Normal dense subspaces of products of “omega 1” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii (see Problem I.5.25 in [2]). One partial positive answer is a theorem attributed to Corson: if Y is a normal dense subspace of a product of separable spaces such that Y \times Y is normal, then Y is collectionwise normal. Another partial positive answer: assuming 2^\omega<2^{\omega_1}, any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). Another partial positive answer to Arkhangelskii’s question is the theorem due to Reznichenko: If C_p(X), which is a dense subspace of the product space \mathbb{R}^X, is normal, then it is collectionwise normal (see Theorem I.5.12 in [2]). In this post, we highlight another partial positive answer to the question posted in [2]. Specifically, we prove the following theorem:

Theorem 1

    Let X=\prod_{\alpha<\omega_1} X_\alpha be a product space where each factor X_\alpha is a separable metric space. Let Y be a dense subspace of X. Then if Y is normal, then Y is collectionwise normal.

Since any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post), it suffices to prove the following theorem:

Theorem 1a

    Let X=\prod_{\alpha<\omega_1} X_\alpha be a product space where each factor X_\alpha is a separable metric space. Let Y be a dense subspace of X. Then if Y is normal, then every closed and discrete subspace of Y is countable, i.e., Y has countable extent.

Arkhangelskii’s question was studied by the author of [3] and [4]. Theorem 1 as presented in this post is essentially the Theorem 1 found in [3]. The proof given in [3] is a beautiful proof. The proof in this post is modeled on the proof in [3] with the exception that all the crucial details are filled in. Theorem 1a (as stated above) is used in [1] to show that the function space C_p(\omega_1+1) contains no dense normal subspace.

It is natural to wonder if Theorem 1 can be generalized to product space of \tau many separable metric factors where \tau is an arbitrary uncountable cardinal. The work of [4] shows that the question at the beginning of this post cannot be answered positively in ZFC. Recall the above mentioned result that assuming 2^\omega<2^{\omega_1}, any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). A theorem in [4] implies that assuming 2^\omega=2^{\omega_1}, for any separable metric space M with at least 2 points, the product of continuum many copies of M contains a normal dense subspace Y that is not collectionwise normal. A side note: for this normal subspace Y, Y \times Y is necessarily not normal (according to Corson’s theorem). Thus [3] and [4] collectively show that Arkhangelskii’s question stated here at the beginning of the post is answered positively (in ZFC) among product spaces of \omega_1 many separable metric factors and that outside of the \omega_1 case, it is impossible to answer the question positively in ZFC.

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Proving Theorem 1a

We use the following lemma. For a proof of this lemma, see the proof for Lemma 1 in this previous post.

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 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.

For any B \subset \omega_1, let \pi_B be the natural projection from the product space X=\prod_{\alpha<\omega_1} X_\alpha into the subproduct space \prod_{\alpha \in B} X_\alpha.

Proof of Theorem 1a
Let Y be a dense subspace of the product space X=\prod_{\alpha<\omega_1} X_\alpha where each factor X_\alpha has a countable base. Suppose that D is an uncountable closed and discrete subset of Y. We then construct a pair of disjoint closed subsets H and K of Y such that for all countable B \subset \omega_1, \pi_B(H) and \pi_B(K) are not separated, specifically \pi_B(H) \cap \overline{\pi_B(K)}\ne \varnothing. Here the closure is taken in the space \pi_B(Y). By Lemma 2, the dense subspace Y of X is not normal.

For each \alpha<\omega_1, let \mathcal{B}_\alpha be a countable base for the space X_\alpha. The standard basic open sets in the product space X are of the form O=\prod_{\alpha<\omega_1} O_\alpha such that

  • each O_\alpha is an open subset of X_\alpha,
  • if O_\alpha \ne X_\alpha, then O_\alpha \in \mathcal{B}_\alpha,
  • O_\alpha=X_\alpha for all but finitely many \alpha<\omega_1.

We use supp(O) to denote the finite set of \alpha such that O_\alpha \ne X_\alpha. Technically we should be working with standard basic open subsets of Y, i.e., sets of the form O \cap Y where O is a standard basic open set as described above. Since Y is dense in the product space, every standard open set contains points of Y. Thus we can simply work with standard basic open sets in the product space as long as we are working with points of Y in the construction.

Let \mathcal{M} be the collection of all standard basic open sets as described above. Since there are only \omega_1 many factors in the product space, \lvert \mathcal{M} \lvert=\omega_1. Recall that D is an uncountable closed and discrete subset of Y. Let \mathcal{M}^* be the following:

    \mathcal{M}^*=\left\{U \in \mathcal{M}: U \cap D \text{ is uncountable }  \right\}

Claim 1. \lvert \mathcal{M}^* \lvert=\omega_1.

First we show that \mathcal{M}^* \ne \varnothing. Let B \subset \omega_1 be countable. Consider these two cases: Case 1. \pi_B(D) is an uncountable subset of \prod_{\alpha \in B} X_\alpha; Case 2. \pi_B(D) is countable.

