Does there exist a meager set of reals M such that every meager set can be covered by countably many translates of M? This is the category analogue of the [following](https://mathoverflow.net/questions/198122/translates-of-null-sets).



Translates of meager sets
No, there is no such set.
The situation for meager sets is dual to that described by Pietro Majer in a comment on [Translates of null sets](https://mathoverflow.net/questions/198122/translates-of-null-sets),



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> "I was vaguely thinking to Hausdorff measures w.r.to gauge functions. One needs to know that, given $N$, there is $\phi=o(t)$ (for $t\rightarrow 0$) such that $H^\phi(N)=0$. So there is still room for a $\psi$, $\phi(t)<\psi(t)<t$ such that there are strict inclusions of the classes of null sets of $H^\phi\subset H^\psi\subset H^1$ (Some close claim is made here <http://en.wikipedia.org/wiki/Hausdorff_measure#Generalizations>)";
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namely here it is not $M$ that has Hausdorff measure 0 but its complement that has positive measure (which is very different as these are not probability measures). I'll claim the following: (\*)


1. There is no bound on how small comeager sets can be: For each gauge function $h$ there is a comeager set $A$ with $H^h(A)=0$.
2. All comeager sets are somewhat large: For each comeager set $A$ there is a gauge function $h$ with $H^h(A)>0$.




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Now, let $M$ be any meager set of reals (potentially very large) and let $A$ be the complement of $M$ (so $A$ is very small). Nevertheless, by (2) $A$ is not that small: we can let $h$ be such that $H^h(A)>0$. Now let $g$ be another dimension function which is sufficiently far from $h$.


By (1) let $B$ be a comeager set with $H^g(B)=0$. Then the complement $N$ of $B$ is too large to be covered by countably many translates of $M$.




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(\*) I apparently proved this for the Cantor space using Kolmogorov complexity. Source: Lecture notes for [MATH 788](http://www.math.cornell.edu/m/Courses/GradCourses/FA06/788) , 2006. At some point my notes say "details missing here, given in class".


