-1
$\begingroup$

A topological space $X$ has Menger's property $\textsf{S}_{\mbox{fin}}(\mathcal{O}, \mathcal{O})$ if, for each sequence of open covers, $\mathcal{U}_1, \mathcal{U}_2, \cdots $, we can select finite sets $\mathcal{F}_1, \mathcal{F}_2, \cdots $ whose union $\bigcup_{n}\mathcal{F}_n$ covers the space.

My question is if the Menger's property is preserved by countable unions, that is, if for each $n\in\omega$, $X_n$ is a Menger space, then $\bigcup_{n\in\omega}X_n$, with the disjoint union topology is a Menger space.

Thanks

$\endgroup$
1
  • 1
    $\begingroup$ Since not everyone is familiar with Menger spaces, you should mention that $F_n\subset U_n$ $\endgroup$ Commented May 8, 2019 at 9:57

1 Answer 1

2
$\begingroup$

(To kick it from the unanswered queue)

Taras Banakh's comment on your crossposted question:

The answer is well-known and is "yes": just divide the sequence of covers into infinitely many parvise disjoint sequences of covers and cover $n$-th set $X_n$ by the union of finite subfamilies from the corresponding subsequences of covers.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .