Let $f(x)$ be uniformly continuous on a bounded open interval $a<x<b$. Show that $f$ is bounded (i.e. $\exists M$ such that $|f(x)|\le M \ \forall x\in (a,b)$).
To be honest, I have no idea how to solve thus problem. I tried to pass directly by the definition of uniformly continuous function and extract something, and I passed by cases distinction ($f$ monotonic or not), but I still can't conclude. Intuitively I see why it is true but can't find a good approach to this problem. If someone could give a hint, I would appreciate it. Thanks in advance
Best Answer
Assume it is not bounded. Then there is a sequence $(x_n)_{n=1}^\infty\subseteq (a,b)$ such that $|f(x_n)|>n$ for all $n\in\mathbb{N}$. Since this sequence is bounded, it must have some Cauchy subsequence $(x_{n_k})$. Now use uniform continuity to show that $f(x_{n_k})$ is also Cauchy (this easily follows from the definition), and this will be a contradiction.