# Can you cover $\mathbb R$ with countably many smol sets?

Say a set $$X\subset\mathbb R$$ is smol if one can't cover $$\mathbb R$$ with countably many translates of $$X$$.

So, for example, null-measure sets are smol. I guess sets of positive measure are not, but I have not been able to prove it yet. Non-measurable sets are a mystery.

I wonder: can you cover $$\mathbb R$$ with countably many smol sets?

This question came as an effort to find a maximal family $$\mathcal F\subset\mathcal P(\mathbb R)$$ such that any countable union over $$\mathcal F$$ is not $$\mathbb R$$. This would give a rough notion of measure (only telling if a set is big or small) to every subset of $$\mathbb R$$.

Yes. By Baire category theorem, sets of first category are smol. And $$\mathbb R$$ can be decomposed as a disjiont union of a measure $$0$$ set and a set of first category (Theorem 1.6 or here). This is exactly the point of the 1st chapter of Measure and Category: There is no uniform notion of being "small".
EDIT: I misunderstood the question. I am explaining how to find non-smol sets, but I don't know about the question of whether or not you can cover $$\Bbb R$$ with smols. This just partially addresses the Op's "guess".
If $$X$$ contains a set of positive measure (it need not be measurable itself) then this is almost true: see here. There is some family of translates whose union's complement has measure zero, so we have "almost" covered all of $$\Bbb R$$. It follows that if $$X$$ contains a set of positive measure, there is a null set $$N$$ such that $$X\cup N$$ is not "smol".
I do not know if literal coverage of all of $$\Bbb R$$ is possible for arbitrary $$X$$ of positive measure.
• Fat Cantor sets are meager, so countably many of them can't cover $\mathbb{R}$. Oct 20, 2023 at 1:50