[net.math] Axiom of Choice

markp@tekmdp.UUCP (Mark Paulin) (10/07/83)

My favorite formulation of the "Axiom of Choice" is:

"The cartesian product of a nonempty family of nonempty sets is nonempty."


The counter-intuitive result mentioned is the so-called "Banach-Tarski Paradox"
which is sometimes stated as, "a sphere the size of a pea may be cut into
finitely many pieces which may be assembled into a life-size statue of Banach."

It has been shown that the axiom of choice is required to prove the existence
of a non-lebesgue-measureable set.

To me, acceptance/rejection of the axiom hinges on the difference between what
*can* be done and what *may* be done.  There may be *no way* to actually find
the cartesian product in the formulation above (or the "choice set" in the
other formulation) but that does not mean (to me) that these sets do not exist,
so I accept the axiom.


Mark Paulin
...tektronix!tekmdp!markp

leichter@yale-com.UUCP (Jerry Leichter) (10/09/83)

I discussed the Banach-Tarski paradox in this newsgroup a couple of months
back.  As I recall, someone also submitted a fairly elementary proof, or
at least an outline of one.  It's actually not at all hard; I saw such a
proof once, years ago, but don't recall it.

Just to make things more definite:  It is possible to cut a (solid) sphere of
arbitrary size into 5 pieces, and reassemble the pieces to from 2 spheres
each of the same size (radius) as the original.  (It turns out that 4 pieces
are sufficient if you start with a sphere missing its center point; the two
new ones will also be missing their center points.)  The "way it works", for
the 4-piece case, is that you get 4 pieces A, B, C and D with the odd property
that A and B are each congruent to A+B - "congruent" in the good old Euclidean
sense; same for C, D and C + D.  I don't remember what the 5th piece does in
the "whole sphere" case.

The proof for 5 pieces is involved.  The simple proof uses 11 pieces.

Before you think this is a way to solve the energy crisis - by multiplying
pieces of coal or whatever - be aware that the pieces are made by "cuts"
something vaguely like:  Put all pieces one of whose coordinates is a
rational number and one of whose ^points^ coordinates is an irrational in
A; etc. - i.e. you have to divide the spheres up literally "point by point".

BTW, Banach-Tarski does NOT work in two dimensions; however, in two and even
in one dimension we have another result:

The existence of a countably additive, translation-invarient measure (an "area"
in two dimensions, a measure of length in one; also, in one, a probability
measure, if defined right) on ALL subsets of the line (or plane) is equivalent
to the axiom of choice.
							-- Jerry
					decvax!yale-comix!leichter leichter@yale

dmn@uvacs.UUCP (10/09/83)

References: tekmdp.2286

Some time ago I attended a talk in which the proof of the Banach-Tarski
theorem was presented.  One step in the partitioning of the pea sphere
involves picking points from the sphere that aren't *close* to each other
in real space (using the Axiom of Choice) and calling this collection
a *piece* of that sphere.  A mathematical point has no volume so that
to my mind the counter-intuitive result follows more from the way this
so called piece is defined than from the fact that the Axiom of Choice
is used in the proof.
  David Nicol (uvax/dmn)