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 is said to have countable chain condition (ccc) if every disjoint collection of open subsets of 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 is a countable dense set in the space and is a countable dense set in the space , then is a countable dense set in the product space . 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 of sets is said to be a Delta-system (or -system) if there is a set such that for every with , we have . When such set exists, it is called the root of the Delta-system . The following is the statement of Delta-system lemma.

**Lemma 1 – Delta-System Lemma**

For every uncountable collection of finite sets, there is an uncountable such that is a -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 be any cardinal that is larger than continuum. Then is a space that has ccc but is not separable.

**Theorem 2**

Suppose that is a family of spaces such that has countable chain condition for every finite . Then has countable chain condition.

**Proof**

In proving the product space having ccc, it suffices to work with basic open sets of the form where for all but finitely many . Let be an uncountable collection of such basic open sets such that open sets in are pairwise disjoint.

For each , let be the finite set such that , and such that if and only if . Let be the set of all such . By Delta-system lemma, there is an uncountable such that is a Delta-system. Let be the root of this Delta-system.

The root of the Delta-system cannot be non-empty. If , then for any and where and , we have , which leads to . Thus .

Let be the collection of all such that . For each , let (i.e. is the projection map). Let be the collection of all where .

We have the following observation:

**Observation**. For any with , implies and implies .

It follows from the above observation that the map is a one-to-one map from into . Then is an uncountable collection of open subsets of , which has ccc by hypothesis. So there exists with . The above observation allows us to define a point , contradicting the assumption that is a pairwise disjoint collection. Thus the entire product space must have countable chain condition.

**Corollary 3**

Suppose that is a family of separable spaces. Then has countable chain condition.

**Proof**

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

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*Reference*

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

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