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

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

# Revisit a lemma dealing with normality in products of separable metric spaces

In this post we continue to discuss a lemma that has been discussed previously in this post. The lemma characterizes the dense normal subspaces of a product of separable metric spaces. The lemma discussed here has been sharpened over the version in the previous post. Two versions of the lemma are given (Lemma 1 and Lemma 2). Any one of these two versions can be used to prove that the $\Sigma$-product of separable metric spaces is normal (see this blog post).

Lemma 1

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 $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$.

The closure in condition 2 and condition 3 is taken in $\pi_B(Y)$. The map $\pi_B$ is the natural projection from the full product space $X=\prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in B} X_\alpha$.

Proof of Lemma 1
$1 \Longrightarrow 2$
Let $H$ and $K$ be disjoint closed subsets of $Y$. Since $Y$ is normal, there exists a continuous function $f: Y \rightarrow [0,1]$ such that $f(H) \subset \left\{0 \right\}$ and $f(H) \subset \left\{1 \right\}$. By Theorem 1 in this previous post, the continuous function $f$ depends on countably many coordinates. This means that there exists a countable $B \subset A$ and there exists a continuous $g:\pi_B(Y) \rightarrow [0,1]$ such that $f= g \circ \pi_B$. The continuity on the full product space is now reduced to the continuity on a countable subproduct. Now $O_H=g^{-1}([0,0.2))$ and $O_K=g^{-1}((0.8,1])$ are disjoint open sets in $\pi_B(Y)$. Since $f= g \circ \pi_B$, it is the case that $\pi_B(H) \subset O_H$ and $\pi_B(K) \subset O_K$. Since $g$ is continuous, we have

$\overline{O_H}=\overline{g^{-1}([0,0.2))} \subset g^{-1}(\overline{[0,0.2)})=g^{-1}([0,0.2]) \ \ \ \ \ \ \ \ (a)$

$\overline{O_K}=\overline{g^{-1}((0.8,1])} \subset g^{-1}(\overline{(0.8,1]})=g^{-1}([0.8,1]) \ \ \ \ \ \ \ \ (b)$

Note that $\overline{\pi_B(H)} \subset \overline{O_H}$ and $\overline{\pi_B(K)} \subset \overline{O_K}$. If $\overline{\pi_B(H)} \cap \overline{\pi_B(K)} \ne \varnothing$, then $g^{-1}([0,0.2]) \cap g^{-1}([0.8,1]) \ne \varnothing$. Thus $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$.

The direction $2 \Longrightarrow 3$ is immediate.

The direction $3 \Longrightarrow 1$ follows from Lemma 1 in this previous post (see the direction $2 \rightarrow 1$ of Lemma 1 in the previous post). $\blacksquare$

The following lemma is another version of Lemma 1 which may be useful in some circumstances. For $B \subset A$, let $\pi_B \times \pi_B$ be the projection map from $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ into $\prod_{\alpha \in B} X_\alpha \times \prod_{\alpha \in B} X_\alpha$ defined by $(\pi_B \times \pi_B)(x,y)=(\pi_B(x),\pi_B(y))$.

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. $Y \times Y$ 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 $(\pi_C \times \pi_C)(H)$ and $(\pi_C \times \pi_C)(K)$ are separated in $\pi_C(Y) \times \pi_C(Y)$.

Proof of Lemma 2
$1 \Longrightarrow 2$
Let $H$ and $K$ be disjoint closed subsets of $Y \times Y$. Since $Y \times Y$ is normal, there exists a continuous function $f: Y \times Y \rightarrow [0,1]$ such that $f(H) \subset \left\{0 \right\}$ and $f(H) \subset \left\{1 \right\}$. By Theorem 2 in this previous post, the continuous function $f$ depends on countably many coordinates. This means that there exists a countable $C \subset A$ and there exists a continuous $g:\pi_C(Y) \times \pi_C(Y) \rightarrow [0,1]$ such that $f= g \circ (\pi_C \times \pi_C)$. Now $O_H=g^{-1}([0,0.2))$ and $O_K=g^{-1}((0.8,1])$ are disjoint open sets in $\pi_C(Y) \times \pi_C(Y)$. Since $f= g \circ (\pi_C \times \pi_C)$, it is the case that $(\pi_C \times \pi_C)(H) \subset O_H$ and $(\pi_C \times \pi_C)(K) \subset O_K$.

Since $g$ is continuous, conditions (a) and (b) in the proof of Lemma 1 also hold here. Note that $\overline{(\pi_C \times \pi_C)(H)} \subset \overline{O_H}$ and $\overline{(\pi_C \times \pi_C)(K)} \subset \overline{O_K}$. It follows that $\overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing$.

The direction $2 \Longrightarrow 3$ is immediate.

$3 \Longrightarrow 1$
Let $H$ and $K$ be disjoint closed subsets of $Y \times Y$. By condition 3, there exists a countable $C \subset A$ such that $F_H=(\pi_C \times \pi_C)(H)$ and $F_K=(\pi_C \times \pi_C)(K)$ are separated in $M=\pi_C(Y) \times \pi_C(Y)$. Note that $\overline{F_H} \cap F_K=\varnothing$ and $F_H \cap \overline{F_K}=\varnothing$. Consider the following subspace of $M$.

$W=M-\overline{F_H} \cap \overline{F_K}$

The space $W$ is an open subspace of $M$. The space $M$ is a subspace of a product of countably many separable metric spaces. Thus both $M$ and $W$ are also second countable and hence normal.

For $L \subset W$, let $Cl_W(L)$ denote the closure of $L$ in the space $W$. Both $Cl_W(F_H)$ and $Cl_W(F_K)$ are disjoint closed subsets of $W$. Let $G_H$ and $G_K$ be disjoint open subsets of $W$ with $Cl_W(F_H) \subset G_H$ and $Cl_W(F_K) \subset G_K$. Then $\pi_B^{-1}(G_H) \cap Y$ and $\pi_B^{-1}(G_K) \cap Y$ are disjoint open subsets of $Y$ separating $H$ and $K$. $\blacksquare$

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Remark

The countable sets in both Lemma 1 and Lemma 2 can be expanded to larger countable sets. For example,

for Lemma 1, for any disjoint closed subsets $H$ and $K$ of $Y$:

1. If for some countable set $B$, $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$, then $\overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing$ for any countable $E \subset A$ with $B \subset E$.
2. If for some countable set $B$, $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$, then $\overline{\pi_E(H)} \cap \pi_E(K)=\pi_E(H) \cap \overline{\pi_E(K)}=\varnothing$ for any countable $E \subset A$ with $B \subset E$.

