If $A$ has measure zero, then is there $x$ such that $A \cap (A-x)=\emptyset$?

Question

If $A\subseteq \mathbb{R}^d$ has Lebesgue measure zero, is it always possible to find an $x$ such that $A \cap (A-x)=\emptyset$?

This is one of those things that seems reasonable, but I'm worried about pathologies. For instance, if $A$ were a group, then the result is obvious. One might consider the group generated by $A$, but this may be too large. E.g. if $A$ is the standard Cantor set on $[0,1]$, then the subgroup generated by $A$ is all of $\mathbb{R}$.

Answer 1

No, it's not always possible. Let $A$ be the union of concentric spheres around the origin, with (all) positive rational radii. This counterexample works in all dimensions greater than $1.$ In $\mathbb{R},$ I am not sure if there is a counterexample (I am guessing yes, but not seeing it).

Answer 2

Let $X$ be a complete metric space and suppose $A,B$ are dense $G_\delta$ subsets of $X.$ Then $A\cap B \ne \emptyset.$ The proof is simple: Suppose $A\cap B = \emptyset,$ dust off DeMorgan's laws, and finish it off with Baire to reach a contradiction.

Now $\mathbb R^d$ is a complete metric space, and it has lots dense $G_\delta$ subsets of measure $0.$ Let $A$ be any one of these. Because translation is a homeomorphism of $\mathbb R^d$ onto itself, every $A-x$ is also a dense $G_\delta$ subset of $\mathbb R^d. $ From the first paragraph we conclude $A\cap (A-x) \ne \emptyset$ for each $x\in \mathbb R^d.$