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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? $\endgroup$ 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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As a modification to Alessandro Codenotti's comment, we can provide a ZFC example of a space (however, it is $T_1$ but non-Hausdorff) that admits non-trivial open covers (open covers $\mathscr U$ of $X$ so that $X \not\in \mathscr U$): consider $\mathbb R$ with its usual topology $\tau$ and a point $\infty \not\in \mathbb R$. Let $X = \mathbb R \cup \{ \infty \}$ have as its topology $$\left\{ U \subseteq X : U \in \tau \vee \left(\infty \in U \wedge X \setminus U \text{ is finite} \right) \right\}.$$ This space is clearly Rothberger and $\mathbb R$, as a subspace of $X$, inherits its usual topology; hence, $\mathbb R$ is a Lindelöf but non-Rothberger subspace of $X$.

If one desires the space to be non-compact, I believe the set $$\left\{ U \subseteq X : U \in \tau \vee \left( \infty \in U \wedge X \setminus U \text{ is countable, closed, and discrete} \right) \right\}$$ produces a non-compact topology on $X$ with the same property as above.

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    $\begingroup$ You could also generalize a bit further to let the complement of U be any closed Rothberger subset (such as any countable set). $\endgroup$ Jul 6, 2023 at 17:53
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    $\begingroup$ I'm finding myself interested in how to obtain a Hausdorff example, or whether there are sufficient conditions for which Lindelöf subspaces must be Rothberger. $\endgroup$ Jul 6, 2023 at 20:49

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