Hausdorff dimension of unit closed ball of $ \mathbb{R}^n$ is $n$ Let's admit as definition of Hausdorff dimension the one given by wikipedia
https://en.wikipedia.org/wiki/Hausdorff_dimension
How to demonstrate dimension of unit closed ball of $\mathbb{R}^n$ is $n$ ?
I've seen in the case of a classical metric space, the dimension of vectorial space is the same as Hausdorff dimension , however , i don't know if i can use any sub-property to treat the case of the unit closed ball.
 A: In this case it is particularly easy, if we use the well-known fact that the Hausdorff top-dimension measure equals the Lebesgue measure. Hence you can show the Lebesgue measure of the $n-$dimensional ball is finite and different than zero, so that to conclude the Hausdorff dimension of the $n-$ball is $n$.
As always, to show a set has a particular (not necessarily integer) Hausdorff dimension, the usual strategy is to divide the proof into two steps: the first one consists in finding one particular cover such that the $n-$dimensional Hausdorff measure is finite. Doing this, we can show the Hausdorff dimension is bounded from above by $n$.
To bound from below, it is usually more difficult, and one needs to either choose a smart cover and show that it is the best one (you have to do this by the definition of Hausdorff dimension, that involves an infimum) or to find a different trick to solve this issue.
A: The inequality $\dim_H \le n$ is easy, covering the ball by cubes of side $1/k$.  We can do it with at most $2k^n$ cubes with side $1/k$ and diameter $2^{1/n} / k$.
For the inequality $\dim_H \ge n$ we can use Lebesgue $n$-dimensional measure (as in the answer of Son Gohan).  Fix a diameter $\epsilon$, any set with diameter ${} \le \epsilon$ is contained in a ball of radius $\epsilon$, so has measure ${} \le C \epsilon^n$ for some constant $C$ (= Lebesgue measure of the unit ball).  If we cover the ball by sets of diameter ${}\le \epsilon$, the sum of their measures is at least $C$.  So the sum of the $n$th power of the diameters of the cover is at least $1$; limit as $\epsilon \to 0$ is ${} \ge 1$.
Another interesting way to do the inequality $\dim H \ge n$. Recall the inequality $\dim_H \ge \dim_\mathrm{ind}$.  If we already know that the unit ball in $\mathbb R^n$ has topological dimension $n$, we can use this to show $\dim_H \ge n$.
