# The relation between Hausdorff dimension of an $n$-manifold and $n$

It is known that for a topological space with different metrics, the Hausdorff dimensions may not be equal in general.

For the case of manifolds, suppose $M$ is a $n$-manifold with a metric(distance), from https://math.stackexchange.com/questions/931628/hausdorf-dimension-of-a-manifold-of-dimension-n, we know that $dim_H M$ and $n$ may not be the same. But whether we can get a relation between $\dim_H M$ and $n$? I mean, whether we can prove that $\dim_H M\leq n$ or $\dim_H M\geq n$?

Unfortunately, I have no idea how we can get it directly from the definition of Hausdorff dimension.

• If you look for a proof of the fact that for a metric space, the Hausdorff dimension is always greater than or equal to the topological dimension, there is a nice book by Juha Heinonen, Lecture notes on analysis on metric spaces 2001. – Changyu Guo Jan 1 '15 at 17:35
• @ChangyuGuo Thank you for the book you recommend. The proof in that book is very clear. However, the definition of topological dimension there is defined for separable metrizable topological spaces. But it is enough for me. – Lewis Zhang Jan 5 '15 at 11:43

In a metrizable topological space, Hausdorff dimension is always larger or equal than the topological (covering) dimension. See Theorem 6.3.10 in Edgar's book "Measure, topology and fractal geometry". In particular, for an $n$-dimensional manifold $M$, if $\rho$ is any metric compatible with the Euclidean topology, then $(M,\rho)$ has Hausdorff dimension at least $n$.
Finally, for any (at least separable) metric space $X$, the topological dimension equals the infimum (which is in fact a minimum) of the Hausdorff dimensions of $(X,\rho)$ where $\rho$ varies among the metrics compatible with the topology of $X$. This is a classical result due to Edward Marczewski.