Axiom of Choice and something I find to not be logical

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The discussion centers on the Banach-Tarski Paradox, which asserts that accepting the Axiom of Choice allows for the division of a sphere into a finite number of non-measurable sets that can be rearranged to form two spheres of the same size. The proof involves advanced concepts from abstract set theory and surgery theory, emphasizing that the intermediate steps utilize sets that lack a defined volume. Consequently, the final configuration does not retain the original volume, challenging conventional notions of measure.

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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?
 
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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
 
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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.
 
Last edited:
Hurkyl said:
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.
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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