Characterization of nowhere differentiable functions Let $N:=\{f\in C([0,1])\vert \text{ f is nowhere differentiable } \}$
and 
$A_n = \{f\in C([0,1]) \vert \exists x\in [0,1]s.t. \forall y\in[0,1]: |f(x)-f(y)|\leq n |x-y|\}$.
Now I have already shown that $\bigcap_{n\in\mathbb{N}}A_n^c \subset N$.
However, I am not sure whether $N=\bigcap_{N\in\mathbb{N}}A_n^c$.
I am thinking that this is not true, because for example the function $f(x)=|x|$ has a bounded slope but is not differentiable at 0.
Can anyone give me an example of a function which is in $N$ but not in $\bigcap_{N\in\mathbb{N}}A_n^c$.
Cheers
 A: First, you probably want $|f(x)-f(y)|\geq n |x-y|$ rather than $|f(x)-f(y)|\leq n |x-y|.$
Second, you're not going to get any reasonable description of the set $N$ in terms of unions and intersections (even iterated transfinitely, any countable ordinal number of times) of "nice sets" because R. Daniel Mauldin showed in 1979 that $N$ is not a Borel subset of the space $C[0,1].$ See his paper The set of continuous nowhere differentiable functions.
Third, the inclusion you have shown is sufficient for what I suspect you want to prove. Specifically, if most continuous functions (Baire category sense) belong to the intersection you are dealing with, then certainly this will be the case of the larger set of continuous and nowhere finitely differentiable functions.
Fourth, the way you want to proceed is to state precisely what you have shown. The fact that it's a stronger result than what you were probably seeking is not a defect.
Finally, for a lot of the lore on this topic see my answer at Generic Elements of a Set.
