Showing a Set is Unbounded Given a Continuous, Unbounded Function Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous, unbounded function. Show that there is a number $t_0$ for which $\{f(n t_o): n$ an integer $\}$ is an unbounded set. 
Here is my attempt at a solution:
Let $f: \mathbb{R} \to \mathbb{R}$ be the continuous, unbounded function with assignment $f(t)=t$ for $t \in \mathbb{R}$. Let $A = \{f(n t_o) : n \in \mathbb{Z}\}$ which is the integer multiples of $f$. 
Suppose $A$ was a bounded set for all $t_0$. If this was true, then we may fit all of $A$ into an open ball with positive radius, call it $B(t,\epsilon)$. However, since $f$ is unbounded and $A$ consists of the set of all integer multiples of $f$, then we cannot contain this entire set within an open ball. Therefore, $A$ is an unbounded set.
For some reason, I feel like if my conclusion did not follow from my argument. Or at least it seems like it took a major detour. Furthermore, my intuition is telling me to somehow incorporate the Baire Category Theorem since this problem is found in the exercises following this statement along with The Contraction Lemma in my textbook. If this is the case, how does one use these tools to solve the problem?
Can somebody please help me prove this? Much thanks in advance for your time and patience. 
 A: The interval $[1,2]$ in $\mathbf{R}$ is a complete metric space. For every positive integer $M$, define a subset $A_M$ of $[1,2]$ by $A_M = \{x\in [1,2]\;|\;\exists n\in\mathbf{Z},|f(nx)|>M\}.$ Since $f$ is continuous, it is clear that $A_M$ is open in $[1,2]$ as it is a union of open sets.
I pretend that $A_M$ is dense in $[1,2]$. If not then you can find an interval $[a,b]$ contained
in $[1,2]$ with $a<b$ which is disjoint from $A_M$. Let $U = \cup_{n \in \mathbf{Z}} n[a,b]$. Then by assumption $f$ is bounded on $U$. On the other hand, $U = a(\cup_{n \in \mathbf{Z}} n[1, b/a])$. Let $d:= b/a - 1$. Now define $V = \cup_{n \in \mathbf{Z}} { n[1, 1+d] }$ so that $U = aV$. Take a positive integer $K$ such that $1/K < d$. So if $n>K$, then $n[1, 1+d] = [n, n+ nd$] contains $[n, n+ Kd]$ which itself contains $[n, n+1]$. It follows that there is a constant $C>0$ such that for any $y \in V$, we have $y< C$. Notice that a number $y$ is in $V$ if and only if $-y$ is in $V$. Therefore, $\mathbf{R}\backslash V$ is in fact a bounded set and so is also $\mathbf{R}\backslash U$. Now $\mathbf{R}\backslash U$ is contained in some compact set $L$ and $f(L)$ is compact as $f$ is continuous, $f(L)$ is therefore also bounded. So $f(\mathbf{R}\backslash U)$ is bounded. From what preceeds, $f$ is bounded on $U$ also. And finally $f$ is bounded which contradicts the hypothesis that $f$ is unbounded.
So $A_M$ is dense in $[1,2]$, for any positive integer M. Let $B = \cap_{M\in\mathbf{N}^*} A_M$. Then, by the Baire's category theorem, $B$ is dense in $[1,2]$ and a fortiori $B$ is non empty and if  $x$ is in $B$ then $\{f(nx)\;|\;n\in\mathbf{Z}\}$ is unbounded.
