# 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 $K$. $\blacksquare$

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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 $H$ 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.