# a question on countable discrete set

Let $X$ be Hausdorff and $C$ is countable discrete in $X$ and $x \in cl(C)$.

Does there exist a subset $D$ of $C$ such that $x \in cl(D)$ and there is a point-finite family $\{U_n\}$of open sets of $X$ such that $D \subset \cup U_n$?

If $X$ is regular, such $D$ does really exist. However I don't know such case that $X$ is Hausdorff. If the answer to the above question is negtive, I want to know what property we add that could make the answer is Yes. (Of course, I don't want to consider that $X$ is regular.)

The reference on this is also very welcome.

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What's a point-finite family? – Rudy the Reindeer Dec 24 '12 at 10:35
I think you also want to specify that each $U_n$ intersects $D$ in a nonempty set. Otherwise, you could take $U_1$ to be an open set containing $x$, take $D = C\cap U_1$, and take $U_n = \emptyset$ for each $n > 1$. Or maybe specify that $\{U_n\}$ should cover $X$? – William Dec 24 '12 at 10:40
@Matt: A family $\mathscr{U}$ of subsets of a set $X$ is point-finite iff each $x\in X$ is contained in at most finitely many members of $\mathscr{U}$. – Brian M. Scott Dec 24 '12 at 11:45

If $C$ is finite, then of course the solution is trivial. Let us assume that $C$ is infinite. Actually, let us assume that there exists a subset $\tilde{C}$ of $C$ consisting of distinct points $\{x_1, x_2, \dots\}$, such that every open neighborhood $U$ of $x$ intersects $\tilde{C}$ in an infinite set. Of course, without loss of generality we may assume that $x\notin \tilde{C}$. Let us construct $\{U_n\}$ inductively as follows.
For $n = 1$, take $U_1$ and $\hat{U}_1$ be disjoint open sets such that $U_1$ contains $x_1$ and $\hat{U}_1$ contains $x$. Assuming that $U_n$ and $\hat{U}_n$ have been constructed, construct $U_{n+1}$ and $\hat{U}_{n+1}$ as follows. Take a point $y$ in $\tilde{C}\cap \hat{U}_{n}$. Let $V$ and $\hat{V}$ be disjoint open sets containing $y$ and $x$, respectively. Let $U_{n+1} = V\cap \hat{U}_n$ and $\hat{U}_{n+1} = \hat{V}\cap \hat{U}_n$.
Finally, set $D = C\bigcap \left(\cup_n U_n\right)$.