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