# (Lower case) sigma-products of separable metric spaces are Lindelof

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $b \in \prod_{\alpha \in A} X_\alpha$, called the base point. The $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ is the following subspace of the product space $X$:

$\Sigma_{\alpha \in A} X_\alpha=\left\{ x \in X: x_\alpha \ne b_\alpha \text{ for at most countably many } \alpha \in A \right\}$

In other words, the space $\Sigma_{\alpha \in A} X_\alpha$ is the subspace of the product space $X=\prod_{\alpha \in A} X_\alpha$ consisting of all points that deviate from the base point on at most countably many coordinates $\alpha \in A$. We also consider the following subspace of $\Sigma_{\alpha \in A} X_\alpha$.

$\sigma=\left\{ x \in \Sigma_{\alpha \in A} X_\alpha: x_\alpha \ne b_\alpha \text{ for at most finitely many } \alpha \in A \right\}$

For convenience , we call $\Sigma_{\alpha \in A} X_\alpha$ the (upper case) Sigma-product (or $\Sigma$-product) of the spaces $X_\alpha$ and we call the space $\sigma$ the (lower case) sigma-product (or $\sigma$-product). Clearly, the space $\sigma$ is a dense subspace of $\Sigma_{\alpha \in A} X_\alpha$. In a previous post, we show that the upper case Sigma-product of separable metric spaces is collectionwise normal. In this post, we show that the (lower case) sigma-product of separable metric spaces is Lindelof. Thus when each factor $X_\alpha$ is a separable metric space with at least two points, the $\Sigma$-product, though not Lindelof, has a dense Lindelof subspace. The (upper case) $\Sigma$-product of separable metric spaces is a handy example of a non-Lindelof space that contains a dense Lindelof subspace.

Naturally, the lower case sigma-product can be further broken down into countably many subspaces. For each integer $n=0,1,2,3,\cdots$, we define $\sigma_n$ as follows:

$\sigma_n=\left\{ x \in \sigma: x_\alpha \ne b_\alpha \text{ for at most } n \text{ many } \alpha \in A \right\}$

Clearly, $\sigma=\bigcup_{n=0}^\infty \sigma_n$. We prove the following theorem. The fact that $\sigma$ is Lindelof will follow as a corollary. Understanding the following proof for Theorem 1 is a matter of keeping straight the notations involving standard basic open sets in the product space $X=\prod_{\alpha \in A} X_\alpha$. We say $V$ is a standard basic open subset of the product space $X$ if $V$ is of the form $V=\prod_{\alpha \in A} V_\alpha$ such that each $V_\alpha$ is an open subset of the factor space $X_\alpha$ and $V_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. The finite set $F$ of all $\alpha \in A$ such that $V_\alpha \ne X_\alpha$ is called the support of the open set $V$.

Theorem 1
Let $\sigma$ be the $\sigma$-product of the separable metrizable spaces $\left\{X_\alpha: \alpha \in A \right\}$. For each $n$, let $\sigma_n$ be defined as above. The product space $\sigma_n \times Y$ is Lindelof for each non-negative integer $n$ and for all separable metric space $Y$.

Proof of Theorem 1
We prove by induction on $n$. Note that $\sigma_0=\left\{b \right\}$, the base point. Clearly $\sigma_0 \times Y$ is Lindelof for all separable metric space $Y$. Suppose the theorem hold for the integer $n$. We show that $\sigma_{n+1} \times Y$ for all separable metric space $Y$. To this end, let $\mathcal{U}$ be an open cover of $\sigma_{n+1} \times Y$ where $Y$ is a separable metric space. Without loss of generality, we assume that each element of $\mathcal{U}$ is of the form $V \times W$ where $V=\prod_{\alpha \in A} V_\alpha$ is a standard basic open subset of the product space $X=\prod_{\alpha \in A} X_\alpha$ and $W$ is an open subset of $Y$.

