I have a real Analysis question to ask: Let $X \subset \mathbb{R}$. Prove that every continuos function $ f: X \rightarrow \mathbb{R}$ is also uniform continuos if and only if $X$ satisfies these properties:
- $X$ is Closed.
- all limit points of $X$ are bounded. e.g $|p| \leq M$ for all limit points $p$ of $X$ and a real positive number $M$.
- There exists $ \delta >0$ such that for every $x,y \in X \backslash [-M , M ]$ with $x \neq y$ it follows that $|x-y| \geq \delta $
My attempt: I wanted to prove the $\Leftarrow$ first. so I suppose the 3 conditions are met. then I wanted to show that $f$ is uniform continuous. so that for every $\epsilon >0$ , there should exist a $ \delta > 0$ such that for every $x,y \in X$, if $|x-y| < \delta$ then $|f(x) - f(y)| < \epsilon$
So I tried to choose a $\delta$ which is mentioned in property 3. Then I had to check 3 cases :
- $x,y \in X \backslash [-M , M ]$ : obviously $|x-y| \geq \delta$ (as mentioned in property 3) so we don't check them anymore.
- $x,y \in [-M , M ]$ : I think I've proved this one correctly. because $[-M , M]$ is close and bounded, then it is compact. since $f$ is continuous , then $f$ is uniform continuous on $[ -M , M]$.
- either $x$ or $y$ lies in $X \backslash [-M , M]$ : I got stuck in this case and I don't know how to proceed.
Also I thought about the $\Rightarrow$ but I couldn't prove it neither.