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

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

Theorem 1

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

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

Theorem 1a

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

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

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

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Proving Theorem 1a

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

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. Then the following conditions are equivalent.

1. $Y$ is normal.
2. For any pair of disjoint closed subsets $H$ and $K$ of $Y$, there exists a countable $B \subset A$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$.
3. For any pair of disjoint closed subsets $H$ and $K$ of $Y$, there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$, meaning that $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Remark

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

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Reference

1. Arkhangelskii, A. V., Normality and dense subspaces, Proc. Amer. Math. Soc., 130 (1), 283-291, 2001.
2. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
3. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
4. Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
5. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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

# Normal dense subspaces of a product of “continuum” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii in [1] (see Problem I.5.25). A partial positive answer is provided by a theorem that is usually attributed to Corson: If $Y$ is a normal dense subspace of a product of separable metric spaces and if $Y \times Y$ is also normal, then $Y$ is collectionwise normal. In this post, using a simple combinatorial argument, we show that any normal dense subspace of a product of continuum many separable metric space is collectionwise normal (see Corollary 4 below), which is a corollary of the following theorem.

Theorem 1
Let $X$ be a normal space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following holds:

• If $Y$ is a closed and discrete subspace of $X$ with $\lvert Y \lvert=\omega_1$, then $Y$ contains a separated subset of cardinality $\omega_1$.

Theorem 1 gives the corollary indicated at the beginning and several other interesting results. The statement $2^\omega<2^{\omega_1}$ means that the cardinality of the power set (the set of all subsets) of $\omega$ is strictly less than the cardinality of the power set of $\omega_1$. Note that the statement $2^\omega<2^{\omega_1}$ follows from the continuum hypothesis (CH), the statement that $2^\omega=\omega_1$. With the assumption $2^\omega<2^{\omega_1}$, Theorem 1 is a theorem that goes beyond ZFC. We also present an alternative to Theorem 1 that removes the assumption $2^\omega<2^{\omega_1}$ (see Theorem 6 below).

A subset $T$ of a space $S$ is a separated set (in $S$) if for each $t \in T$, there is an open subset $O_t$ of $S$ with $t \in O_t$ such that $\left\{O_t: t \in T \right\}$ is a pairwise disjoint collection. First we prove Theorem 1 and then discuss the corollaries.

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Proof of Theorem 1

Suppose $Y$ is a closed and discrete subset of $X$ with $\lvert Y \lvert=\omega_1$ such that no subset of $Y$ of cardinality $\omega_1$ can be separated. We then show that $2^{\omega_1} \le 2^{\omega}$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert \le 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose a maximal disjoint collection $\mathcal{M}_W$ of open sets in $\mathcal{G}_W$. Because no subset of $Y$ of cardinality $\omega_1$ can be separated, each $\mathcal{M}_W$ is countable. If $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that $\lvert \mathcal{P}(Y) \lvert=2^{\omega_1}$. Also note that since $\lvert \mathcal{B} \lvert \le 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert \le 2^\omega$. Thus $2^{\omega_1} \le 2^{\omega}$. $\blacksquare$

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Some Corollaries of Theorem 1

Here’s some corollaries that follow easily from Theorem 1. A space $X$ has the countable chain condition (CCC) if every pairwise disjoint collection of non-empty open subset of $X$ is countable. For convenience, if $X$ has the CCC, we say $X$ is CCC. The following corollaries make use of the fact that any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post).

Corollary 2
Let $X$ be a CCC space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 3
Let $X$ be a CCC space with character $\le 2^\omega$. If CH holds, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 4
Let $X=\prod_{\alpha<2^\omega} X_\alpha$ be a product where each factor $X_\alpha$ is a separable metric space. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $Y$ is a normal dense subspace of $X$, then $Y$ has countable extent.
• If $Y$ is a normal dense subspace of $X$, then $Y$ is collectionwise normal.

Corollary 4 is the result indicated in the title of the post. The product of separable spaces has the CCC. Thus the product space $X$ and any dense subspace of $X$ have the CCC. Because $X$ is a product of continuum many separable metric spaces, $X$ and any subspace of $X$ have characters $\le 2^\omega$. Then Corollary 4 follows from Corollary 2.

When dealing with the topic of normal versus collectionwise normal, it is hard to avoid the connection with the normal Moore space conjecture. Theorem 1 gives the result of F. B. Jones from 1937 (see [3]). We have the following theorem.

