Lets say I need to provide a function that is bounded and continuous on the interval $(1,2)$ but not uniformly continuous on this interval and I have to prove it.
I know the distinction that continuity is defined at a point $a$, whereas uniform continuity is defined on the entire interval. But how do I visualise this? In wikipedia's example, it says something about fitting a point inside a rectangle. Does not-uniformly continuous function revolves around the idea of steepness?
If a function is very steep the $\delta$ cannot catch up to $\epsilon$. So in here, can I say the function f(x) is $\frac{1}{x-1}$? So something like:
This is still count as continuous because $x$ is defined on an open bracket $(1, 2)$ right? and It is not uniformly continuous because of the steepness in the interval?
Can someone verify this? Is this correct? If so how do I prove that it is not uniformly continuous? I am planning to assume it is uniformly continuous and work from the definition. Can someone give me a hint on how to prove this?
Supose you have $f:(a,b) \longrightarrow \mathbb{R}$ that is uniformly continuous and two sequences $(x_n)_n,(y_n)_n \subseteq (a,b)$ such that $|x_n-y_n| \overset{n}{\rightarrow} 0$. Then it's easy to verify that $|f(x_n)-f(y_n)|\overset{n}{\rightarrow} 0$.
So if we take $f:(1,2) \longrightarrow \mathbb{R}$, $f(x)=sin(\frac{1}{x-1})$ and find two sequences $(x_n)_n,(y_n)_n \subseteq (1,2)$ such that $|f(x_n)-f(y_n)|$ doesn't approach $0$ then $f$ can't posibly be uniformly continuous. I think this is one of the easier ways to show what you want.
From that, take $x_n, y_n$ such that $\frac{1}{x_n-1}=\frac{\pi}{2} + 2\pi n,\; \frac{1}{y_n-1}=-\frac{\pi}{2}+2\pi n$.
You will find that $x_n,y_n \rightarrow 1$ and therefore $|x_n-y_n|\rightarrow 0$ but $|f(x_n)-f(y_n)|=2$. So $f$ cannot be uniformly continuous.