From: ah170@FreeNet.Carleton.CA (David Libert)
Subject: Re: Banach-Tarski Construction
Date: 1 May 1999 07:43:46 GMT
Newsgroups: sci.math
Keywords: Outline of construction
"Jonathan W. Hoyle" (jhoyle@kodak.com) writes:
> Is there a resource online or a source I can be directed towards which
> outlines the Banach-Tarski Spherical Decomposition process? I imagine
> that this construction would be reminiscent of, although an order of
> magnitude harder than, the construction of a non-measurable set.
>
> For those who don't know about it, it's a procedure by which a sphere
> can be decomposed into a finite number of subsets, which then can be
> recombined (using only rigid motions) to form a sphere with a different
> volume than it started with. I have also heard it described as a
> process by which two spheres are generated, each of which with the same
> volume as the original.
>
> I understand that the number of subsets can be as low as five, which I
> understand comprises the center point as one set, and the remaining four
> being two sets of two congruent sets. Obviously, at least two of the
> subsets, probably all four, are non-measurable. In addition, I would
> guess that these four subsets are dense within the sphere.
I recall there was an good expository article about this in the
American Mathematical Monthly in the late 1970's. I think it is also
covered in Thomas Jech _The Axiom of Choice_ (a book, maybe (?)
North Holland).
Tarski (maybe co-authored with Banach, I dunno) originally published
a paper called something like A Paradoxical Decomposition of a Sphere.
In the original paper they were decomposing a sphere as opposed to a ball,
ie the sphere being the 2 dimensional surface of the 3 dimensional ball.
If I remember right this involved subsets A, B and C of the sphere which
were disjoint and that had the bizarre property that A is congruent to
B and A is also congruent to B union C. And B is congruent to C, ie it
wasn't the triviality C = empty :-) .
Anyway the construction of these sets involved AC. A while later, maybe
a couple of years, Tarski and possible co-author published the modern
statement we know, namely a three dimensional closed ball can be decomposed
and reassembled by rigid motions (translations and rotations composed)
into two disjoint closed balls each the same radius as the original ball.
The method used the earlier result on spheres. There were steps that
formed subsets of the ball by taking the "cone" generated by the earlier
subset of the sphere, ie exclude the center of the ball but include all
interior points that line up between the center and a point in the special
subset on the surface, also include that special subset of the surface.
The property from the first result was used in showing the second result.
A rumour I heard by personal communication was that Tarski published the
first sphere result hoping it would undermine faith or respect or whatever
for AC. (I think this was in the 1920's). He was disappointed to see it
didn't do so as much as he expected. So he worked on and got and
published the more striking ball version. And it still fell short of his
hopes.
The Banach-Tarski paradox is for me one of the most striking counter-
intuitive AC consequences, but in the end I am firmly in the AC camp
anyway. If any doubts remain I hope the catch-phrase I found can settle
the question of which axiom to adopt for set theory:
the axiom of choice is The Axiom of Choice.
That Monthly paper had further related results from more recent times.
Define an equivalence relation ~ between subsets of 3-space: if you can
get from first to second by finite partition and reassembling the pieces
by separate rigid motions on each piece. Ie so Banach Tarski says one
ball is ~ to two disjoint balls. Define a partial ordering on subsets
of 3-space: A>=B if there is a B' s.t. A ~ B' and B' superset of B.
The author proved a sort of Shroeder-Berstein theorem for ~ and >=.
ie A>=B & B>=A -> A ~ B. As I recall this wasn't very hard and didn't
need AC.
From this result and Tarski's original 1 ball to 2 balls theorem you
can get a powerful generalization. So start with one ball, iterate
the 1->2 ie on the new copies keep splitting them 1 to 2 balls, so
produce as many balls as you want. Then in each ball shave off corners
and extract a cube. So you can get as many of these as you want.
Arrange that these cubes are properly open or closed at various surfaces
so that you can stick them side by side in a three dimensional tiling
so they partition a huge cube, of size you decided originally when you
created the number of balls. So conclude that the original little sphere
is >= any set contained in the big cube.
Under suitable starting premises as I will state momentarily do the
above in two directions and apply "Shroeder-Berstein". So you end up
with this super-generalization of the original Banach-Tarski:
Suppose A1 and A2 are subsets of 3-space s.t. for i=1,2 there are
positive real numbers Ri & ri s.t. Ai is a superset of some open ball
of radius ri (ie the ball can be at any location, not necessarily
centered at the origin) and Ai is a subset of an open ball of radius Ri.
Then: A1 ~ A2.
So ie, that dime in my pocket can be reassembled into all galaxies of
the observable universe with every detail of their shapes.
Those Ai can be anything subject to the premises. So outside of the
little r1 ball A1 could be any weird complicated set, way up in the
Borel hierarchy or whatever. And that could be reassembled from A2
just for example a closed set right at the bottom of the Borel hierachy.
In all this, the real AC power is right back at that first result on
spheres. By comparison as I recall all the later parts were very
concrete constructions.
So I will say the flavour of what I remember of the original sphere
construction. I don't remember it very well and may get this wrong.
I think you have two axis through the sphere. You consider rotations
around one axis by 120 degrees, and rotations around the other axis by
180 degrees. You consider the group of rotations generated by these.
This group is generated by the 120 degree rotation of period 3 and
the 180 degree rotation of period 2. For later in the proof you want
the group to be isomorphic to the free group on two generators, one
of period 3, one of period 2. ie the free group on two generators
modulo the relations saying those generators have respective periods
3 and 2. So no identities hold in this group other than the period
identities and those that follow from that by the pure axioms of
group theory.
You consider different choices of the angle between the two axis
to try to get this condition. There are countably many equations
asserting the equality of some group expression on the generators
to some other such expression. You consider all such equations
which fail in the free group. You must arrange that each of these
equations fails in the desired rotation group. There are countably
many cases.
To handle one such case you choose coordinates and make variables
to express the angles between the two axis. Each of the two generating
rotations can then be written as a matrix with entries depending on
these variables. You write down the relevant equation as an equation
between a suitable product of matrices. Expand out what it says and
it becomes a system of polynomial equations in the base variables.
As a polynomial system it only has finitely many roots. This means
there are only finitely many choices of angle between the axis that
could make that equation mess up the desired freeness property.
There are only countably many cases and each one messes up only
finitely many angles. So at most countably many angles between the
axis ruin the freeness.
In the 1920's when Tarski was doing this Nathan Deeth age 11 hadn't
yet arrived on the scene, so Tarski foolishly believed the work of
Cantor :-) . So with the uncountability of the reals there are angles
left over to be the one we actually use.
So far even this cardinal arithmetic hasn't used AC. Now we will
though. Namely you define some sort of equivalence relation between
points on the sphere (ie the surface) using the group of rotations.
Maybe the orbit relation? That a point can be carried to another by
some rotation in the group.
Anyway, then you use AC, to make a choice set picking a
representative from each equivalence class. Then each point on the
sphere can be uniquely represented as a group element applied to choice
set member. You can represent group elements as (possibly empty)
products of powers of alternating generators, where each power is
less than the respective order of its generator. By the freeness
property distinct representations of this form really do represent
distinct group elements.
Define A, B and C by the leading term in the representation as above,
namely A is elements leading with the 180 degree rotation, B those
leading with 1 power of the 120 degree and C is 2 powers of 120 degree.
Maybe those with the identity group element as representative are in a
fourth set D. Then the 180 degree rotation applied to A carries it
to B union C union D. But the 120 degree rotation applied to A carries
it to B.
Maybe this is actually the construction and the bizarre property is
as just above with that extra D in one case.
Anyway the above I derived is already strange. A, B, C, D are a
partition of the sphere. A 120 degree rotation carries B to C. So A
is congruent to B and congruent to a set that is a superset of two
disjoint copies of B. This is enough to see that A can't be Lebesgue
measurable by the measure on the surface of the sphere. Namely suppose
for contradiction A is Lebesgue measurable. Let m() be this measure.
A congruent to B, so B is measurable with the same measure. On the
other hand A is congruent to a superset of a disjoint union of two
congruent copies of B, so m(A) >= 2*m(B) = 2*m(A). So m(A) = 0.
But A is congruent to B union C union D, so (let u be union)
B u C u D is Leb mble and m(B u C u D) = m(A) = 0. So since A,B,C,D
partitions the sphere, the entire sphere has Leb meas =
m(A) + m(B u C u D) = 0+0, contra.
Anyway once you get the above it would not be too surprising to make
one ball into two. Partition the original ball into A,B,C,D, or rather
the cone version of those and the center. Then a single A can be rotated
to be a B u C u D, and the B can be rotated back to A, so this can be
productive.
Here is another point I recall arising in the proof. Using the sphere
surfaces to induce stuff on the balls, it is convenient to work instead
with the punctured balls: the center deleted. Then the cone construction
makes working in 3-d a closer analogy to working on the sphere surface.
