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}

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