Perfect images of separable metric spaces

Posted on March 28, 2023 by Dan Ma

The Bow-Tie space is the continuous image of a separable metric space and yet is not metrizable (see here). Though taking continuous image can fail to preserve separable metrizability, we show that the perfect image of a separable metric space is a separable metric space. We prove the following theorem, which says that under a perfect map the weight will not increase. The result about separable metric space is a corollary of Theorem 1.

Theorem 1
Let $f:X \longrightarrow Y$ be a perfect map onto the space $Y$ . Then $w(Y) \le w(X)$ , i.e., the weight of $Y$ is no greater than the weight of $X$ .

Proof of Theorem 1

All spaces under consideration are Hausdorff. Let $X$ and $Y$ be spaces. Let $f: \longrightarrow Y$ be a map (or function) from $X$ onto $Y$ . The map $f$ is said to be a closed map if $f(C)$ is closed in $Y$ for any closed subset $C$ of $X$ . The map $f$ is a perfect map if $f$ is continuous, $f$ is a closed map, and $f^{-1}(y)$ is compact for every $y \in Y$ . In words, the last condition is that every point inverse is compact. A point inverse $f^{-1}(y)$ is also referred to as a fiber. Thus, we can say that a perfect map is a continuous closed surjective map with compact fibers. The following lemma is helpful for proving Theorem 1.

Lemma 2
Let $f:X \longrightarrow Y$ be a closed map such that $f(X)=Y$ . Let $V \subset X$ be open. Let $f_*(V)=\{ y \in Y: f^{-1}(y) \subset V \}$ . Then $f_*(V)$ is open in $Y$ and $f_*(V) \subset f(V)$ .

Proof of Lemma 2
We show that $Y \backslash f_*(V)$ is closed in $Y$ . To this end, we show $f(X \backslash V)=Y \backslash f_*(V)$ . Note that $f(X \backslash V)$ is closed in $Y$ since $f$ is a closed map. First, we show $f(X \backslash V) \subset Y \backslash f_*(V)$ . Let $t \in f(X \backslash V)$ . Then $t=f(x)$ for some $x \in X \backslash V$ . Since $x \notin V$ and $x \in f^{-1}(t)$ , we have $f^{-1}(t) \not \subset V$ . This implies that $t \notin f_*(V)$ .

We now show that $Y \backslash f_*(V) \subset f(X \backslash V)$ . Let $z \in Y \backslash f_*(V)$ . Since $z \notin f_*(V)$ , $f^{-1}(z) \not \subset V$ . Choose $x \in f^{-1}(z) \backslash V$ , which implies that $x \in X \backslash V$ . Thus, $z=f(x) \in f(X \backslash V)$ .

To complete the proof of the lemma, we show $f_*(V) \subset f(V)$ . Let $w \in f_*(V)$ . We have $f^{-1}(w) \subset V$ . As a result, $w=f(x)$ for some $x \in V$ . $\square$

Proof of Theorem 1
Let $\mathcal{B}$ be a base for $X$ . We derive a base $\mathcal{B}_1$ for $Y$ such that $\lvert \mathcal{B}_1 \lvert \le \lvert \mathcal{B} \lvert$ , i.e., the cardinality of $\mathcal{B}_1$ is no more than the cardinality of $\mathcal{B}$ . This implies that the minimal cardinality of a base in $Y$ is no more than the minimal cardinality of a base in $X$ , i.e., $w(Y) \le w(X)$ .

We assume that the base $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_1=\{ f_*(B): B \in \mathcal{B} \}$ is a base for $Y$ . Note that $f_*$ is defined in Lemma 2. Let $U \subset Y$ be an open set. Let $y \in U$ . For each $x \in f^{-1}(y)$ , choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$ . Since $f^{-1}(y)$ is compact, there exists finite $F \subset f^{-1}(y)$ such that $f^{-1}(y) \subset \bigcup_{t \in F} B_t=B$ . Note that $B \in \mathcal{B}$ . Since $f^{-1}(y) \subset B$ , we have $y \in f_*(B)$ . We also have $f_*(B) \subset f(B) \subset U$ . Thus, every open subset $U$ of $Y$ is the union of elements of $\mathcal{B}_1$ . This means that $\mathcal{B}_1$ is a base for $Y$ . Theorem 1 is established. $\square$

Corollary 3
Let $f:X \longrightarrow Y$ be a perfect map onto the space $Y$ . Then if $X$ is a separable metric space, then $Y$ is a separable metric space.