In this way you could get a version of the construction making one
punctured ball into two punctured balls.
Then you can make a punctured ball equivalent to a full ball as follows.
Get a circle interior to the ball with the center on the ball on the
circumfrance of the circle. On this circle mark off points at an
infinite sequence of fixed angles going around the circle, where the angle
at each step is an irrational ratio of the full circle, and where the
first point is the center of the ball. So partition the full ball where
one set is these marked points and the other is its compliment. For the
marked points shift the circle forward one step by that angle. On the
compliment keep everything fixed. Then the shift acts on the points
like the Hilbert Hotel and vacates the center and nothing else.
So do that to puncture the center, do the cone construction to make
two punctured balls, and do the reverse construction on each of them to
fill in the two centers.
So I am relying here on the ~ relation being transitive. Earlier I
stated ~ was an equivalence relation. Reflexive and symmetric are
obvious. For transitive take a common refinement. ie the set in the
middle has two partitions corresponding to its two ~ 's. Take a common
refinement of these two partitions. This induces refinements on the
first and third set. Create two new reassemblies based on acting on
the finer pieces by the rigid motion of the enclosing piece. Then the
composition of the rigid motion on each piece makes a reassembly from
the first set to the third.
--
David Libert (ah170@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad." -- David Libert
3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig
==============================================================================
From: ah170@FreeNet.Carleton.CA (David Libert)
Subject: Re: Banach-Tarski Construction
Date: 2 May 1999 08:51:04 GMT
Newsgroups: sci.math
My last article, referenced above, was starting with my hazy memory of
Banach-Tarski and trying to reconstruct it. Alas I see now some parts
are more complicated than I thought.
I realized a difficulty with what I wrote last time and decided I
better make more of an effort to look things up. I did find a brief
description in Thomas Jech's article _The Axiom of Choice_ in
_The Handbook of Mathematic Logic_ ed Jon Barwise, pub North Holland.
In there Jech refers for more detail to his book of the same title which
I referenced last article and which as I guessed then is published also
by North Holland.
Jech's description starts out like mine, the two axis and that group
and finitely many solutions messing up freeness so avoid these.
The first error in my last article was I said that the freeness of the
group guaranteed the representation of points on the sphere are unique.
The problem I overlooked is in a particular orbit with representative m,
given two distinct representations of rotations as suitable products,
the freeness of the group guarantees that the rotations represented r1
and r2 are distinct as group elements, but this does not guarantee
r1(m) != r2(m). ie r1 and r2 being distinct means they must disagree on
some input but it need not be m.
So I haven't ruled out non-uniqueness of the representations behind
the definitions of A,B,C so it is no longer so clear these are pairwise
disjoint.
So here is what we can try. For these cases r1 and r2 are distinct
rotations so r2^-1 o r1 is not the identity rotation so it has exactly
two fixed points. For each pair of representations of rotations as
special form products of generator powers there are at most two points m
creating this problem.
So form the set Q of points m on the sphere such that there are
distinct rotations r1 and r2 in the group with r1(m) = r2(m). There are
only countably many choices of pairs r1 and r2, ie coming from our
countable group.
It can be argued if m in Q and r in the group then r(m) in Q.
Calling the sphere S, S ~ Q is therefore a union of orbits. So do the
corresponding choice set and the corresponding definitions of A,B,C on
S ~ Q instead of on S. Now on S ~ Q it really is behaving with the
uniqueness properties I mistakenly assumed last time for all of S.
So the revised version of what I had last time becomes S can be
partitioned into A,B,C,D,Q where Q is countable and A is congruent to
B,C and (B u C).
My previous mistaken version didn't mention Q, so an equivalent way to
say it is I previously had Q = empty.
As I thought about it I suspected the version I had read before did
have an extra countable set like Q.
One result I had last time is A is not Leb. mble by the measure on S.
This still follows in the new version. The first part copies over and
the change is where I previously had (for m the Leb meas.) that
m(S) = m(A) + m(B u C u D), I now have m(S) = m(A) + m(B u C u D) +
m(Q). But Q is countable so m(Q)=0 so this makes no difference.
It was after I thought of all this I looked up Jech. The first
surprise was he gave a sphere result but he attributed it to Hausdorff
in 1914. He gave the ball result of Banach Tarski as 1924. So what
does this do to my Tarski rumour?
The other strange point is the Hausdorff result he gives is like mine
except it leaves off D. In any event his stated result does have Q
arising in the same way mine did above.
Jech's proof outline is along the lines of my definitions. I don't
see how to avoid having D by the identity group element being a special
case. Jech's outline is so sketchy this point isn't clear.
