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Math Help - Continous, unbounded function in metric spaces

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    Continous, unbounded function in metric spaces

    Let f:R \rightarrow R be a continuous, unbounded function. Show that there is a number t_{0} for which \{f(nt_{0}): n is an integer } is an unbounded set.
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    This is quite a difficult problem. I think the easiest way to attack it is to use a proof by contradiction.

    So suppose that the result is false, and that for each real number t, the set \{f(nt):n\in\mathbb{Z}\} is bounded. For each natural number m, let K_m = \{t\in\mathbb{R}:|f_n(t)|\leqslant m \text{ for all }n\in\mathbb{Z}\}. Then K_m is closed (not too hard to prove), and \bigcup_mK_m = \mathbb{R}.

    Now comes the hard part. Suppose that K_m contains an interval [t_1,t_2], where t_1<t_2. Then for all s\in [t_1,t_2] and for all integers n, |f(ns)|\leqslant m. But as n increases, the intervals [nt_1,nt_2] start to overlap. In fact, this will happen as soon as n is sufficiently large that nt_2>(n+1)t_1, or n>\tfrac{t_1}{t_2-t_1}. But that means that every sufficiently large real number x is in one of the intervals [nt_1,nt_2] and so |f(x)|\leqslant m. It is easy to conclude from that that f is bounded. Since we are told that f is unbounded, it follows that K_m cannot contain an interval. And since K_m is closed, that means that K_m is nowhere dense.

    Thus \{K_m: m\in\mathbb{N}\} is a countable collection of nowhere dense sets whose union is the whole of \mathbb{R}. But that is impossible by the Baire category theorem. That contradiction completes the proof.

    Here is a very similar problem. Let f be a continuous function and suppose that for every t>0 the sequence f(t), f(2t), f(3t), ... tends to 0. Prove that f(x) tends to 0 as x→∞. The proof of that result also relies on the trick of using overlapping intervals and the Baire category theorem. There is a long and careful discussion of that problem in this entry in Tim Gowers's blog.
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