Can someone help me prove this sequence is dense?

Question

Let f:R--> R be continuous, periodic and non constant, with fundamental period p. Show that, if p is irrational, then the sequence (f(n), n=1,2,3.....) is dense in f([0, p]). I was told to use the pigeon hole principle, but I'm really confused.
Thank you

f([0,p]) is the image under f of the closed interval [0, p]. That is, the set I = {f(x) | x is in R}.

Answer 1

The problem is right. The conclusion follows from the following fact: if p is an irrational, then the set A = {m*p + n | m is an integer, n is a positive integer} is dense in R. The proof of this fact can be found doing a search on google, sci.math. It uses the pigeon hole principle.

Let y be an element of f([0, p]). Then, y = f(x) for some x in in [0, p]. Since x is an accumulation point of [0, p] and A is dense in R, x is an accumulation point of A. So, there exists a sequence a_k in A, with pairwise distinct terms, that converges to x. Since a_k is convergent, it is a Cauchy sequence. So, for every eps>0, there is K such that

|a_j - a_k| < eps for every j, k >= K (1).

Since a_k is a sequence in A, it's terms are of the form a_k = m_k * p + n_k, m_k integer, n_k positive integer. For every distinct j and k, a_j <> a_k. Suppose, j, k >=K. If n_j = n_k, them m_j <> m_k, or we'd have a_j = a_k. So,

|a_j - a_k| = |(m_j - m_k)| p.

Since m_j and m_k are distinct integers, then |m_j - m_k| >=1 and |a_j - a_k| = |(m_j - m_k)| p >= p. Therefore, if n_j = n_k, then (1) cannot be satisfied for eps n_k for j<>k, j,k >= K. This means that, for k >= K, the sequence (n_k) has its terms pairwise distinct. This implies that n_k, composed of positive integers, has infinitely many terms.

In virtue of what we have concluded, we can select a subsequence a_i of a_k in which the n_i's form a strict increasing sequence. So, (n_i) is a subsequence of (n) and f(n_i) is a subsequence of f(n). Since a_i converges to x (because so does a_k) and f is continuous, f(a_i) = f(m_i * p + n_i) --> f(x). But since p is the fundamental period of f, f(m_i * p + n_i) = f(n_i) for every i. So, we conclude f(n_i) --> f(x) = y. Since f(n_i) is a subsequence of f(n) and this holds for every y of f([0, p]), it follows every element of f([0, p]) is limit of some subsequence of f(n), that is: f(n) is dense in f([0, p]) , as stated.

Answer 2

One way to simplify this problem:
Instead of directly mapping:
(0, infinity) => f:[0,p] , do it indirectly:
(0, infinity) => [0,p] => f:[0,p] through the mapping:
x => (z = exp(i*2*pi*x/p)) => f(z)
As x runs from 0 to infinity, z will just run around the unit circle in the complex plane. As x traverses distances of p, z will just complete another 2*pi around the circle, coming back to angle 0 again. As long as f(z) is defined to be continuous on the unit circle (so that f(exp(i*2*pi)) = f(exp(0)) ), periodicity in p is automatic.

- OK, the first point then is that when you look at the sequence f(n), the points:
x = 1, 2, 3, 4... map to:
z = exp(i*2*pi*(1/p)), exp(i*2*pi*(2/p)), exp(i*2*pi*(3/p)), exp(i*2*pi*(4/p)), ... and these will wrap around the unit circle.

This is a sequence of infinitely many points, zipping around the unit circle. Because p is stated to be irrational, the sequence of n/p will be dense throughout the unit circle. This is proven in my answer to Steiner's question referenced below. The essence of the proof is that if this sequence were NOT dense in the unit circle, then the sequence of multiples of 1 (in a periodicity of p) would leave a "protected zone" with no multiples of 1, which would lead to the implication of a protected zone from angle 0 to some definite positive angle. This eventually leads to the understanding that p must in fact be rational, which is a contradiction.

So, on these grounds, I take it as proven that the sequence 1, 2, 3... is mapped into a sequence z(1), z(2), z(3)... that is dense on the unit circle.

So now, let's consider a point in the image of
f[0, infinity] = image of [unit circle]. Because the function is continuous, the inverse image of a ball around that point is an open set in the unit circle, centered around the inverse image of that point. There is an open ball centered about the inverse-image point, which lies completely within the inverse image (in the unit circle) of the ball in the range of the mapping f[0, infinity]. Since the points of f(n) are dense in the unit circle, there are points in the ball on the unit circle; and these in turn imply that their mapped-to points in the range will be in the ball centered around the initial point in the range of f[0, infinity].

So this means that points in f[0, infinity] cannot get away from the sequence f(n): any ball around any point in the range will contain members of that sequence.

QED

Answer 3

I think that we could capitilize on the fact that A = {n+mp; n>0, n & m are integers} is a dense set in R in yet another fashion:

Because f is continuous on R, it is also continuous on the closed interval [0, p]. In particular f has both maximum and minimum over [0, p]. Since f is not constant, there exist l and u in [0, p] s.t. l<>u and f(l) is the min and f(u) the max over [0, p]. Because f is also periodic, f(l) and f(u) are the absolute min and max on R respectively. By the intermediate value theorem then, the range of f is [f(l), f(u)].

We will prove that for any x in [0, p] and any 0
Assume that for some x in [0, p] there exists e>0 s.t. B_e(f(x)) doesn't contain any f(n). The continuity of f tells us that there exists an open ball B_d(x) s.t. f[B_d(x)] is contained in B_e(f(x)) . It follows that B_d(x) does not contain any any positive integers e.g. n. **If it does then f(n) is in B_e(f(x)), contary to our hypothesis.**

Now the periodic property comes in handy: because f is periodic with fundamental period p, f(x) = f(x + mp) for any integer m. In particular, f(n) = f(n + mp). What does this tell us? Since A is dense in R, we must have points from A in every interval. Hence, some n + mp is in B_d(x). In other words f(n+mp) is in B_e(f(x)) . But then f(n) is in B_e(f(x)) . And the desired contradiction is obtained.

Originally, I was trying (without success) to prove that for rational 0b, (a+mp, b+mp) contains an integer for some m. Thanks to the comment by Steiner, I altered my strategy.

So, give all your thanks to him! ;D

Answer 4

Thanks, but as it is the problem is wrong
btw, it's I={f(x)|x in [0,p]}