# A factorization theorem for products of separable spaces

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

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

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Factoring a Continuous Map

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

We have the following lemmas.

Lemma 1

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

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

Lemma 1a

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

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

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

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When a Continuous Map Can Be Factored

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

Theorem 1

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

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

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

Lemma 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Another Version

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

Theorem 2

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

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

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

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

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

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

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

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Remarks

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

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Reference

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

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