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It's very well known the embedding theorem for smooth manifolds. I want to know if is there any equivalent result for Lipschitz manifold and $C^1$ manifolds. In particular, I want that the embedding allows us to see the manifold as a closed submanifold of the Euclidean space.

In the $C^\infty$ case, the result says that there exist $f : M \to \mathbb{R}^{k}$ (for sufficiently large $k$) such that $f$ is a smooth embedding. In fact, $f(M)$ is a closed smooth submanifold of the ambient space $\mathbb{R}^{k}$. Furthermore, $M$ is therefore metrizable and each chart constitutes the metric locally and is Bilipschitz homeomorphism (are local diffeomorphism), so the Hausdorff dimension is equal to $\dim M$. If we have an equivalent result for Lipschitz or $C^1$ manifold, we can ensure the same conclusion.

If someone has a different way to attack the problem of the Hausdorff dimension of a $C^1$ or Lipschitz manifold, please comment on it.

JulianDoyle
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  • If I understand the question correctly, for the $C^1$ case, there is the celebrated result stating that, for all $k\ge1$, every $C^k$ manifold is $C^k$-diffeomorphic to some $C^\infty$ manifold and, moreover, $C^k$-diffeomorphic manifolds are $C^k$-diffeomorphic to $C^\infty$-diffeomorphic $C^\infty$ manifolds. So for $C^1$ manifolds you can transform them into $C^\infty$ manifolds and then apply Whitney. –  Oct 21 '20 at 07:54

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