# Countable union of Menger spaces

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 $$\mathcal{F}_1\subseteq\mathcal{U}_1, \mathcal{F}_2\subseteq\mathcal{U}_2, \cdots$$ 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

• Crosspost on MSE (Not receiving answers after 1 hour is not enough for crossposting) – YuiTo Cheng May 8 '19 at 6:53
• 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 subfamies from the corresponding subsequences of covers. – Taras Banakh May 8 '19 at 9:25

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