On a property of regular topology on infinite sets

Question

Let $X$ be an infinite set and $\tau$ be a regular topology on it , then is it true that there exist a sequence $\{x_n\}$ in $X$ and a sequence $\{U_n\}$ of pair wise disjoint open sets in $X$ such that $x_n \in U_n , \forall n \in \mathbb N$ ?

Answer 1

Yes, it’s true, assuming that your regular is my $T_3$, i.e., regular and $T_1$. If $X$ is discrete, just let $\{x_n:n\in\Bbb N\}$ be an infinite set of distinct points of $X$, and for each $n\in\Bbb N$ let $U_n=\{x_n\}$.

Otherwise, there is a $p\in X$ that is not an isolated point. Let $x_0\in X\setminus\{p\}$ be arbitrary. $X$ is $T_1$, so $p$ has an open nbhd $V_0$ such that $x_0\notin V_0$, and $X$ is regular, so $p$ has an open nbhd $W_0$ such that $\operatorname{cl}W_0\subseteq V_0$. Let $U_0=X\setminus\operatorname{cl}W_0$; then $U_0$ is an open nbhd of $x_0$ disjoint from $W_0$.

Since $p$ is not an isolated point, there is an $x_1\in W_0\setminus\{p\}$. $X$ is $T_1$, so $p$ has an open nbhd $G_1$ such that $x_1\notin G_1$; let $V_1=W_0\cap G_1$, use regularity to get an open nbhd $W_1$ of $p$ such that $\operatorname{cl}W_1\subseteq V_1$, and let $U_1=W_0\setminus\operatorname{cl}W_1$; $U_1$ is an open nbhd of $x_1$ disjoint from both $U_0$ and $W_1$.

Keep going. Suppose that we’ve chosen $x_0,\ldots,x_n$, $U_0,\ldots,U_n$, and $W_0,\ldots,W_n$ so that the sets $U_0,\ldots,U_n$ are pairwise disjoint open nbhds of $x_0,\ldots,x_n$, respectively, the sets $W_0,\ldots,W_n$ are open nbhds of $p$, and

$$W_n\cap\bigcup_{k=0}^nU_k=\varnothing\;.$$

Since $p$ is not isolated, there is an $x_{n+1}\in W_n\setminus\{p\}$. $X$ is $T_1$, so $p$ has an open nbhd $G_{n+1}$ such that $x_{n+1}\notin G_{n+1}$; let $V_{n+1}=W_n\cap G_{n+1}$, use regularity to get an open nbhd $W_{n+1}$ of $p$ such that $\operatorname{cl}W_{n+1}\subseteq V_{n+1}$, and let $U_{n+1}=W_n\setminus\operatorname{cl}W_{n+1}$; $U_{n+1}$ is an open nbhd of $x_{n+1}$ disjoint, the sets $U_0,\ldots,U_{n+1}$ are pairwise disjoint, and

$$W_{n+1}\cap\bigcup_{k=0}^{n+1}U_k=\varnothing\;,$$

so the recursive construction goes through to yield points $x_n$ and their pairwise disjoint open nbhds $U_n$ for $n\in\Bbb N$.

You may well find it easier to follow the construction if you draw pictures of the first two or three stages of the recursion.