# Covering $\mathbb R^n$ with disjoint copies of a set with non-zero measure

If $A$ is an open subset of $\mathbb R^n$, with non-zero measure, is it always possible to cover $\mathbb R^n$ with countably many disjoint, scaled, and or rotated copies of $A$?

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Well, if $A$ is open contains a closed interval and its possible cover all the space with copies of that interval. Think in that copies as construction blocks of the whole space. – leo Feb 5 '12 at 1:01
If you start with an open dense set of small positive measure, no two of its translates (/scalings/rotations/images under ambient homeomorphism) will be disjoint. – user83827 Feb 5 '12 at 1:23
I don't see how the edited question has anything to do with AC. – Asaf Karagila Feb 5 '12 at 6:11

It's never possible with the restrictions you set up (except if $A$ is the entire $\mathbb R^n$). If the copies are to be disjoint, no point on the boundary of any of the copies can ever be covered by any of the other ones.

On the other hand, if you relax the conditions such that you don't require the copies to be disjoint, then it's trivially possible: since $A$ is open and nonempty, it contains an open ball, and countably many translated copies of that will cover the entire space.

Edit: The question was modified to allow a measure-zero subset of $\mathbb R^n$ not to be covered. The answer is still no: consider $$A = \{(x,y)\in\mathbb R^2\mid |y|<e^{-|x|} \land x^2+y^2 > \frac14\}$$ Once you place a single copy of this $A$, the central opening in it can never be filled by any copy of $A$ that is disjoint from the first one.

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@HenningMakholm, I think the OP wants to cover the space with non-overlapping sets. And if I recall correctly, it is possible to cover $\mathbb{R}^n$ with countably infinite disjoint open balls. – leo Feb 5 '12 at 1:24