Let $\mathcal{U}_0=\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable subcollection of $\mathcal{U}$ such that $\mathcal{U}_0$ covers $\left\{b \right\} \times Y$. For each $j$, let $U_j=V_j \times W_j$ where $V_j=\prod_{\alpha \in A} V_{j,\alpha}$ is a standard basic open subset of the product space $X$ with $b \in V_j$ and $W_j$ is an open subset of $Y$. For each $j$, let $F_j$ be the support of $V_j$. Note that $\alpha \in F_j$ if and only if $V_{j,\alpha} \ne X_\alpha$. Also for each $\alpha \in F_j$, $b_\alpha \in V_{j,\alpha}$. Furthermore, for each $\alpha \in F_j$, let $V^c_{j,\alpha}=X_\alpha- V_{j,\alpha}$. With all these notations in mind, we define the following open set for each $\beta \in F_j$:

$H_{j,\beta}= \biggl( V^c_{j,\beta} \times \prod_{\alpha \in A, \alpha \ne \beta} X_\alpha \biggr) \times W_j=\biggl( V^c_{j,\beta} \times T_\beta \biggr) \times W_j$

Observe that for each point $y \in \sigma_{n+1}$ such that $y \in V^c_{j,\beta} \times T_\beta$, the point $y$ already deviates from the base point $b$ on one coordinate, namely $\beta$. Thus on the coordinates other than $\beta$, the point $y$ can only deviates from $b$ on at most $n$ many coordinates. Thus $\sigma_{n+1} \cap (V^c_{j,\beta} \times T_\beta)$ is homeomorphic to $V^c_{j,\beta} \times \sigma_n$. Note that $V^c_{j,\beta} \times W_j$ is a separable metric space. By inductive hypothesis, $V^c_{j,\beta} \times \sigma_n \times W_j$ is Lindelof. Thus there are countably many open sets in the open cover $\mathcal{U}$ that covers points of $H_{j,\beta} \cap (\sigma_{n+1} \times W_j)$.

Note that

$\sigma_{n+1} \times Y=\biggl( \bigcup_{j=1}^\infty U_j \cap \sigma_{n+1} \biggr) \cup \biggl( \bigcup \left\{H_{j,\beta} \cap (\sigma_{n+1} \times W_j): j=1,2,3,\cdots, \beta \in F_j \right\} \biggr)$

To see that the left-side is a subset of the right-side, let $t=(x,y) \in \sigma_{n+1} \times Y$. If $t \in U_j$ for some $j$, we are done. Suppose $t \notin U_j$ for all $j$. Observe that $y \in W_j$ for some $j$. Since $t=(x,y) \notin U_j$, $x_\beta \notin V_{j,\beta}$ for some $\beta \in F_j$. Then $t=(x,y) \in H_{j,\beta}$. It is now clear that $t=(x,y) \in H_{j,\beta} \cap (\sigma_{n+1} \times W_j)$. Thus the above set equality is established. Thus one part of $\sigma_{n+1} \times Y$ is covered by countably many open sets in $\mathcal{U}$ while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of $\mathcal{U}$ covers $\sigma_{n+1} \times Y$. $\blacksquare$

Corollary 2
It follows from Theorem 1 that

• If each factor space $X_\alpha$ is a separable metric space, then each $\sigma_n$ is a Lindelof space and that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a Lindelof space.
• If each factor space $X_\alpha$ is a compact separable metric space, then each $\sigma_n$ is a compact space and that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a $\sigma$-compact space.

Proof of Corollary 2
The first bullet point is a clear corollary of Theorem 1. A previous post shows that $\Sigma$-product of compact spaces is countably compact. Thus $\Sigma_{\alpha \in A} X_\alpha$ is a countably compact space if each $X_\alpha$ is compact. Note that each $\sigma_n$ is a closed subset of $\Sigma_{\alpha \in A} X_\alpha$ and is thus countably compact. Being a Lindelof space, each $\sigma_n$ is compact. It follows that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a $\sigma$-compact space. $\blacksquare$

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A non-Lindelof space with a dense Lindelof subspace

Now we put everything together to obtain the example described at the beginning. For each $\alpha \in A$, let $X_\alpha$ be a separable metric space with at least two points. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is collectionwise normal (see this previous post). According to the lemma in this previous post, the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ contains a closed copy of $\omega_1$. Thus the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is not Lindelof. It is clear that the $\sigma$-product is a dense subspace of $\Sigma_{\alpha \in A} X_\alpha$. By Corollary 2, the $\sigma$-product is a Lindelof subspace of $\Sigma_{\alpha \in A} X_\alpha$.

