Sequences proof I'm doing an assignment for my analysis class and need help with the following question:
Let $f$ be a continuous function which is defined on the interval $I = [a, b]$. Assume
that $f$ is not bounded on $I$. Use the fact that $f$ is not bounded to prove that there is a
sequence $(x_n)$ in $I$ for which $|f(x_n)| > n, \forall n \in \mathbb N$
I would really appreciate your help .
Thanks 

If $ f $ is not bounded on $ I $ then, by definition, given any $ C > 0 $ we can find $ x \in I $ such that $ \vert f(x) \vert > C $. In particular, for each $ n \in \mathbb{N} $ we can pick $ x_n \in I $ with $ \vert f(x_n) \vert > n $.
