why is one point set in a first countable $T_1$ space a $G_\delta$? why is one point set in a first countable $T_1$ space a $G_\delta$ set?
I thought, in first countable space, there may be a point $x$ such that a local basis at x does not contain one point set ${x}$..
Thank you for attention in advance.
 A: Let $X$ be a first countable $T_1$ space, and let $x\in X$. Since $X$ is first countable, there is a countable local base $\mathscr{B}= \left\{B_n:n\in \Bbb N \right\}$ at $x$. Let 
$$ G=\bigcap_{B \in \mathscr{B}} B =\bigcap_{n\in\Bbb N}B_n; $$ 
clearly $G$ is a $G_\delta$ set in $X$, and I claim that $G=\{x\}$. Certainly $x\in G$. Suppose that $y\in X$ and $y \neq x$. Let $U=X\setminus\{y\}$; $X$ is $T_1$, so $U$ is open, and clearly $x\in U$. Thus, there is some $B_n\in\mathscr{B}$ such that 
$$ x\in B_n\subseteq U. $$ But then 
$$ G \subseteq B_n\subseteq U, $$
so $y\notin G$. This shows that $x$ is the only point of $X$ lying in $G$ and hence that $G=\{x\}$, so that $\{x\}$ is indeed a $G_\delta$ set in $X$.
A: Let $\{B_n:n\in\Bbb{N}\}$ be a local basis of $x$. We assume that each $B_n$ is open set. If $x\neq y$, then there is a open set $U$ satisfy that $x\in U$ and $y\notin U$. Therefore,
$$\bigcap_{x\in U,\text{$U$ open}} U=\{x\}$$
and you can easily chech that $x\in \bigcap_{n\in\Bbb{N}} B_n\subset \bigcap_{x\in U,\text{$U$ open}} U$. So $\{x\}$ is $G_\delta$.
A: The upper limit in the closed ordinal space [0, $\omega_1$] is not a $G_\delta$ and there is no countable basis at $\omega_1$.
