For separable metric spaces, three fundamental notions of dimension are equivalent: $$ \text{dim }X = \text{Ind }X = \text{ind }X ,$$
Where does the doubling dimension fit into the picture?
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2$\begingroup$ First the first fact is topological (rather to state in term of metrizable spaces). Doubling dimension is a metric property (bilipschitz invariant, not topological). Now the question is certainly too broad: every possible rant about doubling dimension seems to answer the question. $\endgroup$– YCorCommented Apr 8, 2018 at 12:08
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$\begingroup$ So there is no relation between the doubling dimension ddim(X) and the toplogical dimension dim(X)? One can be finite and the other infinite? $\endgroup$– Aryeh KontorovichCommented Apr 8, 2018 at 12:57
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3$\begingroup$ It sounds obvious to construct metric spaces with finite (covering/inductive) dimension and infinite doubling dimension. In the other direction, finite doubling dimension implies finite dimension if I'm correct (e.g., by Assouad's embedding), but I don't know about explicit bounds. $\endgroup$– YCorCommented Apr 8, 2018 at 13:24
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$\begingroup$ I'll be happy to accept the obvious answer -- for both directions. $\endgroup$– Aryeh KontorovichCommented Apr 8, 2018 at 13:32
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1$\begingroup$ It may happen that the topological dimension is equal to the infimum of the doubling dimensions over all metrics generating the topology of the space. $\endgroup$– Taras BanakhCommented Apr 9, 2018 at 7:37
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2 Answers
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In one direction, a rapidly branching tree will have very high doubling dimension, while having topological dimension $0$ (or $1$, if you include the edges). In another direction there is a bound, and this is discussed in the nice paper below (on the first page):
Le Donne, Enrico; Rajala, Tapio, Assouad dimension, Nagata dimension, and uniformly close metric tangents, Indiana Univ. Math. J. 64, No. 1, 21-54 (2015). ZBL1321.54059.
answered Apr 8, 2018 at 16:53
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Let ${\rm dim}_H X$ and ${\rm dim}_d X$ denote the Hausdorff and the doubling dimension respectively. It is easy so see that ${\rm dim}_H X\leq {\rm dim}_d X$. Indeed, if ${\rm dim}_d X=s$, then we can cover a ball of radius $r$ by at most $2^{s}$ balls of radius $r/2.$ Therefore we can cover a ball of radius $1$ by $2^{ks}$ balls of radius $2^{-k}$ so we can estimate the Hausdorff measure $H^s(B(x,1))\leq C 2^{ks}2^{-ks}= C<\infty$ and hence ${\rm dim}_H X\leq s={\rm dim}_d X$. On the other hand by Theorem 8.14 in [1] (the proof is very short) we have that the Hausdorff dimension is greater than equal to the topological dimension so finally we get $$ \text{dim }X = \text{Ind }X = \text{ind }X\leq \text{dim}_H X\leq \text{dim}_d X. $$ As pointed by Igor Rivin the Hausdorff dimension of a rapidly branching tree can be arbitrarily large.
[1] J. Heinonen, Lectures on analysis on metric spaces. Universitext. Springer-Verlag, New York, 2001. MathSciNet review.
answered Apr 8, 2018 at 18:32