It is well known that normality is not preserved by taking products. When nothing is known about the spaces and other than the facts that they are normal spaces, there is not enough to go on for determining whether is normal. In fact even when one factor is a metric space and the other factor is a hereditarily paracompact space, the product can be non-normal (discussed here). This post discusses a productive scenario – the first factor is a normal space and second factor is a metric space with the first factor having the additional property that it is countably compact. In this scenario the product is always normal. This is a well known result in general topology. The goal here is to nail down a proof for use as future reference.

*Main Theorem*

Let be a normal and countably compact space. Then is a normal space for every metric space .

The proof of the main theorem uses the notion of shrinkable open covers.

*Remarks*

The main theorem is a classic result and is often used as motivation for more advanced results for products of normal spaces. Thus we would like to present a clear and complete proof of this classic result for anyone who would like to study the topics of normality (or the lack of) in product spaces. We found that some proofs of this result in the literature are hard to follow. In A. H. Stone’s paper [2], the result is stated in a footnote, stating that “it can be shown that the topological product of a metric space and a normal countably compact space is normal, though not necessarily paracompact”. We had seen several other papers citing [2] as a reference for the result. The Handbook [1] also has a proof (Corollary 4.10 in page 805), which we feel may not be the best proof to learn from. We found a good proof in [3] using the idea of shrinking of open covers.

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**The Notion of Shrinking**

The key to the proof is the notion of shrinkable open covers and shrinking spaces. Let be a space. Let be an open cover of . The open cover of is said to be shrinkable if there is an open cover of such that for each . When this is the case, the open cover is said to be a shrinking of . If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed, e.g. , then a shrinking has the same indexing, e.g. .

A space is a shrinking space if every open cover of is shrinkable. Every open cover of a paracompact space has a locally finite open refinement. With a little bit of rearranging, the locally finite open refinement can be made to be a shrinking (see Theorem 2 here). Thus every paracompact space is a shrinking space. For other spaces, the shrinking phenomenon is limited to certain types of open covers. In a normal space, every finite open cover has a shrinking, as stated in the following theorem.

*Theorem 1*

The following conditions are equivalent.

- The space is normal.
- Every point-finite open cover of is shrinkable.
- Every locally finite open cover of is shrinkable.
- Every finite open cover of is shrinkable.
- Every two-element open cover of is shrinkable.

The hardest direction in the proof is , which is established in this previous post. The directions are immediate. To see , let and be two disjoint closed subsets of . By condition 5, the two-element open cover has a shrinking . Then and . As a result, and . Since the open sets and cover the whole space, and are disjoint open sets. Thus is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space may or may not be shrinkable. It turns out that finding a normal space with an infinite open cover that is not shrinkable is no trivial matter (see Dowker’s theorem in this previous post). However, if an open cover in a normal space point-finite or locally finite, then it is shrinkable.

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**Key Idea**

We now discuss the key idea to the proof of the main theorem. Consider the produce space . Let be an open cover of . Let . The set is stable with respect to the open cover if for each , there is an open set containing such that for some .

Let be a cardinal number (either finite or infinite). A space is a -shrinking space if for each open cover of such that the cardinality of is , then is shrinkable. According to Theorem 1, any normal space is 2-shrinkable.

*Theorem 2*

Let be a cardinal number (either finite or infinite). Let be a -shrinking space. Let be a paracompact space. Suppose that is an open cover of such that the following two conditions are satisfied:

- Each point has an open set containing such that is stable with respect to .
- .

Then is shrinkable.

*Proof of Theorem 2*

Let be any open cover of satisfying the hypothesis. We show that has a shrinking.

For each , obtain the open covers and of as follows. For each , define the following:

Then is an open cover of . Since is -shrinkable, there is an open cover of such that for each .

Now is an open cover of . By the paracompactness of , let be a locally finite open cover of such that for each . For each , define the following:

We claim that is a shrinking of . First it is a cover of . Let . Then for some . There exists such that . Note the following.

This means that . Since , . Thus is an open cover of .

Now we show that is a shrinking of . Let . To show that , let . Let be open in such that and that meets only finitely many , say for . Immediately we have the following relations.

Then it follows that

Thus . This shows that is a shrinking of .

*Remark*

Theorem 2 is the Theorem 3.2 in [3]. Theorem 2 is a formulation of Theorem 3.2 [3] for the purpose of proving Theorem 3 below.

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**Main Theorem**

*Theorem 3 (Main Theorem)*

Let be a normal and countably compact space. Let be a metric space. Then is a normal space.

*Proof of Theorem 3*

Let be a 2-element open cover of . We show that is shrinkable. This would mean that is normal (according to Theorem 1). To show that is shrinkable, we show that the open cover satisfies the two bullet points in Theorem 2.

Fix . Let be a base at the point . Define as follows:

It is clear that is an open cover of . Since is countably compact, choose such that is a cover of . Let . We claim that is stable with respect to . To see this, let . Then for some . By the definition of , there is some open set such that and for some . Furthermore, .

To summarize: for each , there is an open set such that and is stable with respect to the open cover . Thus the first bullet point of Theorem 2 is satisfied. The open cover is a 2-element open cover. Thus the second bullet point of Theorem 2 is satisfied. By Theorem 2, the open cover is shrinkable. Thus is normal.

*Corollary 4*

Let be a normal and pseudocompact space. Let be a metric space. Then is a normal space.

The corollary follows from the fact that any normal and pseudocompact space is countably compact (see here).

*Remarks*

The proof of Theorem 3 actually gives a more general result. Note that the second factor only needs to be paracompact and that every point has a countable base (i.e. first countable). The first factor has to be countably compact. The shrinking requirement for is flexible – if open covers of a certain size for are shrinkable, then open covers of that size for the product are shrinkable. We have the following corollaries.

*Corollary 5*

Let be a -shrinking and countably compact space and let be a paracompact first countable space. Then is a -shrinking space.

*Corollary 6*

Let be a shrinking and countably compact space and let be a paracompact first countable space. Then is a shrinking space.

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**Remarks**

The main theorem (Theorem 3) says that any normal and countably compact space is productively normal with one class of spaces, namely the metric spaces. Thus if one wishes to find a non-normal product space with one factor being countably compact, the other factor must not be a metric space. For example, if , the first uncountable ordinal with the ordered topology, then is always normal for every metric . For non-normal example, is not normal for any compact space with uncountable tightness (see Theorem 1 in this previous post). Another example, is not normal where is the one-point Lindelofication of a discrete space of cardinality (follows from Example 1 and Theorem 7 in this previous post).

Another comment is that normal countably paracompact spaces are examples of Normal P-spaces. K. Morita defined the notion of P-space and he proved that a space is a Normal P-space if and only if is normal for every metric space .

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**Reference**

- Przymusinski T. C.,
*Products of Normal Spaces*, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984. - Stone A. H.,
*Paracompactness and Product Spaces*, Bull. Amer. Math. Soc., Vol. 54, 977-982, 1948. (paper) - Yang L.,
*The Normality in Products with a Countably Compact Factor*, Canad. Math. Bull., Vol. 41 (2), 245-251, 1998. (abstract, paper)

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