Suppose Case 1 is true. Since \prod_{\alpha \in B} X_\alpha is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point y \in \pi_B(D) such that every open neighborhood of y (open in \prod_{\alpha \in B} X_\alpha) contains uncountably many points of \pi_B(D). Thus every standard basic open set U=\prod_{\alpha \in B} U_\alpha, with y \in U, contains uncountably many points of \pi_B(D). Suppose Case 2 is true. There exists one point y \in \pi_B(D) such that y=\pi_B(t) for uncountably many t \in D. Then in either case, every standard basic open set V=\prod_{\alpha<\omega_1} V_\alpha, with supp(V) \subset B and y \in \pi_B(V), contains uncountably many points of D. Any one such V is a member of \mathcal{M}^*.

We can partition the index set \omega_1 into \omega_1 many disjoint countable sets B. Then for each such B, obtain a V \in \mathcal{M}^* in either Case 1 or Case 2. Since supp(V) \subset B, all such open sets V are distinct. Thus Claim 1 is established.

Claim 2.
There exists an uncountable H \subset D such that for each U \in \mathcal{M}^*, U \cap H \ne \varnothing and U \cap (D-H) \ne \varnothing.

Enumerate \mathcal{M}^*=\left\{U_\gamma: \gamma<\omega_1 \right\}. Choose h_0,k_0 \in U_0 \cap D with h_0 \ne k_0. Suppose that for all \beta<\gamma, two points h_\beta,k_\beta are chosen such that h_\beta,k_\beta \in U_\beta \cap D, h_\beta \ne k_\beta and such that h_\beta \notin L_\beta and k_\beta \notin L_\beta where L_\beta=\left\{h_\rho: \rho<\beta \right\} \cup \left\{k_\rho: \rho<\beta \right\}. Then choose h_\gamma,k_\gamma with h_\gamma \ne k_\gamma such that h_\gamma,k_\gamma \in U_\gamma \cap D and h_\gamma \notin L_\gamma and k_\gamma \notin L_\gamma where L_\gamma=\left\{h_\rho: \rho<\gamma \right\} \cup \left\{k_\rho: \rho<\gamma \right\}.

Let H=\left\{h_\gamma: \gamma<\omega_1 \right\} and let K=D-H. Note that K_0=\left\{k_\gamma: \gamma<\omega_1 \right\} \subset K. Based on the inductive process that is used to obtain H and K_0, it is clear that H satisfies Claim 2.

Claim 3.
For each countable B \subset \omega_1, the sets \pi_B(H) and \pi_B(K) are not separated in the space \pi_B(Y).

Let B \subset \omega_1 be countable. Consider the two cases: Case 1. \pi_B(H) is uncountable; Case 2. \pi_B(H) is countable. Suppose Case 1 is true. Since \prod_{\alpha \in B} X_\alpha is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point p \in \pi_B(H) such that every open neighborhood of p (open in \prod_{\alpha \in B} X_\alpha) contains uncountably many points of \pi_B(H). Choose h \in H such that p=\pi_B(h). Then the following statement holds:

  1. For every basic open set U=\prod_{\alpha<\omega_1} U_\alpha with h \in U such that supp(U) \subset B, the open set U contains uncountably many points of H.

Suppose Case 2 is true. There exists some p \in \pi_B(H) such that p=\pi_B(t) for uncountably many t \in H. Choose h \in H such that p=\pi_B(h). Then statement 1 also holds.

In either case, there exists h \in H such that statement 1 holds. The open sets U described in statement 1 are members of \mathcal{M}^*. By Claim 2, the open sets described in statement 1 also contain points of K. Since the open sets described in statement 1 have supports \subset B, the following statement holds:

  1. For every basic open set V=\prod_{\alpha \in B} V_\alpha with \pi_B(h) \in V, the open set V contains points of \pi_B(K).

Statement 2 indicates that \pi_B(h) \in \overline{\pi_B(K)}. Thus \pi_B(h) \in \pi_B(H) \cap \overline{\pi_B(K)}. The closure here can be taken in either \prod_{\alpha \in B} X_\alpha or \pi_B(Y) (to apply Lemma 2, we only need the latter). Thus Claim 3 is established.

Claim 3 is the negation of condition 3 of Lemma 2. Therefore Y is not normal. \blacksquare

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Remark

The proof of Theorem 1a, though a proof in ZFC only, clearly relies on the fact that the product space is a product of \omega_1 many factors. For example, in the inductive step in the proof of Claim 2, it is always possible to pick a pair of points not chosen previously. This is because the previously chosen points form a countable set and each open set in \mathcal{M}^* contains \omega_1 many points of the closed and discrete set D. With the “\omega versus \omega_1” situation, at each step, there are always points not previously chosen. When more than \omega_1 many factors are involved, there may be no such guarantee in the inductive process.

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Reference

  1. Arkhangelskii, A. V., Normality and dense subspaces, Proc. Amer. Math. Soc., 130 (1), 283-291, 2001.
  2. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  3. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
  4. Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
  5. 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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