Hausdorff dimension

Remarks

If $\mathcal{H}^s(A) < \infty$ and $t > s$, then $\mathcal{H}^t(A) = 0$. So $\mathcal{H}^s(A) = \infty$ for $s < \dim_H A$ and $\mathcal{H}^s(A) = 0$ for $s > \dim_H A$. At $s = \dim_H A$ the value can be $0$, positive and finite, or $\infty$. Equivalently, $\dim_H A = \sup \set{s \ge 0 : \mathcal{H}^s(A) = \infty}$, with $\sup \emptyset = 0$.

Example: the middle-thirds Cantor set $C$ has $\dim_H C = \log 2 / \log 3$. At the $k$-th step, $C$ is covered by $2^k$ intervals of length $3^{-k}$, so $\mathcal{H}^s_{3^{-k}}(C) \le 2^k 3^{-ks}$, which equals $1$ for $s = \log 2 / \log 3$. This gives $\dim_H C \le \log 2 / \log 3$; the matching lower bound needs a separate argument.

A dense set can have dimension $0$: $\Q$ is dense in $\R$, yet $\dim_H \Q = 0$. Each point has $\mathcal{H}^s$ measure $0$ for $s > 0$, and $\mathcal{H}^s$ is countably subadditive.

For subsets of $\R^n$, one has $\dim_H A \le n$, and every nonempty open set has dimension $n$.