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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.

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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,

"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)";

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 $H^h(A)>0$.


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 $M$.


(*) I apparently proved this for the Cantor space using Kolmogorov complexity. Source: Lecture notes for MATH 788 , 2006. At some point my notes say "details missing here, given in class".

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  • $\begingroup$ Thanks! Can you tell me if your notes are available online? $\endgroup$ – Ashutosh Mar 1 '15 at 13:54
  • $\begingroup$ No, although I am considering to work on putting some of them up. I wonder if (1) and (2) appear somewhere else. $\endgroup$ – Bjørn Kjos-Hanssen Mar 1 '15 at 15:43
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    $\begingroup$ (1) seems easy: For each $\epsilon > 0$, choose an open set $U_{\epsilon}$ that contains all rationals and has $h$-Hausdroff measure $< \epsilon$. (2) follows from Theorem 35 in "C. A. Rogers, Hausdorff measures" where it is shown that every perfect set has positive $h$-Hausdorff measure for some $h$. I am interested in seeing how you use Kolmogorov complexity for (2). $\endgroup$ – Ashutosh Mar 1 '15 at 17:25
  • $\begingroup$ Here are the notes: overleaf.com/read/xtdkyvymdsrp $\endgroup$ – Bjørn Kjos-Hanssen Mar 1 '15 at 18:57

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