TITLE: Hausdorff Measure and Hausdorff Dimension
QUESTION [9 upvotes]: Could someone explain the intuition behund the Hausdorff Measure and Hausdorff Dimension?
The Hausdorff Measure is defined as the following:   
Let $(X,d)$ be a metric space. $\forall S \subset X$, let $\operatorname{diam} U$ denote the diameter, that is 
$$\operatorname{diam} U = \sup \{ \rho(x,y) : x,y \in U \} \,\,\,\,\, \operatorname{diam} \emptyset = 0 $$
Let $S$ be any subset of $X$ and $\delta > 0 $ a real number. Define
$$ H^{d}_{\delta} (S) := \inf \{ \sum_{i=1}^{\infty} (\operatorname{diam} U_i )^d : \bigcup_{i=1}^{\infty} U_i \supseteq S, \operatorname{diam} U_i < \delta \}$$
What does $\rho(x,y)$ denote? What is meant by the diameter of a set? I'm just trying to understand the intuition behind this definition.
The Hausdorff Dimension is defined as the following:    
Let $X$  be a metric space. If $S \subset X$ and $d \in \mathbb{R}^+$, the Hausdorff content is defined as 
$$C^{d}_H (S) := \inf \{ \sum_{i} r_i^d : \, \, \text{there is a cover of $S$ by balls of radii} \, \, r_i > 0 \} $$
Then the Hausdorff Dimension is defined as
$$\operatorname{dim}_{H} (X) := \inf \{d \geq 0 : C^d_H = 0\}$$
Could someone explain the intuition behind these definitions?

REPLY [3 votes]: Let me try to explain, the way I understand, the intuition behind the definition of Hausdorff dimension.  As user103254 said, $\rho =d$.  Now let us look at some examples.  
As always, intuition is derived from $\mathbb{R}$, $\mathbb{R}^2$ and $\mathbb{R}^3$.  For simplicity, I would use cubes of sides $r_i$ rather than balls.  A cube in $\mathbb{R}$ is an interval, a cube in $\mathbb{R}^2$ is a square, a cube in $\mathbb{R}^3$ is the usual cube and so on.  Also, I would assume that all cubes are of same side-length.  Thus the sum of the radii is nothing but the number of cubes multiplied by the side-length.  First look at the interval [0,1].  If we want to cover it with intervals of size $\frac{1}{n}$ the best we can hope to do is to cover it with $n$ closed intervals of length $\frac{1}{n}$.  Similarly you need $n^2$ squares of sides $\frac{1}{n}$ to cover $[0,1]\times [0,1]$, $n^3$ cubes of sides $\frac{1}{n}$ to cover $[0,1]\times [0,1]\times [0,1]$ and so on.  This can also be seen from the fact that the volume of a $d$-dimensional cube of side $\frac{1}{n}$ is intuitively $\frac{1}{n^d}$ where as the volume of $[0,1]^d$ is $1$.  If the size of the box is fixed, then higher the dimension the more boxes you would require to cover.
Thus, the sum of radii in the definition of $d$-dimensional content of $[0,1]^k$ in this special case will look like $n^k \times \frac{1}{n^d} = n^{k-d}$.  As $n$ tends to infinity, $n^{d-k}$ tend to zero if $d \geq k$ and infinity otherwise.  So naively the dimension of $[0,1]^k$ is $k$.  I say naively because while calculating the infinum we could have taken covers of different radii.  Of course we did not even take balls we took cubes.  But I guess the idea is clear.  
This I think is the basic intuition.  Hope it was helpful.