TITLE: If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded?
QUESTION [5 upvotes]: 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!

REPLY [6 votes]: There is a $\delta >0$ such that for all $x,y\in (a,b)$ with $|x-y|\leq \delta$ we have $|f(x)-f(y)|\leq 1$. Let $p=\min (\delta,\frac{b-a}{3})$. Then $f$ is continuous on $I=[a+p,b-p]$. Hence $f$ is bounded on $I$. In addition, $f$ is bounded by $|f(a+p)|+|f(b-p)|+1$ on $(a-b)-I$. This means $f$ is bounded on $(a,b)$.