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

Question

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 $\mathbb R$.

Answer 1

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".

Answer 2

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.