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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.

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  • $\begingroup$ 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 $1/2$? $\endgroup$ – GEdgar Jan 11 at 12:33
  • $\begingroup$ @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. $\endgroup$ – Emilio Martinez Jan 11 at 13:32

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