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}

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