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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

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    $\begingroup$ Crosspost on MSE (Not receiving answers after 1 hour is not enough for crossposting) $\endgroup$ – YuiTo Cheng May 8 '19 at 6:53
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    $\begingroup$ 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. $\endgroup$ – Taras Banakh May 8 '19 at 9:25
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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.

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