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:)


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 $k<n$ so that $y_k\in U_k\subseteq\operatorname{cl}U_k\subseteq X\setminus Y_k$. Then $\bigcup_{k<n}\operatorname{cl}U_k\cup Y_n$ is a closed set not containing $y_n$, so there is an open set $U_n$ such that $y_n\in U_n\subseteq\operatorname{cl}U_n\subseteq X\setminus Y_n$, and the construction goes through to produce such open sets $U_n$ for each $n\in\omega$. Clearly these $U_n$ separate the points of $Y$.


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 $X$.

  • $\begingroup$ Your counterexample is not regular. If the space $X$ is regular, then what will happen? $\endgroup$ – Paul Jul 4 '12 at 12:35
  • 2
    $\begingroup$ @John: I don't see in the question an assumption of regularity. Please edit the question to include all relevant information. $\endgroup$ – Asaf Karagila Jul 4 '12 at 14:33

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