A collection of quotes, fortunes, anecdotes, and quips. Get new quotes everyday on facebook, twitter, and tumblr.

john milnor quotes

Related Tags math

Mandelbrot made quite good computer pictures, which seemed to show
a number of isolated "islands" (in the Mandelbrot set M).
Therefore, he conjectured that the set M has many distinct connected
components.
(The editors of the journal thought that his islands were specks of
dirt, and carefully removed them from the pictures).
-- John Milnor, in Dynamics in one complex variable, 1991


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