Sigma-products of separable metric spaces are collectionwise normal

Let \prod_{\alpha \in \omega_1} \mathbb{R}=\mathbb{R}^{\omega_1} be the Cartesian product of \omega_1 many copies of the real line. This product product space is not normal since it contains \prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1} as a closed subspace. The subspace of \mathbb{R}^{\omega_1} consisting of points which have at most countably many non-zero coordinates is collectionwise normal. Such spaces are called \Sigma-products. In this post, we show that the \Sigma-product of separable and metrizable spaces is always collectionwise normal. To place the result proved in this post in a historical context, see the comments at the end of the post.

Consider the product space X=\prod_{\alpha \in A} X_\alpha. Let a \in X. The \Sigma-product of the spaces \left\{X_\alpha \right\}_{\alpha \in A} about the base point a is the following subspace of X:

    \Sigma_{\alpha \in A} X_\alpha(a)=\left\{x \in X: x_\alpha \ne a_\alpha \text{ for at most countably many } \alpha \in A \right\}

When the base point a is understood, we denote the space by \Sigma_{\alpha \in A} X_\alpha. First we want to eliminate cases that are not interesting. If the index set A is countable, then the \Sigma-product is simply the Cartesian product. We assume that the index set A is uncountable. If all but countably many of the factors consist of only one point, then the \Sigma-product is also the Cartesian product. So we assume that each X_\alpha has at least 2 points. When these two assumptions are made, the resulting \Sigma-products are called proper.

The collectionwise normality of \Sigma_{\alpha \in A} X_\alpha is accomplished in two steps. First, \Sigma_{\alpha \in A} X_\alpha is shown to be normal if each factor X_\alpha is a separable metric space (Theorem 1). Secondly, observe that normality in \Sigma-product is countably productive, i.e., if Y=\Sigma_{\alpha \in A} X_\alpha is normal, then Y^\omega is also normal (Theorem 2). Then the collectionwise normality of \Sigma_{\alpha \in A} X_\alpha follows from a theorem attributed to Corson. We have the following theorems.

Theorem 1

    For each \alpha \in A, let X_\alpha be a separable and metrizable space. Then \Sigma_{\alpha \in A} X_\alpha is a normal space.

Theorem 2

    For each \alpha \in A, let X_\alpha be a separable and metrizable space. Let Y=\Sigma_{\alpha \in A} X_\alpha. Then Y^\omega is a normal space.

Theorem 3 (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.

    For a proof of Corson’s theorem, see this post.

The above three theorems lead to the following theorem.

Theorem 4

    For each \alpha \in A, let X_\alpha be a separable and metrizable space. Then \Sigma_{\alpha \in A} X_\alpha is a collectionwise normal space.

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Proofs

Before proving the theorems, let’s set some notations. For each B \subset A, \pi_B is the natural projection from \prod_{\alpha \in A} X_\alpha into \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.

The following lemma is used (for a proof, see Lemma 1 in this post).

Lemma 5

    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.

Proof of Theorem 1

Let X=\prod_{\alpha \in A} X_\alpha be the product space in question. Let Y=\Sigma_{\alpha \in A} X_\alpha be defined using the base point b \in X. Note that Y is dense in the product space X=\prod_{\alpha \in A} X_\alpha. In light of Lemma 5, to show Y is normal, it suffices to show that for each 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. Let H and K be disjoint closed subsets of Y.

Before building up to a countable set B, let’s set some notation that will be used along the way. For each y \in Y, let S(y) denote the set of all \alpha \in A such that y_\alpha \ne b_\alpha. For any set T \subset Y, let S(T)=\bigcup_{y \in T} S(y).

