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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?
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    By definition a zero-dimensional manifold is locally homeomorphic to the space consisting of a single point. Thus a zero-dimensional manifold consists of isolated points. In the Cantor set (fat or standard) there is a nonisolated point so it's not a zero-dimensional manifold.2011-10-19
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    I just thought that none of the Cantor sets are manifolds, so it may not make sense to talk about diffeomorphisms between them.2011-10-19
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    Is there then, a definition of diffeomorphism that makes sense between non-manifolds?2011-10-19
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    There is. You can define a smooth function on any set C in R^n as a function that extends to a smooth one on some open neighbourhood U containing C. Then a diffemorphism between your sets is just a smooth function with a smooth inverse. Also, I believe the answer to be no, because your smooth function would take a zero-measure subset to a non-zero one.2011-10-19
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    \int_a^b f(x)\, dx is the TeX code for $\int_a^b f(x)\, dx $2011-10-19
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    Piotr: I think absolutely continuous functions preserve measure, but I don't know if this is true for smooth ones, and is definitely not true for continuous alone, with the function mapping the middle-third Cantor set into [0,1].2011-10-19
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    Jim: thanks; unfortunately, you took away my excuse for not editing, and now I am just lazy :).2011-10-19
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    @Gary: smooth functions are continuously differentiable, and hence (locally) Lipschitz-continuous, and hence absolutely continuous.2011-10-19
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    @Willie: I may have mixed up my terms; diffeomorphism here originally only meant that there is a differentiable map with a differentiable inverse, so I think the issue is more difficult. I think if we have at least $C^1$, then, by compactness, f would be Lipschitz, and , as you said, absolutely continuous. But I don't think differentiable with differentiable inverse is enough to conclude a.c.2011-10-26
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    But I maybe wrong....2011-10-26
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    @Gary: when people say smooth (with no other qualifiers), usually $C^\infty$ is meant. My comment was in response to what you wrote on Oct 19 at 19:22. If you actually meant only differentiable but not necessarily continuously differentiable, then you are right that the argument outlined in my comment won't apply.2011-10-29

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