[math-fun] Banach-Tarski vs scissors
Banach-Tarski "paradoxes" can include, e.g. cutting a ball into a finite # peices, then reassembling it into two balls (i.e. twice the volume). In order to accomplish this, the"pieces" are non-measurable sets. This is far nastier than the square-circle "paradox" Plouffe mentioned. The main "use" of Banach-Tarski and their successors seems to be in demonstrating that the foundations of mathematics are a dangerous and nasty place, and maybe you shouldn't really believe some of those ZFC axioms... One starts to feel some sympathy with Brouwer and the "constructivists"... In contrast, the phrase "scissors congruence" is usually used to mean something far tamer and also more practical -- one cuts A into a finite number of pieces using "scissors" (i.e. piecewise-smooth boundaries, everything measurable) which then are re-assembled into B. This was already studied by Bolyai who showed that any two polygons with the same area, were scissors congruent. However, Dehn showed in one of the first Hilbert problem solves, that a regular tetrahedron and cube (same volume) are NOT scissors congruent. It is fairly well understood now what is scissors congruent to what in all low dimensions (say, up to 4) and there is at least one entire book giving amazing scissors congruences. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
----- Banach-Tarski "paradoxes" can include, e.g. cutting a ball into a finite # pieces, then reassembling it into two balls (i.e. twice the volume). In order to accomplish this, the "pieces" are non-measurable sets. This is far nastier than the square-circle "paradox" Plouffe mentioned. The main "use" of Banach-Tarski and their successors seems to be in demonstrating that the foundations of mathematics are a dangerous and nasty place, and maybe you shouldn't really believe some of those ZFC axioms... One starts to feel some sympathy with Brouwer and the "constructivists"... ----- After learning that the Axiom of Choice is equivalent to "The cartesian product of nonempty sets is nonempty", I decided that any counterintuitive consequences of AC were problems with my intuition, not with AC. ----- In contrast, the phrase "scissors congruence" is usually used to mean something far tamer and also more practical -- one cuts A into a finite number of pieces using "scissors" (i.e. piecewise-smooth boundaries, everything measurable) which then are re-assembled into B. ----- I thought scissors congruence normally refers just to decomposition of two spaces into (finitely many) topological closed disks with disjoint interiors, where the collection of disks in one case is isometric to the collection of disks in the other case. ----- This was already studied by Bolyai who showed that any two polygons with the same area, were scissors congruent. ----- In fact, congruent by dissection into finitely many convex polygons. --- However, Dehn showed in one of the first Hilbert problem solves, that a regular tetrahedron and cube (same volume) are NOT scissors congruent. ----- I thought Dehn proved only that they could not each be dissected into the same finite set of polyhedra (up to isometries), analogous to the Bolyai dissections. Incidentally, Sydler found an exact algebraic invariant for when two polyhedra can be dissected into each other. ----- It is fairly well understood now what is scissors congruent to what in all low dimensions (say, up to 4) and there is at least one entire book giving amazing scissors congruences. ----- Is there a proof that a 4-dimensional cube and 4-dimensional ball (sphere + interior) of the same volume are not scissors congruent? --Dan
On Sun, Apr 27, 2014 at 1:56 PM, Dan Asimov <dasimov@earthlink.net> wrote:
After learning that the Axiom of Choice is equivalent to "The cartesian product of nonempty sets is nonempty", I decided that any counterintuitive consequences of AC were problems with my intuition, not with AC.
If that's the case, I don't see why it's not constructive: in order to demonstrate that the sets you're taking the product over are nonempty, you have to construct at least one element of each. Once you've got the elements, constructing the list is trivial. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
On 28/04/2014 00:31, Mike Stay wrote:
If that's the case, I don't see why it's not constructive: in order to demonstrate that the sets you're taking the product over are nonempty, you have to construct at least one element of each.
Nope. You can prove a set is nonempty without constructing any elements.
Once you've got the elements, constructing the list is trivial.
Nope. If what you constructed was, say, an unordered pair of undistinguished elements of each set, getting an element in the cartesian product requires a set that performs the function of choosing one element from each pair. (If you have infinitely many pairs of shoes, you don't need AC to get one shoe from each pair, because you can e.g. take the left shoe from each. But if you have infinitely many pairs of socks, you do need AC because there is no distinction between left and right socks.) -- g
On Sun, Apr 27, 2014 at 5:48 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 28/04/2014 00:31, Mike Stay wrote:
If that's the case, I don't see why it's not constructive: in order to demonstrate that the sets you're taking the product over are nonempty, you have to construct at least one element of each.
Nope. You can prove a set is nonempty without constructing any elements.
How?
Once you've got the elements, constructing the list is trivial.
Nope. If what you constructed was, say, an unordered pair of undistinguished elements of each set, getting an element in the cartesian product requires a set that performs the function of choosing one element from each pair.
(If you have infinitely many pairs of shoes, you don't need AC to get one shoe from each pair, because you can e.g. take the left shoe from each. But if you have infinitely many pairs of socks, you do need AC because there is no distinction between left and right socks.)
OK, that makes sense. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
There's the old example of whether the set X of [irrational pairs (a,b) such that a^b is rational] is nonempty. Set a = b = sqrt(2). If (a,b) is not in X (i.e., a^b is irrational), then setting c = a^b and d = sqrt(2) shows that (c,d) is in X. --Dan Mike S. wrote: ----- Adam G. wrote: ----- . . . You can prove a set is nonempty without constructing any elements. ----- How? -----
participants (4)
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Dan Asimov -
Gareth McCaughan -
Mike Stay -
Warren D Smith