It is straightforward to verify these facts. For the sake of completeness, we verify condition 2. Suppose that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$ for some countable $B \subset A$. Let $E \subset A$ be countable with $B \subset E$. We show $\overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing$. Suppose $x \in \overline{\pi_E(H)} \cap \overline{\pi_E(K)}$. Then $\pi_B(x) \notin \overline{\pi_B(H)} \cap \overline{\pi_B(K)}$. Choose some standard basic open set $O=\prod_{\alpha \in B} O_\alpha$ with $\pi_B(x) \in O$ such that $O \cap \overline{\pi_B(H)}=\varnothing$ and $O \cap \overline{\pi_B(K)}=\varnothing$. Consider $O_1=\prod_{\alpha \in E} O_\alpha$ such that $O_\alpha=X_\alpha$ for all $\alpha \in C-B$. Clearly $x \in O_1$. Then there exist $h \in H$ and $k \in K$ such that $\pi_E(h) \in O_1$ and $\pi_E(h) \in O_1$. It follows that $\pi_B(h) \in O_1$ and $\pi_B(h) \in O$, a contradiction. Thus $\overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing$.

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$\copyright \ 2014 \text{ by Dan Ma}$
Revised 3/31/2014.

# A factorization theorem for products of separable spaces

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $f:X \rightarrow T$ be a continuous function where $T$ is a topological space. In this post, we discuss what it means for the continuous function $f$ to depend on countably many coordinates and then discuss some conditions that we can impose on the product space and on the range space $T$ to ensure that every continuous $f$ defined on the product space will depend on countably many coordinates. This notion of a continuous function depending on countably many coordinates is equivalent to factoring the continuous function into the composition of a projection map and a continuous function defined on a countable subproduct (see Lemma 1 below).

Let’s set some notation about the product space we work with in this post. Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $f:X \rightarrow T$ be continuous. For any $B \subset A$, $\pi_B$ is the natural projection from the full product space $X=\prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in B} X_\alpha$. Standard basic open sets of $X=\prod_{\alpha \in A} X_\alpha$ are of the form $\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is open in $X_\alpha$ and that $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. We use $supp(\prod_{\alpha \in A} O_\alpha)$ to denote the finite set of $\alpha \in A$ where $O_\alpha \ne X_\alpha$.

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Factoring a Continuous Map

The function $f$ is said to depend on countably many coordinates if there exists a countable set $B \subset A$ such that for any $x,y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$. Suppose $f$ is instead defined on a subspace $Y$ of $X$. The function $f$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x,y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.

We have the following lemmas.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $f:X \rightarrow T$ be continuous. Then the following are equivalent.

1. There exists a countable $B \subset A$ such that for any $x,y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.
2. There exists a countable $B \subset A$ such that $f=g \circ \pi_B$ where $g: \prod_{\alpha \in B} X_\alpha \rightarrow T$ is continuous.

Lemma 1a

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $Y$ be a dense subspace of $X$. Let $f:Y \rightarrow T$ be continuous. Then the following are equivalent.

1. There exists a countable $B \subset A$ such that for any $x,y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.
2. There exists a countable $B \subset A$ such that $f=g \circ \pi_B$ where $g: \pi_B(Y) \rightarrow T$ is continuous.

It is straightforward to verify Lemma 1 and Lemma 1a. We use condition 1 to define what it means for a function to be dependent on countably many coordinates. Both lemmas indicate that either condition is a valid definition. These two lemmas also indicate why the notion being discussed can be called a factorization notion.

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When a Continuous Map Can Be Factored

We discuss some conditions that we can place on the product space $X=\prod_{\alpha \in A} X_\alpha$ and on the range space $T$ so that any continuous map depends on countably many coordinates. We prove the following theorem.

Theorem 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space such that each factor $X_\alpha$ is a separable space. Let $T$ be a second countable space (i.e. having a countable base). Then for any dense subspace $Y$ of $X$, any continuous function $f:Y \rightarrow T$ depends on countably many coordinates, i.e., either one of the conditions in Lemma 1a holds.

Before stating the main theorem, we need one more lemma. Let $W \subset X=\prod_{\alpha \in A} X_\alpha$. The set $W$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x \in W$ and for any $y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in W$.

When we try to determine whether a function $f:Y \rightarrow T$, where $Y \subset X$, can be factored, we will need to decide whether a set $W \subset Y$ depends on countably many coordinates. Let $W \subset Y \subset X=\prod_{\alpha \in A} X_\alpha$. The set $W$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x \in W$ and for any $y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in W$. We have the following lemma.

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space with the countable chain condition. Let $Y$ be a dense subspace of $X$.

1. Let $U$ be an open subset of $X$. Then $\overline{U}$ depends on countably many coordinates.
2. Let $W$ be an open subset of $Y$. Then $\overline{W}$ depends on countably many coordinates (closure in $Y$).

Proof of Lemma 2
Proof of Part 1
Let $U \subset X$ be open. Let $\mathcal{B}$ be a collection of pairwise disjoint open subsets of the open set $U \subset X$ such that $\mathcal{B}$ is maximal with this property, i.e., if you throw one more open set into $\mathcal{B}$, it will be no longer pairwise disjoint. Let $V=\bigcup \mathcal{B}$. Since $\mathcal{B}$ is maximal, $\overline{V}=\overline{U}$. Since $X$ has the countable chain condition, $\mathcal{B}$ is countable.

Let $B=\bigcup \left\{supp(O): O \in \mathcal{B} \right\}$. The set $B$ is a countable subset of $A$ since $B$ is the union of countably many finite sets. We have the following claims.

Claim 1
The open set $V$ depends on the coordinates in $B$.

Let $x \in V$ and $y \in X$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show that $y \in V$. Firstly, $x \in O$ for some $O \in \mathcal{B}$. It follows that $x_\alpha=y_\alpha$ for all $\alpha \in supp(O)$. Thus $y \in O \subset V$. This completes the proof of Claim 1.

Claim 2
The set $\overline{V}$ depends on the coordinates in $B$.

