# Separating points of closed countable discrete subspace of a $T_4$ space with pairwise disjoint open sets

Let $$X$$ be a $$T_4$$ space (i.e.it is $$T_1$$ and every pair of disjoint closed sets can be separated by disjoint open sets). Suppose that $$D=\{d_n\}_{n\in\mathbb N}$$ is a countable, closed and discrete subspace of $$X$$. Can we always find a family $$\{U_n\}_{n\in\mathbb N}$$ of pairwise disjoint open sets of $$X$$ such that $$d_n\in U_n$$ for each $$n$$? I know this is possible if $$D$$ is finite, but I would like to know if this happens as well if $$D$$ is infinite. This question is inspired by what happens with the subspace of integers on the real line.

• It seems that $\{d_n\}$ and $D\setminus \{ d_n\}$ are both closed? Jan 30, 2022 at 16:44
• Clearly yes these are both closed. Jan 30, 2022 at 16:45

It is true: the map $$d_n\to n$$ is continuous from $$D$$ to $$\mathbb{R}$$. By the Tietze-Urysohn theorem it has a continuous extension $$f:X\to\mathbb{R}$$. Now let $$U_n=f^{-1}\bigl[(n-\frac13,n+\frac13)\bigr]$$ for all $$n$$.

Yes, you can do it as follows, and it suffices that $$D$$ is discrete and $$X$$ is $$T_3$$.

First, chose open $$U_1$$ with $$U_1\cap D = \{d_1\}$$.

Let $$V_1$$ be such that $$d_1\in V_1\subseteq \overline{V_1}\subseteq U_1$$ which we can find since $$X$$ is regular.

Next, let $$U_2$$ be such that $$U_2\cap D = \{d_2\}$$ and $$U_2\cap \overline{V_1} = \emptyset$$. Again find $$V_2$$ from regularity with desired properties.

Continue this process to find a sequence of sets $$V_n$$ with $$d_n\in V_n\subseteq \overline{V_n}\subseteq U_n$$ and $$U_{n+1}\cap (\bigcup_{i=1}^n \overline{V_i}) = \emptyset$$.

Then $$\{V_n:n\in \mathbb{N}\}$$ is a pairwise disjoint family of open sets such that $$d_n\in V_n$$.