If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded? If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded?
($a$ and $b$ being finite numbers).
I tried proving and disproving it. Couldn't find an example for a non-bounded image. 
Is there any basic proof or counter example for any of the cases?
Thanks a million!

Let $\delta$ be so that $|x-y|< \delta \Rightarrow |f(x)-f(y)|<1$. Now prove that there exists a finite set $F$ so that 
$$(a,b) \subset \cup_{z \in F} (z-\frac{\delta}{2}, z+\frac{\delta}{2} ) \,.$$
Now, what can you say about $|f(x)|$ and $\max\{f(z)|z \in F \}+1\,.$?
P.S. The existence of the finite set is really the fact that $(0,1)$ is precompact in the topology of $\mathbb R$. So this solution is not really different than Ittay's...
