The standard proofs can be found here:
Every function $f: \mathbb{N} \to \mathbb{R}$ is continuous?
But I want to see how this could be proved using directly the definition of continuity of real functions:
$ \forall x,y \;\forall \epsilon > 0 \; \exists \delta > 0 $ such that $ |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$
I tried $\delta$ = 1 and $\delta=\frac{\epsilon}{2}$ but didn't work for me.
Thanks.
Best Answer
Actually, $\delta=1$ works. For natural numbers, $|x-y|<1$ means that $x=y$ and $f(x) = f(y)$, so $|f(x)-f(y)|=0<\epsilon$.