# Hausdorff dimension of a smooth manifold

I read a book about fractal stating that without proof: every $m$-dimension $(m<n)$ smooth manifold $M$ in $\mathbb{R}^n$ has Hausdorff dimension $m$. How can we prove it?

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Let $d < m < e$. It suffices to show $0 = \mathcal{H}^d(U) < \mathcal{H}^e(U) = \infty$ for one smooth chart $(U,\phi)$ around each point $p \in M$, since $M$ can be covered with countably many of these ($M$ is Lindelöf!). But for each point we can find some small neighborhood $U$ which can be written as graph of a smooth function having its domain in an $m$-dimensional affine subspace of $\mathbb{R}^n$ and mapping into the orthogonal complement of that subspace. But such graphs have finite $m$-dimensional Hausdorff-measure as Lipschitz-transformations of bounded subsets of $m$-dimensional affine spaces.