When the spaces and are normal spaces, the product space is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors and can have no influence on the normality of the product . The dynamics in the other direction are totally different. When the product is hereditarily normal, the two factors and are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.

A subset of a space is said to be a -set in if is the intersection of countably many open subsets of . A space is perfectly normal if it is normal and that every closed subset of is a -set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product is a very strong property.

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**Katetov’s theorems**

*Theorem 1*

If is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

- The factor is perfectly normal.
- Every countable and infinite subset of the factor is closed.

*Proof of Theorem 1*

The strategy we use is to define a subspace of that is not normal after assuming that none of the two conditions is true. So assume that has a closed subspace that is not a -set and assume that is an infinite subset of that is not closed. Let be a limit point of such that . The candidate for a non-normal subspace of is:

Note that is an open subspace of since it is the result of subtracting a closed set from . The following are the two closed sets that demonstrate that is not normal.

It is clear that and are closed subsets of . Let and be open subsets of such that and . We show that . To this end, define for each . It follows that for each , . Furthermore each is an open subspace of . Thus . Since is not a -set in , there must exist such that . Then and .

Since is open in the product , choose open sets and such that and . With , there exists some such that . First, . Since , . Thus . This completes the proof that the subspace is not normal and that is not hereditarily normal.

Let’s see what happens in Theorem 1 when both factors are compact. If both and are compact and if is hereditarily normal, then both and must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.

*Theorem 2*

If and are compact and is hereditarily normal, then both and are perfectly normal.

Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.

*Theorem 3*

Let be a compact space. If is hereditarily normal, then is metrizable.

*Proof of Theorem 3*

Suppose that is hereditarily normal. By Theorem 2, the compact spaces and are perfectly normal. In particular, the following subset of is a -set in .

The set is said to be the diagonal of the space . It is a well known result that any compact space whose diagonal is a -set in the square is metrizable (discussed here).

The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of . This was indeed posted by Katetov himself. This leads to the discussion in the next post.

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**Reference**

- Engelking R.,
*General Topology*, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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