Subspace of regular countably compact $X$ consisting of all points at which $X$ is weakly countably tight is $G_\delta$ 
In the last two lines, why did the author say $X_t$ is a $G_\delta$-subset of $X$? What's his intention?
 A: First, $\mathscr{B}_0=\{B\in\mathscr{B}:B\cap X_t\ne\varnothing\}$ is countable, since the base $\mathscr{B}$ is point-countable at each point of $X_t$ and $X_t$ is separable. Let $U$ be any open nbhd of $X_t$; for each $x\in X_t$ there is a $B_x\in\mathscr{B}_0$ such that $x\in B_x\subseteq U$. $X_t$ is compact, so there is a finite $F(U)\subseteq X_t$ such that $X_t\subseteq\bigcup_{x\in F(U)}B_x\subseteq U$. Then
$$\mathscr{G}=\left\{\bigcup_{x\in F(U)}B_x:U\text{ is an open nbhd of }X_t\right\}$$
is a countable family of open sets whose intersection is $X_t$:


*

*$\mathscr{G}$ is countable, since $\mathscr{B}_0$ has only countably many finite subsets, and every $G\in\mathscr{G}$ is a union of finitely many members of $\mathscr{B}_0$.  

*Clearly $X_t\subseteq\bigcap\mathscr{G}$.  

*If $x\in X\setminus X_t$, then $X\setminus\{x\}$ is an open nbhd of $X_t$, so $X_t\subseteq G\subseteq X\setminus\{x\}$ for some $G\in\mathscr{G}$.


For each $G\in\mathscr{G}$ let $\mathscr{B}_G=\big\{\{x\}:x\in X\setminus G\big\}$; then $\mathscr{B}_G$ is discrete in $X$. Clearly $\{B\}$ is a discrete collection for each $B\in\mathscr{B}_0$. Thus, the base $\mathscr{B}_0\cup\big\{\{x\}:x\in X\setminus X_t\big\}$ is $\sigma$-discrete: the countably many discrete subcollections are the singletons $\{B\}$ for $B\in\mathscr{B}_0$ and the families $\mathscr{B}_G$ for $G\in\mathscr{G}$. We need to know that $X_t$ is a $G_\delta$-set in $X$ in order to break up $\big\{\{x\}:x\in X\setminus X_t\big\}$ into countably many discrete subcollections.
