Let be the closed unit interval . In 1951, Dowker proved that for normal spaces , is normal if and only if is countably paracompact. A space is countably paracompact if every countable open cover of has a locally finite open refinement. Dowker’s theorem is a fundamental result on products of normal spaces. With this theorem, a question was raised about the existence of a normal but not countably paracompact space (such a space became known as a Dowker space). For a detailed discussion on Dowker spaces, see the survey paper [Rudin]. The focus here is on presenting a proof for the Dowker’s theorem, laying the groundwork for future discussion.

An open cover of a space is shrinkable if there exists an open cover such that for each . The open cover is called a shrinking of . We use the following lemma in proving Dowker’s theorem. Go here to see a proof of this lemma.

**Lemma**

A space is normal if and only if every point-finite open cover of is shrinkable.

**Dowker’s Theorem**

For a normal space , the following conditions are equivalent:

- is a countably paracompact space.
- If is an open cover of , then there is a locally finite open refinement such that for each .
- is normal for any compact metrizable space .
- is normal.
- For each sequence of closed sets such that and , there exists open sets such that .

**Proof**

Suppose is countably paracompact. Let be an open cover of . Then has a locally open refinement . For each , let . Note that is still locally finite (thus point-finite). By the lemma, it has a shrinking .

Let be a compact metrizable space. Let be a base of such that it is closed under finite unions. Let be two disjoint closed subsets of . The goal here is to find an open set such that and misses .

For each , consider the following:

.

For each , let be defined by:

and .

Claim 1.

Each is an element of some .

Both and are compact. Thus we can find some such that and .

Claim 2.

Each is open in .

Fix . For each , . Choose open set such that and misses . The set of all is a cover of . We can find a finite subcover covering , say . Let .

For each , . Choose open set such that and misses . The set of all is a cover of . We can find a finite subcover covering , say . Let .

Now, let . Clearly . To show that , pick . We want to show and . Suppose we have and . Then for some . As a result, . This would mean that . But this contradicts with . So we have . On the other hand, suppose we have and . Then for some . As a result, . This means that . This contradicts with . So we have . It follows that and is open in .

Claim 3.

We can find an open set such that and misses .

Let . Note that and misses . The open set being constructed will satisfiy . The open cover has a locally finite open refinement such that for each . Now define . Note that .

We also have . We have a closure preserving situation because the open sets are from a locally finite collection. Continue the derivation and we have:

.

is obvious.

Let be closed sets with empty intersection. We show that can be expanded by open sets such that and .

Choose and a sequence of distinct points converging to . Let and . These are disjoint closed sets. Since is normal, we can find open set such that and misses .

Let . Note that each is open in . Also, . We want to show that . Let . The point and . There exist open and open such that and . Then for some . Note that . Thus . It follows that .

Let be an open cover of . Let . Let . Each is closed and . So there exist open sets such that .

Since is normal, there exists open such that . It follows that . Let .

Claim. is an open cover of . Furthermore, .

Let . Since , for some . Then and . To show the second half of the claim, let . Then for each . This means and for each . Then is an open set containing that contains no point of . Thus we have .

Let . We show that is a locally finite open refinement of . Clearly . To show that the open sets form a cover, let . Choose least such that . Then for each , . This means and for each . It follows that for each . This implies . To see that the open sets form a locally finite collection, note that each belong to some . The open set misses for all . Thus can only meet for . We just prove that is countably paracompact.

* Note*. In , all we need is that the factor has a non-trivial convergent sequence. That is, if is normal and has a non-trivial convergent sequence, then satisfies condition .

**Reference**

[Dowker] Dowker, C. H. [1951], *On Countably Paracompact Spaces*, Canad. J. Math. 3, 219-224.

[Rudin] Rudin, M. E., [1984], *Dowker Spaces, Handbook of Set-Theoretic Topology* (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 761-780.

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Revised November 27, 2016