# One way to get CCC spaces that are not separable

A space $X$ is said to have countable chain condition (ccc) if every disjoint collection of open subsets of $X$ is countable. For convenience, we refer to spaces that have countable chain condition as ccc spaces. It is easy to verify that separable spaces are ccc spaces. We present a specific way of generating spaces that always have ccc but are not separable. These spaces are the sigma products of separable spaces.

The product of separable spaces always have ccc (see Product of Spaces with Countable Chain Condition). However, the product of separable spaces is not separable when the number of factors is greater than continuum. Thus one way to get an example of ccc but not separable space is to take the product of more than continuum many separable spaces. For example, if $c$ is the cardinality of continuum, $\left\{0,1 \right\}^{2^c}$, the product of $2^c$ many copies of $\left\{0,1 \right\}$, is a space that has ccc but is not separable.

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Another example is obtained from taking sigma product of separable spaces. Let $\left\{X_\alpha: \alpha \in A \right\}$ be a collection of spaces where $A$ is some index set. Consider the product space $\prod \limits_{\alpha \in A} X_\alpha$. Fix a point $p$ in the product. The sigma-product about the point $p$ is denoted by $\sum \limits_{\alpha \in A} X_\alpha$ and is the following subspace of the product space $\prod \limits_{\alpha \in A} X_\alpha$:

$\sum \limits_{\alpha \in A} X_\alpha=\left\{x \in \prod \limits_{\alpha \in A} X_\alpha: x(\alpha) \ne p(\alpha) \text{ for at most countably many } \alpha \right\}$

To obtain the desired example, let $A$ be an uncountable index set and let each $X_\alpha$ be separable. The product space $\prod \limits_{\alpha \in A} X_\alpha$ has ccc. Note that $\sum \limits_{\alpha \in A} X_\alpha$ is always dense in the product space $\prod \limits_{\alpha \in A} X_\alpha$. Thus the sigma-product $\sum \limits_{\alpha \in A} X_\alpha$ has ccc since it is a dense subspace of a ccc space. On the other hand, $\sum \limits_{\alpha \in A} X_\alpha$ is never separable as long as there are uncountably many spaces $X_\alpha$.

As specific example, take $X_\alpha=\left\{0,1 \right\}$ for each $\alpha < \omega_1$ and let the fixed point $p$ be such that $p(\alpha)=0$ for all $\alpha < \omega_1$. The resulting $\sum \limits_{\alpha < \omega_1} \left\{0,1 \right\}$ is a ccc space that is not separable. Of course, $\sum \limits_{\alpha < \omega_1} X_\alpha$ in this case is the set of all $x \in \left\{0,1 \right\}^{\omega_1}$ such that $x(\alpha) \ne 0$ for at most countably many $\alpha < \omega_1$.

One interesting note about the sigma-product $\sum \limits_{\alpha < \omega_1} \left\{0,1 \right\}$ is that the overall product space $\left\{0,1 \right\}^{\omega_1}$ is an example of a separable but not hereditarily separable space. Another interesting point is that $\sum \limits_{\alpha < \omega_1} \left\{0,1 \right\}$ is a countably compact non-compact space (see A note about sigma-product of compact spaces).

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Previous discussion of CCC spaces in this blog:

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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