Let $x \in \overline{V}$ and $y \in X$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show $y \in \overline{V}$. To this end, let $O=\prod_{\alpha \in A} O_\alpha$ be a standard basic open set with $y \in O$. The goal is to find some $q \in O \cap V$. Define $G=\prod_{\alpha \in A} G_\alpha$ such that $G_\alpha=O_\alpha$ for all $\alpha \in B$ and $G_\alpha=X_\alpha$ for all $\alpha \in A-B$. Then $x \in G$. Since $x \in \overline{V}$, there exists some $p \in V \cap G$. Define $q$ such that $q_\alpha=p_\alpha$ for all $\alpha \in B$ and $q_\alpha=y_\alpha$ for all $\alpha \in A-B$. Since $supp(V) \subset B$, $q \in V$. On the other hand, $q \in O$. This completes the proof of Claim 2.

As noted above, $\overline{V}=\overline{U}$. Thus $\overline{U}$ depends on countably many coordinates, namely the coordinates in the set $B$. This completes the proof of Part 1.

Proof of Part 2
For any $S \subset X$, let $\overline{S}$ denote the closure of $S$ in $Y$. Let $Cl_X(S)$ denote the closure of $S$ in $X$. Let $W \subset Y$ be open. Let $W_1$ be open in $X$ such that $W=W_1 \cap Y$. By Part 1, $Cl_X(W_1)$ depends on countably many coordinates, say the coordinates in the countable set $B \subset A$. This means that for any $x \in Cl_X(W_1)$ and for any $y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in Cl_X(W_1)$. Thus for any $x \in \overline{W}$ and for any $y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in Cl_X(W_1)$. If we have $y \in \overline{W}$, then we are done. So we only need to show that if $y \in Y$ and $y \in Cl_X(W_1)$, then $y \in \overline{W}$. This is why we need to assume $Y$ is dense in $X$.

Let $y \in Y$ and $y \in Cl_X(W_1)$. Let $C$ be an open subset of $Y$ with $y \in C$. There exists an open subset $D$ of $X$ such that $C=D \cap Y$. Then $D \cap Cl_X(W_1) \ne \varnothing$. Note that $D \cap W_1$ is open and $D \cap W_1 \ne \varnothing$. Since $Y$ is dense in $X$, $D \cap W_1$ must contain points of $Y$. These points of $Y$ are also points of $W$. Thus $C$ contains points of $W$. It follows that $y \in \overline{W}$. This concludes the proof of Part 2. $\blacksquare$

Proof of Theorem 1
Let $Y$ be a dense subspace of $X=\prod_{\alpha \in A} X_\alpha$. Let $f:Y \rightarrow T$ be continuous. Let $\mathcal{M}$ be a countable base for the separable metrizable space $T$. By Lemma 2 Part 2, for each $M \in \mathcal{M}$, $\overline{f^{-1}(M)}$ depends on countably many coordinates, say the countable set $B_M$. Let $B=\bigcup_{M \in \mathcal{M}} B_M$.

We claim that $B$ is a countable set of coordinates we need. Let $x,y \in Y$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show that $f(x)=f(y)$. Suppose $f(x) \ne f(y)$. Choose $\left\{M_1,M_2,M_3,\cdots \right\} \subset \mathcal{M}$ such that

• $\left\{f(x) \right\}=\bigcap_{j=1}^\infty M_j=\bigcap_{j=1}^\infty \overline{M_j}$
• $\overline{M_{j+1}} \subset M_j$ for each $j$

This is possible since $T$ is a second countable space. Then $f(y) \notin \overline{M_{k}}$ for some $k$. Furthermore, $y \notin f^{-1}(\overline{M_{k}})$. Since $f$ is continuous, $\overline{f^{-1}(M_{k})} \subset f^{-1}(\overline{M_{k}})$. Therefore, $y \notin \overline{f^{-1}(M_{k})}$. On the other hand, $\overline{f^{-1}(M_{k})}$ depends on the countably many coordinates in $B_{M_k}$. We assume above that $x_\alpha=y_\alpha$ for all $\alpha \in B$. Thus $x_\alpha=y_\alpha$ for all $\alpha \in B_{M_k}$. This means that $y \in \overline{f^{-1}(M_{k})}$, a contradiction. It must be that case that $f(x)=f(y)$. $\blacksquare$

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Another Version

We state another version of Theorem 1 that will be useful in some situations.

Theorem 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space such that each factor $X_\alpha$ is a separable space. Let $T$ be a second countable space. Let $Y$ be a dense subspace of $X$. Let $f:Y \times Y \rightarrow T$ be any continuous function. Then the function $f$ depends on countably many coordinates, which means either one of the following two conditions:

1. There exists a countable set $C \subset A$ such that for any $(x,y),(p,q) \in Y \times Y$, if $x_\alpha=p_\alpha$ and $y_\alpha=q_\alpha$ for all $\alpha \in C$, then $f(x,y)=f(p,q)$.
2. There exists a countable set $C \subset A$ and there exists a continuous $g:\pi_C(Y) \times \pi_C(Y) \rightarrow T$ such that $f=g \circ (\pi_C \times \pi_C)$.

The map $\pi_C \times \pi_C$ is the projection map from $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in C} X_\alpha \times \prod_{\alpha \in B} X_\alpha$ defined by $(\pi_C \times \pi_C)(x,y)=(\pi_C(x),\pi_C(y))$. In Theorem 2, we only need to consider $\pi_C \times \pi_C$ being defined on the subspace $Y \times Y$.

Theorem 2 follows from Theorem 1. It is only a matter of fitting Theorem 2 in the framework of Theorem 1. Note that the product $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ is identical to the product $\prod_{\alpha \in A \cup A^*} X_\alpha$ where $A^*$ is a disjoint copy of the index set $A$. For $(x,y) \in \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$, let $x \times y \in \prod_{\alpha \in A \cup A^*} X_\alpha$ be defined by $(x \times y)_\alpha=x_\alpha$ for all $\alpha \in A$ and $(x \times y)_\alpha=y_\alpha$ for all $\alpha \in A^*$.

With the identification of $(x,y)$ with $x \times y$, we have a setting that fits Theorem 1. The product $\prod_{\alpha \in A \cup A^*} X_\alpha$ is also a product of separable spaces. The set $Y \times Y$ is a dense subspace of the product $\prod_{\alpha \in A \cup A^*} X_\alpha$. In this new setting, we view a point in $Y \times Y$ as $x \times y$. The map $f:Y \times Y \rightarrow T$ is still a continuous map. We can now apply Theorem 1.

