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Edit: According to comment of Andre Henriques we revise the question: In this question a fractal is a metric space whose topological and Hausdorff dimensions are different. So we would like that this possible counter example comes from a Riemannian structure not merely from a metric space view point.

Is there an example of an analytic manifold $M$ with two real analytic Riemanian metrics $g_1,g_2$ such that $M$ has a compact subset which is a fractal set with respect to $g_1$ but is not a fractal set with respect to $g_2$?

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  • $\begingroup$ Under any mathematically formal definition of "fractal", I would expect the answer to be "no". But please: do provide a definition of the word "fractal". Without a definition, it is not possible to answer the question, because the terms in the question are not well defined. $\endgroup$
    – André Henriques
    Commented May 18, 2019 at 19:27
  • $\begingroup$ @AndréHenriques Thanks for your attention to my question. A fractal is a set whose topological dimension is not equal to its Hausdorff dimension. $\endgroup$
    – Ali Taghavi
    Commented May 18, 2019 at 19:31
  • $\begingroup$ @AndréHenriques Now I revise the question. $\endgroup$
    – Ali Taghavi
    Commented May 18, 2019 at 20:06
  • $\begingroup$ @AndréHenriques Why do you expect the answer is "no"? $\endgroup$
    – Ali Taghavi
    Commented May 18, 2019 at 21:08
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    $\begingroup$ I expect the answer to be "no" because Hausdorff dimension is invariant under diffeomorphisms. $\endgroup$
    – André Henriques
    Commented May 18, 2019 at 22:05

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The answer is no. Any two Riemannian metrics when restricted to a compact set are bi-Lipschitz equivalent and bi-LIpschitz homeomorphism preserves the Hausdorff dimension.

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$
    – Ali Taghavi
    Commented May 19, 2019 at 21:02

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