A topological space $X$ satisfies the ccc property if for every family of open sets $\{U_i\}_{i\in\mathcal{I}}$ with empty intersection then $\mathcal{I}$ is countable. In exercise III.2.20 of Kunen's book Set Theory we hace the following characterization of the ccc property.

A space $X$ satisfies the ccc property iff there is no an increasing family of open sets $\{U_\alpha\}_{\alpha<\omega_1}$ such that for all $\alpha < \beta< \omega_1$ we have that $U_\beta\setminus\overline{U_\alpha}\neq \emptyset$.

Could anybody help me with the right-left implication? Thanks in advance.

  • $\begingroup$ I’m pretty sure that you’ve left out part of the statement of the result. How are $\alpha$ and $\beta$ related, and how are they quantified? $\endgroup$ – Brian M. Scott Dec 4 '15 at 13:51
  • $\begingroup$ Oh, for sure... $\beta>\alpha$. Thanks for your appreciation! $\endgroup$ – Ergonvi Dec 4 '15 at 14:00

So suppose $X$ satisfies the second condition and let $(U_i)_{i\in I}$ be a pairwise disjoint collection of open sets. By the well ordering theorem we may suppose that $I = \gamma$ is an ordinal, for $\alpha < \gamma$ define $$ V_\alpha = \bigcup_{\alpha' < \alpha} U_{\alpha'} $$ Let $\alpha < \beta$. $U_\alpha \subseteq \bar V_\beta$, but as $U_\alpha$ is disjoint form all $U_{\alpha'}$, $\alpha' < \alpha$, we have $V_\alpha \subseteq X \setminus U_\alpha$, hence, $\bar V_\alpha \subseteq X \setminus U_\alpha$. So $U_\alpha \subseteq \bar V_{\beta}\setminus \bar V_\alpha$. As $U_\alpha$ is non-empty, $\bar V_\beta \setminus \bar V_\alpha\ne \emptyset$. By the given condition $\gamma$ is countable, hence, so is $I$.

  • $\begingroup$ Thank you @martini! But probably, you would say $U_\alpha\subset V_\beta$ when $\alpha<\beta$ instead of $U_\alpha\subset \overline{V_\beta}$. I think that it's enough to ensure that $V_\beta\setminus \overline{V_\alpha}\neq \emptyset$. $\endgroup$ – Ergonvi Dec 4 '15 at 14:31

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