Let $B \subset A \cup A^*$ be a countable set such that for all $x \times y,p \times q \in Y \times Y$, if $(x \times y)_\alpha=(p \times q)_\alpha$ for all $\alpha \in B$, then $f(x \times y)=f(p \times q)$. Specifically, if $x_\alpha=p_\alpha$ for all $\alpha \in B \cap A$ and $y_\alpha=q_\alpha$ for all $\alpha \in B \cap A^*$, then $f(x,y)=f(p,q)$.

Choose a countable set $C \subset A$ such that $B \cap A \subset C$ and $B \cap A^* \subset C^*$. Here, $C^*$ is the copy of $C$ in $A^*$. We claim that $C$ is a countable set we need in condition 1 of Theorem 2. Let $(x,y),(p,q) \in Y \times Y$ such that $x_\alpha=p_\alpha$ and $y_\alpha=q_\alpha$ for all $\alpha \in C$. This implies that $x_\alpha=p_\alpha$ for all $\alpha \in B \cap A$ and $y_\alpha=q_\alpha$ for all $\alpha \in B \cap A^*$. Then $f(x,y)=f(p,q)$. Thus condition 1 of Theorem 2 holds. It is also straightforward to verify that Condition 1 and Condition 2 are equivalent.

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Remarks

The notion of factorizing a continuous map defined on a product space is an old topic. Theorem 1 discussed in this post is based on Theorem 4 found in [6]. Theorem 4 found in [6] is to factor continuous maps defined on a product of separable spaces. Theorem 1 in this post is modified to consider continuous maps defined on a dense subspace of a product of separable spaces. This modification will make it more useful. The references listed below represent a small sample of papers or books that have involves theorems of factoring functions defined on products. The work in [3] and [5] have more systematic treatment.

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Reference

1. Brandenburg H., Husek M., On mappings from products into developable spaces, Topology Appl., 26, 229-238, 1987.
2. Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Engelking R., On functions defined on Cartesian products, Fund. Math., 59, 221-231, 1966.
4. Keesling J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
5. Noble N., Ulmer M., Factoring functions on Cartesian products, Trans. Amer. Math. Soc., 163, 329-339, 1972.
6. Ross K. A., Stone A. H., Products of separable spaces, Amer. Math. Monthly, 71, 398-403, 1964.

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

# A lemma dealing with normality in products of separable metric spaces

In this post we prove a lemma that is a great tool for working with product spaces of separable metrizable spaces. As an application of the lemma, we give an alternative proof for showing the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$ where each $X_\alpha$ is a separable and metrizable space. The lemma we discuss here is a tool that can shed some light on normality of dense subspaces of the product space $X$. The lemma is stated in two equivalent forms (Lemma 1 and Lemma 2).

Before stating the lemmas, let’s fix some notations. For any $B \subset A$, the map $\pi_B$ is the natural projection from the full product $X=\prod_{\alpha \in A} X_\alpha$ to the subproduct $\prod_{\alpha \in B} X_\alpha$. 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 $O_\alpha \ne X_\alpha$.

Given a space $W$, and given $F,G \subset W$, the sets $F$ and $G$ are separated if $F \cap \overline{G}=\varnothing=\overline{F} \cap G$.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. For any sets $H,K \subset Y$, the following two conditions are equivalent:

1. There exist disjoint open subsets $U$ and $V$ of $Y$ such that $H \subset U$ and $K \subset V$.
2. There exists a countable $B \subset A$ such that the sets $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$.

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 $Y$ is normal if and only if for each 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 $\pi_B(Y)$.

If Lemma 1 holds, it is clear that Lemma 2 holds. We prove Lemma 1. The lemmas indicate that to separate disjoint sets in the full product, it suffices to separate in a countable subproduct. In this sense normality in dense subspaces of a product of separable metrizable spaces only depends on countably many coordinates.

This lemma seems to have been around for a long time. We cannot find any reference of this lemma in Engelking’s topology textbook (see [4]). We found three references. One is Corson’s paper (see [3]), in which the lemma is mentioned in relation to the non-normality of $\mathbb{N}^{\omega_1}$ and is attributed to a paper of M. Bockstein in 1948. Another is a paper of Baturov (see [2]), in which the lemma is used to prove a theorem about normality in dense subspace of $M^{\omega_1}$ where $M$ is a separable metric space. In [2] the lemma is attributed to Uspenskii. Another reference is Arkhangelskii’s book on function space (see Lemma I.6.1 on p. 43 in [1]) where the lemma is used in proving some facts about normality in function spaces $C_p(X)$.

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Proof of Lemma 1

$1 \Longrightarrow 2$
Let $U$ and $V$ be disjoint open subsets of $Y$ with $H \subset U$ and $K \subset V$. Let $U_1$ and $V_1$ be open subsets of $X$ such that $U=U_1 \cap Y$ and $V=V_1 \cap Y$. Since $Y$ is dense in $X$, $U_1 \cap V_1=\varnothing$.

Let $\mathcal{U}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $U_1$. Let $\mathcal{V}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $V_1$. These two collections can be obtained using a Zorn lemma argument. The product space $X$ has the countable chain condition since it is a product of separable spaces. So both $\mathcal{U}$ and $\mathcal{V}$ are countable. Let $B$ be the union of finite sets each one of which is a $supp(O)$ where $O \in \mathcal{U} \cup \mathcal{V}$. The set $B$ is countable too.

Let $U^*=\cup \mathcal{U}$ and $V^*=\cup \mathcal{V}$. Note that $U^* \cap V^*=\varnothing$. We have the following observations:

$\pi^{-1}_B(\pi_B(U^*))=U^* \subset U_1$ and $\pi^{-1}_B(\pi_B(V^*))=V^* \subset V_1$

The above observations lead to the following observations:

$\pi^{-1}_B(\pi_B(U^*)) \cap \pi^{-1}_B(\pi_B(V^*)) \subset U_1 \cap V_1=\varnothing$

implying that $\pi_B(U^*) \cap \pi_B(V^*)=\varnothing$. Both $\pi_B(U^*)$ and $\pi_B(V^*)$ are open subsets of $\pi_B(X)$ and are dense in $\pi_B(X)$, respectively.

We claim that $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$. Suppose that $y \in \pi_B(U_1) \cap \pi_B(V_1)$. Then $\pi_B(V_1)$ contains a point of $\pi_B(U^*)$, say $t$. With $t \in \pi_B(U^*)$, $t=\pi_B(q)$ for some $q \in O$ where $O \in \mathcal{U}$. Note that $supp(O) \subset B$. Thus $\pi^{-1}_B(\pi_B(q))=\pi^{-1}_B(t)=O \subset U_1$. On the other hand, $t \in \pi_B(V_1)$ implies that $t=\pi_B(w)$ for some $w \in V_1$. It follows that $w \in U_1 \cap V_1$, a contradiction. Therefore $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$.

