Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.