Comment About Lemma 2
A perfect map is not necessarily an open map. If the perfect map $f$ in Theorem 1 is an open map, then Lemma 2 is not needed and $\mathcal{B}_1=\{ f(B): B \in \mathcal{B} \}$ would be a base for $Y$ . However, we cannot assume $f$ is an open map simply because it is a perfect map. To see this, let $X=\mathbb{R}$ be the real line with the usual topology. Collapse the closed interval $[1,2]$ to one point called $p$ . The resulting quotient space is $Y$ where $Y=(-\infty,1) \cup \{ p \} \cup (2,\infty)$ . In $Y$ , the open neighborhoods of points in $(-\infty,1) \cup (2,\infty)$ are the usual Euclidean neighborhoods. The open neighborhoods of the point $p$ are the usual Euclidean open sets containing the interval $[1,2]$ . The resulting quotient map $f$ is an identity map on $(-\infty,1) \cup (2,\infty)$ and it maps points in $[1,2]$ to the point $p$ . It can be verified that $f$ is a perfect map. For the open set $V=(1,2)$ , $f(V)=\{ p \}$ , which is not open in $Y$ . For the open set $V=(0,1.5)$ , $f(V)=(0,1) \cup \{ p \}$ , which is not open in $Y$ . Lemma 2 says that for any open $X \subset$ , $f(V)$ may not be open but has an open subset $f_*(V)$ if $f(V)$ has non-empty interior. The interior sets $f_*(V)$ can work as a base in $Y$ .

Invariant and Inverse Invariant

Let $\mathcal{P}$ be a property of topological spaces. We say that $\mathcal{P}$ is an invariant of the perfect maps or that $\mathcal{P}$ is invariant under the perfect maps if the property $\mathcal{P}$ is preserved by perfect maps, i.e., for each perfect map $f:X \longrightarrow Y$ where $Y=f(X)$ , if the space $X$ has $\mathcal{P}$ , so does $Y$ . On the other hand, $\mathcal{P}$ is an inverse invariant of the perfect maps if this holds: for each perfect map $f:X \longrightarrow Y$ where $Y=f(X)$ , if $Y$ has $\mathcal{P}$ , so does $X$ .

The notions invariant and inverse invariant defined here are for perfect maps. In general, the notions are much broader and can be defined in relation to any class of continuous maps. For example, we know that the continuous image of a separable space is separable. We can say that separability is an invariant of the continuous maps or that separability is invariant under continuous maps. We can now restate Theorem 1 and Corollary 3 as follows.

Theorem 4…..Restatement of Theorem 1
The property “weight $\le \mathcal{K}$ ” is an invariant of the perfect maps.

Corollary 5
The second axiom of countability is invariant under the perfect maps, but is not an invariant of the continuous maps.

The Bow-Tie space is the continuous image of a separable space but cannot have a countable base (see here). In light of Theorem 1, the continuous map that maps a separable metric space to the Bow-Tie space (shown here) cannot be a perfect map. With respect to that map, the upper half plane in the domain (the separable metric space) is closed but its continuous image in the Bow-Tie space is open and not closed.

$\text{ }$

Dan Ma Bow-Tie space
Daniel Ma Bow-Tie space

Dan Ma perfect map
Daniel Ma perfect map

Dan Ma separable metric space
Daniel Ma separable metric space

Dan Ma topology
Daniel Ma topology

$\copyright$ 2023 – Dan Ma

Revised March 31, 2023
Revised March 12, 2024

Sigma-products of separable metric spaces are monolithic

Posted on June 3, 2014 by Dan Ma

Let $\Sigma(\kappa)$ be the $\Sigma$ -product of $\kappa$ many copies of the real lines where $\kappa$ is any infinite cardinal number. Any compact space that can be embedded in $\Sigma(\kappa)$ for some $\kappa$ is said to be a Corson compact space. Corson compact spaces play an important role in functional analysis. Corson compact spaces are also very interesting from a topological point of view. Some of the properties of Corson compact spaces are inherited (as subspaces) from the $\Sigma$ -product $\Sigma(\kappa)$ . One such property is the property that the $\Sigma$ -product $\Sigma(\kappa)$ is monolithic, which implies that the closure of any countable subspace of $\Sigma(\kappa)$ is metrizable.