We have $\pi_B(H) \subset \pi_B(U) \subset \pi_B(U_1)$ and $\pi_B(K) \subset \pi_B(V) \subset \pi_B(V_1)$. This implies that $\overline{\pi_B(H)} \cap \pi_B(K)=\varnothing$ and $\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$ (closure in $\pi_B(X)$). Then $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$ as well. This concludes the proof for the $1 \Longrightarrow 2$ direction.

$2 \Longrightarrow 1$
Suppose that $B \subset A$ is countable such that $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$. Note that $\pi_B(H) \subset \pi_B(Y)$ and $\pi_B(K) \subset \pi_B(Y)$. Then we have the following:

$\pi_B(H) \cup \pi_B(K) \subset \pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)}) \ \ \ \ \text{closures in } \pi_B(Y)$

Consider $W=\pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)})$. The space $W$ is an open subspace of $\pi_B(Y)$. Furthermore, $\pi_B(Y)$ is a subspace of $\prod_{\alpha \in B} X_\alpha$, which is a separable and metrizable space. Thus the space $W$ is metrizable and hence normal.

For $L \subset W$, let $Cl_W(L)$ denote the closure of $L$ in the space $W$. Note that $Cl_W(\pi_B(H))$ and $Cl_W(\pi_B(K))$ are disjoint and closed sets in $W$. Let $G_H$ and $G_K$ be disjoint open subsets of $W$ such that $Cl_W(\pi_B(H)) \subset G_H$ and $Cl_W(\pi_B(K)) \subset G_K$. Then $\pi^{-1}_B(G_H) \cap Y$ and $\pi^{-1}_B(G_K) \cap Y$ are disjoint open subsets of $Y$ such that $H \subset \pi^{-1}_B(G_H) \cap Y$ and $K \subset \pi^{-1}_B(G_H) \cap Y$. $\blacksquare$

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Remark

The proof of Lemma 1 does not need the full strength of separable metric in each factor of the product space. The above proof only makes two assumptions about the product space: the product space $X=\prod_{\alpha \in A} X_\alpha$ has the countable chain condition (CCC) and that any countable subproduct is normal, i.e., $\prod_{\alpha \in B} X_\alpha$ is normal for any countable $B \subset A$.

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Example

As an application of the above lemma, we give another proof of the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers. See this post for a version of A. H. Stone’s original proof.

Let $\mathbb{N}$ be the set of all nonnegative integers and let $\omega_1$ be the first uncountable ordinal (i.e. the set of all countable ordinals). We provide an alternative proof that $\mathbb{N}^{\omega_1}$ is not normal. In A. H. Stone’s proof, the following disjoint closed sets cannot be separated in $\mathbb{N}^{\omega_1}$:

$H=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 0, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

$K=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 1, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

We can also use Lemma 1 to show that $H$ and $K$ cannot be separated. Note that for each countable $B \subset \omega_1$, the sets $\pi_B(H)$ and $\pi_B(K)$ have non-empty intersection. Hence they cannot be separated in $\pi_B(\mathbb{N}^{\omega_1})$. By Lemma 1, $H$ and $K$ cannot be separated in the full product space $\mathbb{N}^{\omega_1}$.

To see that $\pi_B(H) \cap \pi_B(K) \ne \varnothing$, choose a function $g:\omega_1 \rightarrow \mathbb{N}$ such that $g^{-1}(0) \cap B=\varnothing$. Let $g_B:B \rightarrow \mathbb{N}$ be defined by $g_B(\alpha)=g(\alpha)$ for all $\alpha \in B$. Then $g_B \in \pi_B(H) \cap \pi_B(K)$.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
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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$\copyright \ 2014 \text{ by Dan Ma}$

# The normality of the product of the first uncountable ordinal with a compact factor

The product of a normal space with a compact space needs not be normal. For example, the product space $\omega_1 \times (\omega_1+1)$ is not normal where $\omega_1$ is the first uncountable ordinal with the order topology and $\omega_1+1$ is the immediate successor of $\omega_1$ (see this post). However, $\omega_1 \times I$ is normal where $I=[0,1]$ is the unit interval with the usual topology. The topological story here is that $I$ has countable tightness while the compact space $\omega_1+1$ does not. In this post, we prove the following theorem:

Theorem 1

Let $Y$ be an infinite compact space. Then the following conditions are equivalent:

1. The product space $\omega_1 \times Y$ is normal.
2. $Y$ has countable tightness, i.e., $t(Y)=\omega$.

Theorem 1 is a special case of the theorem found in [4]. The proof for the direction of countable tightness of $Y$ implies $\omega_1 \times Y$ is normal given in [4] relies on a theorem in another source. In this post we attempt to fill in some of the gaps. For the direction $2 \Longrightarrow 1$, we give a complete proof. For the direction $1 \Longrightarrow 2$, we essentially give the same proof as in [4], proving it by using a series of lemmas (stated below).

The authors in [2] studied the normality of $X \times \omega_1$ where $X$ is not necessarily compact. The necessary definitions are given below. All spaces are at least Hausdorff.

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Definitions and Lemmas

Let $X$ be a topological space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal number $\kappa$ such that for any $A \subset X$ and for any $x \in \overline{A}$, there exists a $B \subset A$ such that $x \in \overline{B}$ and $\lvert B \lvert \le \kappa$. When $t(X)=\omega$, we say $X$ has countable tightness or is countably tight. When $t(X)>\omega$, we say $X$ has uncountably tightness or is uncountably tight. An handy example of a space with uncountably tightness is $\omega_1+1=\omega_1 \cup \left\{\omega_1 \right\}$. This space has uncountable tightness at the point $\omega_1$. All first countable spaces and all Frechet spaces have countable tightness. The concept of countable tightness and tightness in general are discussed in more details here.

A sequence $\left\{x_\alpha: \alpha<\tau \right\}$ of points of a space $X$ is said to be a free sequence if for each $\alpha<\tau$, $\overline{\left\{x_\beta: \beta<\alpha \right\}} \cap \overline{\left\{x_\beta: \beta \ge \alpha \right\}}=\varnothing$. When a free sequence is indexed by the cardinal number $\tau$, the free sequence is said to have length $\tau$. The cardinal function $F(X)$ is the least infinite cardinal $\kappa$ such that if $\left\{x_\alpha \in X: \alpha<\tau \right\}$ is a free sequence of length $\tau$, then $\tau \le \kappa$. The concept of tightness was introduced by Arkhangelskii and he proved that $t(X)=F(X)$ (see p. 15 of [3]). This fact implies the following lemma.

