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?
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 $s \in [0,+\infty)$, $$ \mathcal H_\delta^s(A) = \inf \varnothing = +\infty . $$
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 $$ \dim A = \sup\{s : \mathcal H^s(A) > 0\} = +\infty. $$