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A space $X$ is said to be Rothberger if for each sequence $(\mathcal{U}_n)$ of open covers of $X$ there exists a sequence $(U_n)$ such that for each $n$ $U_n\in\mathcal{U}_n$ and $\{U_n : n\in\mathbb{N}\}$ is an open cover of $X$.

Give an example of a Rothberger space $X$ which has a Lindelöf subspace $Y$ that is not Rothberger.

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    $\begingroup$ This is phrased as an imperative, and such questions tend to look like homework. If this is not homework, then please edit to remove the imperative, and talk more about the context in which it arose. // Also, your question should be in the body of the post as well as the subject. I have edited accordingly. $\endgroup$
    – LSpice
    Mar 27, 2021 at 16:22
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    $\begingroup$ Maybe I'm missing something obvious here, but without assumptions on $X$ cannot you just take $Y$ to be your favourite Lindelöf but not rothberger space and $X=Y\sqcup\{\ast\}$ where the only open set containing the new point is the whole space? $\endgroup$
    – Alessandro Codenotti
    Mar 27, 2021 at 18:16

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A canonical example for a Rothberger space is, assuming CH, the union of a scale and the "rationals" (in $P(\mathbb N)$ this is the set of finite sets). This set is concentrated on the rationals, and hence Rothberger. Removing the rationals, you end up with a set that is not even Menger.

To understand my terminology, I refer you, for example, to my paper Menger's and Hurewicz's Problems: Solutions from ``The Book'' and refinements, Contemporary Mathematics 533 (2011).

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