# A theorem attributed to Corson

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 $\Sigma$-product of separable metric spaces (see this blog post). We prove the following theorem.

Theorem 1 (Corson’s Theorem)

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. If $Y \times Y$ is normal, then $Y$ 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 re-stated: any dense normal subspace $Y$ of a product of separable metric spaces must be collectionwise normal if one additional condition is satisfied: the square of $Y$ is also normal. Thus we have the following theorem:

Theorem 1a

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a normal and dense subspace of $X$. If $Y \times Y$ is normal, then $Y$ is collectionwise normal.

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A Brief Background Discussion

A space $S$ is said to be collectionwise normal if for any discrete collection $\mathcal{A}$ of closed subsets of $S$, there exists a pairwise disjoint collection $\mathcal{U}$ of open subsets of $S$ such that for each $A \in \mathcal{A}$, there is exactly one $U \in \mathcal{U}$ such that $A \subset U$. 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 $S$ be a normal space. If all closed and discrete subsets of $S$ are countable, then $S$ is collectionwise normal.

Proof of Theorem 2
We first establish the following lemma.

Lemma 2a
Let $L$ be a normal space. Let $\left\{C_1,C_2,C_3,\cdots \right\}$ be a countable discrete collection of closed subsets of $L$. Then there exists a pairwise disjoint collection $\left\{O_1,O_2,O_3,\cdots \right\}$ of open subsets of $L$ such that $C_j \subset O_j$ for each $j$.

Proof of Lemma 2a
For each $j$, choose disjoint open subsets $U_j$ and $V_j$ such that $C_j \subset U_j$ and $\bigcup_{n \ne j} C_n \subset V_j$. Let $O_1=U_1$. For each $j>1$, let $O_j=U_j \cap \bigcap_{n \le j-1} V_n$. It follows that $\left\{O_1,O_2,O_3,\cdots \right\}$ is pairwise disjoint such that $C_j \subset O_j$ for each $j$. This completes the proof for Lemma 2a.

Suppose that all closed and discrete subsets of the normal space $S$ are countable. It follows that any discrete collection of closed subsets of $S$ is countable. Then the collectionwise normality of $S$ follows from Lemma 2a. $\blacksquare$

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Proving Corson’s Theorem

If $Y \times Y$ is normal, then clearly $Y$ is normal. In light of Theorem 2, to show $Y$ is collectionwise normal, it suffices to show that every closed and discrete subspace of $Y$ is countable. Thus Theorem 1 is established by combining Theorem 2 and the following theorem.

Theorem 3

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. If $Y \times Y$ is normal, then every closed and discrete subspace of $Y$ is countable.

Before proving Theorem 3, we state one more lemma that is needed. For $C \subset A$, $\pi_C$ is the natural projection map from $X=\prod_{\alpha \in A} X_\alpha$ into $\prod_{\alpha \in C} X_\alpha$. The map $\pi_C \times \pi_C$ refers to the projection map from $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ into $\prod_{\alpha \in C} X_\alpha \times \prod_{\alpha \in C} X_\alpha$ defined by $(\pi_C \times \pi_C)(x,y)=(\pi_C(x),\pi_C(y))$.

Lemma 4

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 $H_1=(\pi_C \times \pi_C)(H)$ and $K_1=(\pi_C \times \pi_C)(K)$ are separated in $\pi_C(Y) \times \pi_C(Y)$, meaning that $H_1 \cap \overline{K_1}=\varnothing=\overline{H_1} \cap K_1$.

Lemma 4 deals with the dense subspace $Y$ of $X=\prod_{\alpha \in A} X_\alpha$ and the dense subspace $Y \times Y$ of $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$. So the map $\pi_C$ should be restricted to $Y$ and the map $\pi_C \times \pi_C$ is restricted to $Y \times Y$. For a proof of Lemma 4, see the proof of Lemma 2 in this previous post.