Lemma 2

Let $X$ be compact. If $t(X) \ge \tau$, then there exists a free sequence $\left\{x_\alpha \in X: \alpha<\tau \right\}$ of length $\tau$.

A proof of Lemma 2 can be found here.

The proof of the direction $1 \Longrightarrow 2$ also uses the following lemmas.

Lemma 3

For any compact space $Y$, $\beta (\omega_1 \times Y)=(\omega_1+1) \times Y$.

Lemma 4

Let $X$ be a normal space. For every pair $H$ and $K$ of disjoint closed subsets of $X$, $H$ and $K$ have disjoint closures in $\beta X$.

For Lemma 3, see 3.12.20(c) on p. 237 of [1]. For Lemma 4, see 3.6.4 on p. 173 of [1].

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Proof of Theorem 1

$1 \Longrightarrow 2$
Let $X=\omega_1 \times Y$. Suppose that $X$ is normal. Suppose that $Y$ has uncountable tightness, i.e., $t(Y) \ge \omega_1$. By Lemma 2, there exists a free sequence $\left\{y_\alpha \in Y: \alpha<\omega_1 \right\}$. For each $\beta<\omega_1$, let $C_\beta=\left\{y_\alpha: \alpha>\beta \right\}$. Then the collection $\left\{\overline{C_\beta}: \beta<\omega_1 \right\}$ has the finite intersection property. Since $Y$ is compact, $\bigcap_{\beta<\omega_1} \overline{C_\beta} \ne \varnothing$. Let $p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}$. Consider the following closed subsets of $X=\omega_1 \times Y$.

$H=\overline{\left\{(\alpha,y_\alpha): \alpha<\omega_1 \right\}}$
$K=\left\{(\alpha,p): \alpha<\omega_1 \right\}$

We claim that $H \cap K=\varnothing$. Suppose that $(\alpha,p) \in H \cap K$. Either $p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}}$ or $p \in \overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}$. The latter case is not possible. Note that $[0,\alpha] \times Y$ is an open set containing $(\alpha,p)$. This open set cannot contain points of the form $(\delta,p)$ where $\delta \ge \alpha+1$. So the first case $p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}}$ must hold. Since $p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}$, $p \in \overline{C_\alpha}=\overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}$, a contradiction. So $H$ and $K$ are disjoint closed subsets of $X=\omega_1 \times Y$.

Now consider $\beta X$, the Stone-Cech compactification of $X=\omega_1 \times Y$. By Lemma 3, $\beta X=\beta (\omega_1 \times Y)=(\omega_1+1) \times Y$. Let $H^*=\overline{H}$ and $K^*=\overline{K}$ (closures in $\beta X$). We claim that $(\omega_1,p) \in H^* \cap K^*$. Let $O=(\theta,\omega_1] \times V$ be an open set in $\beta X$ with $(\omega_1,p) \in O$. Note that $p \in \overline{C_\theta}=\left\{y_\delta: \delta>\theta \right\}$. Thus $V \cap \overline{C_\theta} \ne \varnothing$. Choose $\delta>\theta$ such that $y_\delta \in V$. We have $(\delta,y_\delta) \in (\theta,\omega_1] \times V$ and $(\delta,y_\delta) \in H^*$. On the other hand, $(\delta,p) \in K^*$. Thus $(\omega_1,p) \in H^* \cap K^*$, a contradiction. Since $X=\omega_1 \times Y$ is normal, Lemma 4 indicates that $H$ and $K$ should have disjoint closures in $\beta X=(\omega_1+1) \times Y$. Thus $Y$ has countable tightness.

$2 \Longrightarrow 1$
Suppose $t(Y)=\omega$. Let $H$ and $K$ be disjoint closed subsets of $\omega_1 \times Y$. The following series of claims will complete the proof:

Claim 1
For each $y \in Y$, there exists an $\alpha<\omega_1$ such that either $W_{H,y} \subset \alpha+1$ or $W_{K,y} \subset \alpha+1$ where

$W_{H,y}=\left\{\delta<\omega_1: (\delta,y) \in H \right\}$
$W_{K,y}=\left\{\delta<\omega_1: (\delta,y) \in K \right\}$

Proof of Claim 1
Let $y \in Y$. The set $V=\omega_1 \times \left\{y \right\}$ is a copy of $\omega_1$. It is a known fact that in $\omega_1$, there cannot be two disjoint closed and unbounded sets. Let $V_H=V \cap H$ and $V_K=V \cap K$. If $V_H \ne \varnothing$ and $V_K \ne \varnothing$, they cannot be both unbounded in $V$. Thus the claim follows if both $V_H \ne \varnothing$ and $V_K \ne \varnothing$. Now suppose only one of $V_H$ and $V_K$ is non-empty. If the one that is non-empty is bounded, then the claim follows. Suppose the one that is non-empty is unbounded, say $V_K$. Then $W_{H,y}=\varnothing$ and the claim follows.

Claim 2
For each $y \in Y$, there exists an $\alpha<\omega_1$ and there exists an open set $O_y \subset Y$ with $y \in O_y$ such that one and only one of the following holds:

$H \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (1)$
$K \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (2)$

Proof of Claim 2
Let $y \in Y$. Let $\alpha<\omega_1$ be as in Claim 1. Assume that $W_{H,y} \subset \alpha+1$. We want to show that there exists an open set $O_y \subset Y$ with $y \in O_y$ such that (1) holds. Suppose that for each open $O \subset Y$ with $y \in O$, there is a $q \in \overline{O}$ and there exists $\delta_q>\alpha$ such that $(\delta_q,q) \in H$. Let $S$ be the set of all such points $q$. Then $y \in \overline{S}$. Since $Y$ has countable tightness, there exists countable $T \subset S$ such that $y \in \overline{T}$. Since $T$ is countable, choose $\gamma >\omega_1$ such that $\alpha<\delta_q<\gamma$ for all $q \in T$. Note that $[\alpha,\gamma] \times \left\{y \right\}$ does not contain points of $H$ since $W_{H,y} \subset \alpha+1$. For each $\theta \in [\alpha,\gamma]$, the point $(\theta,y)$ has an open neighborhood that contains no point of $H$. Since $[\alpha,\gamma] \times \left\{y \right\}$ is compact, finitely many of these neighborhoods cover $[\alpha,\gamma] \times \left\{y \right\}$. Let these finitely many open neighborhoods be $M_i \times N_i$ where $i=1,\cdots,m$. Let $N=\bigcap_{i=1}^m N_i$. Then $y \in N$ and $N$ would contain a point of $T$, say $q$. Then $(\delta_q,q) \in M_i \times N_i$ for some $i$, a contradiction. Note that $(\delta_q,q)$ is a point of $H$. Thus there exists an open $O_y \subset Y$ with $y \in O_y$ such that (1) holds. This completes the proof of Claim 2.

