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.
Best Answer
I'm back on the topic after four years. The answer is no. Here's an example of a $F_\sigma$ set of dimension $1$ such that its relative complement to every $G_\delta$ that contains it has dimension $1$ too: $\mathbb{Q} \times [0, 1]$. Will explain if requested, but it's not difficult, it uses that $\mathbb{Q}$ is not a $G_\delta$.