Revisit a lemma dealing with normality in products of separable metric spaces

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 \Sigma-product of separable metric spaces is normal (see this blog post).

Lemma 1

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \pi_B(H) and \pi_B(K) are separated in \pi_B(Y), meaning that \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing.

The closure in condition 2 and condition 3 is taken in \pi_B(Y). The map \pi_B is the natural projection from the full product space X=\prod_{\alpha \in A} X_\alpha into the subproduct \prod_{\alpha \in B} X_\alpha.

Proof of Lemma 1
1 \Longrightarrow 2
Let H and K be disjoint closed subsets of Y. Since Y is normal, there exists a continuous function f: Y \rightarrow [0,1] such that f(H) \subset \left\{0 \right\} and f(H) \subset \left\{1 \right\}. By Theorem 1 in this previous post, the continuous function f depends on countably many coordinates. This means that there exists a countable B \subset A and there exists a continuous g:\pi_B(Y) \rightarrow [0,1] such that f= g \circ \pi_B. The continuity on the full product space is now reduced to the continuity on a countable subproduct. Now O_H=g^{-1}([0,0.2)) and O_K=g^{-1}((0.8,1]) are disjoint open sets in \pi_B(Y). Since f= g \circ \pi_B, it is the case that \pi_B(H) \subset O_H and \pi_B(K) \subset O_K. Since g is continuous, we have

    \overline{O_H}=\overline{g^{-1}([0,0.2))} \subset g^{-1}(\overline{[0,0.2)})=g^{-1}([0,0.2]) \ \ \ \ \ \ \ \ (a)

    \overline{O_K}=\overline{g^{-1}((0.8,1])} \subset g^{-1}(\overline{(0.8,1]})=g^{-1}([0.8,1]) \ \ \ \ \ \ \ \ (b)

Note that \overline{\pi_B(H)} \subset \overline{O_H} and \overline{\pi_B(K)} \subset \overline{O_K}. If \overline{\pi_B(H)} \cap \overline{\pi_B(K)} \ne \varnothing, then g^{-1}([0,0.2]) \cap g^{-1}([0.8,1]) \ne \varnothing. Thus \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.

The direction 2 \Longrightarrow 3 is immediate.

The direction 3 \Longrightarrow 1 follows from Lemma 1 in this previous post (see the direction 2 \rightarrow 1 of Lemma 1 in the previous post). \blacksquare

The following lemma is another version of Lemma 1 which may be useful in some circumstances. For B \subset A, let \pi_B \times \pi_B be the projection map from \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha into \prod_{\alpha \in B} X_\alpha \times \prod_{\alpha \in B} X_\alpha defined by (\pi_B \times \pi_B)(x,y)=(\pi_B(x),\pi_B(y)).

Lemma 2

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y \times Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y \times Y, there exists a countable C \subset A such that \overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y \times Y, there exists a countable C \subset A such that (\pi_C \times \pi_C)(H) and (\pi_C \times \pi_C)(K) are separated in \pi_C(Y) \times \pi_C(Y).

Proof of Lemma 2
1 \Longrightarrow 2
Let H and K be disjoint closed subsets of Y \times Y. Since Y \times Y is normal, there exists a continuous function f: Y \times Y \rightarrow [0,1] such that f(H) \subset \left\{0 \right\} and f(H) \subset \left\{1 \right\}. By Theorem 2 in this previous post, the continuous function f depends on countably many coordinates. This means that there exists a countable C \subset A and there exists a continuous g:\pi_C(Y) \times \pi_C(Y) \rightarrow [0,1] such that f= g \circ (\pi_C \times \pi_C). Now O_H=g^{-1}([0,0.2)) and O_K=g^{-1}((0.8,1]) are disjoint open sets in \pi_C(Y) \times \pi_C(Y). Since f= g \circ (\pi_C \times \pi_C), it is the case that (\pi_C \times \pi_C)(H) \subset O_H and (\pi_C \times \pi_C)(K) \subset O_K.

Since g is continuous, conditions (a) and (b) in the proof of Lemma 1 also hold here. Note that \overline{(\pi_C \times \pi_C)(H)} \subset \overline{O_H} and \overline{(\pi_C \times \pi_C)(K)} \subset \overline{O_K}. It follows that \overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing.

The direction 2 \Longrightarrow 3 is immediate.

3 \Longrightarrow 1
Let H and K be disjoint closed subsets of Y \times Y. By condition 3, there exists a countable C \subset A such that F_H=(\pi_C \times \pi_C)(H) and F_K=(\pi_C \times \pi_C)(K) are separated in M=\pi_C(Y) \times \pi_C(Y). Note that \overline{F_H} \cap F_K=\varnothing and F_H \cap \overline{F_K}=\varnothing. Consider the following subspace of M.

    W=M-\overline{F_H} \cap \overline{F_K}

The space W is an open subspace of M. The space M is a subspace of a product of countably many separable metric spaces. Thus both M and W are also second countable and hence normal.

For L \subset W, let Cl_W(L) denote the closure of L in the space W. Both Cl_W(F_H) and Cl_W(F_K) are disjoint closed subsets of W. Let G_H and G_K be disjoint open subsets of W with Cl_W(F_H) \subset G_H and Cl_W(F_K) \subset G_K. Then \pi_B^{-1}(G_H) \cap Y and \pi_B^{-1}(G_K) \cap Y are disjoint open subsets of Y separating H and K. \blacksquare

____________________________________________________________________

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 H and K of Y:

    1. If for some countable set B, \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing, then \overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing for any countable E \subset A with B \subset E.
    2. If for some countable set B, \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing, then \overline{\pi_E(H)} \cap \pi_E(K)=\pi_E(H) \cap \overline{\pi_E(K)}=\varnothing for any countable E \subset A with B \subset E.

It is straightforward to verify these facts. For the sake of completeness, we verify condition 2. Suppose that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing for some countable B \subset A. Let E \subset A be countable with B \subset E. We show \overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing. Suppose x \in \overline{\pi_E(H)} \cap \overline{\pi_E(K)}. Then \pi_B(x) \notin \overline{\pi_B(H)} \cap \overline{\pi_B(K)}. Choose some standard basic open set O=\prod_{\alpha \in B} O_\alpha with \pi_B(x) \in O such that O \cap \overline{\pi_B(H)}=\varnothing and O \cap \overline{\pi_B(K)}=\varnothing. Consider O_1=\prod_{\alpha \in E} O_\alpha such that O_\alpha=X_\alpha for all \alpha \in C-B. Clearly x \in O_1. Then there exist h \in H and k \in K such that \pi_E(h) \in O_1 and \pi_E(h) \in O_1. It follows that \pi_B(h) \in O_1 and \pi_B(h) \in O, a contradiction. Thus \overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing.

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\copyright \ 2014 \text{ by Dan Ma}
Revised 3/31/2014.

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