Let $(X,d)$ be a separable metric space and $A\subseteq X$ be compact.

Since every finite set is doubling then, the collection $\mathcal{A}$ of doubling subsets of $A$ cannot be empty. My initial question was, how large can the following be $$ \sup_{\tilde{A}\in \mathcal{A}}\,d(A,\tilde{A}) ? $$ Where $d(A,\tilde{A})$ denotes the Hausdorff distance between $A$ and $\tilde{A}$.

However, I have a feeling that the above quantity is $0$ if $A$ is not doubling by the separability of $X$.

So then, let me refine my question: For any $\tilde{A}\subseteq A$ and any $r>0$ let $N(\tilde{A},r)$ denote the smallest number of balls of radius $r$ required to cover $\tilde{A}$?

For any $C,d>0$ how large can the following be: $$ \begin{aligned} \sup_{\tilde{A}\in \mathcal{A}}& \,d(A,\tilde{A})\\ \mbox{s.t.}\, & N(\tilde{A},r) \leq C(|A|/r)^d \mbox{for all $0<r\leq |\tilde{A}|$} \end{aligned} $$

Alternatively, is there a way to quantify how far a metric space is from being doubling?

  • $\begingroup$ Does $d(A, \bar{A})$ denote Hausdorff distance, or something else? $\endgroup$ May 11, 2022 at 16:57
  • $\begingroup$ @NateEldredge Yes indeed, I incorporated the clarification. $\endgroup$ May 16, 2022 at 7:32


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