Using specific factor spaces, if each $X_\alpha=\mathbb{R}$ with the usual topology, then $\Sigma_{\alpha<\omega_1} X_\alpha$ is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each $X_\alpha=[0,1]$ with the usual topology, then $\Sigma_{\alpha<\omega_1} X_\alpha$ is a non-Lindelof space with a dense $\sigma$-compact subspace. Another example of a non-Lindelof space with a dense Lindelof subspace is given In this previous post (see Example 1).

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

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

# Looking for another closed and discrete subspace of a product space

Let $\omega_1$ be the first uncountable ordinal. In a previous post called Looking for a closed and discrete subspace of a product space, it was shown that the product space $\mathbb{R}^c$, the product of continuum many copies of the real line $\mathbb{R}$, contains a closed and discrete subset of cardinality continuum. This example shows that a product space of uncountably many copies of a “nice” space is “big and wide” enough to hide uncountable closed and discrete sets even when the product space is separable. This post reinforces this same fact by showing that $\mathbb{R}^{\omega_1}$ contains a closed and discrete subset of cardinality $\omega_1$. It follows that for any uncountable cardinal $\tau$, the product space $\mathbb{R}^\tau$ contains an uncountable closed and discrete subset, i.e., the product of uncountably many copies of the real line $\mathbb{R}$ has uncountable extent.

Let $\omega$ be the first infinite ordinal, i.e., the set of all nonnegative integers. Consider $\omega^{\omega_1}$, the product of $\omega_1$ many copies of $\omega$ with the discrete topology. Since $\omega^{\omega_1}$ is a closed subspace of $\mathbb{R}^{\omega_1}$, it suffices to show that $\omega^{\omega_1}$ has an uncountable closed and discrete subset.

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The Construction

We now construct an uncountable closed and discrete subset of $\omega^{\omega_1}$. Let $\delta$ be an infinite ordinal such that $\omega<\delta<\omega_1$. Let $W=\left\{\alpha: \delta \le \alpha<\omega_1 \right\}$. For each $\alpha \in W$, let $Y_\alpha=\left\{\beta<\omega_1: \beta<\alpha \right\}$. We can also use interval notations: $W=[\delta,\omega_1)$ and $Y_\alpha=[0,\alpha)$. Consider $Y_\alpha$ as a space with the discrete topology. Then it is clear that $\omega^{\omega_1}$ is homeomorphic to the product space $\prod_{\alpha \in W} Y_\alpha$. Thus the focus is now on finding an uncountable closed and discrete subset of $\prod_{\alpha \in W} Y_\alpha$.

One interesting fact about the space $\prod_{\alpha \in W} Y_\alpha$ is that every function $f \in \prod_{\alpha \in W} Y_\alpha$ is a pressing down function. That is, for every $f \in \prod_{\alpha \in W} Y_\alpha$, $f(\alpha)<\alpha$ for all $\alpha \in W$. Note that $f$ is defined on $W$, a closed and unbounded subset of $\omega_1$ (hence a stationary set). It follows that for each $f \in \prod_{\alpha \in W} Y_\alpha$, there is a stationary set $S \subset W$ and there exists $\rho<\omega_1$ such that $f(\alpha)=\rho$ for all $\alpha \in S$. This fact is called the pressing down lemma and will be used below. See this post for more information about the pressing down lemma.

For each $\gamma \in W$, let $h_\gamma: Y_{\gamma+1} \rightarrow \delta$ be a one-to-one function. For each $\gamma \in W$, define $t_\gamma \in \prod_{\alpha \in W} Y_\alpha$ as follows:

$t_\gamma(\alpha) = \begin{cases} h_\gamma(\alpha) & \mbox{if } \delta \le \alpha \le \gamma \\ \gamma & \mbox{if } \gamma < \alpha <\omega_1 \end{cases}$

Note that each $t_\gamma$ is a pressing down function. Thus each $t_\gamma \in \prod_{\alpha \in W} Y_\alpha$. Let $T=\left\{t_\gamma: \gamma \in W \right\}$. Clearly $t_\gamma \ne t_\mu$ if $\gamma \ne \mu$. Thus $T$ has cardinality $\omega_1$. We claim that $T$ is a closed and discrete subset of $\prod_{\alpha \in W} Y_\alpha$. It suffices to show that for each $f \in \prod_{\alpha \in W} Y_\alpha$, there exists an open set $V$ with $f \in V$ such that $V$ contains at most one $t_\gamma$.