Theorem 5
If $2^\omega<2^{\omega_1}$, then every separable normal Moore space is metrizable.

Though this was not how Jones proved it in [3], Theorem 5 is a corollary of Corollary 2. By Corollary 2, any separable normal Moore space is collectionwise normal. It is well known that collectionwise normal Moore space is metrizable (Bing’s metrization theorem, see Theorem 5.4.1 in [2]).

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A ZFC Theorem

We now prove a result that is similar to Corollary 2 but uses no set-theory beyond the Zermelo–Fraenkel set theory plus axiom of choice (abbreviated by ZFC). Of course the conclusion is not as strong. Even though the assumption $2^\omega<2^{\omega_1}$ is removed in Theorem 6, note the similarity between the proof of Theorem 1 and the proof of Theorem 6.

Theorem 6
Let $X$ be a CCC space with character $\le 2^\omega$. Then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ has cardinality less than continuum.

Proof of Theorem 6
Let $X$ be a normal CCC space with character $\le 2^\omega$. Let $Y$ be a closed and discrete subset of $X$. We show that $\lvert Y \lvert < 2^\omega$. Suppose that $\lvert Y \lvert = 2^\omega$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert = 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose $\mathcal{M}_W \subset \mathcal{G}_W$ such that $\mathcal{M}_W$ is a maximal disjoint collection. Since $X$ is CCC, $\mathcal{M}_W$ is countable. It is clear that if $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that since $\lvert \mathcal{B} \lvert = 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert = 2^\omega$. Thus $\lvert \mathcal{P}(Y) \lvert \le 2^{\omega}$. However, $Y$ is assumed to be of cardinality continuum. Then $\lvert \mathcal{P}(Y) \lvert>2^{\omega_1}$, leading to a contradiction. Thus it must be the case that $\lvert Y \lvert < 2^\omega$. $\blacksquare$

With Theorem 6, Corollary 3 still holds. Theorem 5 removes the set-theoretic assumption of $2^\omega<2^{\omega_1}$. As a result, the upper bound for cardinalities of closed and discrete sets is (at least potentially) higher.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Jones, F. B., Concerning normal and completely normal spaces, Bull. Amer. Math. Soc., 43, 671-677, 1937.

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

# One theorem about normality of Cp(X)

Assuming that the function space $C_p(X)$ is normal, what can be said about the domain space $X$? In this post, we prove a theorem that yields a corollary that for any normal space $X$, if $C_p(X)$ is normal, then $X$ has countable extent (i.e. every closed and discrete subset of $X$ is countable). Thus the normality of the function space limits the size of a closed and discrete subset of the domain space. It then follows that for any metric space $X$, if $C_p(X)$ is normal, $X$ has is second countable (i.e. having a countable base). Another immediate, but slightly less obvious, corollary is that for any $X$ that is a normal Moore space, if $C_p(X)$ is normal, then $X$ is metrizable.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

Let $X$ be a space. Let $Y \subset X$. Let $\pi_Y$ be the natural projection from the product space $\mathbb{R}^X$ into the product space $\mathbb{R}^Y$. Specifically, if $f \in \mathbb{R}^X$, then $\pi_Y(f)=f \upharpoonright Y$, i.e., the function $f$ restricted to $Y$. In the discussion below, $\pi_Y$ is defined just on $C_p(X)$, i.e., $\pi_Y$ is the natural projection from $C_p(X)$ into $C_p(Y)$. It is always the case that $\pi_Y(C_p(X)) \subset C_p(Y)$. It is not necessarily the case that $\pi_Y(C_p(X))=C_p(Y)$. However, if $X$ is a normal space and $Y$ is closed in $X$, then $\pi_Y(C_p(X))=C_p(Y)$ and $\pi_Y$ is the natural projection from $C_p(X)$ onto $C_p(Y)$. We prove the following theorem.

Theorem 1

Suppose that $C_p(X)$ is a normal space. Let $Y$ be a closed subspace of $X$. Then $\pi_Y(C_p(X))$ is a normal space.

Theorem 1 is found in [1] (see Theorem I.6.2). In proving Theorem 1, we need the following lemma.

Lemma 2

Let $T=\prod_{\alpha \in A} T_\alpha$ be a product of separable metrizable spaces. Let $S$ be a dense subspace of $T$. Then the following conditions are equivalent.

