A simple question from someone new to the field:


In a metric space, the [Hausdorff dimension](http://en.wikipedia.org/wiki/Hausdorff_dimension) of a subset is defined by covering the subset with $\epsilon$-balls and looking at how the number of required balls grows as a power or $\epsilon$ in the limit $\epsilon \to 0$.


My question is this: If my space is $\mathbb{R}^n$ and the metric $d\_p$ is the metric induced by the $p$-norm, $p \in (1, \infty]$, does the Hausdorff dimension of a subset $A \subseteq \mathbb{R}^n$ depend on the choice of $p$?


(My initial thought was that every $d\_p$-ball has the same $d\_q$-dimension for all $q$ (namely $n$), so that I believe all the dimensions should coincide, but I'm sure I'm overlooking something.)


Thanks!



Does the Hausdorff dimension depend on the L^p-norm?
Let $B\_p$ denote the 1-ball with centre 0 with respect to the
$l^p$ norm. For any $p$ and $q$ there is a number $N$ such that $B\_p$
is covered by $N$ translates of $B\_q$. Then any $\epsilon$-ball in
the $l^p$ norm is covered by $N$ $\epsilon$-balls in the $l^q$ norm.
Thus within a constant factor, the number of $\epsilon$-balls
required to cover a set in the $l^p$ and $l^q$ norms is the same.
This constant factor won't affect the asymptotic power in the
number of $\epsilon$-balls required to cover a given set,
so the Hausdorff dimension in both cases is the same.