In my last article I recalled the sphere result got the ball result by
a cone construction but I didn't remember the details. Tonight before I
realized my Q problem I thought about this and came up with a solution
for the ball result based on last article's version without Q. I don't
yet see how to do this based on the correct sphere result with Q.
Jech claims there is a solution based on his version with Q and no D.
For me D wasn't the problem. Jech gives no details.
Maybe Jech's book is more detailed than the article. Or else that
Monthly article if anyone can track it down.
--
David Libert (ah170@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad." -- David Libert
3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig
==============================================================================
From: ah170@FreeNet.Carleton.CA (David Libert)
Subject: Re: Banach-Tarski Construction
Date: 3 May 1999 03:34:07 GMT
Newsgroups: sci.math
Since my last article in this thread I have found how to use the
latest versions of the sphere decomposition to get the basic ball
result. Recall from the last article my most recent corrected version
of the sphere decomposition had A,B,C,D,Q while the Hausdorff result
as quoted by Jech in the Handbook had A,B,C,Q . I still don't see
how to get that Hausdorff version for the sphere without D. Since
last article I have found how to get the basic ball result using either
my or Hausdorff's version of the sphere decomposition.
I will correct an error from my last post summarizing my A,B,C,D,Q
sphere decomposition. I wrote:
> So the revised version of what I had last time becomes S can be
>partitioned into A,B,C,D,Q where Q is countable and A is congruent to
>B,C and (B u C).
That should be A is congruent to B, C and (B u C u D) .
By contrast the Hausdorff result is S can be partitioned into A,B,C,Q
where Q is countable and A is congruent to B, C and (B u C) .
Also Jech's book is titled _The Axiom of Choice_ , but his Handbook
article is titled _About the Axiom of Choice_ . Last article I wrote
that both book and article were titled _The Axiom of Choice_ .
Last article I claimed without details that Q is a union of orbits. I
will prove this now. Recall Q is the set of those points s on the sphere
S s.t. there exist r1 and r2 distinct elements of the group s.t.
r1(s) = r2(s) . For such s,r1,r2 if r = (r2^-1 o r1) then r is not
the identity rotation and r is in the group and r(s) = s. Conversely
if r is in the group and not the identity and r(s) = s then
r1 = identity rotation and r2 = r are distinct group elements with
r1(s) = r2(s). So we can equivalently define Q as points fixed by some
nonidentity group element.
So suppose s is in Q, namely r1 in the group is not the identity
rotation and r1(s)=s and consider r2(s) in the same orbit as s for some
group element r2. Let r3 = r2 o r1 o r2^-1 . Since r1 != identity
therefore r1 o r2^-1 != r2^-1. So r2 o r1 o r2^-1 != identity, ie
r3 != identity. r3(r2(s)) = r2 o r1 o r2^-1 o r2 (s) = r2 o r1 (s)
= r2(r1(s)) = r2(s). So r3 fixes r2(s) so r2(s) in Q.
So Q is a union of orbits as claimed.
With this claim my last two articles sketched how to get my sphere
decomposition result with A,B,C,D,Q. I will now use this result to
show how to reassemble one punctured ball as two punctured balls of the
same radius as the starting ball. My first article discussed how to
improve this to the corresponding result with the centers undeleted
and then to the supergeneralization with bounded sets of nonempty
interior. After discussing this punctured ball doubling based on my
A,B,C,D,Q I will more briefly indicate how to get the same doubling
from Hausdorff's A,B,C,Q.
So assuming my A,B,C,D,Q and start with a closed punctured ball.
Consider those 5 subsets of the spherical surface of the ball. For
each such subset consider the corresponding induced cone set: namely
union to the original surface subset the interior points on radii
joining surface subset points to the center point of the ball. Do
not include the center point of the ball. Since we are dealing with the
punctured ball where the center point that would otherwise overlap
between the cone sets has been deleted, we have the 5 cone sets induced
partition the punctured ball. The congruences I claimed were all based
on rotations around the ball center, so each original congruence on
the surface subsets corresponds to a congruence on the induced cone
sets. So I will refer to these induced cone subsets of the punctured
ball by the same variable letters I originally used for the
corresponding surface subsets, ie A,B,C,D,Q.
Call the punctured ball P. So P - Q is partitioned by A,B,C,D.
B and C are each congruent to A, so we can assemble by rigid motions
a copy of P - Q from three copies of A and one copy of D.
