that . An earlier result shows that there exists an such that . I claim that . If not, then there exists such that , which implies that , for some . But then by identity (3) of Lemma 1

which contradicts the minimality of . This completes the proof of the claim.

Define sets and by

We have the trivial inclusions . After much head banging against the wall, I claim that no intersects both and when . Indeed, suppose and , where , then there are sets nontrivially intersecting and , respectively. Recall the observation that if , then the union is contained in some set . Therefore there are sets and . Since , we obtain that there is a set containing their union. But by construction, and are both nonempty, which is a contradiction.

I claim that is closed, despite being the possibly uncountable union of closed sets. If , then any open neighborhood intersects intersects for some . Be careful: if we change the neighborhood , the index may change, so we can’t yet conclude that ; we need to show that the is unique. Replacing with a smaller neighborhood, if necessary, we may assume that . But then since no set in meets two distinct , the neighborhood can intersect at most one set . Hence, .

Define open sets by

We complete the proof of Stone’s theorem by showing that is the desired locally finite refinement.

First, I claim that . It suffices to prove the inclusion . If , then there exists and such that . Taking smaller, if necessary, we may assume that is the minimal integer such that . Then and , which implies that .

Second, I claim that refines, though not necessarily star refines, . Indeed, by Lemma 1,

Third, I claim that is locally finite. For , we know that for some and some . Hence, , which satisfies for any . Observe that if , then the star refinement assumption of the implies that for any ,

for some . If meets and , where , then meets both and , which is a contradiction. We conclude that is an open neighborhood of which at most intersects .