How can we show this sequence is dense in this set?

Question

Let f:R --> R be a continuous and periodic function such that it's fundamental period p is irrational. Let x_n be an increasing sequence of positive numbers such that x_n --> oo and (x_(n+1) - x_n) --> 0. Show that f(x_n) is dense in f([0, p]) .

Answer 1

I'm not sure what the irrationality of p has to do with anything--red herring, or am I missing something?

Anyway, here's my solution. Given Y in f(0,p) and ε>0, we want to show that there is an n in N such that |Y-f(x_n)| < ε.

Choose any X in f-inverse of Y. By continuity of f at X, we can choose δ > 0 such that |f(X)-f(x)|< ε whenever |X-x|< δ. Note that this same delta works for ANY X+kp (in f-inverse of Y), k in N, by periodicity. And, since |x_(n+1)-x_n| →0, we can choose M so that |x_(n+1) - x_n| < δ for all n ≥ M.

Now, from {X+kp: k in N}, we may choose an element X' > x_M. Then choose n so that x_n < X' ≤ x_(n+1) (which is possible since x_n→∞ monotonically). Note that n≥M, which means that |x_n - X'| < δ, so that |f(x_n) - f(X')| = |f(x_n) - Y| < ε, as desired.

QED.


*** ADDENDUM:

Apparently there was a bit of confusion because there are several definitions of "dense." The above proof addresses the standard topological definition, namely that a set Y is dense in X if every open set in X contains an element of Y.

An alternate definition, particular to sequences in a metric space, is that a sequence {x_n} is dense in a set X if for each x in X there is a subsequence {x_n_k} of {x_n} converging to x.

To address this second definition with my proof, simply associate with each k in N an x_n_k corresponding to ε=1/k in the proof above. Then |Y-f(x_n_k)| < 1/k for each k in N, so {f(x_n_k)} →Y as desired.

(Note that in general, this idea of choosing ε≈1/k shows the equivalence of the two definitions in the case of metric spaces.)
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Answer 2

Since the n+1th iteration is larger than its predecessor (i.e. it's increasing), we know that in any arbitrarily small subset of the real line we have a point x_n.

Basically, you could argue that you can create any arbitrarily small segment of the real line on [0,p] and have a point in the sequence fall into it, thus making it dense.

I know that's pretty hand-wavey, but I'm a little too far removed from my college analysis courses to delve too deeply into this. I know that what I'm saying is essentially correct though.