thomas (12/01/82)
This result is due to Hausdorff, and is taken from the appendix to his
book "Grundzuege der Mengenlehre" (1914). Most of the material here is
directly translated, although paraphrasing has been done in spots. It
gets a little hairy in the middle, so you can skip if you want. The
banger is at the end.
We will show that it is impossible to define an area function on the
surface of a sphere K which satisfies the conditions
i) if A and B are disjoint sets, then
area(A+B) = area(A)+area(B)
ii) if A and B are congruent, then area(A) = area(B)
and which is always positive. The proof rests on the remarkable fact
that a half-sphere and a third-sphere can be congruent, or that (with a
countable set ignored), we can divide K into three sets A, B, C, which
are mutually congruent but also congruent to B+C.
Any countable set must have measure 0.
A sketch of the procedure is as follows:
Let Q and R be rotations of the sphere, Q is a rotation by pi
and R is a rotation by 2*pi/3 about separate axes. We can show
that Q and R are completely independent for some set of axes.
Then we partition the surface of the sphere into (an infinite
number of) sets, each of which is generated by taking a single
point and applying all possible combinations of Q and R.
[These are equivalence classes.] Next, we pick one point from
each set and declare them all to be members of the set A. Then
we can inductively classify all the rest of the points by
saying that points in sets A, B, C go under the rotation Q into
the sets B, A, A, and under the rotation R into B, C, A. Thus,
we have defined our 3 sets so that B+C is congruent (via the
rotation R) to A, and A is congruent to B is congruent to C
(via the rotation Q). A formal proof follows.
We can define a group G of rotations which is generated by Q and R. If
we let Q, R, and R**2 (denoted by R2) be primitive factors, then we can
order the elements of the group by their factors:
G = { 1 | Q, R, R2 | QR, QR2, RQ, R2Q | .... }
Now, we shall show that we can choose the axes of rotation so that,
except for the fact that Q**2 = R**3 = 1, there is no relation between
Q and R.
The product of 2 or more factors must be of one of the four
forms:
(1) Q R**m1 Q R**m2...Q R**mn
(2) R**m1 Q R**m2...Q R**mn Q
(3) Q R**m1 Q R**m2...Q R**mn Q
(4) R**m1 Q R**m2...Q R**ms
where m1 ... mn are either 1 or 2. Now we shall show that two
products are equal if and only if they are formally identical.
The relationship x = y implies that x * y**-1 = 1. If x and y
are not formally identical, then a product formally different
from 1 must be equal to 1. We will show that this is not
possible with the proper choice of axes. It is easy to see
that we need only consider the case when a product of form (1)
is equal to 1, since if, for example, we had a product of form
(2), we need only multiply fore and aft by R to put it into
form (1). Let us denote this product as P.
Put a set of coordinate axes through the center of the sphere
such that the R axis is the z axis and the Q axis is in the xz
plane. Denote the angle between the axes by t/2 and set
l = cos(2*pi/3) = -1/2, m = sin(2*pi/3) = sqrt(3)/2
then we have
R(x,y,z) -> (xl-ym, xm+yl, z)
Q(x,y,z) -> (-x*cos(t)+z*sin(t), -y, x*sin(t)+z*cos(t))
QR(x,y,z)-> (-xl*cos(t)+ym+zl*sin(t),
-xm*cos(t)-yl+zm*sin(t),
x *sin(t) +z *cos(t))
R**2 is obtained from R by substituting -m for m.
Now, P is a product of n factors of the form QR or QR**2, and
P' = PRQ or P" = PRQ**2 is a product of n+1 such factors. The
point (0, 0, 1) goes into (x, y, z) by P, and by P' into
(x', y', z'), (and by P" ...) so that the transformation QR (or
QR**2) goes between these points. We claim that x/sin(t) and
y/sin(t) are polynomials of degree n-1 in cos(t), and that z is
a polynomial of degree n in cos(t):
x = sin(t) * (a*cos**n-1(t) + ... )
y = sin(t) * (b*cos**n-1(t) + ... )
z = c*cos**n(t) + ...
for n=1 this is obvious, since the point (0,0,1) goes to the
point (l*sin(t), +/-m*sin(t), cos(t)) by QR or QR**2. Now
assume (here comes an inductive proof) it for n, it easily
follows that
x' = sin(t) * (a'*cos**n(t) + ...)
y' = sin(t) * (b'*cos**n(t) + ...)
z' = c'*cos**n+1(t) + ...
where
a' = l*(c-a)
b' = +/-m*(c-a)
c' = c-a
thus
c' - a' = (1-l) * (c-a) = 3/2 * (c-a)
and
c - a = (3/2) ** n
So the z coordinate of the point (1, 0, 0) goes by an n-factor
product P into
z = (3/2)**n-1 * cos**n(t) + ...
which, for some t is not 1 for all n (this is easily seen by
taking cos(t) to be a transcendental number, i.e., a number
which is not the solution of any polynomial. There are
uncountably many of these, so it should be easy to find one.)
End of proof that Q and R are independent.
Now, we separate G into 3 classes A, B and C such that given P in G,
one of P and PQ belongs to A and the other to B+C, and out of
P, PR, and PR**2 one belongs to A, one to B and one to C. We
will do this by the following recursive procedure: once we
have all the products with n factors separated, we will
determine the separation of the n+1 factor products as follows:
if P ends in R or R**2, and P is in A, B or C, then PQ is in B,
A or A, respectively. If P ends in Q, and P is in A, B or C,
then PR is in B, C or A and PR**2 is in C, A or B respectively.
Thus we get (assigning 1 to class A):
A| 1 | | RQ, R**2Q, QR**2 | QRQ | ...
B| | Q, R | | QR**2Q, RQR, R**2QR | ...
C| | R**2 | QR | RQR**2, R**2QR**2 | ...
Now, let Q be the set of all the fixed points of members of G.
Q is obviously countable (G is countable, and each member has
only 2 fixed points). Let P = K - Q (sorry about the
duplication of letters here, but I don't have Greek, script,
etc. on my terminal). A point x in P goes by the elements of G
into x, xQ, xR, ... (all distinct). Call this set Px - it is
an equivalence class of K. Choose one point from each
equivalence class [Axiom of Choice!] and call this set M. Then
P = M + MQ + MR + ...
***HERE IT IS****
Now, using the separation of G, we can separate P into 3 sets
A, B and C which we will identify with the classes A, B and C
above, such that
P = A + B + C
A = M + MRQ + MR**2Q + ...
B = MQ + MR + ...
C = MR**2 + MQR + ...
and by construction
AQ = B+C, AR = B, AR**2 = C
and therefore (== means congruent)
A == B == C == B+C.
If we assign to P an area of 1, then
area(A) = area(B) = area(C) = 1/3 (since A == B == C)
and
area(A) = area(B+C) = 1/2 (since A == B+C)
Take heart, though, area still works fine in 2-space.
References:
Hausdorff, F. "Grundzuege der Mengenlehre", (1914) pp 469-472
For more on area see
Banach, S. "Sur le probleme de mesure", Fund Math IV, 1923?
(near the beginning)