# Another characterization about CCC spaces

In this post, we present another characterization about spaces with the countable chain condition (CCC spaces for short). The theorem presented here (Theorem 1 below) will provide more insight about CCC spaces and should be useful in proving theorems about CCC spaces. This characterization will make it easy to see that CCC spaces are weakly Lindelof.

This post can be considered a continuation of an earlier post, which discusses a different characterization of CCC spaces.

All spaces under consideration are at least $T_1$ and regular. A space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of making a statement or stating a result, if $X$ has the CCC, we also say that $X$ is a CCC space or $X$ is CCC. We prove the following theorem.

Theorem 1

Let $X$ be a space. Then the following conditions are equivalent.

1. The space has the CCC.
2. For any collection $\mathcal{U}$ of non-empty open subsets of $X$, there exists a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}$.

Proof of Theorem 1
$1 \Longrightarrow 2$
Suppose that condition 2 does not hold. Then there exists a collection $\mathcal{U}$ of non-empty open subsets of $X$ such that for any countable $\mathcal{V} \subset \mathcal{U}$, there exists a point $x \in \bigcup \mathcal{U}=Y$ such that $x \notin \overline{\bigcup \mathcal{V}}$. From the collection $\mathcal{U}$, performing a transfinite inductive process, we will generate an uncountable collection of pairwise disjoint non-empty open subsets of $X$.

Choose some $U_0 \in \mathcal{U}$. Choose some $x_0 \in Y-\overline{U_0}$. For $\alpha < \omega_1$, suppose that the following have been chosen

$\left\{x_\beta \in Y: \beta<\alpha \right\}$

$\left\{U_\beta \in \mathcal{U}: \beta<\alpha \right\}$

such that for each $\beta<\alpha$, $x_\beta \notin \overline{\bigcup \limits_{\gamma < \beta} U_\gamma}$ and $x_\beta \in U_\beta$. Then by the assumption about $\mathcal{U}$, there exists $x_\alpha \in Y$ such that $x_\alpha \notin \overline{\bigcup \limits_{\gamma < \alpha} U_\gamma}$. Now choose some $U_\alpha \in \mathcal{U}$ such that $x_\alpha \in U_\alpha$. The inductive process is completed.

For each $\alpha<\omega_1$, let $O_\alpha=U_\alpha-\overline{\bigcup \limits_{\gamma < \alpha} U_\gamma}$. Clearly $O_\alpha \ne \varnothing$ since $x_\alpha \in O_\alpha$. For $\beta<\alpha<\omega_1$, we have $O_\beta \cap O_\alpha=\varnothing$. With the non-empty open sets $O_\alpha$ being pairwise disjoint, we conclude that $X$ does not have the CCC.

$2 \Longrightarrow 1$
This is the easier direction. Suppose $X$ is not CCC. Let $\mathcal{W}=\left\{W_\alpha: \alpha<\omega_1 \right\}$ be a collection of pairwise disjoint non-empty open subsets of $X$. It is clear that for any countable $\mathcal{V} \subset \mathcal{W}$, the closure $\overline{\bigcup \mathcal{V}}$ has to miss some $W_\alpha$ (e.g. choose some $W_\alpha \notin \mathcal{V}$). Thus condition 2 does not hold. $\blacksquare$

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Weakly Lindelof Spaces

With Theorem 1, the CCC property looks like a covering property. Let $X$ be a CCC space. Let $\mathcal{U}$ be an open cover of $X$. By Theorem 1, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $X=\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}$ (in other words, $\bigcup \mathcal{V}$ is dense in $X$). So any CCC space $X$ satisfies the following covering property:

For any open cover $\mathcal{U}$ of $X$, there exists a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense in $X$.

Any space satisfying the above property is called a weakly Lindelof space. Any CCC space is weakly Lindelof. On the other hand, the weakly Lindelof property is strictly weaker than CCC. For a further discussion, see the next post called Weakly Lindelof spaces.

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$\copyright \ 2014 \text{ by Dan Ma}$