To start, choose \gamma \in A and let A_1=\left\{\gamma \right\}. Consider \pi_{A_1}(H) and \pi_{A_1}(K). They are subsets of \prod_{\alpha \in A_1} X_\alpha, which is a hereditarily separable space. Choose a countable D_1 \subset \pi_{A_1}(H) and a countable E_1 \subset \pi_{A_1}(K) such that \overline{D_1} = \pi_{A_1}(H) and \overline{E_1} = \pi_{A_1}(K). For each u \in D_1, choose f(u) \in H such that \pi_{A_1}(f(u))=u. For each v \in E_1, choose g(v) \in H such that \pi_{A_1}(g(v))=v. Let H_1 and K_1 be defined by:

    H_1=\left\{f(u): u \in D_1 \right\}
    K_1=\left\{g(v): v \in E_1 \right\}

Clearly \pi_{A_1}(H)=\overline{\pi_{A_1}(H_1)} and \pi_{A_1}(K)=\overline{\pi_{A_1}(K_1)}. Let A_2=A_1 \cup S(H_1) \cup S(K_1).

Now perform the next step inductive process. Consider \pi_{A_2}(H) and \pi_{A_2}(K). As before, we can find countable dense subsets of these 2 sets. Choose a countable D_2 \subset \pi_{A_2}(H) and a countable E_2 \subset \pi_{A_2}(K) such that \overline{D_2} = \pi_{A_2}(H) and \overline{E_2} = \pi_{A_2}(K). For each u \in D_2, choose f(u) \in H such that \pi_{A_2}(f(u))=u. For each v \in E_2, choose g(v) \in H such that \pi_{A_2}(g(v))=v. Let H_2 and K_2 be defined by:

    H_2=\left\{f(u): u \in D_2 \right\} \cup H_1
    K_2=\left\{g(v): v \in E_2 \right\} \cup K_1

Clearly \pi_{A_2}(H) \subset \overline{\pi_{A_2}(H_2)} and \pi_{A_2}(K) \subset \overline{\pi_{A_2}(K_2)}. To prepare for the next step, let A_3=A_1 \cup A_2 \cup S(H_2) \cup S(K_2).

Continue the inductive process and when completed, the following sequences are obtained:

  • a sequence of countable sets A_1 \subset A_2 \subset A_3 \subset \cdots \subset A
  • a sequence of countable sets H_1 \subset H_2 \subset H_3 \subset \cdots \subset H
  • a sequence of countable sets K_1 \subset K_2 \subset K_3 \subset \cdots \subset K

such that

  • \pi_{A_j}(H) \subset \overline{\pi_{A_j}(H_j)} and \pi_{A_j}(K) \subset \overline{\pi_{A_j}(K_j)} for each j
  • A_{j+1}=(\bigcup_{i \le j} A_i) \cup S(H_j) \cup S(K_j) for each j

Let B=\bigcup_{j=1}^\infty A_j, H^*=\bigcup_{j=1}^\infty H_j and K^*=\bigcup_{j=1}^\infty K_j. We have the following claims.

Claim 1
\pi_B(H) \subset \overline{\pi_B(H^*)} and \pi_B(K) \subset \overline{\pi_B(K^*)}.

Claim 2
\overline{\pi_B(H^*)} \cap \overline{\pi_B(K^*)}=\varnothing.

Proof of Claim 1
It suffices to show one of the set inclusions. We show \pi_B(H) \subset \overline{\pi_B(H^*)}. Let h \in H. We need to show that \pi_B(h) is a limit point of \pi_B(H^*). To this end, let V=\prod_{\alpha \in B} V_\alpha be a standard basic open set containing \pi_B(h). Then supp(V) \subset A_j for some j. Let V_j=\prod_{\alpha \in A_j} V_\alpha. Then \pi_{A_j}(h) \in V_j. Since \pi_{A_j}(H) \subset \overline{\pi_{A_j}(H_j)}, there is some t \in H_j such that \pi_{A_j}(t) \in V_j. It follows that \pi_B(t) \in V. Thus every open set containing \pi_B(h) contains a point of \pi_B(H^*).

