Give an example of a continuous function $f: \mathbb R\to \mathbb R$ which is not uniformly continuous on $(1,3)$. I'm not looking for someone to do this for me or anything but I feel completely stuck. I can't even find a jumping off point.
I understand what qualifies a function as continuous and uniformly continuous. What I'm having trouble with (and I feel silly for admitting it) is being able to product a function that is not uniformly continuous at the given interval, (1,3).
I've done the internet searches but I'm having trouble relating what I'm finding to my given interval. I'm sorry if that doesn't narrow it down enough, but feeling stuck from the very start doesn't give me a whole lot to shave off.
Best Answer
Someone has given you an impossible task. If a function $f$ with domain $\mathbb{R}$ is continuous, then $f$ is continuous on the compact interval $[1,3]$, and a function that is continuous on $[1,3]$ is uniformly continuous on $[1,3]$, hence it is uniformly continuous on any subset of $[1,3]$, such as $(1,3)$.
The fact that if $f$ is continuous on $[1,3]$, then it must be uniformly continuous on $[1,3]$, is not too hard to prove. One way to prove it is by contradiction. Assume $f$ is continuous but not uniformly continuous on $[1,3]$, use the definition of uniform continuity and the compactness of $[1,3]$, and eventually you will get a contradiction.
Undoubtedly this is proven in scores of real analysis textbooks, but I don't know of one offhand, and I don't know if proof by contradiction is the most popular or the best way to do it.