From our original P we have three copies of A and one of D. (Also
a Q but set that to the side for the moment). A is congruent to
B u C u D, ie 2 copies of A and one of D. So one copy of A can be
split to 2 copies of A and one new copy of D. When we do this to one
A copy in a collection we augment the population by one net new A copy
and one new D copy.
So do this splitting operation six times in iteration and grow our
original collection of 3 A copies and one D copy into 9 A copies
and 7 D copies. These can be reassembled into 3 disjoint copies of
P - Q, with left over on the side 4 copies of D and that original
one copy of Q.
I need an extra Q copy. I will harvest that from the third P - Q
copy, that's why I made three P~Q copies instead of two. Namely
consider how Q is a subset of the original P. Consider an axis through
P, not necessarily one of the two axis we used in setting up the
original rotation group. Namely consider an axis disjoint from Q,
possible since Q^S (= Q intersect S, S being the surface of the sphere)
is countable and there are continuum many choices of axis. Consider
various choices of rotation to perform around this axis.
Consider the image of Q under such a rotation r1: r1"Q a subset of P.
If Q and r1"Q are not disjoint there are points q1 and q2 in Q^S s.t.
q1 = r1(q2). If r2 is another rotation around the same axis then
q1 != r2(q2). This is because the axis was chosen disjoint from Q,
so q2 is not along the common axis of rotation of r1 and r2. So
r1(q2) = r2(q2) would imply r1 = r2.
So for each rotation r1 around the chosen axis which makes Q and
r1"Q overlap by AC we can pick an ordered pair of Q^S members q1 and q2
with q1 != r1(q1). For r1, r2 distinct rotations about that axis both
making overlaps of ri"Q with Q, the pair of Q points chosen for q1 is
a distinct pair from that chosen for r2 by the above comments. There
are only countably many pairs of Q elements, so there are only
countably many choices of angles r around the special axis with
Q and r"Q overlapping.
There are continuum many rotations around that axis so pick one r
not among those countably many, so Q and r"Q are disjoint. So r"Q
is a subset of P - Q. So P - Q contains another congruent copy of Q.
So in that third copy of P - Q pull out that extra copy of Q. We
also have the first two copies we created of P - Q, and the left over
copy of Q from the original P. So combine by rigid motions the two
copies of P - Q and the two copies of Q to produce two copies of P.
The waste products we have off to side from constructing these two
copies of P are the third copy which by now is a P copy with two
disjoint copies of Q removed, and also those extra 4 unused copies of
D we had to create as a side effect of all the A copies we had to make.
I have been describing this construction in a series of stages each
involving some reassembly. So I have been describing a sequence of ~
relations between stages. I discussed in the first article how ~ is
transitive, so this staged description induces an actual ~ style single
reassembly.
From one copy of P we have reassembled two copies of P and waste
products. So P >= 2 disjoint copies of P.
On the other hand one copy of P is a subset of 2 disjoint copies of
P, so two disjoin copiss of P >= one copy of P.
So apply "Shroeder Berstein" to these two directions of >= and get
one copy of P ~ two disjoint copies of P.
So that is the main core of the Banach Tarski construction. My first
post covered the extensions of this punctured ball version.
That was Banach Tarski based on my A,B,C,D,Q sphere decomposition.
To do the corresponding thing from the claimed Hausdorff A,B,C,Q
version, you have the original P-Q with three copies of A. One copy
of A can now be split into two copies of A. You do the splitting
operation six times, growing three copies of A to nine copies of A.
In this version you have less waste products, ie the extra D copies.
Nine copies of A become three copies of P-Q, so everything is similar.
You still have some waste product from the third P~Q copy, and so need
Shroeder Berstein.
I will now discuss the proof of the "Shroeder Berstein" theorem for
~ and >=. Recall the usual proof of the regular Shroeder Berstein
theorem of pure set theory. So suppose A and B are sets and we have
injections f:A -> B and g:B -> A. We seek a bijection h:A -> B.
We define sequences of subsets. A0 = A, B0 = B. Define inductively
A(n+1) = g"Bn, B(n+1) = f"An. ie that notation means A(n+1) is the
image of Bn under g etc. Define
Aw = "Aomega" = intersection (n>=0) An and
Bw = intersection (n>=0) Bn . For n non-negative finite define
A'n = An ~ A(n+1) & B'n = Bn ~ B(n+1) . Then
A = union {A'n| 0<= n <= w} and {A'n| 0<= n <= w} are pairwise
disjoint. Similarly B is the disjoint union of {B'n|0<=n<=w}.
(Note these include the case n=w).