Claim 3
For each $y \in Y$, there exists an $\alpha<\omega_1$ and there exists an open set $O_y \subset Y$ with $y \in O_y$ such that there are disjoint open subsets $Q_H$ and $Q_K$ of $\omega_1 \times \overline{O_y}$ with $H \cap (\omega_1 \times \overline{O_y}) \subset Q_H$ and $K \cap (\omega_1 \times \overline{O_y}) \subset Q_K$.

Proof of Claim 3
Let $y \in Y$. Let $\alpha$ and $O_y$ be as in Claim 2. Assume (1) in the statement of Claim 2 holds. Note that $(\alpha+1) \times \overline{O_y}$ is a product of two compact spaces and is thus compact (and normal). Let $R_{H,y}$ and $R_{K,y}$ be disjoint open sets in $(\alpha+1) \times \overline{O_y}$ such that $H \cap (\alpha+1) \times \overline{O_y} \subset R_{H,y}$ and $K \cap (\alpha+1) \times \overline{O_y} \subset R_{K,y}$. Note that $[\alpha+1,\omega_1) \times \overline{O_y}$ contains no points of $H$. Then $Q_{H,y}=R_{H,y}$ and $Q_{K,y}=R_{K,y} \cup [\alpha+1,\omega_1) \times \overline{O_y}$ are the desired open sets. This completes the proof of Claim 3.

To make the rest of the proof easier to see, we prove the following claim , which is a general fact that is cleaner to work with. Claim 4 describes precisely (in a topological way) what is happening at this point in the proof.

Claim 4
Let $Z$ be a space. Let $C$ and $D$ be disjoint closed subsets of $Z$. Suppose that $\left\{U_1,U_2,\cdots,U_m \right\}$ is a collection of open subsets of $Z$ covering $C \cup D$ such that for each $i=1,2,\cdots,m$, only one of the following holds:

$C \cap \overline{U_i} \ne \varnothing \text{ and } D \cap \overline{U_i}=\varnothing$
$C \cap \overline{U_i} = \varnothing \text{ and } D \cap \overline{U_i} \ne \varnothing$

Then there exist disjoint open subsets of $Z$ separating $C$ and $D$.

Proof of Claim 4
Let $U_C=\cup \left\{U_i: \overline{U_i} \cap C \ne \varnothing \right\}$ and $U_D=\cup \left\{U_i: \overline{U_i} \cap D \ne \varnothing \right\}$. Note that $\overline{U_C}=\cup \left\{\overline{U_i}: \overline{U_i} \cap C \ne \varnothing \right\}$. Likewise, $\overline{U_D}=\cup \left\{\overline{U_i}: \overline{U_i} \cap D \ne \varnothing \right\}$. Let $V_C=U_C-\overline{U_D}$ and $V_D=U_D-\overline{U_C}$. Then $V_C$ and $V_D$ are disjoint open sets. Furthermore, $C \subset V_C$ and $D \subset V_D$. This completes the proof of Claim 4.

Now back to the proof of Theorem 1. For each $y \in Y$, let $O_y$, $Q_{H,y}$ and $Q_{K,y}$ be as in Claim 3. Since $Y$ is compact, there exists $\left\{y_1,y_2,\cdots,y_n \right\} \subset Y$ such that $\left\{O_{y_1},O_{y_2},\cdots,O_{y_n} \right\}$ is a cover of $Y$. For each $i=1,\cdots,n$, let $L_i=Q_{H,y_i} \cap (\omega_1 \times O_y)$ and $M_i=Q_{K,y_i} \cap (\omega_1 \times O_y)$. Note that both $L_i$ and $M_i$ are open in $\omega_1 \times Y$. To apply Claim 4, rearrange the open sets $L_i$ and $M_i$ and re-label them as $U_1,U_2,\cdots,U_m$. By letting $Z=\omega_1 \times Y$, $C=H$ and $D=K$, the open sets $U_i$ satisfy Claim 4. Tracing the $U_i$ to $L_j$ or $M_j$ and then to $Q_{H,y_j}$ and $Q_{K,y_j}$, it is clear that the two conditions in Claim 4 are satisfied:

$H \cap \overline{U_i} \ne \varnothing \text{ and } K \cap \overline{U_i}=\varnothing$
$H \cap \overline{U_i} = \varnothing \text{ and } K \cap \overline{U_i} \ne \varnothing$

Then by Claim 4, the disjoint closed sets $H$ and $K$ can be separated by two disjoint open subsets of $\omega_1 \times Y$. $\blacksquare$

The theorem proved in [4] is essentially the statement that for any compact space $Y$, the product $\kappa^+ \times Y$ is normal if and only $t(Y) \le \kappa$. Here $\kappa^+$ is the first ordinal of the next cardinal that is greater than $\kappa$.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Gruenhage, G., Nogura, T., Purisch, S., Normality of $X \times \omega_1$, Topology and its Appl., 39, 263-275, 1991.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.

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$\copyright \ 2014 - 2015 \text{ by Dan Ma}$

# When all powers of a space are normal

It follows from the Tychonoff Theorem that if a topological space $X$ is a compact Hausdorff space, then the product space $X^\tau$ is normal for any cardinal number $\tau$. The converse is also true. If $X^\tau$ is normal for any cardinal number $\tau$, then $X$ must be compact. This is a theorem that is due to Noble (see [5]). The proof in [5] is a corollary resulted from a long chain of previous results of Noble and others. Many authors had produced simpler and more direct proofs of Noble’s theorem (e.g. [2], [3] and [4]). All these more direct proofs make use of the fact that the product space $\omega^{\omega_1}$ is not normal (due to A. H. Stone). All of them except [2] make use of other strong topological results in order to derive Noble’s theorem. In [2], Engelking established Noble’s theorem by an elementary proof. In this post, we present the proof in [2] in full details. Noble’s theorem is also given in Engelking’s textbook as an exercise (see 3.12.15 on p. 233 in [1]).

