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 nonzero 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.
Theorem 1

For each , let be a separable and metrizable space. Then is a normal space.
Theorem 2

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.
Theorem 4

For each , let be a separable and metrizable space. Then is a collectionwise normal space.
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Proofs
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).
Lemma 5

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
such that
 and for each
 for each
Let , and . We have the following claims.
Claim 1
and .
Claim 2
.
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.
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A Brief History
The notion of products was introduced by Corson in [1] 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 [4] showed that the product of separable metric spaces is normal. In 1977, Gulko [2] and Rudin [6] independently proved the product of metric spaces is normal. In 1978, Kombarov [3] generalized Gulko and Rudin’s result by showing that any product of paracompact pspaces is collectionwise normal if and only if all spaces have countable tightness. A useful resource is Przymusinski’s chapter in the Handbook of SetTheoretic Topology [5], which has a section on products.
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Reference
 Corson H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785796, 1959.
 Gulko S. P., On the properties of sets lying in products, Dokl. Acad. Nauk. SSSR, 237, 505508, 1977 (in Russian).
 Kombarov A. P., On the tightness and normality of products, Dokl. Acad. Nauk. SSSR, 239, 775778, 1978 (in Russian).
 Kombarov A. P., Malyhin V. I., On products, Soviet Math. Dokl., 14 (6), 19801983, 1973.
 Przymusinski, T. C., Products of Normal Spaces, Handbook of SetTheoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781826, 1984.
 Rudin M. E., Book Review, Bull. Amer. Math. Soc., 84, 271272, 1978.
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