# A lemma dealing with normality in products of separable metric spaces

In this post we prove a lemma that is a great tool for working with product spaces of separable metrizable spaces. As an application of the lemma, we give an alternative proof for showing the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$ where each $X_\alpha$ is a separable and metrizable space. The lemma we discuss here is a tool that can shed some light on normality of dense subspaces of the product space $X$. The lemma is stated in two equivalent forms (Lemma 1 and Lemma 2).

Before stating the lemmas, let’s fix some notations. For any $B \subset A$, the map $\pi_B$ is the natural projection from the full product $X=\prod_{\alpha \in A} X_\alpha$ to the subproduct $\prod_{\alpha \in B} X_\alpha$. The standard basic open sets in the product space $X=\prod_{\alpha \in A} X_\alpha$ are of the form $\prod_{\alpha \in A} O_\alpha$ where $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. We use $supp(\prod_{\alpha \in A} O_\alpha)$ to denote the set of finitely many $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

Given a space $W$, and given $F,G \subset W$, the sets $F$ and $G$ are separated if $F \cap \overline{G}=\varnothing=\overline{F} \cap G$.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. For any sets $H,K \subset Y$, the following two conditions are equivalent:

1. There exist disjoint open subsets $U$ and $V$ of $Y$ such that $H \subset U$ and $K \subset V$.
2. There exists a countable $B \subset A$ such that the sets $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$.

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 $Y$ is normal if and only if for each 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 $\pi_B(Y)$.

If Lemma 1 holds, it is clear that Lemma 2 holds. We prove Lemma 1. The lemmas indicate that to separate disjoint sets in the full product, it suffices to separate in a countable subproduct. In this sense normality in dense subspaces of a product of separable metrizable spaces only depends on countably many coordinates.

This lemma seems to have been around for a long time. We cannot find any reference of this lemma in Engelking’s topology textbook (see [4]). We found three references. One is Corson’s paper (see [3]), in which the lemma is mentioned in relation to the non-normality of $\mathbb{N}^{\omega_1}$ and is attributed to a paper of M. Bockstein in 1948. Another is a paper of Baturov (see [2]), in which the lemma is used to prove a theorem about normality in dense subspace of $M^{\omega_1}$ where $M$ is a separable metric space. In [2] the lemma is attributed to Uspenskii. Another reference is Arkhangelskii’s book on function space (see Lemma I.6.1 on p. 43 in [1]) where the lemma is used in proving some facts about normality in function spaces $C_p(X)$.

____________________________________________________________________

Proof of Lemma 1

$1 \Longrightarrow 2$
Let $U$ and $V$ be disjoint open subsets of $Y$ with $H \subset U$ and $K \subset V$. Let $U_1$ and $V_1$ be open subsets of $X$ such that $U=U_1 \cap Y$ and $V=V_1 \cap Y$. Since $Y$ is dense in $X$, $U_1 \cap V_1=\varnothing$.

Let $\mathcal{U}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $U_1$. Let $\mathcal{V}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $V_1$. These two collections can be obtained using a Zorn lemma argument. The product space $X$ has the countable chain condition since it is a product of separable spaces. So both $\mathcal{U}$ and $\mathcal{V}$ are countable. Let $B$ be the union of finite sets each one of which is a $supp(O)$ where $O \in \mathcal{U} \cup \mathcal{V}$. The set $B$ is countable too.

Let $U^*=\cup \mathcal{U}$ and $V^*=\cup \mathcal{V}$. Note that $U^* \cap V^*=\varnothing$. We have the following observations:

$\pi^{-1}_B(\pi_B(U^*))=U^* \subset U_1$ and $\pi^{-1}_B(\pi_B(V^*))=V^* \subset V_1$

The above observations lead to the following observations:

$\pi^{-1}_B(\pi_B(U^*)) \cap \pi^{-1}_B(\pi_B(V^*)) \subset U_1 \cap V_1=\varnothing$

implying that $\pi_B(U^*) \cap \pi_B(V^*)=\varnothing$. Both $\pi_B(U^*)$ and $\pi_B(V^*)$ are open subsets of $\pi_B(X)$ and are dense in $\pi_B(X)$, respectively.

