Let be the first uncountable ordinal. A separable space is one that has a countable dense subset. Any separable space satisfies this property:

For every collection of nonempty open subsets of , there is an uncountable such that .
Any space that has this property is said to have caliber . Spaces that have caliber have the countable chain condition (abbreviated by CCC, which means that there is no uncountable pairwise disjoint collection of open subsets of the space). So we have the following implications:
The chain condition “caliber ” is an interesting one. Some of the spaces that have the CCC actually have caliber . For example, separable spaces and products of separable spaces have caliber . Thus the product space (for any uncountable cardinal ) not only has the countable chain condition. It has the stronger property of having caliber , which may make certain proof easier to do. In this post we also provide examples to show that none of the above implications is reversible.
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Delta System Lemma
In proving that product of separable spaces has caliber , the Delta system lemma is used. A collection of sets is said to be a Deltasystem (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 Deltasystem . The following is the statement of Deltasystem lemma.
Lemma 1 – DeltaSystem Lemma
For every uncountable collection of finite sets, there is an uncountable such that is a system.
The statement of Deltasystem lemma presented here is a special case for a general version (see [2], page 49).
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Products of Spaces with Caliber
Theorem 2 below shows that whenever “caliber ” is preserved by taking product with any finite number of factors, “caliber ” 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 caliber . The proof of Theorem 2 is similar to the one stating 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 (see the previous post Product of Spaces with Countable Chain Condition).
Theorem 2
Suppose that is a family of spaces such that has caliber for every finite . Then has caliber .
Proof
In proving the product space having caliber , it suffices to work with basic open sets of the form where for all but finitely many . Let be an uncountable collection of such nonempty open sets. Our plan is to find an uncountable such that .
For each , let be the finite set such that if and only if . Consider . By the Deltasystem lemma, there is an uncountable such that is a Deltasystem. Let be the root of this Deltasystem.
Consider the case that the root of the Deltasystem is empty. Then for any and where , we have . For each , we have and we choose in . Then we can define an in such that extends for all . Then in this case let and we have . So we move onto the case that .
Now assume . Let . For each where , let be (i.e. is a projection map). Let .
Consider two cases. One is that is countable. The second is that is uncountable. Suppose is countable. Then there is an uncountable such that for all . Then fix and choose . Then is extendable to some in such that for all . Thus we have .
Now assume is uncountable. By assumption has caliber . Then there is an uncountable such that . Choose . As in the previous case, is extendable to some in such that for all . Thus we have .
Corollary 3
Suppose that is a family of separable spaces. Then has caliber .
Proof
This follows directly from Theorem 2. Note that the product of finitely many separable is separable (hence has caliber ).
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Examples
We now show that the following implications are not reversible.
To get a space with caliber that is not separable, consider the product of many copies of where is a cardinal number greater than continuum. Since is a product of separable spaces, it has caliber according to Corollary 3. It is well known that the product of more than continuum many separable spaces is not separable (see Product of Separable Spaces).
To get a space with the CCC that does not have caliber . Consider the subspace of the product space ( many copies of ) where is the set of all such that for at most countably many . Note that is a space with the CCC since it is a product of separable spaces. Furthermore is a dense subspace of . The property of CCC is hereditary with respect to dense subsets. Thus has the countable chain condition. Here's a discussion of the countable chain condition.
To see that does not have caliber , look at the collection of open sets where each . Note that each belongs to at most countably many . Thus for any uncountable , .
The example shows that dense subspace of a space with caliber need not have caliber .
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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, NorthHolland, Amsterdam, 1980.
 Willard, S., General Topology, AddisonWesley Publishing Company, 1970.
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