**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)