Proof of Claim 2
Suppose that x \in \overline{\pi_B(H^*)} \cap \overline{\pi_B(K^*)}. Define y \in Y=\Sigma_{\alpha \in A} X_\alpha such that y_\alpha=x_\alpha for all \alpha \in B and y_\alpha=b_\alpha for all \alpha \in A-B. It follows that y \in \overline{H} \cap \overline{K}=H \cap K, which is a contradiction since H and K are disjoint closed sets. To see that y \in H \cap K, let W=\prod_{\alpha \in A} W_\alpha be a standard basic open set containing y. Let W_1=\prod_{\alpha \in B} W_\alpha. Since x \in W_1, there exist h \in H^* and k \in K^* such that \pi_B(h) \in W_1 and \pi_B(k) \in W_1. Note that the supports S(h) \subset B and S(k) \subset B. For the coordinates outside of B, both h and k agree with the base point b and hence with y. Thus h \in W and k \in W. We have just shown that every open set containing y contains a point of H and a point of K. This means that y \in H \cap K, a contradiction. This completes the proof of Claim 2.

Both Claim 1 and Claim 2 imply that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing. By Lemma 5, Y=\Sigma_{\alpha \in A} X_\alpha is normal. \blacksquare

Proof of Theorem 2

Let Y=\Sigma_{\alpha \in A} X_\alpha be the \Sigma-product about the base point b \in \prod_{\alpha \in A} X_\alpha. The following countable product

    \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a)

is a product of separable metric spaces. So any \Sigma-product that can be defined within the product space (a) is normal (by Theorem 1). In particular, consider the \Sigma-product defined about the base point c=(b, b, b, \cdots) (countably many coordinates). Denote this \Sigma-product by T. Observe that T is homeomorphic to the following countable product of Y=\Sigma_{\alpha \in A} X_\alpha:

    \Sigma_{\alpha \in A} X_\alpha \times \Sigma_{\alpha \in A} X_\alpha \times \Sigma_{\alpha \in A} X_\alpha \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (b)

Thus T can be identified with Y^\omega. We can conclude that Y^\omega is normal. \blacksquare

Proof of Theorem 4

Let Y=\Sigma_{\alpha \in A} X_\alpha be the \Sigma-product of the separable metric spaces X_\alpha. By Theorem 1, Y is normal. By Theorem 2, Y^\omega is normal. In particular, Y \times Y is normal. Clearly, Y is a dense subspace of the product space X=\prod_{\alpha \in A} X_\alpha. By Corson’s theorem (Theorem 3), Y is collectionwise normal. \blacksquare

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A Brief History

The notion of \Sigma-products was introduced by Corson in [1] where he proved that the \Sigma-product of complete metric spaces is normal. Corson then asked whether the \Sigma-product of copies of the rationals is normal. In 1973, Kombarov and Malyhin [4] showed that the \Sigma-product of separable metric spaces is normal. In 1977, Gulko [2] and Rudin [6] independently proved the \Sigma-product of metric spaces is normal. In 1978, Kombarov [3] generalized Gulko and Rudin’s result by showing that any \Sigma-product of paracompact p-spaces \left\{X_\alpha: \alpha \in A \right\} is collectionwise normal if and only if all spaces X_\alpha have countable tightness. A useful resource is Przymusinski’s chapter in the Handbook of Set-Theoretic Topology [5], which has a section on \Sigma-products.

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Reference

  1. Corson H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
  2. Gulko S. P., On the properties of sets lying in \Sigma-products, Dokl. Acad. Nauk. SSSR, 237, 505-508, 1977 (in Russian).
  3. Kombarov A. P., On the tightness and normality of \Sigma-products, Dokl. Acad. Nauk. SSSR, 239, 775-778, 1978 (in Russian).
  4. Kombarov A. P., Malyhin V. I., On \Sigma-products, Soviet Math. Dokl., 14 (6), 1980-1983, 1973.
  5. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
  6. Rudin M. E., Book Review, Bull. Amer. Math. Soc., 84, 271-272, 1978.

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\copyright \ 2014 \text{ by Dan Ma}

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