[alogo] Bolzano Weierstrass theorem application

Here we consider the subset of points of [0,a]x[0,a], of the form:
S = {(t-n*a, beta*t-m*a), for t real, m, n integers} Apply the Bolzano-Weierstrass theorem to prove, that if beta is irrational, then the points of S are dense in the square .
This means, given (t0, t1) in this square and e>0, one can find (t, m, n), t: real and m, n: integers, such that both inequalities hold: |t-t0-a*m| < e, and |beta*t-t1-a*n| < e.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]

By replacing t = t0+a*m in the second inequality, it suffices to show that there is a tripple (t,m,n) such that |beta*(t0+a*m)-t1-a*n| < e. Equivalently a*|(beta*t0-t1)/a + beta*m-n| < e.
Setting t2 = (beta*t0-t1)/a, this amounts to show that for any real t2 in the interval [0,1], there are m, n such that |t2 +beta*m-n| < e.

This can be proved by applying the Bolzano-Weierstrass theorem on the set T of points of the interval I=[0,1], of the form:
T = {t2+beta*m-n, with variable integers m,n}. Note that T is non-empty, since for each choice of m, the real number t2+beta*m has, for general m, a fractional part u and an integer part v. Thus, taking n=v we obtain a number in T. Besides two numbers of T cannot be equal, since this would imply: beta*m-n = beta*m'-n' ==> beta = (n-n')/(m-m'), and beta would be a rational number.

Thus, T is an infinite set of I, and by the thoerem referred above has a sequence {si} converging to some point s0 of [0,1]. By applying the Cauchy criterium, we can find then i,j such that |si-sj| < e.
Considering the form of the elements of T, we can assume that d = si-sj = beta*m-n, with 0 < b*m-n < e. Divide then [0,1] in intervals of length d and look for the place of t2 inside these subintervals of length d. Certainly t2 is contained in some of them i.e. there is an integer k, such that k*d <= t2 < (k+1)*d. Thus |t2-k*d| < d < e. But k*d = beta*k*m- k*n = beta*m' - n' has the required form and we have found m', n' such that |t2 - beta*m' - n'| < e .

Bolzano-Weierstrass theorem: Every bounded set in an euclidean space has an accumulation point.
Cauchy-Criterium: If a sequence {si} converges, then for every e>0 there is an integer N, such that for every i, j>N: |si - sj| < e.
The above figure illustrates the assertion proved. The line defined parametrically by t*(1,beta), when intercepts the boundary of the square at a point (a,x) or (x,a), it is continued in the same direction from the point lying on the opposite side i.e from (0,x) or (x,0). By repeating this procedure we create the above image representing a part of set S. If beta is irrational and we continue drawing, then the green lines will cover the whole area of the square.
Note that a similar argument prooves that for an angle w, with w/pi irrational the multiples n*w define a dense set on the unit circle.

Return to Gallery

Produced with EucliDraw©