This came as a discussion of this question Local isometry and Hausdorff dimension . I have looked into this answer Hausdorff Dimension of a manifold of dimension n? but I understood it vaguely. I am thinking in this way; suppose $M$ is of dimension $n$ and $p\in M$. If $(U,\phi)$ be a local chart around $p$ and $\phi:U\to\mathbb{R}^m$ local chart, cover $M$ be countably many this $U_i$ and using Lindelof property extract a countable subcover and then using countable stability of Hausdorff dimension can we say $dim_{H}(M)=dim_{H}(U_i)=dim_{H}(\mathbb{R}^m)=m$ or I am missing something crucial steps. The thing I am not sure about is whether local diffeomorphism preserve Hausdorff dimension which I used to claim that $dim_{H}(U_i)=dim_{H}(\mathbb{R}^m)$. In fact in the answer I mentioned above, it's written that diffeomorphisms are bi-Lipschitz, I am not sure how does this follow that diffeomorphisms are Bi-Lipschitz?
Please let me know if my thinking is correct or not. Or it's needed to proceed like as they did. In that case, it would be helpful if anyone could elaborate it bit more Mathematically. Also are there any references to make it precise mathematically? I had looked into Mattila's book but did't find this. Any insight would be very much helpful.
It suffices to show that local charts are locally bi-Lipschitz, as Hausdorff dimension is clearly preserved under bi-Lipschitz equivalence, and the Hausdorff dimension coincides with the topological dimension in $\mathbb R^n$.
This mostly follows immediately from continuity of the Riemannian metric, but we have to be a little careful about our choice of neighborhood, since path lengths being comparable by a given factor in some neighborhood need not force distances to be comparable by the same factor in the same neighborhood.
To see that charts are locally bi-Lipschitz, suppose $U\subseteq \mathbb R^n$ and $\phi\colon U\to M$ is a local chart for $M$. Then in any ball $B=B(x,\epsilon)$ with $\overline{B}\subseteq U$, there is by continuity of the Riemannian metric some constant $C$ such that for $y\in \overline{B}$ and $v\in T_y(\mathbb R^n)$ we have $$\frac{1}{C}\|v\| \leq \|\phi_*v\|\leq C \|v\|,$$ from which it follows that every smooth path $\gamma\colon I\to \overline{B}$ satisfies
$$\frac{1}{C}\ell(\gamma)\leq \ell(\phi\circ\gamma) \leq C\ell(\gamma).$$
Now suppose $p,q\in \frac{1}{2} B:=B\left(x,\frac{\epsilon}{2}\right)$.
On the one hand, if $\gamma$ is a parametrized segment between $p$ and $q$, we have $$d(\phi(p),\phi(q))\leq \ell(\phi\circ\gamma) \leq C \ell(\gamma) = Cd(p,q).$$
On the other hand, suppose $\gamma\colon I\to M$ joins $\phi(p)$ and $\phi(q)$. Then either $\gamma$ lies completely in $\phi(B)$, so that $$d(p,q)\leq \ell(\phi^{-1}\circ\gamma)\leq C\ell(\gamma),$$ or $\gamma$ leaves $\phi(B)$, and therefore contains two subcurves $\gamma_1,\gamma_2\colon I\to \phi(\overline{B})$ each joining $\partial\phi(B)$ to $\partial\phi\left(\frac{1}{2}B\right)$, whereby $$d(p,q)\leq \epsilon \leq \ell(\phi^{-1}\circ \gamma_1)+\ell(\phi^{-1}\circ\gamma_2)\leq C(\ell(\gamma_1)+\ell(\gamma_2))\leq C\ell(\gamma).$$
Since $d(p,q)\leq C\ell(\gamma)$ for all smooth paths joining $\phi(p)$ to $\phi(q)$, we have $d(p,q)\leq Cd(\phi(p),\phi(q))$.