# Hausdorff dimension of compact set

Let $$(X, d)$$ be a metric space and $$S \subset X$$. Let $$\DeclareMathOperator{\diam}{diam}\diam(S)$$ denote diameter of $$S$$, that is $$\diam(S) = \sup \{ d(x, y) \colon \: x, y \in S \}$$. Let $$\delta > 0$$, $$s \ge 0$$ and define $$H_{\delta}^s(S) = \inf \left\{ \sum_{i=1}^{\infty}(\diam \, U_i)^s \colon \enspace S \subset \bigcup_{i=1}^{\infty} U_i \: \land \: \diam U_i < \delta \right\}.$$

Now, let $$H^s(S) = \lim_{\delta \to 0^+} H^s_{\delta}(S)$$. Then $$H^s$$ is a outer (Hausdorff) measure. (Hausdorff) dimension of set $$S$$ can be defined by $$\dim_{\mathrm{Haus}}(S) = \inf \{ s \ge 0 \colon \: H^s(S) = 0 \}$$.

My questions are:

1. If $$S$$ is a compact subset of $$X$$, is it possible that $$\dim_{Haus}(S) = \infty$$?
2. If in 1. it is possible, then, is there a way to define what would the (outer) measure of such set be? (Is there a way of generalizing outer Hausdorff measure to infinite-dimensional sets?)
• The Hilbert cube has infinite Hausdorff dimension. Jul 13, 2019 at 20:25

Yes to both questions. For 1., observe that $$[0,1]^{\mathbb N}$$ is compact by Tychonoff's theorem and is infinite dimensional. For 2., this can be achieved by using the generalization of Hausdorff measure described at the wikipedia page, where instead of $$(\textrm{diam} U_i)^s$$ we use $$\phi(\textrm{diam} U_i)$$ for an appropriate gauge function $$\phi$$. When $$\phi$$ grows superpolynomially, you can get non-trivial Hausdorff measures defined for infinite dimensional metric spaces.