> Banach-Tarski Yes, but in that construction, not all the sets are Lebesgue measurable. It's a curious fact, already true just in one dimension, i.e., the

> Banach-Tarski Yes, but in that construction, not all the sets are Lebesgue measurable. It's a curious fact, already true just in one dimension, i.e., the reals, that we can't have a nice way to assign a length, area, or measure to all the subsets. Instead there have to be some subsets to which we can't assign a length, etc. Here the usual proof on the reals does use the axiom of choice! With measurable sets, which already can be wildly bizarre, can't do anything like Banach-Tarski. So, suddenly in this thread, instead of hackers, we are all turning into undergraduate pure math majors!