Define C1 = ( union {A'n|0<=nB is a bijection from A to B.
These definitions and the proof that h is a bijection don't need AC.
The above is general set theory, not especially related to the finite
partitioning and reassembly in ~. The above is a sharper statement of
the result of the usual set theoretic construction than is usually
summarized in the statement of Shroeder Berstein. Normally the
statement of set theoretic Shroeder Berstein just says a bijection
exists. The sharper statement expresses the bijection as a union of
restrictions of f and g^-1.
Now we turn to the desired "Shroeder Berstein" for ~ and >=. That
is, assume A and B are subsets of 3-space and A>=B and B>=A, and we
want to prove A~B.
From B>=A we can include A in a superset A' and we can partitition
and reassemble A' to B. So we have an injection carrying points of A
to points of B by including in A' and reassembling to B. Similarly we
have an injection B->A from A>=B. Call these f:A->B and g:B->A
and get C1, C2 and h as above.
A is included in A', and A' has its partition corresponding to to the
equialence A' ~ B. This inclusion and partition of a superset induces
a partition on A. The reassmbly from A' to B is rigid motions on the
respective partition elements. We get that f:A->B is respective rigid
motions on the parition elements on A we just defined.
Similarly g:B->A is respective rigid motions on pieces of a partition
on B. There is an associated partition on A to this g, namely the
partition on A from the B' ~ A for the B' superset of B.
So we have two partitions on A, one associated to each direction of
injection. Form a finite partition on A, a common refinement of these
two. Let this be {Pi| 0* Pi is a rigid motion,
mapping the (part of B that injects by g into Pi) into Pi.
Define Si,j = Pi intersection Cj 0**B.
Consider h|Si,1 for some 0**= 's. Again the crude upper bound is a product of the
respective starting partition sizes. Finally the "Shroeder Berstein"
proof depended on further splitting the partition by C1, C2, ie
basically taking a further common refinement of the previously
constructed refinement partition with the partition C1, C2 of A. Or
to restate it more clearly, we are finally producing a triple common
refinement of the two ~ partitions behind the starting hypothesis and
the C1,C2 partition from the set theoretic Shroeder Berstein.
The question of sharpening those upper bounds beyond the crude ones
I just mentioned may in cases be difficult because it might depend on
details of what the choice set was on the orbits. We just got that
choice set from AC and we know little more about it.
The prinicples I just gave should be sufficient to work through the
proofs and determine a crude upper bound on the partition sizes. I
am sure this would produce a big number, ie loosely speaking
exponential in the number of steps in the construction with those
steps of multiplying subpart upper bounds.
Jonathan mentioned that 5 pieces may suffice for reassembling 1 ball
to 2 balls. I think I have read this too. Getting this must involve
some further ideas beyond what I have done here.
--
David Libert (ah170@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad." -- David Libert
3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig
==============================================================================
From: ah170@FreeNet.Carleton.CA (David Libert)
Subject: Re: Banach-Tarski Construction
Date: 5 May 1999 06:43:14 GMT
Newsgroups: sci.math
Maybe the Banach Tarski paradox is that an original proof attempt at
the Banach Tarski paradox expands out to an infinite sequence of
corrections.
I previously posted to this thread correcting the earlier attempt by
providing Q a countable set of exceptional points to be excluded from
the basic definition of A,B,C,D . Now I have realized I had an error
in one of the stated congruence relations of the sphere decomposition,
namely an error in how D should figure into one of the congruences.
This error does not affect the Hausdorff version which has no D.
I will correct this error and the later results building on the
corrected revision in this post. Since my previous concise description
of the congruences concealed an error I will spell out in more detail
this time the proofs of the congruence relations and the underlying
definitions of the partition sets.
To start we are reviewing the sphere result involving A,B,C,D,Q .
Let G be our countable group of rotations of the sphere S, G being
the group generated by r180 and r120, these being rotations of
180 and 120 degrees respectively around two respective distinct axis.
An orbit O of G is a subset of S s.t. each pair of points in O are
accessible from each other by G elements, and there is no G
accesibility between O and S-O. The sphere S is partitioned by the
orbits of G.
Our definitions of A,B,C,D will be based on divisions performed
inside various orbits, with parallel definitions running in different
orbits. Within an orbit O, once you pick a base point s0 in O, you
can refer to any point s1 of O by referring to a G element g s.t.
s1 = g(s0). So you can use G to easily label O and discuss it.
