# On the Menger property and the Alexandroff duplicate

Recall that a space $$X$$ is Menger if for each sequence $$(\mathcal{U}_n)_{n\in\omega}$$ of open covers of $$X$$, there is a sequence $$(\mathcal{V}_n)_{n\in\omega}$$ such that, for each $$n\in \omega$$, $$\mathcal{V}_n$$ is a finite subcollection of $$\mathcal{U}_n$$ and $$\bigcup\{\mathcal{V}_n:n\in\omega\}$$ is an open cover of $$X$$.

On the other hand, given a space $$X$$, the Alexandroff duplicate $$A(X)$$ is defined as follows: The underlying set is $$X\times\{0,1\}$$, each point of $$X\times\{1\}$$ is isolated and a basic open neighbourhood for a point $$\langle x, 0 \rangle \in X\times\{0\}$$ is a set of the form $$U\times\{0, 1\}\setminus\{\langle x, 1\rangle\}$$ where $$U$$ is an open neighbourhood of $$x$$ in $$X$$.

It turns out that for several topological properties $$\mathcal{P}$$, it occurs that a space $$X$$ has the property $$\mathcal{P}$$ if and only if its Alexandroff duplicate $$A(X)$$ has the property $$\mathcal{P}$$. Examples of such properties are compactness, Lindelofness, normality, countably paracompactness, etc. I have found in some papers that, for the Menger property is well-known that this fact also occurs; that is, a space $$X$$ is Menger if and only if its Alexandroff duplicate $$A(X)$$ is Menger. The problem is that I have not been able to find a reference where this fact has been showed for the first time. Does anyone knows any reference to be cited where this result had been showed? I agree that showing this fact is not dificult.

Thanks for any help!

Every closed subset of a Menger set is Menger. Thus if $$A(X)$$ is Menger, then its closed subset $$X\times\{0\}\cong X$$ is Menger.
Suppose $$X$$ is Menger. Given basic open covers $$\mathcal U_n$$ of $$A(X)$$ for $$n<\omega$$, let $$\mathcal F_n'$$ be a finite subcover of $$\{\pi_0 (U):U\in\mathcal U_n\}$$ such that $$\bigcup_{n<\omega}\mathcal F_n'\supseteq X$$. So each $$U\in\mathcal U_n$$ is of the form $$(V_U\times\{0,1\})\setminus\{\langle x_U,1\rangle\}$$. So let $$\mathcal F_n$$ be a subcover of $$\mathcal U_n$$ that covers $$\bigcup\{(V_U\times\{0,1\})\setminus\{\langle x_U,1\rangle\}: U\in\mathcal F_n\}\cup\{\langle x_U,1\rangle:U\in\mathcal F_n'\}=\bigcup\{V_U\times\{0,1\}:U\in\mathcal F_n'\}.$$
Then since $$\bigcup_{n<\omega}\mathcal F_n'\supseteq X$$, we have $$\bigcup_{n<\omega}\bigcup\{V_U\times\{0,1\}:U\in\mathcal F_n'\}\supseteq X\times\{0,1\}=A(X)$$.