$G_\delta$ singletons in compact Hausdorff and first countability

Given a compact Hausdorff space $X$, if $a \in X$ is a $G_\delta$ singleton (i.e., $\{ a \}$ is a $G_\delta$ set), then there is a countable local base at $a$.

The point $a$ can be written as countable intersection of open sets let's call $U_{n}$.

How can I use the compactness? What is the next step for proof? Could you give me hint please?

Solution 1:

$\newcommand{\cl}{\operatorname{cl}}$Suppose that $X$ is a compact Hausdorff space, $a\in X$, and $\{a\}=\bigcap_{n\in\Bbb N}U_n$, where each $U_n$ is open in $X$. Let $V_0=U_0$. A compact Hausdorff space is regular, so there is an open $V_1$ such that $$a\in V_1\subseteq\cl V_1\subseteq V_0\cap U_1\;.$$ Continue in this fashion: given $V_n$, let $V_{n+1}$ be an open set such that $$a\in V_{n+1}\subseteq\cl V_{n+1}\subseteq V_n\cap U_{n+1}\;.$$ Then $\{a\}=\bigcap_{n\in\Bbb N}V_n=\bigcap_{n\in\Bbb N}\cl V_n$, and $$V_0\supseteq\cl V_1\subseteq V_1\supseteq\cl V_2\supseteq V_2\supseteq\ldots~\;.$$

For each $n\in\Bbb N$ let $W_n=X\setminus\cl V_n$, and let $\mathscr{W}=\{W_n:n\in\Bbb N\}$; then $\mathscr{W}$ is an open cover of $X\setminus\{a\}$, and $W_0\subseteq W_1\subseteq W_2\subseteq\ldots~$.

Now suppose that $U$ is an open set containing $a$, and let $F=X\setminus U$; $a\notin F$, so $\mathscr{W}=\{W_n:n\in\Bbb N\}$ is an open cover of $F$. Moreover, $F$ is closed and therefore compact, so a finite subfamily of $\mathscr{W}$ covers $F$. Use this to show that there is an $n\in\Bbb N$ such that $a\in V_n\subseteq U$ and conclude that $\{V_n:n\in\Bbb N\}$ is a base at $a$.