Conference rusure::math

Well, sort of. The theorem is certainly non-intuitive and is frequently cited
as an example of the ways in which our intuition fails us when dealing with
infinite sets. The 'pieces' in the B-T theorem are dense sets bounded by
the surface of the sphere, and may occupy the same finite region of the sphere
while being disjoint. To give an example, consider the set of all fractions
in {0,1} with terminating binary expansions. It is interleaved with its
complement, the set of all nonterminating fractions, but both are dense in
the interval. In E3 it gets substantially more complex. There subsets may
map into 'larger' or 'smaller' sets, so intuitive concepts of volume break
down completely.

Lynn

The branch of mathematics called MEASURE THEORY can be used to get a better
grip on problems like these as well as the following interesting question...

	what is the integral between 0 and 1 of the function

		F(X) = 0 if X is rational,
		       1 if X is irrational

/Bevin

There are a couple of ways of looking at what the Banach-Tarski Paradox is
sayin.  One is that it indicates that our notion of "congruence" does not
extend very well to "weird" sets in E3.  This enters because congruence is
fundamental to the notion of "equivalence by finite disection", which is what
B-T is talking about.  Let's formalize a bit:

Def:  Let A, B be subsets of E3.  We say A and B are CONGRUENT if there is
a rigid motion M such that M(A) = B.

Def.  A function M:E3 -> E3 is a RIGID MOTION if it can be represented as
a finite number of compositions of translations and rotations.

F is a TRANSLATION if there is a vector v such that, for all x, F(x) = x+v.
F is a ROTATION is there is a 3 x 3 matrix M of determinant + or - 1 such
that, for all x, F(x) = Mx.

(The fact that this is a rotation, you'll have to take on faith.  Actually, I've
fudged a bit by allowing determinant -1, which allows "reflecting" rotations.
Rotations are usually taken to allow only +1 as the determinant, and then you
add in reflection - x -> -x - as a third primitive.)

If you think about this a while, you'll see that it really captures what you
would want "congruence" to mean - I can take one of them sets, move it around,
perhaps rotate it, perhaps reflect it, and lay it exactly over the other set.

It's easy to show that translations, rotations, and reflections are all inver-
tible, jence so are rigid motions.  From this it follows that congruence is
an equivalence relation.

Now, the funny thing about congruence is that it is much more complicated than
you think.  In particular, it is fairly easy to construct bounded sets A and B
in E3 with the following weird property:  A is congruent to B; but also A is
congruent to A union B.  (A and B are disjoint.)  You might want to compare
this to the fact that it is possible for an infinite set to be placed in 1-1
correspondence with a proper subset of itself, something that cannot happen with
finite sets.  (In fact, this has been taken as a DEFINITION of an infinite set.)

A "weak" form of the B-T paradox can be shown using a varient of this construc-
tion.  Take a solid sphere less its center.  It's possible to divide this
object into four pieces, A, B, C and D, such that A ~ B ~ A+B, C ~ D ~ C+D,
where "~" means "is congruent to" and "+" is union.  A, B, C and D are pair-
wise disjoint.  Now take A and rigidly move it until it overlaps A+B, and
C until it overlaps C+D; joining these two, we get A+B+C+D, which is the ori-
ginal sphere-less-center - but we still have the B and D pieces of the original
sphere left over.  We can do the same thing with them, building another sphere-
less-center.  QED.

The 5th piece in the full B-T theorem is needed to take care of the center
point.  This is tricky to do; you have to start over again from scratch.  There
is a much simpler construction that cuts the sphere up into 11 pieces.  I
remember seeing this one many years ago and understanding it, but I no longer
remember anything about the proof.  I also don't have a reference; one would
be nice.
							-- Jerry

Re .3:

> Now, the funny thing about congruence is that it is much more complicated than
> you think.  In particular, it is fairly easy to construct bounded sets A and B
> in E3 with the following weird property:  A is congruent to B; but also A is
> congruent to A union B.  (A and B are disjoint.)

I would like to see such sets.  I understand one-to-one functions between an
infinite set and one of its proper subsets, but I get the impression such
functions "distort" the set while they transform it -- which rigid motions
cannot do.


				-- edp

Are such sets constructed via an infinite number of steps such as
used to create Cantor dust?  I have an inkling that if one can
create a set of measure zero on the interval [0,1] with the same
power as the continuum on that interval it should be possible to
construct the sets in E3 you mentioned but see no obvious way to do
it.

- Jim

Let O be the origin. Let A be the point (1,0).
Let T be the transformation that rotates the plane
counterclockwise about O through 1 radian.

Let Tn be the transformation T applied n times.  T0 is the identity.

Let S be the union of Ti applied to A as i ranges from 0 to infinity.

S is an infinite bounded set.  It is infinite because pi is
incommensurable with 1 so no TiA ever coincides with A again.

Let S'=TS, i.e. rotate S through 1 radian to get S'.
No point of S moves into A (again since pi is irrational).
Thus S' does not contain A.  It is easy to see that S=S' union {A}.

Thus S is bounded, and is congruent to a proper subset of itself.
(S and S' are congruent.)

Re .6:

Thanks, that makes things fall into place.  Can anyone now show two disjoint
sets, A and B, such that A, B, and A union B are all congruent to each other?
It would be nice if the sets had a cardinality greater than that of the
integers but were all contained in a circle.


				-- edp

Re .6:

Thanks, that makes things fall into place.  Can anyone now show two disjoint
sets, A and B, such that A, B, and A union B are all congruent to each other?
It would be nice if the sets had a cardinality greater than that of the
integers but were all contained in a circle.


				-- edp

re .8;
The B-T theorem does not operate in two dimensions, where mapping is more
rational; it holds only for three or more dimensions. There is an informal
description of the problem in "Some Unsolved Problems in Plane Geometry"
by Victor Klee, MATHEMATICS MAGAZINE Vol 52 #3, May 1979 (Thanks to STAN
for a copy.) - Lynn Yarbrough

The closest analogue to B-T in two dimensions is:  There does not exists a
non-trivial countably-additive translation-invarient measure on all subsets
of the plane.  (The B-T theorem essentially replaces "countably additive" by
"finitely additive" in 3 dimensions.)

To parse out what this all means:

m is a measure on some collection of subsets of (say) the plane if, for all
S in the collection, m(S) >= 0; m(null set) = 0; m(S+S') m(S) + m(S'), where
S intersect S' = null and all the sets are in the collections.  (Actually, one
normally defines m over an algebra of sets, so the qualifications aren't
needed.)  m is nontrivial if there is some set with non-zero measure.  (Often
on allows infinite measures, in which case nontrivial means some set with
non-zero finite measure.)

m is translation-invarient if m(S) = m(S+a) where S+a = {s+a | s in S}.

m is countably-additive if, for any pairwise-disjoint countable collection of
sets S1, S2, ..., m(union Si) = sum m(Si).

The construction that demonstrates this is similar to Stan's construction above.
I think I gave it in some other note in this file.

Interestingly enough, this theorem is equivalent to the axiom of choice!

							-- Jerry