Proof of Theorem 3
Let $T=\left\{t_\alpha: \alpha<\omega_1 \right\}$ be an uncountable closed and discrete subset of $Y$. We define two disjoint closed subsets $H$ and $K$ of $Y \times Y$ such that for each countable set $C \subset A$, $(\pi_C \times \pi_C)(H) \cap \overline{(\pi_C \times \pi_C)(K)} \ne \varnothing$. By Lemma 4, $Y \times Y$ is not normal. Consider the following two subsets of $Y \times Y$:

$H=\left\{(t_\alpha,t_\alpha): \alpha<\omega_1 \right\}$

$K=\left\{(t_\delta,t_\rho): \delta,\rho<\omega_1 \text{ and } \delta \ne \rho \right\}$

Clearly $H$ and $K$ are disjoint. It is clear that $H$ is a closed subset of $Y \times Y$. Because $T$ is closed and discrete in $Y$, $K$ is a closed subset of $Y \times Y$. Thus $H$ and $K$ are disjoint closed subsets of $Y \times Y$.

Let $C \subset A$ be countable. Note that $(\pi_C \times \pi_C)(H)$ and $(\pi_C \times \pi_C)(K)$ are:

$(\pi_C \times \pi_C)(H)=\left\{(\pi_C(w_\alpha),\pi_C(w_\alpha)): \alpha<\omega_1 \right\}$

$(\pi_C \times \pi_C)(K)=\left\{(\pi_C(w_\delta),\pi_C(w_\rho)): \delta,\rho<\omega_1, \delta \ne \rho \right\}$

Consider $P=\left\{\pi_C(w_\alpha): \alpha<\omega_1 \right\}$. Clearly, $P \times P=(\pi_C \times \pi_C)(H)$. We consider two cases: $P$ is uncountable or $P$ is countable.

Case 1: $P$ is uncountable.
Note that $\prod_{\alpha \in C} X_\alpha$ is the product of countably many separable metric spaces and is therefore a hereditarily Lindelof space. As a subspace of $\prod_{\alpha \in C} X_\alpha$, $\pi_C(Y)$ is also hereditarily Lindelof. Since $P$ is an uncountable subspace of $\pi_C(Y)$, there must exist a point $p \in P$ such that every open set (open in $\pi_C(Y)$) containing $p$ must contain uncountably many points of $P$. Note that $(p,p) \in (\pi_C \times \pi_C)(H)$.

Let $O$ be an open subset of $\pi_C(Y)$ with $p \in O$. Then there exist $\gamma, \rho<\omega_1$ with $\gamma \ne \rho$ such that $\pi_B(w_\gamma) \in O$ and $\pi_B(w_\rho) \in O$. Then $(\pi_B(w_\gamma), \pi_B(w_\gamma))$ is a point of $(\pi_C \times \pi_C)(H)$ that is in $O \times O$. The point $(\pi_B(w_\gamma), \pi_B(w_\delta))$ is a point of $(\pi_C \times \pi_C)(K)$ that is in $O \times O$. This means that $(p,p) \in \overline{(\pi_C \times \pi_C)(K)}$. Thus we have $(\pi_C \times \pi_C)(H) \cap \overline{(\pi_C \times \pi_C)(K)} \ne \varnothing$.

Case 2: $P$ is countable.
Then there exists $p \in P$ such that $p=\pi_C(w_\alpha)$ for uncountably many $\alpha$. Choose $\gamma, \rho<\omega_1$ such that $\gamma \ne \rho$ and $p=\pi_C(w_\gamma)$ and $p=\pi_C(w_\rho)$. Then $(p,p) \in (\pi_C \times \pi_C)(H)$ and $(p,p) \in (\pi_C \times \pi_C)(K)$.

In either case, we can say that $(\pi_C \times \pi_C)(H) \cap \overline{(\pi_C \times \pi_C)(K)} \ne \varnothing$. By Lemma 4, $Y \times Y$ is not normal. $\blacksquare$

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Reference

1. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
2. Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
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.

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