Let $f \in \prod_{\alpha \in W} Y_\alpha$. As discussed above, there is a stationary set $S \subset W$ and there exists $\rho<\omega_1$ such that $f(\alpha)=\rho$ for all $\alpha \in S$. In particular, choose $\mu, \lambda \in S$ such that $\mu \ne \lambda$. Thus $f(\mu)=f(\lambda)=\rho$. Let $V$ be the open set defined by:

$V=\left\{g \in \prod_{\alpha \in W} Y_\alpha: g(\mu)=g(\lambda)=\rho \right\}$

Clearly, $f \in V$. We show that if $t_\gamma \in V$, then $\gamma=\rho$. Suppose $t_\gamma \in V$. Then $t_\gamma(\mu)=t_\gamma(\lambda)=\rho$. Consider two cases: Case 1: $\delta \le \mu, \lambda \le \gamma$; Case 2: one of $\mu$ and $\lambda>\gamma$. The definition of $t_\gamma$ indicates that $t_\gamma=h_\gamma$ on the interval $[\delta, \gamma]$. Note that $h_\gamma$ is a one-to-one function. Since $t_\gamma(\mu)=t_\gamma(\lambda)$, it cannot be that $\mu, \lambda \in [\delta, \gamma]$, i.e., Case 1 is not possible. Thus Case 2 holds, say $\mu>\gamma$. Then by definition, $t_\gamma(\mu)=\gamma$. Putting everything together, $\gamma=t_\gamma(\mu)=t_\gamma(\lambda)=\rho$. Thus $V \cap T \subset \left\{t_\rho \right\}$. This concludes the proof that the set $T$ is a closed and discrete subset of $\prod_{\alpha \in W} Y_\alpha$. $\blacksquare$

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

# Sigma-products of first countable spaces

A product space is never first countable if there are uncountably many factors. For example, $\prod_{\alpha < \omega_1}\mathbb{R}=\mathbb{R}^{\omega_1}$ is not first countable. In fact any dense subspace of $\mathbb{R}^{\omega_1}$ is not first countable. In particular, the subspace of $\mathbb{R}^{\omega_1}$ consisting of points which have at most countably many non-zero coordinates is not first countable. This subspace is called the $\Sigma$-product of $\omega_1$ many copies of the real line $\mathbb{R}$ and is denoted by $\Sigma_{\alpha<\omega_1} \mathbb{R}$. However, this $\Sigma$-product is a Frechet space (or a Frechet-Urysohn space). In this post, we show that the $\Sigma$-product of first countable spaces is a Frechet space.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $a \in X$. Consider the following subspace of $X$:

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

The above subspace of $X$ is called the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ about the base point $a$. When the base point is understood, we simply say the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ and use the notation $\Sigma_{\alpha \in A} X_\alpha$ to denote the space.

For each $y \in \Sigma_{\alpha \in A} X_\alpha$, define $S(y)$ to be the set of all $\alpha \in A$ such that $y_\alpha \ne a_\alpha$, i.e., the support of the point $y$. Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set $O=\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is an open subset of $X_\alpha$. The support of $O$, denoted by $supp(O)$ is the finite set of all $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

A space $Y$ is said to be first countable if there exists a countable local base at each point in $Y$. A space $Y$ is said to be a Frechet space if for each $y \in Y$ and for each $M \subset Y$, if $y \in \overline{M}$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $M$ such that the sequence converges to $y$. Frechet spaces also go by the name of Frechet-Urysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.

Theorem 1

Suppose each factor $X_\alpha$ is a first countable space. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is a Frechet space.

Proof of Theorem 1
Let $\Sigma=\Sigma_{\alpha \in A} X_\alpha$. Let $M \subset \Sigma$ and let $x \in \overline{M}$. We proceed to define a sequence of points $t_n \in M$ such that the sequence $t_n$ converges to $x$. For each $\alpha \in A$, choose a countable local base $\left\{B_{\alpha,j}: j=1,2,3,\cdots \right\}$ at the point $x_\alpha \in X_\alpha$. Assume that $B_{\alpha,1} \supset B_{\alpha,2} \supset B_{\alpha,3} \supset \cdots$. Then enumerate the countable set $S(x)$ by $S(x)=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\cdots \right\}$. Let $C_1=\left\{\beta_{1,1} \right\}$. The following set $O_1$ is an open subset of $\Sigma$.

