# 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.

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What paper is this from? –  Arthur Fischer Aug 7 '12 at 9:31
What do you mean by "understanding this sentence"? Is there a problem with understanding the statement itself or only its proof? –  t.b. Aug 7 '12 at 9:32
@ArthurFischer See the link in the question:) –  Paul Aug 7 '12 at 9:40
@t.b. Thanks for your link and your kindful reminding. The problem is the statement itself. –  Paul Aug 7 '12 at 9:42

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.

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Thanks stefan for your answer. I will take some time to take it up. –  Paul Aug 7 '12 at 10:27
Your answer is very clear for me and very helpful. One thing I don't understand what is why we can assume that the $U\in \mathcal{U}$ are actually open subsets of $\beta X$? Because $X$ is open in $\beta X$? –  Paul Aug 8 '12 at 11:26
Sorry that this was not clear. Start with an open cover $\mathcal U$ of $X$. Here $\mathcal U$ is a family of open subsets of $X$. Now embed $X$ into $\beta X$. Since $X$ is a subspace of $\beta X$, for every open subset set $U$ of $X$ there is an open subset $V$ of $\beta X$ such that $V\cap X=U$. It follows that we can "blow up" each $U\in\mathcal U$ to an open subset of $\beta X$ whose intersection with $X$ is just $U$. Now I have a family of open subsets of $\beta X$ whose union contains $X$. I call this family also an open cover of $X$. –  Stefan Geschke Aug 8 '12 at 14:41
Now I see. It is abtained by "blow up". Thanks again. –  Paul Aug 9 '12 at 11:50