Axiom of Choice and something I find to not be logical

[Ed Quanta]

I heard something along the lines of when you accept the axiom of choice as true, you can then prove using some abstract set theory that by dividing a sphere, you can divide it and then put it together so that it is bigger than it originally was????

Is the math behind this proof difficult? And is this true?

[quartodeciman]

from Kuro5hin - Layman's Guide to the Banach-Tarski Paradox --->
http://www.kuro5hin.org/story/2003/5/23/134430/275

search "Banach-Tarski" for more stuff.

Feynman said phooey about B-T --->
http://www.ams.org/new-in-math/mathdigest/200112-choice.html

[Hurkyl]

IIRC, the proof uses surgery theory.

One of the main things to emphasize about the construction is that its intermediate steps involve sets that are not measurable. All of the clever work is done with sets for which you cannot define volume, so there isn't any reason to expect that you have the original volume when you're done.


And, incidentally, the big point about the construction is that it only uses five pieces. It's a trivial exercise to prove that you can rearrange all of the points in one sphere to form two spheres of the same size, if you do it one point at a time.

[selfAdjoint]

IIRC, the proof uses surgery theory.

One of the main things to emphasize about the construction is that its intermediate steps involve sets that are not measurable. All of the clever work is done with sets for which you cannot define volume, so there isn't any reason to expect that you have the original volume when you're done.


And, incidentally, the big point about the construction is that it only uses three pieces. It's a trivial exercise to prove that you can rearrange all of the points in one sphere to form two spheres of the same size, if you do it one point at a time.


The "not measureable" subsets are disjoint and add up to the whole original ball. Therefore by linearity of measure their total measure is the original volume, even though that can't be allocated to them in any way.