In this post, we prove a theorem that is attributed to Corson. It had been reported in the literature (see [1] and [2] for two instances) and on the Internet that this theorem can be deduced from a paper of Corson [3]. Instead of having an indirect proof, we give a full proof of this theorem. One application of this theorem is that we can use it to show the collectionwise normality of a product of separable metric spaces (see this blog post). We prove the following theorem.
Theorem 1 (Corson’s Theorem)

Let be a product of separable metrizable spaces. Let be a dense subspace of . If is normal, then is collectionwise normal.
Another way to state this theorem is through the angle of finding normal spaces that are collectionwise normal. The above theorem can be restated: any dense normal subspace of a product of separable metric spaces must be collectionwise normal if one additional condition is satisfied: the square of is also normal. Thus we have the following theorem:
Theorem 1a

Let be a product of separable metrizable spaces. Let be a normal and dense subspace of . If is normal, then is collectionwise normal.
____________________________________________________________________
A Brief Background Discussion
A space is said to be collectionwise normal if for any discrete collection of closed subsets of , there exists a pairwise disjoint collection of open subsets of such that for each , there is exactly one such that . Here’s some previous posts on the definitions and a background discussion on collectionwise normality.
There is one circumstance where normality implies collectionwise normality. If all closed and discrete subsets of a normal space are countable, then it is collectionwise normal. We have the following theorem.
Theorem 2

Let be a normal space. If all closed and discrete subsets of are countable, then is collectionwise normal.
Proof of Theorem 2
We first establish the following lemma.
Lemma 2a
Let be a normal space. Let be a countable discrete collection of closed subsets of . Then there exists a pairwise disjoint collection of open subsets of such that for each .
Proof of Lemma 2a
For each , choose disjoint open subsets and such that and . Let . For each , let . It follows that is pairwise disjoint such that for each . This completes the proof for Lemma 2a.
Suppose that all closed and discrete subsets of the normal space are countable. It follows that any discrete collection of closed subsets of is countable. Then the collectionwise normality of follows from Lemma 2a.
____________________________________________________________________
Proving Corson’s Theorem
If is normal, then clearly is normal. In light of Theorem 2, to show is collectionwise normal, it suffices to show that every closed and discrete subspace of is countable. Thus Theorem 1 is established by combining Theorem 2 and the following theorem.
Theorem 3

Let be a product of separable metrizable spaces. Let be a dense subspace of . If is normal, then every closed and discrete subspace of is countable.
Before proving Theorem 3, we state one more lemma that is needed. For , is the natural projection map from into . The map refers to the projection map from into defined by .
Lemma 4

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 .
Lemma 4 deals with the dense subspace of and the dense subspace of . So the map should be restricted to and the map is restricted to . For a proof of Lemma 4, see the proof of Lemma 2 in this previous post.
Proof of Theorem 3
Let be an uncountable closed and discrete subset of . We define two disjoint closed subsets and of such that for each countable set , . By Lemma 4, is not normal. Consider the following two subsets of :
Clearly and are disjoint. It is clear that is a closed subset of . Because is closed and discrete in , is a closed subset of . Thus and are disjoint closed subsets of .
Let be countable. Note that and are:
Consider . Clearly, . We consider two cases: is uncountable or is countable.
Case 1: is uncountable.
Note that is the product of countably many separable metric spaces and is therefore a hereditarily Lindelof space. As a subspace of , is also hereditarily Lindelof. Since is an uncountable subspace of , there must exist a point such that every open set (open in ) containing must contain uncountably many points of . Note that .
Let be an open subset of with . Then there exist with such that and . Then is a point of that is in . The point is a point of that is in . This means that . Thus we have .
Case 2: is countable.
Then there exists such that for uncountably many . Choose such that and and . Then and .
In either case, we can say that . By Lemma 4, is not normal.
____________________________________________________________________
Reference
 Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111116, 1990.
 Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374383, 2007.
 Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785796, 1959.
 Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
____________________________________________________________________
Pingback: Normality in Cp(X)  Dan Ma's Topology Blog