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THEOREM (A. Katok) There exists a measurable set E of area one in 
the unit square (0,1) x [0,1] together with a family of disjoint 
smooth real analytic curves G(y) which fill out this square, so that 
each curve G(y) intersects E in at most one single point. 
PROOF. Define for p in (0,1) the piecewise linear map T on [0,1] 
by T(x)=x/p for x in A=[0,p) and f(x)=(x-p)/(1-p) for x in B=[p,1). 
It is easy to see that T is measure-preserving. Denote by T^n(x) the
n'th iterate of the map, that is T^n(x)=T(T^(n-1)(x)). For fixed p, code
x by an infinite sequence b(n)=0 if x(n)=T^n(x) in A and b(n)=1 else. 
In terms of this coding, T corresponds to the shift map. By the strong
law of large numbers, for Lebesgue almost every x in [0,1], the 
frequency of 1's in the associated symbol space is defined and equal
to (1-p). Let E be the subset of (p,x) in (0,1) x [0,1] such that
the frequency of 1's is equal to 1-p. It is a measurable set. Because
the intersection of E with each line {p} x [0,1] has full Lebesgue
measure, Fubini's theorem implies that E has Lebesgue area 1. 
For x in [0,1] define y(p,x) = sum b(n) 2^n, where b(n) is the coding
of x. The sets G(y) = { (p,x) |  y(p,x)=y } are disjoint and each G(y) 
is a smooth real analytic curve. 
[Proof: set p(0)=p,p(1)=1-p. From x(n)=b(n) p(0)+x(n+1) p(b(n)) follows
  x=x(p,y)=p(0)(b(1)+p(b(1))(b(2)+p(b(2))(b(3)+...) ...) ...)
          =p(0)(b(1)+b(2) p(b(1))                 +b(3) p(b(1)) p(b(2)) 
                    +b(4) p(b(1)) p(b(2)) p(b(3)) +...)
  Set p(0)=p=(1+t)/2, p(1)=1-p=(1-t)/2. If |t| x(p,y). Since a given symbol 
sequence b(n) can have at most one limiting frequency 
lim (b(1)+ ... + b(n))/n = 1-p, it follows that each G(y) can intersect 
E in at most a single point (p,x(p,y)). 
                           -- John Milnor, Mathem. Intelligencer, Vol 19, 1997