A question on the Lindelöf property and the Stone–Čech compactification This question is from a paper I'm reading. I cannot understand this sentence in the poof of Theorem 1. It says:

Suppose $X$ is not Lindelöf. Then there exists a compactum $C\subset \beta X \setminus X$ such that any $G_\delta$-set in $\beta X$ containing $C$ meets $X$. ($X$ is Tychonoff.)

Could anybody help me understand this sentence? Thanks ahead:)
See the paper by Raushan Z. Buzyakova, On absolutely submetrizable spaces, Comment. Math. Univ. Carolin. 47, 3 (2006) 483–490. Also available at DML-CZ.
 A: I think that there should be a comma after Lindelöf.  Other than that, all I can do is paraphrase the statement, hoping that this helps.  
Suppose $X$ is not Lindelöf.  In particular, $X$ is not compact.  Since $X$ is Tychonoff, $X$ has a Stone-Cech compactification $\beta X$. 
As usual, we consider $X$ to be a subset of $\beta X$ and can now form the Stone-Cech remainder $\beta X\setminus X$.  The remainder is nonempty since $X$ is not compact.
Now the claim is the following:  There is a compact set $C\subseteq\beta X\setminus X$ with the following property:
For all $G_\delta$-sets $A\subseteq\beta X$ with $C\subseteq A$, $A\cap X\not=\emptyset$.
A $G_\delta$-set is an intersection of countably many open sets.
I hope this clarifies something.
You can get this compact set as follows:  Let $\mathcal U$ be an open cover of $X$ without a countable subcover.  This is possible since $X$ is not Lindelöf.  We can assume that the $U\in\mathcal U$ are actually open subsets of $\beta X$ so that no countable subcollection of $\mathcal U$ covers $X$.
$\mathcal U$ is not a cover of $\beta X$ since in this case, by compactness of $\beta X$, finitely many elements of $\mathcal U$ would already cover $\beta X$ and in particular $X$.  
It follows that the compact set $C=\beta X\setminus\bigcup\mathcal U$ is nonempty.
Let $A$ be a $G_\delta$ subset of $\beta X$ with $C\subseteq A$.
Since $A$ is $G_\delta$, $\beta X\setminus A$ is the union of a countable family $\mathcal B$ of closed sets.  Each $B\in\mathcal B$ is compact and disjoint from $C$.  Since $C=\beta X\setminus\bigcup\mathcal U$, each $B\in\mathcal B$ is covered by $\mathcal U$ and hence by finitely many elements of $\mathcal U$.
It follows that $\bigcup\mathcal
B$ is covered by countably many elements of $\mathcal U$.
But $X$ is not covered by countably many elements of $U$.  It follows that there is $x\in X\setminus\bigcup\mathcal B=A\cap X$.
So $A$ meets $X$.  This finishes the proof.
