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?)
  • 1
    $\begingroup$ The Hilbert cube has infinite Hausdorff dimension. $\endgroup$
    – Lee Mosher
    Jul 13, 2019 at 20:25

1 Answer 1


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.


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