Previous blog posts on $\Sigma$ -products:

A previous blog post on monolithic spaces: A short note on monolithic spaces. A listing of other blog posts on Corson compact spaces is given at the end of this post.

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Defining Sigma-product

Let $\kappa$ be an infinite cardinal number. For each $\alpha<\kappa$ , let $X_\alpha$ be a topological space. Let $b \in \prod_{\alpha<\kappa} X_\alpha$ . The $\Sigma$ -product of the spaces $X_\alpha$ about the base point $b$ is defined as follows:

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

If each $X_\alpha=\mathbb{R}$ and if the base point $b$ is such that $b_\alpha=0$ for all $\alpha<\kappa$ , then we use the notation $\Sigma(\kappa)$ for $\Sigma_{\alpha<\kappa} X_\alpha$ , i.e., $\Sigma(\kappa)$ is defined as follows:

$\Sigma(\kappa)=\left\{x \in \mathbb{R}^\kappa: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \kappa \right\}$

A compact space is said to be a Corson compact space if it can be embedded in the $\Sigma$ -product $\Sigma(\kappa)$ for some infinite cardinal $\kappa$ .

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Monolithic Spaces

A space $X$ is monolithic if for every subspace $Y$ of $X$ , the density of $Y$ equals the network weight of $Y$ , i.e., $d(Y)=nw(Y)$ . A space $X$ is strongly monolithic if for every subspace $Y$ of $X$ , the density of $Y$ equals the weight of $Y$ , i.e., $d(Y)=w(Y)$ . See the previous post called A short note on monolithic spaces.

The proof of the fact that $\Sigma$ -product of separable metrizable spaces is monolithic can be worked out quite easily from definitions. Interested readers are invited to walk through the proof. For the sake of completeness, we prove the following theorem.

Theorem 1
Suppose that for each $\alpha<\kappa$ , $X_\alpha$ is a separable metric space. Then the $\Sigma$ -product $\Sigma_{\alpha<\kappa} X_\alpha$ is strongly monolithic.

Proof of Theorem 1
Let $b$ be the base point of the $\Sigma$ -product $X=\Sigma_{\alpha<\kappa} X_\alpha$ . For each $x \in X$ , let $S(x)$ be the support of the point $x$ , i.e., the set of all $\alpha<\kappa$ such that $x_\alpha \ne b_\alpha$ . Let Y be a subspace of $X$ . We show that $d(Y)=w(Y)$ .

Let $T=\left\{t_\delta: \delta<\tau \right\}$ be a dense subspace of $Y$ such that $d(Y)=\lvert T \lvert=\tau$ . Note that $\overline{T}=Y$ (closure is taken in $Y$ ). Let $S=\bigcup_{\delta<\tau} S(t_\delta)$ . Clearly $\lvert S \lvert \le \tau$ . Consider the following subspace of $X$ :

$X(S)=\left\{x \in X: S(x) \subset S \right\}$

It is clear that $X(S)$ is a closed subspace of $X$ . Since $T \subset X(S)$ , the closure of $T$ (closure in $X$ or in $Y$ ) is a subspace of $X(S)$ . Thus $Y \subset X(S)$ . Note that $\overline{T}=Y \subset X(S)$ . Since each $X_\alpha$ has a countable base, the product space $\prod_{\alpha<\tau} X_\alpha$ has a base of cardinality $\tau$ . Thus $\prod_{\alpha<\tau} X_\alpha$ has weight $\le \tau$ . Since $X(S) \subset \prod_{\alpha<\tau} X_\alpha$ , both $Y$ and $X(S)$ have weights $\le \tau$ . We have $w(Y) \le d(Y)=\tau$ . Note that $d(Y) \le w(Y)$ always holds. Therefore $d(Y)=w(Y)$ . $\blacksquare$

Corollary 2
For any infinite cardinal $\kappa$ , the $\Sigma$ -product $\Sigma(\kappa)$ is strongly monolithic.