This depends on getting started by picking the base point s0, which
can't be done in any obvious definable way. So you pick s0
arbitrarily. In order to know that such procdures on each orbit O
can be combined to give an overall construction handling infinitely
many orbits you use AC to pick the s0 for of the O's to be handled.
This is actually the only place the Banach Tarski construction uses
AC.
For some orbits O the labeling method of g(s0) can be ambiguous:
possibly g1(s0) = g2(s0) for distinct g1, g2 in G. We don't want
this complication, so all the orbits where this can happen anywhere
in the orbits are thrown in as subsets of Q. I previously argued
that Q is countable. We will be continuing by partitioning S-Q
into A,B,C,D . As indicated above these definitions will work in
parallel on various orbits which are subsets of S-Q. I previously
argued that Q is a union of orbits, so working now in S-Q we have
full orbits as subsets. By throwing away Q we now have for each
remaining orbit O under consideration the labelling of O elements
by G elements is unique.
So with AC and Q, if we partition G we can use this to partition
all of the remaining orbits O subset of S-P. This is way the proof
proceeds.
G is generated by r180 and r120 of respective periods 2 and 3.
So G elements can be represented by words consisting of intermixed
blocks of r180's and r120's. Each block of r180 can only contain
exactly one r180 copy. A block of r120 may be one copy of r120 or
two copies. The periods of r180 and r120 are the reason for these
block sizes. There are no restrictions on whether the number of
blocks is even or odd or whether the first block is of r180 or r120.
Also the empty word is allowed (the identity element of the group,
th identity rotation).
In general for some 180 and 120 degree rotation groups involving
distinct axis it would be possible for this labelling of group
elements by words to be non-unique: some group elements may have
more than one word label. But the freeness property of G makes
the labelling unique, this is what freeness means. We choose the
angle between the axis to have this freeness property for G.
So with AC, Q and G freeness it has been arranged that S-Q can be
partitioned by parititioning the set of words I have just described.
This is how the proof proceeds.
Namely the words are classified by their leading (= leftmost) block.
A corresponds to r180 block, B to one r120, C to two r120's. This
leaves the empty word left over which has no leading block: that is
what D is. So as a subset of words D has only one element: the empty
word. A, B and C each correspond to a countable infinity of words.
So D corresponds to a one element subset of G: the identity element.
So in each O orbit D corresponds to only one element, namely s0.
In O A,B,C correspond to a countable infinity of elements.
On the other hand, in S-Q there are continuum many orbits O
contained. (ie Q is countable so S-Q is continuum size. Each orbit
O is countable so there are continuum many orbits O subset of S-Q).
So the actual D subset of S-Q is still continuum sized, even though
each subcase of defining parts of D (ie subcases by orbits O), makes
D look small (size = 1).
For r1 a rotation in G, r1 acts on points in S-Q by moving them around
inside their respective orbits. (ie basically by definition of orbit).
In an orbit O subset of S-Q r1 moving on points corresponds to a certain
function on the words labeling points. Namely we consider the group
G to be the usual order of composition of function, where the function
applied last in the composition is the left operand of the group
operation. So r1 acting on points corresponds to prepending r1 to a
word. Namely r1 as a G element is also represented by a word as
described, so prepend this to a word labelling a point r2(s0) in O.
Such prepending may not produce a legal word as I described them,
because the rightmost block for the word for r1 may juxtapose with the
leftmost block of the word for r2 to violate the block size rules.
(ie r180 juxtaposed to r180 for example). For such words cancel down
using the periodicities of r180 and r120, ie two r180's disappear,
three r120's disapear. This cancellation process may have to be
iterated, but in the end produces a legal word.
The real cases of interest for the proof are r1 = r180 and
r1 = r120 so for these at most one cancellation suffices.
With this background we want to check the congruences relating to
A,B,C,D . Last time I claimed that B is congruent to C. This was
correct, and is the easiest to see. B elements are labeled by words
with leading block r120. When you apply r120 to such points you
prepend r120 to the word producing an r120^2 block, hence in C.
On the other hand any C element with an r120^2 leading block can
have its word produced by prepending r120 to an r120 block, so
such C elements are r120 applied to a B element.
We conclude that r120 carries B exactly to C, so B and C are
congruent as claimed last article.
Next consider a slightly trickier case which was correct last time.
Consider a point in A, so represented by a word with leading block
r180. Apply r180 to this, which corresponds to prepending r180. The
leading block becomes r180^2, which cancels to disappear. So in
effect we lose the leading block r180 of the original point's word.
