# Could the closed discrete subset of the first countable regular space have arbitrary small closed nbhd?

Suppose $X$ is any first countable regular space, and $D$ is the countable closed discrete subset of $X$, then could $D$ have arbitary small closed nbhd, i.e., for any open set $U$ of $X$ which contains $D$, $D$ has an small open set $V$, which satisfies that $\overline{V}\subset U$?

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What do you mean by an arbitrary small nhood in this case? Is $X$ is a metric space? – T. Eskin Feb 19 '13 at 8:21
$X$ is not metric space. Arbitrary small closed nbhd means for any open set $U$ of $X$ which contains $D$, $D$ has an small open set $V$, which satisfies that $\overline{V}\subset U$. – Paul Feb 19 '13 at 8:26
I see now. However, the word "small" has no mathematical relevance, and is quite misleading too, unless it carries some actual content through a definition. – T. Eskin Feb 19 '13 at 8:37

Here’s a counterexample.

Let $\mathscr{A}$ be a maximal family of almost disjoint subsets of $\omega$; i.e., if $A,B\in\mathscr{A}$ and $A\ne B$, then $A\cap B$ is finite. Let $X=\mathscr{A}\cup\omega$. Points of $\omega$ are isolated. Basic open nbhds of $A\in\mathscr{A}$ are sets of the form

$$B_n(A)=\{A\}\cup\{k\in A:k\ge n\}\;.$$ (This is a Mrówka $\Psi$-space.) Clearly $X$ is a first countable, zero-dimensional Tikhonov space, and $\mathscr{A}$ is a closed, discrete set in $X$.

Fix a countably infinite $D=\{A_n:n\in\omega\}\subseteq\mathscr{A}$, and let $U=D\cup\omega$. Let $V$ be any open nbhd of $D$ such that $V\subseteq U$. There is a function $n:\omega\to\omega$ such that $B_{n(k)}(A_k)\subseteq U$ for each $k\in\omega$ and the sets $B_{n(k)}(A_k)$ are pairwise disjoint. Let $S\subseteq\omega$ be such that $|S\cap B_{n(k)}(A_k)|=1$ for each $k\in\omega$; by the maximality of the family $\mathscr{A}$ there is an $A\in\mathscr{A}$ such that $A\cap S$ is infinite, and it follows immediately that $A\in\operatorname{cl}V$. Thus, $D$ has no open nbhd whose closure is contained in $U$.

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Sorry. I can't follow you immediately. Why $A \in clV$? and why does $D$ have no open nbhd whose closure is contained in $U$? – Paul Feb 20 '13 at 0:58
@Paul: $A$ has infinite intersection with $T$, so $B_n(A)$ has infinite intersection with $T$ for every $n\in\omega$. But $T\subseteq V$, so every open nbhd of $A$ (as a point of $X$) meets $V$. Thus, $A\in\operatorname{cl}V$, and clearly $A\notin U$. – Brian M. Scott Feb 20 '13 at 1:05
Thanks. I get it. – Paul Feb 20 '13 at 1:22
@Paul: You’re welcome. – Brian M. Scott Feb 20 '13 at 1:25