$O_1=\biggl(\prod_{\alpha \in C_1} B_{\alpha,1} \times \prod_{\alpha \in A-C_1} X_\alpha \biggr) \cap \Sigma$

Note that $O_1$ is an open set containing $x$. Choose $t_2 \in O_1 \cap M$. Enumerate the support $S(t_2)$ by $S(t_2)=\left\{\beta_{2,1},\beta_{2,2},\beta_{2,3},\cdots \right\}$. Form the finite set $C_2$ by picking the first two points of $S(x)$ and the first two points of $S(t_2)$, i.e., $C_2=\left\{\beta_{1,1},\beta_{1,2},\beta_{2,1},\beta_{2,2} \right\}$. Then form the following open subset of $\Sigma$.

$O_2=\biggl(\prod_{\alpha \in C_2} B_{\alpha,2} \times \prod_{\alpha \in A-C_2} X_\alpha \biggr) \cap \Sigma$

Choose $t_3 \in O_2 \cap M$. Enumerate the support $S(t_3)$ by $S(t_3)=\left\{\beta_{3,1},\beta_{3,2},\beta_{3,3},\cdots \right\}$. Then let $C_3=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\ \beta_{2,1},\beta_{2,2},\beta_{2,3},\ \beta_{3,1},\beta_{3,2},\beta_{3,3} \right\}$, i.e., picking the first three points of $S(x)$, the first three points of $S(t_2)$ and the first three points of $S(t_3)$. Now, form the following open subset of $\Sigma$.

$O_3=\biggl(\prod_{\alpha \in C_3} B_{\alpha,3} \times \prod_{\alpha \in A-C_3} X_\alpha \biggr) \cap \Sigma$

Choose $t_4 \in O_2 \cap M$. Let this inductive process continue and we would obtain a sequence $t_2,t_3,t_4,\cdots$ of points of $M$. We claim that the sequence converges to $x$. Before we prove the claim, let’s make a few observations about the inductive process of defining $t_2,t_3,t_4,\cdots$. Let $C=\bigcup_{j=1}^\infty C_j$.

• Each $C_j$ is the support of the open set $O_j$.
• The sequence of open sets $O_j$ is decreasing, i.e., $O_1 \supset O_2 \supset O_3 \supset \cdots$. Thus for each integer $j$, we have $t_k \in O_j$ for all $k \ge j$.
• The support of the point $x$ is contained in $C$, i.e., $S(x) \subset C$.
• The support of the each $t_j$ is contained in $C$, i.e., $S(t_j) \subset C$.
• In fact, $C=S(x) \cup S(t_2) \cup S(t_3) \cup \cdots$.
• The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets $O_j$.
• Consequently, for each $j$, $x_\alpha=(t_j)_\alpha=a_\alpha$ for each $\alpha \in A-C$. In other words, $x$ and each $t_j$ agree (and agree with the base point $a$) on the coordinates outside of the countable set $C$.

Let $U=\prod_{\alpha \in A} U_\alpha$ be a standard open set in the product space $X=\prod_{\alpha \in A} X_\alpha$ such that $x \in U$. Let $U^*=U \cap \Sigma$. We show that for some $n$, $t_j \in U^*$ for all $j \ge n$.

Let $F=supp(U)$ be the support of $U$. Let $F_1=F \cap C$ and $F_2=F \cap (A-C)$. Consider the following open set:

$U^{**}=\biggl(\prod_{\alpha \in C} U_\alpha \times \prod_{\alpha \in A-C} X_\alpha \biggr) \cap \Sigma$

Note that $supp(U^{**})=F_1$. For each $\alpha \in F_1$, choose $B_{\alpha,k(\alpha)} \subset U_\alpha$. Let $m$ be the maximum of all $k(\alpha)$ where $\alpha \in F_1$. Then $B_{\alpha,m} \subset U_\alpha$ for each $\alpha \in F_1$. Choose a positive integer $p$ such that:

$F_1 \subset W=\left\{\beta_{i,j}: i \le p \text{ and } j \le p \right\}$

Let $n=\text{max}(m,p)$. It follows that there exists some $n$ such that $O_n \subset U^{**}$. Then $t_j \in U^{**}$ for all $j \ge n$. It is also the case that $t_j \in U^{*}$ for all $j \ge n$. This is because $x=t_j$ on the coordinates not in $C$. $\blacksquare$

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