>differentiable everywhere and continuous nowhere>either the concept of differentiable is bullshit>or the density of irrationals in rationals (and vice versa) is bullshit>effectively reals are bullshit Pick your poison.
>>16928918>differentiable everywhereUh, no. It's literally an easy exercise in calc 1 to show that differentiability implies continuity.
It is nowhere differentiable. I am going to be charitable and assume you meant to say it's not Riemann integrable.
>>16928918>differentiable everywhere and continuous nowherethis doesn't even need a proof to disprove because you're just violating a definition
>>16928995That differentiable functions are continuous is a theorem, not a premise.Let [math] \left(\mathbf F,+,\times,\tau_{\bf F}\right) [/math] be a topological field, [math] \left(X,+,\cdot,\tau\right) [/math] be a TVS over it, [math]D[/math] be a subset of [math]\bf F[/math], [math]x[/math] be a point in [math]D[/math] and [math] f:D\longrightarrow X[/math] be have a derivative [math] \delta [/math] at [math]x[/math] (in a non-Hausdorff, derivatives aren't unique).Let [math]W[/math] be a neighborhood of [math]\bf 0[/math]. Choose [math] V\in\mathcal N\left(\mathbf 0\right) [/math] such that [math] V+V\subseteq W [/math]. By boundedness of singletons, there exists [math] U_1\in\mathcal N\left(0\right) [/math] such that [math] U_1\delta \subseteq V[/math]. There also exist [math] U_2 \in \mathcal N\left(0\right) [/math] and [math] U_\delta \in \mathcal N\left(\mathbf 0\right) [/math] such that [math] U_2U_\delta \subseteq V[/math].By differentiability, there exists [math] U_2\in\mathcal N\left(0\right) [/math] such that[eqn] \forall y\in D \cap \left(x + U_2\right), \frac{f\left(y\right) - f\left(x\right)}{y - x}\in \delta + U_\delta.[/eqn]Put [math] U_x=U_1\cap U_2 [/math]. Then for all [math] y \in D \cap \left(x + U_x\right) \setminus\left\{x\right\} [/math],[eqn] f\left(y\right) - f\left(x\right) = \left(y - x\right) \frac{f\left(y\right) - f\left(x\right)}{y - x} \in U_x \left(\delta + U_\delta\right) \subseteq U_x\delta + U_xU_\delta \subseteq U_1\delta + U_2U_\delta \subseteq V+V \subseteq W. [/eqn]And [math] f\left(x\right) \in f\left(x\right) + W[/math]. It follows that [math] \exists U\in\mathcal N\left(x\right), f\left(U\right) \subseteq f\left(x\right) + W [/math]. Since this holds for all such [math]W[/math], we conclude that [math]f[/math] is continuous at [math]x[/math].