Before proceeding to the main theorem, let’s set up some notation for working with the product space $\prod_{\alpha \in S} X_\alpha$. For $x \in \prod_{\alpha \in S} X_\alpha$, the $\alpha^{th}$ coordinate of $x$ is denoted by $x_\alpha$ or $(x)_\alpha$. For $M \subset S$, the map $P_M: \prod_{\alpha \in S} X_\alpha \rightarrow \prod_{\alpha \in M} X_\alpha$ is the natural projection map. In the product space $\prod_{\alpha \in S} X_\alpha$, standard basic open sets are of the form $\prod_{\alpha \in S} O_\alpha$ where $O_\alpha= X_\alpha$ for all but finitely many $\alpha$. We use $supp(\prod_{\alpha \in S} O_\alpha)$ to denote the set of the finite number of $\alpha \in S$ where $O_\alpha \ne X_\alpha$.

Noble’s Theorem

If each power of a space $X$ is normal, then $X$ is compact.

Proof
Suppose that $X^\tau$ is normal for all cardinal numbers $\tau$. Suppose that $X$ is not compact. Then there exists a collection $\mathcal{H}=\left\{H_\alpha: \alpha \in S \right\}$ of closed subsets of $X$ such that $\mathcal{H}$ has the finite intersection property but has empty intersection. Let $H=\prod_{\alpha \in S} H_\alpha$, which is a subspace of the product space $\prod_{\alpha \in S} X_\alpha$ where each $X_\alpha=X$. We can also denote the product space $\prod_{\alpha \in S} X_\alpha$ by $X^\tau$ where $\tau=\lvert S \lvert$.

Let $K=\left\{k \in X^{\lvert S \lvert}: \forall \ \beta, \gamma \in S, \ k_\beta=k_\gamma \right\}$. Note that $K$ is commonly referred to as the diagonal of the product space in question. Both $H$ and $K$ are closed sets in the product space $X^\tau$. Because the collection $\mathcal{H}$ has empty intersection, $H$ and $K$ are disjoint closed sets. Since $X^\tau$ is normal, there exist disjoint open subsets $U$ and $V$ of $X^\tau$ such that $H \subset U$ and $K \subset V$.

Let $x_1 \in H$. Let $O_1$ be a basic standard open set with $x \in O_1 \subset U$. Let $S_1=supp(O_1)$. Then we have $P^{-1}_{S_1}(P_{S_1}(x_1)) \subset U$. Since $\mathcal{H}$ has the finite intersection property, choose $a_1 \in \bigcap_{\alpha \in S_1} H_\alpha$. Then define $x_2 \in H$ such that $(x_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(x_2)_\alpha=(x_1)_\alpha$ for all $\alpha \in S-S_1$.

Let $O_2$ be a basic standard open set with $x_2 \in O_2 \subset U$. Let $S_2=supp(O_2)$. By making $O_2$ a smaller open set if necessary, we can have $S_1 \subset S_2$. Then we have $P^{-1}_{S_2}(P_{S_2}(x_2)) \subset U$. Choose $a_2 \in \bigcap_{\alpha \in S_2} H_\alpha$. Then define $x_3 \in H$ such that $(x_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(x_3)_\alpha=(x_2)_\alpha$ for all $\alpha \in S-S_2$.

After this inductive process is completed, we can obtain:

• a sequence $x_1,x_2,x_3,\cdots$ of points of $H=\prod_{\alpha \in S} H_\alpha$,
• a sequence $S_1 \subset S_2 \subset S_3, \subset \cdots$ of finite subsets of the index set $S$,
• a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for each $n \ge 2$, $(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$.

By A. H. Stone’s theorem (Theorem 5 in [6]), $X$ cannot contain a closed copy of $\mathbb{N}$ (the space of the positive integers with the discrete topology). A proof that $\mathbb{N}^{\omega_1}$ is also found in this post. Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$. Either $A$ is infinite or finite.

Case 1
Assume that $A$ is an infinite set. Then $A$ has a limit point $a$, meaning that every open subset of $X$ containing $a$ contains some $a_n$ different from $a$. For each $n \ge 2$, define $y_n \in \prod_{\alpha \in S} X_\alpha$ such that

• $(y_n)_\alpha=(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$
• $(y_n)_\alpha=a$ for all $\alpha \in S-S_n$

It is the case that $y_n \in P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$ for all $n$, since $y_n$ agrees with $x_n$ on the finite set $S_n$. Let $t \in K$ such that $t_\alpha=a$ for all $\alpha \in S$. It follows that $t$ is a limit point of $\left\{y_2,y_3,y_4,\cdots \right\}$. Thus $t \in \overline{U}$. Since $t \in K \subset V$, the open set $V$ would have to contain points of $U$. But $U$ and $V$ are supposed to be disjoint open subsets of the product space $\prod_{\alpha \in S} X_\alpha$. Thus we have a contradiction.

Case 2
Assume that $A$ is a finite set. Then for some $m$, $a_j=a_m$ for all $j \ge m$. For each $n \ge 2$, define $y_n \in \prod_{\alpha \in S} X_\alpha$ such that

• $(y_n)_\alpha=(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$
• $(y_n)_\alpha=a_m$ for all $\alpha \in S-S_n$

Let $t \in K$ such that $t_\alpha=a_m$ for all $\alpha \in S$. Then $y_n=t$ for all $n \ge m+1$. As in Case 1, $y_n \in P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$ for all $n$, implying that $t \in K \cap U$. This is a contradiction, since $K \subset V$ and $U$ and $V$ are supposed to be disjoint.

Both cases lead to a contradiction. Thus if all powers of $X$ is normal, $X$ must be compact. This completes the proof of the theorem.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Engelking, R., An elementary proof of Noble’s theorem on normality of powers, Comment. Math. Univ. Carolinae, 29.4, 677-678, 1988.
3. Franklin, S. P., Walker, R. C., Normalit of powers implies compactness, Proc. Amer. Math. Soc., 36, 295-296, 1972.
4. Keesling, J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
5. Noble, N., Products with closed projections, II, Trans. Amer. Math. Soc., 160, 169-183, 1971.
6. Ross, K. A., Stone, A. H. Products of separable spaces, Amer. Math. Monthly, 71, 398-403, 1964.

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