# Not having c.c.c. implies existence of sequence of open sets with strictly increasing closures?

Suppose a topological space $X$ does not satisfy the countable chain condition (by that I mean that there exists an uncountable family of pairwise disjoint non-empty open subsets of $X$). Does it follow that there is a sequence of open sets such that their closures form a strictly increasing sequence with respect to inclusion?

• Do you want the sequence to have uncountable length? – user642796 Mar 21 '14 at 10:24
• No, by a sequence I mean a countable sequence. – Bach Mar 21 '14 at 10:25

Suppose you have any space $X$ in which there are pairwise disjoint nonempty open sets $\{ U_i : i \in \mathbb{N} \}$. I claim that $\langle \bigcup_{i=1}^n U_i \rangle_{n =1}^\infty$ is as required:
Given $n \geq 1$, note that $U_{n+1} \cap \bigcup_{i=1}^n U_i = \varnothing$, and so $U_{n+1} \cap \overline{ \bigcup_{i=1}^n U_i } = \varnothing$ (since $U_{n+1}$ is open, it is a witness for each $x \in U_{n+1}$ that $x \notin \overline{ \bigcup_{i=1}^n U_i }$), which implies that $\overline{ \bigcup_{i=1}^n U_i } \subsetneq \overline{ \bigcup_{i=1}^{n+1} U_i }$.