# Infinite Hausdorff dimension in discrete metric spaces

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?

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 $$\dim A = \sup\{s : \mathcal H^s(A) > 0\} = +\infty.$$
• 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. – GEdgar Feb 11 at 15:48
• 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. – Moishe Kohan Feb 11 at 16:04