A topological space $X$ has [Menger's property](https://en.wikipedia.org/wiki/Menger_space) $\textsf{S}\_{\mbox{fin}}(\mathcal{O}, \mathcal{O})$ 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



Countable union of Menger spaces
Copying @Taras Banakh's answer from the comments. The answer is correct and complete in my view.


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.


