Let M ⊂ R, ξ ∈ M and f : M → C be continous. Let ξ be an accumulation point of M. Prove that f(ξ) is an accumulation point of f(M) or f is differntiable in ξ with f '(ξ) =0.
I really need help with this proof. I am struggeling for 3 hours on this proof but i didn't find something useful.
Thank you for your help.
If $f(\xi)$ is an accumulation point of $f(M)$, nothing is to show.
If $f(\xi)$ is not an accumulation point of $f(M)$, there is $\epsilon>0$ such that $f(M)\cap (f(\xi)-\epsilon,f(\xi)+\epsilon) = \{f(\xi)\}$. For each sequence $(x_n)_{n\in\mathbb N}$ in $M$ with $x_n\to\xi$, we have $f(x_n)\in f(M)\cap (f(\xi)-\epsilon,f(\xi)+\epsilon) = \{f(\xi)\}$ for sufficiently large $n$ due to continuity. Hence $f(x_n) = f(\xi)$ for sufficiently large $n$. This yields $$\lim_{x\to \xi}\frac{f(x)-f(\xi)}{x-\xi} = 0.$$
Let me remark that we are using a slightly unusual definition of differentiability here. Usually one demands in inner point to differentiate. Here, we only demand an accumulation point.