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

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.

• First, let's do the case of $\mathbb Q$ in $\mathbb R$. The obvious attempt give us a $G_\delta$ set $G \supset \mathbb Q$ with Lebesgue measure $0$. But that is far from Hausdorff dimension $0$. Can we even get a $G_\delta$ set $G \supset \mathbb Q$ with Hausdorff dimension $1/2$? – GEdgar Jan 11 at 12:33
• @GEdgar According to the linked result, G could be chosen so it has dimension zero, which admittedly sounds strange to me for a dense $G_\delta$, but I find no error. – Emilio Martinez Jan 11 at 13:32