1. $S$ is normal.
2. For any pair of disjoint closed subsets $H$ and $K$ of $S$, 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 $S$, there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(S)$, meaning that $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$.

For a proof of Lemma 2, see Lemma 1 in this previous post.

Proof of Theorem 1
Note that $\pi_Y(C_p(X))$ is a dense subspace of $\mathbb{R}^Y$. Let $H$ and $K$ be disjoint closed subsets of $\pi_Y(C_p(X))$. To show $\pi_Y(C_p(X))$ is normal, by Lemma 2, we only need to produce a countable $B \subset Y$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$. The closure here is taken in $\pi_B(\pi_Y(C_p(X)))$.

Let $H_1=\pi_Y^{-1}(H)$ and $K_1=\pi_Y^{-1}(K)$. Both $H_1$ and $K_1$ are closed subsets of $C_p(X)$. By Lemma 2, there exists some countable $C \subset X$ such that $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$. The closure here is taken in $\pi_C(C_p(X))$. According to the remark at the end of this previous post, for any countable $D \subset X$ such that $C \subset D$, $\overline{\pi_D(H_1)} \cap \overline{\pi_D(K_1)}=\varnothing$. In other words, the countable set $C$ can be enlarged and the conclusion of the lemma still holds. With this observation in mind, we can assume that $C \cap Y \ne \varnothing$. If not, we can always throw countably many points of $Y$ into $C$ and still have $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$.

Let $B=C \cap Y$. We claim that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing$. The closure here is taken $\pi_B(C_p(X))$. Suppose that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)} \ne \varnothing$. Choose $f \in C_p(X)$ such that $f \upharpoonright B \in \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}$. It follows that $f \upharpoonright C \in \overline{\pi_C(H_1)}$. To see this, let $f \upharpoonright C \in U=\prod_{x \in C} U_x$ where $U$ is a standard basic open set. Let $F$ be the support of $U$, i.e., the finite set of $x \in C$ such that $U_x \ne \mathbb{R}$. Let $F_1=F \cap Y$ and $F_2=F \cap (X-Y)$. Let $U^*=\prod_{x \in B} U_x$. Note that $f \upharpoonright B \in U^*$. Since $f \upharpoonright B \in \overline{\pi_B(H_1)}$, there is some $g \in H_1$ such that $\pi_B(g) \in U^*$. Note that $F_1$ is the support of $U^*$.

Because the space $X$ is completely regular, there is a $h \in C_p(X)$ such that $h(x)=0$ for all $x \in Y$ and $h(x)=f(x)-g(x)$ for all $x \in F_2$. Let $w=h+g$. Since $w \upharpoonright Y=g \upharpoonright Y$, $w \in H_1$. Note that $w=g$ on $Y$, hence on $F_1$ and that $w=f$ on $F_2$. Thus $w \upharpoonright C \in U$. Since $U$ is an arbitrary open set containing $f \upharpoonright C$, it follows that $f \upharpoonright C \in \overline{\pi_C(H_1)}$. By a similar argument, it can be shown that $f \upharpoonright C \in \overline{\pi_C(K_1)}$. This is a contradiction since $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$. Therefore the claim that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing$ is true, with the closure being taken in $\pi_B(C_p(X))$.

Because $B \subset Y$, observe that $\pi_B(H_1)=\pi_B(H)$ and $\pi_B(K_1)=\pi_B(K)$. Furthermore, $\pi_B(\pi_Y(C_p(X)))=\pi_B(C_p(X))$. Thus we can claim that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$, with the closure being taken in $\pi_B(\pi_Y(C_p(X)))$. By Lemma 2, $\pi_Y(C_p(X))$ is normal. $\blacksquare$

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Some Corollaries

Corollary 3

Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e., every closed and discrete subset of $X$ is countable.

Proof of Corollary 3
Let $Y$ be a closed and discrete subset of $X$. We show that $Y$ must be countable. Since $Y$ is closed and $X$ is normal, $\pi_Y(C_p(X))=C_p(Y)$. By Theorem 1, $C_p(Y)$ is normal. Since $Y$ is discrete, $C_p(Y)=\mathbb{R}^Y$. If $Y$ is uncountable, $\mathbb{R}^Y$ is not normal. Thus $Y$ must be countable. $\blacksquare$

Corollary 4

Let $X$ be a metrizable space. If $C_p(X)$ is normal, then $X$ has a countable base.

