# Smallest doubling subset of a set in a metric space

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

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

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