If the topological space $X$ has CCC (= countable chain condition ) with given a countable closed discreted subspace $Y$ of $X$, could we seperate the points in $Y$ by countable disjoint open sets in $X$? Thanks for any help:)
On Countable chain condition
2 Answers
As Arthur Fischer points out, the result is false as stated.
If you assume that the space is regular, the result is true even without the countable chain condition. Let $Y=\{y_k:k\in\omega\}$. For $n\in\omega$ let $Y_n=\{y_k:k>n\}$; each $Y_n$ is closed.
Since $X$ is regular, there is an open set $U_0$ such that $y_0\in U_0\subseteq\operatorname{cl}U_0\subseteq X\setminus Y_0$. Suppose that $n\in\omega$ and we’ve chosen open sets $U_k$ for $k
Consider the space $X = \omega \cup \{ * \}$ where the non-empty open sets are of the form $A \cup \{ * \}$ where $A \subseteq \omega$. Clearly $X$ has the c.c.c. (even more, any two nonempty open sets meet) and $\omega$ is a closed discrete subspace of $X$. However the points of $\omega$ cannot be separated by pairwise disjoint open subsets of $X$.
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2@John: I don't see in the question an assumption of regularity. Please edit the question to include all relevant information. – 2012-07-04