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!


1 Answer 1


The reference is here, provided you cannot find anything more classical.

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 $\bigcup_{n<\omega}\bigcup\{V_U\times\{0,1\}:U\in\mathcal F_n'\}\supseteq X\times\{0,1\}=A(X)$.

  • 1
    $\begingroup$ Your link points to this very question $\endgroup$ Commented Dec 4, 2022 at 4:59
  • $\begingroup$ Thank you for adding the proof @stevenclontz. This is basically the same way I had thought of it. However, I was looking for some paper (if it exists) where this fact had been proved or pointed out for the first time. Do you know any classical paper mentioning/proving this fact? $\endgroup$
    – J. Casas
    Commented Dec 4, 2022 at 19:11

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