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A simple question from someone new to the field:

In a metric space, the 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!

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    $\begingroup$ Hausdorff dimension is a bi-Lipschitz invariant, and all norms of $\mathbb{r}^n$ are equivalent, so the Hausdorff dimension does not depend upon the norm. $\endgroup$
    – Benoît Kloeckner
    Commented Nov 3, 2010 at 11:30

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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.

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  • $\begingroup$ It is better to say that $(A,\ell^p)$ is bi-Lipschitz equivalent to $(A,\ell^q)$ and therefore their Hausdorff dimensions coincide. $\endgroup$
    – Anton Petrunin
    Commented Nov 3, 2010 at 18:25

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