Corollary 3
Any Corson compact space is strongly monolithic.

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Blog posts on Corson compact spaces

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

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

Posted on April 21, 2014 by Dan Ma

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

Posted on April 19, 2014 by Dan Ma

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

$X=\prod_{\alpha<\omega_1} X_\alpha$

$X_\alpha$

$Y$

$X$

$Y$

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

$X=\prod_{\alpha<\omega_1} X_\alpha$

$X_\alpha$

$Y$

$X$

$Y$

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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

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

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:

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

Arkhangelskii, A. V., Normality and dense subspaces, Proc. Amer. Math. Soc., 130 (1), 283-291, 2001.
Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
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

Posted on April 18, 2014 by Dan Ma

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

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

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

Cp(X) where X is a separable metric space

Posted on April 3, 2014 by Dan Ma

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

Posted on March 28, 2014 by Dan Ma

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

$\alpha \in A$

$X_\alpha$

$\Sigma_{\alpha \in A} X_\alpha$

Theorem 2

$\alpha \in A$

$X_\alpha$

$Y=\Sigma_{\alpha \in A} X_\alpha$

$Y^\omega$

Theorem 3 (Corson’s Theorem)

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y \times Y$

$Y$

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

The above three theorems lead to the following theorem.

Theorem 4

$\alpha \in A$

$X_\alpha$

$\Sigma_{\alpha \in A} X_\alpha$

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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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

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

Corson H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
Gulko S. P., On the properties of sets lying in $\Sigma$ -products, Dokl. Acad. Nauk. SSSR, 237, 505-508, 1977 (in Russian).
Kombarov A. P., On the tightness and normality of $\Sigma$ -products, Dokl. Acad. Nauk. SSSR, 239, 775-778, 1978 (in Russian).
Kombarov A. P., Malyhin V. I., On $\Sigma$ -products, Soviet Math. Dokl., 14 (6), 1980-1983, 1973.
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.
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

Posted on March 26, 2014 by Dan Ma

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)

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y \times Y$

$Y$

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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y \times Y$

$Y$

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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

$S$

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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y \times Y$

$Y$

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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y \times Y$ is normal.
For any pair of disjoint closed subsets $H$ and $K$ of $Y \times Y$ , there exists a countable $C \subset A$ such that $\overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing$ .
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

Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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

Posted on March 25, 2014 by Dan Ma

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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

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

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y \times Y$ is normal.
For any pair of disjoint closed subsets $H$ and $K$ of $Y \times Y$ , there exists a countable $C \subset A$ such that $\overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing$ .
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,

$H$

$K$

$Y$

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 $B \subset E$ .
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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A factorization theorem for products of separable spaces

Posted on March 23, 2014 by Dan Ma

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

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

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Factoring a Continuous Map

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

We have the following lemmas.

Lemma 1

$X=\prod_{\alpha \in A} X_\alpha$

$T$

$f:X \rightarrow T$

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

Lemma 1a

$X=\prod_{\alpha \in A} X_\alpha$

$T$

$Y$

$X$

$f:Y \rightarrow T$

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

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

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When a Continuous Map Can Be Factored

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

Theorem 1

$X=\prod_{\alpha \in A} X_\alpha$

$X_\alpha$

$T$

$Y$

$X$

$f:Y \rightarrow T$

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

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

Lemma 2

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Another Version

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

Theorem 2

$X=\prod_{\alpha \in A} X_\alpha$

$X_\alpha$

$T$

$Y$

$X$

$f:Y \times Y \rightarrow T$

$f$

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

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

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

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

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

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

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Remarks

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

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Reference

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

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Dan Ma's Topology Blog

Expository Articles In General Topology

Tag Archives: Separable metrizable space

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