Proof of Corollary 4
Note that in any metrizable space, the weight equals the extent. By Corollary 3, $X$ has countable extent and thus has countable base. $\blacksquare$

Corollary 5

Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ is collectionwise normal.

Proof of Corollary 5
Any normal space with countable extent is collectionwise normal. See Theorem 2 in this previous post. $\blacksquare$

Corollary 6

Let $X$ be a normal Moore space. If $C_p(X)$ is normal, then $X$ is metrizable.

Proof of Corollary 6
Suppose $C_p(X)$ is normal. By Theorem 1, $X$ has countable extent. By Corollary 5, $X$ is collectionwise normal. According to Bing’s metrization theorem, any collectionwise normal Moore space is metrizable (see [2] Theorem 5.4.1 in page 329). $\blacksquare$

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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

# Cp(X) where X is a separable metric space

Let $\tau$ be an uncountable cardinal. Let $\prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau}$ be the Cartesian product of $\tau$ many copies of the real line. This product space is not normal since it contains $\prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1}$ as a closed subspace. However, there are dense subspaces of $\mathbb{R}^{\tau}$ are normal. For example, the $\Sigma$-product of $\tau$ copies of the real line is normal, i.e., the subspace of $\mathbb{R}^{\tau}$ consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of $\mathbb{R}^{\tau}$ that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space $C_p(X)$ where $X$ is a separable metrizable space.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

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$C_p(X)$ when $X$ is a separable metric space

In the remainder of the post, $X$ denotes a separable metrizable space. Then, $C_p(X)$ is more than normal. The function space $C_p(X)$ has the following properties:

• normal,
• Lindelof (hence paracompact and collectionwise normal),
• hereditarily Lindelof (hence hereditarily normal),
• hereditarily separable,
• perfectly normal.

All such properties stem from the fact that $C_p(X)$ has a countable network whenever $X$ is a separable metrizable space.

Let $L$ be a topological space. A collection $\mathcal{N}$ of subsets of $L$ is said to be a network for $L$ if for each $x \in L$ and for each open $O \subset L$ with $x \in O$, there exists some $A \in \mathcal{N}$ such that $x \in A \subset O$. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for $C_p(X)$, let $\mathcal{B}$ be a countable base for the domain space $X$. For each $B \subset \mathcal{B}$ and for any open interval $(a,b)$ in the real line with rational endpoints, consider the following set:

$[B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}$

There are only countably many sets of the form $[B,(a,b)]$. Let $\mathcal{N}$ be the collection of sets, each of which is the intersection of finitely many sets of the form $[B,(a,b)]$. Then $\mathcal{N}$ is a network for the function space $C_p(X)$. To see this, let $f \in O$ where $O=\bigcap_{x \in F} [x,O_x]$ is a basic open set in $C_p(X)$ where $F \subset X$ is finite and each $O_x$ is an open interval with rational endpoints. For each point $x \in F$, choose $B_x \in \mathcal{B}$ with $x \in B_x$ such that $f(B_x) \subset O_x$. Clearly $f \in \bigcap_{x \in F} \ [B_x,O_x]$. It follows that $\bigcap_{x \in F} \ [B_x,O_x] \subset O$.

Examples include $C_p(\mathbb{R})$, $C_p([0,1])$ and $C_p(\mathbb{R}^\omega)$. All three can be considered subspaces of the product space $\mathbb{R}^c$ where $c$ is the cardinality of the continuum. This is true for any separable metrizable $X$. Note that any separable metrizable $X$ can be embedded in the product space $\mathbb{R}^\omega$. The product space $\mathbb{R}^\omega$ has cardinality $c$. Thus the cardinality of any separable metrizable space $X$ is at most continuum. So $C_p(X)$ is the subspace of a product space of $\le$ continuum many copies of the real lines, hence can be regarded as a subspace of $\mathbb{R}^c$.

A space $L$ has countable extent if every closed and discrete subset of $L$ is countable. The $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ of the separable metric spaces $\left\{X_\alpha: \alpha \in A \right\}$ is a dense and normal subspace of the product space $\prod_{\alpha \in A} X_\alpha$. The normal space $\Sigma_{\alpha \in A} X_\alpha$ has countable extent (hence collectionwise normal). The examples of $C_p(X)$ discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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

# 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 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 $\lvert A \lvert \le \kappa$. 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.

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

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

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

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