Hausdorff dimension of the relative complement of a set contained in a $G_\delta$

142 Views Asked by At

We have the following result:

Every set is contained in a $G_\delta$ set of the same Hausdorff dimension

I was wondering how tight can this inclusion be made, complement-wise. Is true that:

Let $A \subseteq \mathbb{R}^n$ be an analytic set. Then there exists $A \subseteq G \in G_\delta$ such that $dim_\mathcal{H}(G \setminus A) = 0$

If not, what if $A$ is required to be a Borel set or even $F_\sigma$? Any advice on finding a counterexample? And any comments about dimensions of relative complements are appreciated.