Here is the question:
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be any function. Show that $f$ is continuous at $x$ if and only if for every $n\in \mathbb{N}$ there exists $\delta >0$ so that $|f(a)-f(b)|<\frac{1}{n}$ for all $a,b \in (x-\delta,x+\delta)$.
Here are my thoughts:
-->: Suppose $f$ is continuous at $x$. Then for all positive $\epsilon$ there is a positive $\delta$ so that if $|x-x_{0}|<\delta$ then $|f(x)-f(x_{0})|<\epsilon$. Okay so I know that $a$ and $b$ are within $\delta$ of $x$ and this means that $|f(x)-f(a)|< \epsilon$ and $|f(x)-f(b)|< \epsilon$. I don't know where to go from here. In particular, I am not sure where the $|f(a)-f(b)|<\frac{1}{n}$ is coming from. I know that $a,b$ are in the same neighborhood of $x$, but that doesn't necessarily mean that $f$ is continuous at $a$, right? Any guidance would be appreciated. Thank you.