Let us recall that a topological space $X$ has the *[Rothberger property](https://en.wikipedia.org/wiki/Rothberger_space)* if for any sequence $(\mathcal U\_n)\_{n\in\omega}$ of open covers of $X$ there exists a sequence $(U\_n)\_{n\in\omega}\in\prod\_{n\in\omega}\mathcal U\_n$ such that $X=\bigcup\_{n\in\omega}U\_n$.


I am interested in the finitary version of the Rothberger property.



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> **Definition.** A topological space $X$ is defined to have a *finitary Rothberger property* if for any sequence $(\mathcal U\_n)\_{n\in\omega}$ of finite open covers of $X$ there exists a a sequence $(U\_n)\_{n\in\omega}\in\prod\_{n\in\omega}\mathcal U\_n$ such that $X=\bigcup\_{n\in\omega}U\_n$.
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It is clear that each topological space $X$ with the Rothberger property has the finitary Rothberger property.


On the other hand, each subspace $X\subset \mathbb R$ with the finitary Rothberger property has strong measure zero and hence is countable if the Borel conjecture is true.



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> **Question 1.** Is there a (necessarily consistent) example of a topological space that has the finitary Rothberger property but fails to have the Rothberger property?
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By a [result of Fremlin and Miller](https://www.math.wisc.edu/~miller/res/hmr.pdf), a metrizable space $X$ has the Rothberger property if and only if $X$ has strong measure zero with respect to any metric $d$ generating the topology of $X$. The latter means that for any sequence of positive real numbers $(\varepsilon\_n)\_{n\in\omega}$ there exists a cover $\{U\_n\}\_{n\in\omega}$ of $X$ such that each set $U\_n$ has diameter $<\varepsilon\_n$ with respect to the metric $d$. 


By analogy it can be shown that a metrizable space $X$ has the finitary Rothberger propery if and only if $X$ has strong measure zero with respect to any totally bounded metric generating the topology of $X$.


**Added in Edit.** After the answer of Boaz Tsaban I looked at the Handbook's article ``[Special subsets of the real line](http://www.math.wisc.edu/~miller/res/special.pdf)" by Arnold Miller and found that my question is equivalent to the question of Rothberger: *Is every $C'$ set a $C''$ set*, which was still open in 1984, but has been answered in negative by Fremlin and Miller in 1985 and eventually published in [1988](https://www.math.wisc.edu/~miller/res/hmr.pdf).



Rothberger property for finite covers
I hope I'm not messing things up in the following attempted answer.


Just for history, I seem to remember that the property you mention is called C' by Rothberger (and his main property is called C''). C alone stands for strong measure zero. So, if I'm not misquoting, your property is well known. It appears in some later works, too, but I believe that it wasn't studied much.


You say that, for example in the Cantor space, C' is just C. So, your question there reduces to the nonequivalence of C'' and C, which is well known, and there are many ways to get that. To start with, the critical cardinalities are not the same. But there are also nontrivial examples, e.g. using CH. I believe you can find some in my paper with Machura on the Baer-Specker group [The combinatorics of the Baer-Specker group, Israel Journal of Mathematics 168 (2008), 125-151], but surely there are much, much earlier examples, perhaps in Fremlin-Miller already.


