In this post we continue to discuss a lemma that has been discussed previously in this post. The lemma characterizes the dense normal subspaces of a product of separable metric spaces. The lemma discussed here has been sharpened over the version in the previous post. Two versions of the lemma are given (Lemma 1 and Lemma 2). Any one of these two versions can be used to prove that the product of separable metric spaces is normal (see this blog post).
Lemma 1

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 .
The closure in condition 2 and condition 3 is taken in . The map is the natural projection from the full product space into the subproduct .
Proof of Lemma 1
Let and be disjoint closed subsets of . Since is normal, there exists a continuous function such that and . By Theorem 1 in this previous post, the continuous function depends on countably many coordinates. This means that there exists a countable and there exists a continuous such that . The continuity on the full product space is now reduced to the continuity on a countable subproduct. Now and are disjoint open sets in . Since , it is the case that and . Since is continuous, we have
Note that and . If , then . Thus .
The direction is immediate.
The direction follows from Lemma 1 in this previous post (see the direction of Lemma 1 in the previous post).
The following lemma is another version of Lemma 1 which may be useful in some circumstances. For , let be the projection map from into defined by .
Lemma 2

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 .
Proof of Lemma 2
Let and be disjoint closed subsets of . Since is normal, there exists a continuous function such that and . By Theorem 2 in this previous post, the continuous function depends on countably many coordinates. This means that there exists a countable and there exists a continuous such that . Now and are disjoint open sets in . Since , it is the case that and .
Since is continuous, conditions (a) and (b) in the proof of Lemma 1 also hold here. Note that and . It follows that .
The direction is immediate.
Let and be disjoint closed subsets of . By condition 3, there exists a countable such that and are separated in . Note that and . Consider the following subspace of .
The space is an open subspace of . The space is a subspace of a product of countably many separable metric spaces. Thus both and are also second countable and hence normal.
For , let denote the closure of in the space . Both and are disjoint closed subsets of . Let and be disjoint open subsets of with and . Then and are disjoint open subsets of separating and .
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Remark
The countable sets in both Lemma 1 and Lemma 2 can be expanded to larger countable sets. For example,

for Lemma 1, for any disjoint closed subsets and of :
 If for some countable set , , then for any countable with .
 If for some countable set , , then for any countable with .
It is straightforward to verify these facts. For the sake of completeness, we verify condition 2. Suppose that for some countable . Let be countable with . We show . Suppose . Then . Choose some standard basic open set with such that and . Consider such that for all . Clearly . Then there exist and such that and . It follows that and , a contradiction. Thus .
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Revised 3/31/2014.