If there was a block after the original leading r180, it must have
been r120 or r120^2. In these cases the original A point went by r180
into B u C. On the other hand if nothing followed the original r180
block, ie if the original word was just one block of r180, then on
another r180 prepend and cancellation we get the empty word, ie into
D. So such A points travel by r180 to D. We conclude r180 carries
A into B u C u D.
Given any element r(s0) of B u C u D, the word for r does not have
leading block r180. If you prepend r180 to this word there is no
cancellation, so the resulting word has a leading r180 block. So
r180(r(s0)) in A. Then r180(r180(r(s0))) = r180 o r180 o r (s0)
= r(s0). (Last = since r180 o r180 = identity). So our arbibrary
B u C u D element is in the image of A under r180.
We conclude r180 carries A to exactly B u C u D, so A is
congruent to B u C u D, as claimed last article.
So we arrive at the third claimed congruence from previous articles,
which was the error. I claimed r120 carries A exactly to B, showing
A and B are congruent.
Consider an A point r(s0). So the leading block of the r word is
r180 (by definition of A). Act on this by r120 and you prepend an r120.
This doesn't cancel against r180 so we have one leading block of r120
so we are in B, as I claimed.
Now consider D. Points of D are represented by the empty word.
Prepend r120 and get r120 again, also putting us into B. So r120 also
carries D into B. Since r120 is 1-1, the image of A under r120 can't
fill B. This contradicts my previous article claim that r120 carries
A to B exactly.
So far we have r120 carries A u D into B. On the other hand
consider an element r(s0) of B. r must have a word representation with
leading block r120 (by definition of B). Following this leading block
there must be either be an r180 block or nothing. So r120^-1(r(s0))
is either in A or D respectively. So any B member has an r120 preimage
in A u D, so r120 carries A u D exactly to B.
So in the last article where I claimed r120 carries A exactly to B
I should have claimed r120 carries A u D exactly to B.
So my revised congruence claims are: A u D is congruent to B and C,
and A is congruent to B u C u D.
Previously I erroneously claimed A is congruent to B, C and
(B u C u D).
From this erroneous claim I had two arguments, one that A is
non-Lebesgue measurable by the measure on the sphere S, and the other
the Banach Tarski construction doubling a punctured ball. Both of these
have to be revised by the change to the congruence claim.
Regarding the first one on sphere Lebesgue measure, the new version I
will prove now from the corrected congruence claim will be that B is
not Lebesgue measurable by m the Lebesgue measure on the sphere S.
Suppose for contradiction B is Leb. measurable. B is congruent to
A u D and A is congruent to B u C u D, ie a superset of two disjoint
congruent copies of B (ie B and C). So m(B) >= 2m(B). So m(B)=0.
A u D is congruent to B, so A is included in a measurable set of
measure 0, so A has outer measure 0, so A is measurable of measure 0.
A is congruent to B u C u D, so (B u C u D) is measurable of measure
0. S is partitioned by A,B,C,D,Q so
m(S) = m(A) + m(B u C u D) + m(Q) . We found the first two are 0.
m(Q) = 0 since Q is countable. So m(S) = 0, contra.
Next the Banach Tarski doubling. Begin with P a punctured sphere.
P-Q has one copy of A, two of B and one of D. (ie B congruent to C).
Given a B copy we can split it into an A and D copy. We can then split
that A into two copies of B and a D. So from one B we can produce two
B's and two D's. Given a mixed collection of copies of A, B and D,
if we perform this splitting on one B copy, we increase the count of
B copies by one (net increase), and the count of D copies by 2.
So starting from 1 A, 2 B's and 1 D from the original P-Q,
iteratively expand the collection by splitting a B six times, producing
1 A, 8 B's and 13 D's. Split two of the B's again each into an A and a
D, producing 3 A's, 6 B's and 15 D's. Combine these into 3 (P-Q)'s,
leaving a leftover of 12 D's. Continue as in the previous articles.
--
David Libert (ah170@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad." -- David Libert
3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig
==============================================================================
From: Martin GOLDSTERN
Subject: Re: Infinite cardinals question
Date: 19 Dec 1999 20:03:18 GMT
Newsgroups: sci.math
John Savard wrote:
:> [Banach-Tarski] is a paradox, a surprising fact.
:> It does not need to be "resolved".
: Well, an explicit construction for decomposing a three-dimensional
: cube into one of a different size *would* be nice.
What does "explicit construction" mean? Recursive? Borel?
First order definable?
We know (i.e., we can prove with the appropriate set-theoretic axioms)
that there is no such construction.
Martin.Goldstern@tuwien.ac.at
*