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?
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?
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.
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.