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- From: columbus@strident.think.com (Michael Weiss)
- Newsgroups: sci.math,sci.physics
- Subject: Banach-Tarski paradox (was: Bayes' theorem and QM)
- Date: 30 Dec 92 17:03:54
- Organization: Thinking Machines Corporation, Cambridge MA, USA
- Lines: 55
- Message-ID: <COLUMBUS.92Dec30170354@strident.think.com>
- References: <1992Dec24.101452.16194@oracorp.com>
- <1ht1arINNf8a@chnews.intel.com>
- NNTP-Posting-Host: strident.think.com
- In-reply-to: bhoughto@sedona.intel.com's message of 30 Dec 1992 20:37:47 GMT
-
-
- Description of the Banach-Tarski paradox by Daryl McCullough omitted.
-
- Blair P. Houghton replies:
-
- I'll buy the explanation (conditioned on the conjecture
- that such a situation might exist, which as yet isn't
- anything more than conjecture), but not the example.
-
- The two spheres thus formed do indeed have the same volume
- sum as the original sphere, but each has less volume than
- the original, or else they weren't composed of a finite
- number of pieces[*], or else when constructed they contained
- gaps, internally.
-
- That, or their "volume" is fractal, being the effective
- volume of a convoluted surface, and therefore isn't
- actually of order 3 but of some order less than 3 and
- greater than 2. [...]
-
- I know it sounds unbelievable. The proof relies on (and in
- fact requires) the axiom of choice, so that's one way around it.
-
- But if you do accept the "usual axioms of mathematics" (i.e., ZFC,
- Zermelo-Frankel set theory with the axiom of choice), then you're stuck
- with the BT paradox, which is indeed a theorem of ZFC.
-
- You can find a more extensive discussion of the BT theorem in the FAQ for
- sci.math. Here is a statement of one form of the result.
-
- Let S, S1, and S2 be three spheres of radius 1. Let B be the set of
- all points on or inside S; ditto for B1, B2. Assume S, S1, and S2 are
- placed so that B, B1, and B2 are pairwise disjoint.
-
- The sets B, B1, and B2 can each be partitioned into a finite number of
- subsets: (here, "+" stands for disjoint union)
-
- B = P_1 + ... + P_m + P_{m+1} + ... + P_{m+n}
- B1 = Q_1 + ... + Q_m
- B2 = Q_{m+1} + ... + Q_{m+n}
-
- such that P_i is congruent to Q_i for i=1 to m+n. "Congruent" means
- that there is a rigid motion taking P_i onto Q_i.
-
- One says that B and B1+B2 are "equivalent by finite decomposition". There
- are no gaps in B1 or B2--- they each consist of *all* points inside or on
- their respective spheres. There are indeed a finite number of pieces in
- each partition (i.e., m and n are finite--- in fact I think m and n can be
- chosen rather small, something like 20 or so.)
-
- Note that the statement of the BT theorem does not mention the word
- "volume". B, B1, and B2 are really quite ordinary point sets, and I don't
- believe there is any reasonable sense in which one could say they have
- "fractal volume". Of course, the P's and Q's are a different story...
-
-