1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ It seems that $\{d_n\}$ and $D\setminus \{ d_n\}$ are both closed? $\endgroup$ Commented Jan 30, 2022 at 16:44
  • $\begingroup$ Clearly yes these are both closed. $\endgroup$ Commented Jan 30, 2022 at 16:45

2 Answers 2

2
$\begingroup$

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 $n$.

$\endgroup$
1
$\begingroup$

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 $d_n\in V_n$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .