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Let $(X,d)$ be a metric space with finite Assouad dimension $0<C_X$. It seems intuitive to me that if $\emptyset \subset Y\subseteq X$ then $Y$ is also doubling and its Assouad dimension, denoted here by $C_Y$, should satisfy $C_Y\leq c C_X$ (where $c$ is some absolute constant independent of $X$ and of $Y$).

Is this true, and if so where can I find this fact?

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  • $\begingroup$ Just to make sure that we are on the same page: what do you mean by the doubling constant? $\endgroup$
    – Piotr Hajlasz
    Commented Aug 3, 2021 at 13:04
  • $\begingroup$ Oh, I mean the Assouad dimension to be precise. $\endgroup$
    – SetValued_Michael
    Commented Aug 3, 2021 at 13:04
  • $\begingroup$ That is a very different question :) $\endgroup$
    – Piotr Hajlasz
    Commented Aug 3, 2021 at 13:09
  • $\begingroup$ @PiotrHajlasz Why so, I thought the doubling dimension is just $\log_2(c_X)$ where $c_X$ is the doubling constant of $X$ (in the sense of that $c_X$ is the smallest natural number for which every ball in $X$ of radius $r>0$ can be covered by $c_X$ balls of radius $\frac{r}{2}$.) $\endgroup$
    – SetValued_Michael
    Commented Aug 3, 2021 at 13:26

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This is Lemma 9.6(i) in J. C. Robinson, Dimensions, embeddings, and attractors. Cambridge Tracts in Mathematics, 186. Cambridge University Press, Cambridge, 2011.

In the proof the author says "it is obvious". I am no longer sure if it is obvious (perhaps it is) since I see a potential issue: if $Y\subset X$, then every ball in $Y$ is a restriction of a ball from $X$, but if $B$ is a ball in $X$ not centered at $Y$, then $B\cap Y$ is not a ball in $Y$ so there are less balls in $Y$ to cover. Is it an issue? I haven't checked (I still did not have my morning coffee).

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  • $\begingroup$ Just out of curiosity, where could I find the analogous result for the doubling constant (as defined in the comments above)? $\endgroup$
    – SetValued_Michael
    Commented Aug 3, 2021 at 13:58
  • $\begingroup$ @SetValued_Michael The difficulty is similar to proving that a subset of a separable space is separable: the countable and a dense set may have an empty intersection with the subset. From second afternoon coffee I guess you are in Europe. Do I know you or - if you are a graduate student - do I know your advisor? I believe, I know (at least by name) most of the people working with the Assouad dimension. $\endgroup$
    – Piotr Hajlasz
    Commented Aug 3, 2021 at 13:59
  • $\begingroup$ Oops I deleted the previous comment. Anyway, likely not as I am not in metric space theory but rather in applied mathematics (and have been using this condition to get some nice bounds :) ) ...I came across this question when reading this paper: arxiv.org/abs/1801.07533 $\endgroup$
    – SetValued_Michael
    Commented Aug 3, 2021 at 14:06
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Given $y\in Y$ and $r>0$, using twice the doubling property of $X$, cover $B(y,r)$ with $C_X^2$ balls of radius $\frac{r}{4}$ centered in $X$. Inside every such ball take any $y\in Y$ (if there are any) and center there a ball of radius $\frac{r}{2}$. This gives a cover of the original ball.

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  • $\begingroup$ Yes, but you get a quite bad estimate for the constant. Can you get the same constant for the subset or at least $C_X+\epsilon$? I haven't thought about it. $\endgroup$
    – Piotr Hajlasz
    Commented Aug 3, 2021 at 20:37
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    $\begingroup$ Good example. That explains everything. $\endgroup$
    – Piotr Hajlasz
    Commented Aug 3, 2021 at 21:41
  • $\begingroup$ @PiotrHajlasz I do think there may be some bad example, where the constant increases. The following example came to mind: $\mathbb{R}^d$ with the sup norm has doubling constant $2^d$ (at least if you consider closed balls...) but the subset given by all points whose coordinates belong to $\{-1,0,1\}$ should have constant $3^d$ (consider the ball centered at 0 with radius slightly above 1) $\endgroup$
    – Del
    Commented Aug 3, 2021 at 21:44
  • $\begingroup$ Sorry I deleted by mistake my previous comment trying to edit it. It was a particular case of the last one I wrote $\endgroup$
    – Del
    Commented Aug 3, 2021 at 21:46
  • $\begingroup$ As this example made me somewhat nervous (the Assouad dimension is not monotone???) let me say what allayed this: If one defines the Assouad dimension in a more careful way (as is done in Assouad's original paper, or in two equivalent ways in and after Def. 10.15 of Heinonen's Lectures on Analysis on Metric Spaces), then the dimension is indeed monotone. But it seems that then it does not necessarily exactly agree with log_2(doubling constant). $\endgroup$
    – user142382
    Commented Aug 5, 2021 at 14:37

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