Question about Čech-Stone compactification Let $\beta X$ be the Čech-Stone compactification of $X$ and $p\in \beta X\setminus X$. Is it true that $\{p\}$ can not be a $G_\delta$ set ?
 A: Yes, it’s true.
$\newcommand{\cl}{\operatorname{cl}}$Let $p\in\beta X\setminus X$, and suppose that $\{p\}$ is a $G_\delta$ in $\beta X$. Then there is a continuous $f:\beta X\to[0,1]$ such that $f(p)=0$ and $f(x)\ne 0$ for $x\in\beta X\setminus\{p\}$. Since $X$ is dense in $\beta X$ we can find points $x_n\in X$ for $n\in\omega$ such that $f(x_{n+1})<f(x_n)\le 2^{-n}$ for each $n\in\omega$. Let $D=\{x_n:n\in\omega\}$, let $H=\{x_{2n}:n\in\omega\}$, and let $K=D\setminus H$. There is a continuous $g:(0,1]\to[0,1]$ such that $g(\alpha)=0$ for $\alpha\in f[H]$ and $g(\alpha)=1$ for $\alpha\in f[K]$, so the function $g\circ f$ functionally separates $H$ and $K$. It follows that $\cl_{\beta X}H\cap\cl_{\beta X}K=\varnothing$, so without loss of generality $p\notin\cl_{\beta X}H$. $H$ is closed and discrete in $X$, so there must be some $q\in(\cl_{\beta X}H)\setminus H$. But then $f(q)=0$, since $f$ is continous, and $q=p$, a contradiction.
This argument can be elaborated slightly to show that in fact $\cl_{\beta X}D$ is homeomorphic to $\beta\omega$ and hence that a non-empty, closed $G_\delta$ in $\beta X\setminus X$ must have cardinality at least $|\beta\omega\setminus\omega|=2^{2^\omega}=2^{\mathfrak c}$.
A: It certainly can happen. Let X = w1, the least uncountable ordinal, with the order topology. The identity embedding of X into w1+1 is the minimal,and maximal, compactification of X. If F is a countable non-empty family of neighborhoods of the POINT  p=w1 in the space w1+1 , the complement of each member of F is a compact, hence countable, subset of X. So the common intersection of F is uncountable. 
