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I was searching for a metric space that has infinite Hausdorff dimenion . I stumbled upon the example of $\mathbb{R}$ with discrete metric. $\mathbb{R}$ should then have infinite dimension but I cannot understand why.

In the answer to this question it is stated that "If your discrete metric space is countable, its Hausdorff dimension is also 0; if it’s uncountable, its Hausdorff dimension is $\infty$"

If you consider a covering of a set $A \subset \bigcup A_k$ in a discrete metric space, where all covering sets have diameter smaller than 1 ($diam(A_k) < 1$), the only possible covering is the covering where the covering sets only contain one element. $A_k= \{a\}$ The diameter of a set containing one point is obviously zero. $diam(A_k) =0$

Now if I consider such a covering of $\mathbb{Q}$, the sum of the diameters of the covering sets to the power of $s < \infty$ is zero. $\sum diam(A_k)^s =0$. Therefore the Hausdorff dimension would be zero as well. I would assume the same for a covering of $\mathbb{R}$. Does it have infinite Hausdorff dimension because the covering would be uncountable? Or is every set that does not have an countable $\delta$-cover of infinite Hausdorff dimension? If so can some one explain to me why?

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For the definition of Hausdorff measure, you may use only countable covers. Uncountable covers are not allowed.

Let $A$ be an uncountable set in your discrete space. So if $0 < \delta < 1$, there is no cover of $A$ by sets of diameter ${} < \delta$. Then for any $s \in [0,+\infty)$, $$ \mathcal H_\delta^s(A) = \inf \varnothing = +\infty . $$

The infimum (greatest lower bound) of the empty set is $+\infty$ because every real number is a lower bound of $\varnothing$.

Thus the $s$-dimensional Hausdorff measure is $$ \mathcal H^s(A) = \lim_{\delta \to 0} H_\delta^s(A) = +\infty . $$ This is true for any $s \in (0,+\infty)$, so the Hausdorff dimension is $$ \dim A = \sup\{s : \mathcal H^s(A) > 0\} = +\infty. $$

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  • $\begingroup$ Very true. In principle, when defining the Hausdorff dimension, one could have allowed uncountable covers and used sums over uncountable index sets, but this is not how it is usually done. $\endgroup$ – Moishe Kohan Feb 11 at 15:44
  • $\begingroup$ You can allow uncountable covers by balls $B_r(x) = \{y : d(x,y) < r\}$ where $r>0$. And use sums $\sum_i r_i^s$. Uncountable sums of positive numbers always give you $+\infty$. If you allow sets of diameter $0$ in your covers, and sum $\sum_i(\mathrm{diam}\; A_i)^s$, then every set in every metric space has an uncountable cover by sets of diameter $0$. Not much use. $\endgroup$ – GEdgar Feb 11 at 15:48
  • $\begingroup$ What I had in mind was using covers by open sets of diameter $\le \delta$. However, the only novelty of this definition is that every space with discrete metric will have measure $0$, while for separable metric spaces, this would be the usual definition. $\endgroup$ – Moishe Kohan Feb 11 at 16:04

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