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 $A$?
Covering $\mathbb R^n$ with disjoint copies of a set with non-zero measure
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0I don't see how the edited question has anything to do with AC. – 2012-02-05
1 Answers
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|
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1@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. – 2012-02-05