All: I know any two Cantor sets; "fat" , and "Standard"(middle-third) are homeomorphic to each other. Still, are they diffeomorphic to each other? I think yes, since they are both $0$-dimensional manifolds (###), and any two $0$-dimensional manifolds are diffeomorphic to each other. Still, I saw an argument somewhere where the claim is that the two are not diffeomorphic.
The argument is along the lines that, for $C$ the characteristic function of the standard Cantor set integrates to $0$ , since $C$ has (Lebesgue) measure zero, but , if $g$ where a diffeomorphism into a fat Cantor set C', then: $ f(g(x))$ is the indicator function for C', so its integral is positive.
And (my apologies, I don't remember the Tex for integral and I don't have enough points to look at someone else's edit ; if someone could please let me know )
By the chain rule, the change-of-variable \int_0^1 f(g(x))g'(x)dx should equal $\int_a^b f(x)dx$ but g'(x)>0 and $f(g(x))>0$ . So the change-of-variable is contradicted by the assumption of the existence of the diffeomorphism $g$ between $C$ and C'.
Is this right?
(###)EDIT: I realized after posting --simultaneously with "Lost in Math"* , that the Cantor sets {C} are not 0-dimensional manifolds (for one thing, C has no isolated points). The problem then becomes, as someone posted in the comments, one of deciding if there is a differentiable map $f:[0,1]\rightarrow [0,1]$ taking C to C' with a differentiable inverse.
- I mean, who isn't, right?