# Product of Spaces with Countable Chain Condition

A topological space is said to be separable if it has a countable dense subset. In any separable space, there cannot exist uncountably many pairwise disjoint open subsets (if that is the case, any dense set will have to be uncountable). A topological space $X$ is said to have countable chain condition (ccc) if every disjoint collection of open subsets of $X$ is countable. Thus any separable space is a space with ccc. We take a look at what happen when we take product of spaces with these two properties.

The product of two separable spaces is always a separable space. If $A$ is a countable dense set in the space $X$ and $B$ is a countable dense set in the space $Y$, then $A \times B$ is a countable dense set in the product space $X \times Y$. It follows that the product of finitely many separable spaces is separable. When the number of factors is infinite, the cardinality of the continuum is the dividing line. The product of separable spaces is sometimes separable (when the number of factors is less than or equal to continuum) and is sometimes not separable (when the number of factors is greater than continuum). For a discussion of this result, see Product of Separable Spaces in this blog, or see [1] and [3].

Is the product of two ccc spaces a space with ccc? It turns out that this question is independent of ZFC. That is, this question cannot be answered on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC (see [2], page 50). However it can be proven in ZFC that the product of any number of separable spaces has countable chain condition. In proving this result, Delta-system lemma is used.

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Delta-System Lemma

A collection $\mathcal{D}$ of sets is said to be a Delta-system (or $\Delta$-system) if there is a set $D$ such that for every $A,B \in \mathcal{D}$ with $A \ne B$, we have $A \cap B = D$. When such set $D$ exists, it is called the root of the Delta-system $\mathcal{D}$. The following is the statement of Delta-system lemma.

Lemma 1 – Delta-System Lemma
For every uncountable collection $\mathcal{A}$ of finite sets, there is an uncountable $\mathcal{D} \subset \mathcal{A}$ such that $\mathcal{D}$ is a $\Delta$-system.

The statement of Delta-system lemma presented here is a special case (see [2], page 49) for the general version.

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Products of CCC Spaces

Even though the question of whether the product of two ccc spaces has ccc cannot be answered just within ZFC, we have a theorem that can gives us quite a bit of clarity. We have an amazing result that whenever the product of two ccc spaces has ccc, the product of any number of ccc spaces has ccc. Note that when countable chain condition is preserved by taking product with two factors, ccc is preserved by taking product with any finite number of factors. The following theorem shows that whenever ccc is preserved by taking product with any finite number of factors, ccc is preserved by taking the product of any number of factors. As a corollary, we have the result that the product of any number of separable spaces has ccc.

By manipulating the number of factors, we can easily obtain a space that has ccc but is not separable. For example, let $\mathcal{K}$ be any cardinal that is larger than continuum. Then $\left\{0,1\right\}^\mathcal{K}$ is a space that has ccc but is not separable.

Theorem 2
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of spaces such that $\prod \limits_{\alpha \in F} X_\alpha$ has countable chain condition for every finite $F \subset T$. Then $\prod \limits_{\alpha \in T} X_\alpha$ has countable chain condition.

Proof
In proving the product space $\prod \limits_{\alpha \in T} X_\alpha$ having ccc, it suffices to work with basic open sets of the form $\prod \limits_{\alpha \in T} U_\alpha$ where $U_\alpha=X_\alpha$ for all but finitely many $\alpha \in T$. Let $\mathcal{U}$ be an uncountable collection of such basic open sets such that open sets in $\mathcal{U}$ are pairwise disjoint.

For each $G = \prod \limits_{\alpha \in T} U_\alpha \in \mathcal{U}$, let $A(G)$ be the finite set such that $A(G) \subset T$, and such that $\alpha \in A(G)$ if and only if $U_\alpha \ne X_\alpha$. Let $\mathcal{A}$ be the set of all such $A(G)$. By Delta-system lemma, there is an uncountable $\mathcal{D} \subset \mathcal{A}$ such that $\mathcal{D}$ is a Delta-system. Let $D$ be the root of this Delta-system.

The root of the Delta-system cannot be non-empty. If $D = \varnothing$, then for any $A(G_1) \in \mathcal{D}$ and $A(G_2) \in \mathcal{D}$ where $A(G_1) \ne A(G_2)$ and $G_1,G_2 \in \mathcal{U}$, we have $A(G_1) \cap A(G_2) = \varnothing$, which leads to $G_1 \cap G_2 \ne \varnothing$. Thus $D \ne \varnothing$.

Let $\mathcal{U}^*$ be the collection of all $G \in \mathcal{U}$ such that $A(G) \in \mathcal{D}$. For each $G = \prod \limits_{\alpha \in T} U_\alpha \in \mathcal{U}^*$, let $p(G)=\prod \limits_{\alpha \in D} U_\alpha$ (i.e. $p$ is the projection map). Let $\mathcal{U}^{**}$ be the collection of all $p(G)$ where $G \in \mathcal{U}^*$.

We have the following observation:

• Observation. For any $A(G_1),A(G_2) \in \mathcal{D}$ with $A(G_1) \ne A(G_2)$, $\alpha \in A(G_1)-D$ implies $\alpha \notin A(G_2)-D$ and $\alpha \in A(G_2)-D$ implies $\alpha \notin A(G_1)-D$.

It follows from the above observation that the map $p$ is a one-to-one map from $\mathcal{U}^*$ into $\mathcal{U}^{**}$. Then $\mathcal{U}^{**}$ is an uncountable collection of open subsets of $\prod \limits_{\alpha \in D} X_\alpha$, which has ccc by hypothesis. So there exists $p(G_1),p(G_2) \in \mathcal{U}^{**}$ with $p(G_1) \cap p(G_2) \ne \varnothing$. The above observation allows us to define a point $x \in G_1 \cap G_2$, contradicting the assumption that $\mathcal{U}$ is a pairwise disjoint collection. Thus the entire product space $\prod \limits_{\alpha \in T} X_\alpha$ must have countable chain condition. $\blacksquare$

Corollary 3
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of separable spaces. Then $\prod \limits_{\alpha \in T} X_\alpha$ has countable chain condition.

Proof
This follows directly from Theorem 2. Note that the product of finitely many separable is separable (hence has ccc). $\blacksquare$

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ 2012$

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