We claim that $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$. Suppose that $y \in \pi_B(U_1) \cap \pi_B(V_1)$. Then $\pi_B(V_1)$ contains a point of $\pi_B(U^*)$, say $t$. With $t \in \pi_B(U^*)$, $t=\pi_B(q)$ for some $q \in O$ where $O \in \mathcal{U}$. Note that $supp(O) \subset B$. Thus $\pi^{-1}_B(\pi_B(q))=\pi^{-1}_B(t)=O \subset U_1$. On the other hand, $t \in \pi_B(V_1)$ implies that $t=\pi_B(w)$ for some $w \in V_1$. It follows that $w \in U_1 \cap V_1$, a contradiction. Therefore $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$.

We have $\pi_B(H) \subset \pi_B(U) \subset \pi_B(U_1)$ and $\pi_B(K) \subset \pi_B(V) \subset \pi_B(V_1)$. This implies that $\overline{\pi_B(H)} \cap \pi_B(K)=\varnothing$ and $\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$ (closure in $\pi_B(X)$). Then $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$ as well. This concludes the proof for the $1 \Longrightarrow 2$ direction.

$2 \Longrightarrow 1$
Suppose that $B \subset A$ is countable such that $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$. Note that $\pi_B(H) \subset \pi_B(Y)$ and $\pi_B(K) \subset \pi_B(Y)$. Then we have the following:

$\pi_B(H) \cup \pi_B(K) \subset \pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)}) \ \ \ \ \text{closures in } \pi_B(Y)$

Consider $W=\pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)})$. The space $W$ is an open subspace of $\pi_B(Y)$. Furthermore, $\pi_B(Y)$ is a subspace of $\prod_{\alpha \in B} X_\alpha$, which is a separable and metrizable space. Thus the space $W$ is metrizable and hence normal.

For $L \subset W$, let $Cl_W(L)$ denote the closure of $L$ in the space $W$. Note that $Cl_W(\pi_B(H))$ and $Cl_W(\pi_B(K))$ are disjoint and closed sets in $W$. Let $G_H$ and $G_K$ be disjoint open subsets of $W$ such that $Cl_W(\pi_B(H)) \subset G_H$ and $Cl_W(\pi_B(K)) \subset G_K$. Then $\pi^{-1}_B(G_H) \cap Y$ and $\pi^{-1}_B(G_K) \cap Y$ are disjoint open subsets of $Y$ such that $H \subset \pi^{-1}_B(G_H) \cap Y$ and $K \subset \pi^{-1}_B(G_H) \cap Y$. $\blacksquare$

____________________________________________________________________

Remark

The proof of Lemma 1 does not need the full strength of separable metric in each factor of the product space. The above proof only makes two assumptions about the product space: the product space $X=\prod_{\alpha \in A} X_\alpha$ has the countable chain condition (CCC) and that any countable subproduct is normal, i.e., $\prod_{\alpha \in B} X_\alpha$ is normal for any countable $B \subset A$.

____________________________________________________________________

Example

As an application of the above lemma, we give another proof of the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers. See this post for a version of A. H. Stone’s original proof.

Let $\mathbb{N}$ be the set of all nonnegative integers and let $\omega_1$ be the first uncountable ordinal (i.e. the set of all countable ordinals). We provide an alternative proof that $\mathbb{N}^{\omega_1}$ is not normal. In A. H. Stone’s proof, the following disjoint closed sets cannot be separated in $\mathbb{N}^{\omega_1}$:

$H=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 0, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

$K=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 1, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

We can also use Lemma 1 to show that $H$ and $K$ cannot be separated. Note that for each countable $B \subset \omega_1$, the sets $\pi_B(H)$ and $\pi_B(K)$ have non-empty intersection. Hence they cannot be separated in $\pi_B(\mathbb{N}^{\omega_1})$. By Lemma 1, $H$ and $K$ cannot be separated in the full product space $\mathbb{N}^{\omega_1}$.

To see that $\pi_B(H) \cap \pi_B(K) \ne \varnothing$, choose a function $g:\omega_1 \rightarrow \mathbb{N}$ such that $g^{-1}(0) \cap B=\varnothing$. Let $g_B:B \rightarrow \mathbb{N}$ be defined by $g_B(\alpha)=g(\alpha)$ for all $\alpha \in B$. Then $g_B \in \pi_B(H) \cap \pi_B(K)$.

____________________________________________________________________

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

____________________________________________________________________

$\copyright \ 2014 \text{ by Dan Ma}$