Return to Answer

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".

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".

No, there is no such set.

First, I'll claim the following: (*)

1. There is no bound on how small comeager sets can be: For each generalized dimension 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 dimension 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, while teaching MATH 788. At some point my notes say "details missing here, given in class".

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".

No. I'll claim that the following can be proved: (*)

1. There is no bound on how small comeager sets can be: For each generalized dimension 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 dimension function $h$ with $H^h(A)>0$.

Thus, 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 in November 2006 using Kolmogorov complexity, while teaching the grad course Math 788 at Cornell. At some point my notes say "details missing here, given in class"...)

No, there is no such set.

First, I'll claim the following: (*)

1. There is no bound on how small comeager sets can be: For each generalized dimension 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 dimension 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, while teaching MATH 788. At some point my notes say "details missing here, given in class".