# Thread: Continous, unbounded function in metric spaces

1. ## 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.

2. 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. 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.