The Hausdorff dimension $\dim(A)$ of $A \subseteq \mathbb{R}^d$ is the unique number $\alpha$ such that the $\beta$-dimensional Hausdorff measure $H_\beta(A)=0$ for $\beta>\alpha$ and $H_\beta(A)=\infty$ for $\beta<\alpha$.
What is $\dim(\mathbb{R}^n)$?
For every $\delta,\beta>0$ and every (countable) cover $\mathfrak{A}$ of $\mathbb{R}^n$ where $\forall A\in\mathfrak{A}. \text{diam}(A)>\delta$ we get $\sum\limits_{A\in\mathfrak{A}}\text{diam}(A)^\beta=\infty$ which would imply $\dim(\mathbb{R}^n)=0$ while Wikipedia states $\dim(\mathbb{R}^n)=n$.
So where did i go wrong?
Thanks, Takirion
I'm not an expert on this but I think I figured it out for $\mathbb R^1$ at least. Consider Covering $\mathbb R^1$ with balls (intervals) $I_n$ of length $\frac{1}{n}$. We know this is possible because $\sum_{n=1}^\infty\frac{1}{n}$ diverges so the intervals can cover the whole line. Now, by the $p$-series test or the integral test, $\sum_{n=1}^\infty \text{diam}(I_n)^\beta=\sum_{n=1}^\infty\frac{1}{n^\beta}$ converges for all $\beta\leq1$ and diverges for $\beta\geq 1$.
Can you finish the argument from here? And generalize it to $n$ using boxes of volume $1/n$ instead?