2
$\begingroup$

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.

  • 0
    @t.b. Thanks for your link and your kindful reminding. The problem is the statement itself.2012-08-07

1 Answers 1

3

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.

  • 0
    Now I see. It is abtained by "blow up". Thanks again.2012-08-09