Axiom of Choice and something I find to not be logical

  • Context: Graduate 
  • Thread starter Thread starter Ed Quanta
  • Start date Start date
  • Tags Tags
    Axiom Choice
Click For Summary

Discussion Overview

The discussion centers around the Axiom of Choice and its implications, particularly in relation to the Banach-Tarski Paradox. Participants explore the mathematical concepts involved, the nature of the proof, and the logical implications of dividing a sphere into parts that can be rearranged to form a larger sphere.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions the logic behind the Axiom of Choice, specifically regarding the Banach-Tarski Paradox and whether the mathematics involved is difficult.
  • Another participant provides a reference to a layman's guide on the Banach-Tarski Paradox and mentions Richard Feynman's skepticism about it.
  • Several participants note that the proof involves surgery theory and emphasizes that the intermediate steps use sets that are not measurable, which complicates the notion of volume preservation.
  • There is a claim that the construction of the paradox only requires five pieces, while another participant asserts it only requires three pieces, highlighting a potential disagreement on the specifics of the proof.
  • One participant mentions that it is possible to rearrange points in a sphere to form two spheres of the same size, suggesting a triviality in the rearrangement process when done point by point.
  • Discussion includes the idea that the "not measurable" subsets are disjoint and sum to the original volume, despite the inability to allocate volume to them in a conventional sense.

Areas of Agreement / Disagreement

Participants express differing views on the specifics of the proof, particularly regarding the number of pieces involved in the construction. There is also a general uncertainty about the implications of the Axiom of Choice and the nature of the sets used in the proof.

Contextual Notes

Limitations include the dependence on the definitions of measurability and volume, as well as unresolved mathematical steps regarding the proof's construction.

Ed Quanta
Messages
296
Reaction score
0
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?
 
Physics news on Phys.org
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
 
Last edited by a moderator:
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 97 ·
4
Replies
97
Views
13K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 27 ·
Replies
27
Views
6K
  • · Replies 64 ·
3
Replies
64
Views
2K