Let be the Cartesian product of many copies of the real line. This product product space is not normal since it contains as a closed subspace. The subspace of consisting of points which have at most countably many non-zero coordinates is collectionwise normal. Such spaces are called -products. In this post, we show that the -product of separable and metrizable spaces is always collectionwise normal. To place the result proved in this post in a historical context, see the comments at the end of the post.
Consider the product space . Let . The -product of the spaces about the base point is the following subspace of :
When the base point is understood, we denote the space by . First we want to eliminate cases that are not interesting. If the index set is countable, then the -product is simply the Cartesian product. We assume that the index set is uncountable. If all but countably many of the factors consist of only one point, then the -product is also the Cartesian product. So we assume that each has at least 2 points. When these two assumptions are made, the resulting -products are called proper.
The collectionwise normality of is accomplished in two steps. First, is shown to be normal if each factor is a separable metric space (Theorem 1). Secondly, observe that normality in -product is countably productive, i.e., if is normal, then is also normal (Theorem 2). Then the collectionwise normality of follows from a theorem attributed to Corson. We have the following theorems.
For each , let be a separable and metrizable space. Then is a normal space.
For each , let be a separable and metrizable space. Let . Then is a normal space.
Theorem 3 (Corson’s Theorem)
Let be a product of separable metrizable spaces. Let be a dense subspace of . If is normal, then is collectionwise normal.
For a proof of Corson’s theorem, see this post.
The above three theorems lead to the following theorem.
For each , let be a separable and metrizable space. Then is a collectionwise normal space.
Before proving the theorems, let’s set some notations. For each , is the natural projection from into . The standard basic open sets in the product space are of the form where for all but finitely many . We use to denote the set of finitely many such that .
The following lemma is used (for a proof, see Lemma 1 in this post).
Let be a product of separable metrizable spaces. Let be a dense subspace of . Then the following conditions are equivalent.
- is normal.
- For any pair of disjoint closed subsets and of , there exists a countable such that .
- For any pair of disjoint closed subsets and of , there exists a countable such that and are separated in , meaning that .
Proof of Theorem 1
Let be the product space in question. Let be defined using the base point . Note that is dense in the product space . In light of Lemma 5, to show is normal, it suffices to show that for each pair of disjoint closed subsets and of , there exists a countable such that . Let and be disjoint closed subsets of .
Before building up to a countable set , let’s set some notation that will be used along the way. For each , let denote the set of all such that . For any set , let .
To start, choose and let . Consider and . They are subsets of , which is a hereditarily separable space. Choose a countable and a countable such that and . For each , choose such that . For each , choose such that . Let and be defined by:
Clearly and . Let .
Now perform the next step inductive process. Consider and . As before, we can find countable dense subsets of these 2 sets. Choose a countable and a countable such that and . For each , choose such that . For each , choose such that . Let and be defined by:
Clearly and . To prepare for the next step, let .
Continue the inductive process and when completed, the following sequences are obtained:
- a sequence of countable sets
- a sequence of countable sets
- a sequence of countable sets
- and for each
- for each
Let , and . We have the following claims.
Proof of Claim 1
It suffices to show one of the set inclusions. We show . Let . We need to show that is a limit point of . To this end, let be a standard basic open set containing . Then for some . Let . Then . Since , there is some such that . It follows that . Thus every open set containing contains a point of .
Proof of Claim 2
Suppose that . Define such that for all and for all . It follows that , which is a contradiction since and are disjoint closed sets. To see that , let be a standard basic open set containing . Let . Since , there exist and such that and . Note that the supports and . For the coordinates outside of , both and agree with the base point and hence with . Thus and . We have just shown that every open set containing contains a point of and a point of . This means that , a contradiction. This completes the proof of Claim 2.
Both Claim 1 and Claim 2 imply that . By Lemma 5, is normal.
Proof of Theorem 2
Let be the -product about the base point . The following countable product
is a product of separable metric spaces. So any -product that can be defined within the product space (a) is normal (by Theorem 1). In particular, consider the -product defined about the base point (countably many coordinates). Denote this -product by . Observe that is homeomorphic to the following countable product of :
Thus can be identified with . We can conclude that is normal.
Proof of Theorem 4
Let be the -product of the separable metric spaces . By Theorem 1, is normal. By Theorem 2, is normal. In particular, is normal. Clearly, is a dense subspace of the product space . By Corson’s theorem (Theorem 3), is collectionwise normal.
A Brief History
The notion of -products was introduced by Corson in  where he proved that the -product of complete metric spaces is normal. Corson then asked whether the -product of copies of the rationals is normal. In 1973, Kombarov and Malyhin  showed that the -product of separable metric spaces is normal. In 1977, Gulko  and Rudin  independently proved the -product of metric spaces is normal. In 1978, Kombarov  generalized Gulko and Rudin’s result by showing that any -product of paracompact p-spaces is collectionwise normal if and only if all spaces have countable tightness. A useful resource is Przymusinski’s chapter in the Handbook of Set-Theoretic Topology , which has a section on -products.
- Corson H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
- Gulko S. P., On the properties of sets lying in -products, Dokl. Acad. Nauk. SSSR, 237, 505-508, 1977 (in Russian).
- Kombarov A. P., On the tightness and normality of -products, Dokl. Acad. Nauk. SSSR, 239, 775-778, 1978 (in Russian).
- Kombarov A. P., Malyhin V. I., On -products, Soviet Math. Dokl., 14 (6), 1980-1983, 1973.
- 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.
- Rudin M. E., Book Review, Bull. Amer. Math. Soc., 84, 271-272, 1978.