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Recall the definition of the Hausdorff dimension of a subset $A$ of a metric space $\mathcal{M}$. (I'm not especially wedded to the Hausdorff dimension per se and I'd be interested in answers for any other dimension notion which similarly "works" over all metric spaces, but I'm focusing on Hausdorff here to keep things reasonably concrete.)

Every (isometry class of) compact metric space(s) corresponds to a point in the Gromov-Hausdorff space $\mathcal{GH}$ (see also these lecture notes of Tuzhilin). Consequently we can define, for an isometry-invariant class $\mathbb{K}$ of compact metric spaces, the value $\mathsf{dim}_{GH}(\mathbb{K})$ to be the Hausdorff dimension of the subset of $\mathcal{GH}$ corresponding to $\mathbb{K}$. Now at first glance this doesn't seem too interesting - every natural example I can think of winds up yielding either $\infty$ or $0$. Basically, $\mathcal{GH}$ has so many "degrees of freedom" that in order for a class of compact metric spaces to not have infinite dimension, it needs to be so sparse that it winds up having dimension $0$.

I'm curious if there's an actual result behind this. Vaguely inspired by $0$-$1$-laws in finite model theory, the following concrete question occurred to me which I haven't been able to make headway on:

Given a metric space $\mathcal{X}=(X,d)$, let $Struc_\mathcal{X}$ be the two-sorted first-order structure $(X,\mathbb{R};d,0,1,+,\times,<)$ consisting of the points of $X$, the ordered field of real numbers, and the metric $d$ connecting them. For $\varphi$ a sentence in the signature $\Sigma=\{d,0,1,+,\times,<\}$, let $\hat{\varphi}$ be the class of compact metric spaces $\mathcal{X}=(X,d)$ such that $Struc_\mathcal{X}\models\varphi$.

Question: Is there a first-order $\Sigma$-sentence $\varphi$ such that $\mathsf{dim}_{GH}(\hat{\varphi})\not\in\{0,\infty\}$?

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Claim. There exists an isometric embedding $f: [0,\infty)\to {\mathcal G}{\mathcal H}$ whose image is the set $S$ of metric spaces consisting of one or two points.

Proof. Start with the metric space $M_1$ given by $\{x,y\}, d_1(x,y)=2$. For each $t>0$ define $M_t=(M_1, td_1)$. For $t=0$ let $M_0$ be the singleton. Thus, we get a map $f: t\mapsto M_t, t\ge 0$.

It is a pleasant exercise, which is a mild generalization of my answer here, to see that $f$ is an isometric embedding from the half-line (with the standard metric) to the Gromov-Hausdorff space
${\mathcal G}{\mathcal H}$. Clearly, the image of $f$ is exactly $S$. qed

Thus, $S\subset {\mathcal G}{\mathcal H}$ has Hausdorff dimension 1. At the same time, the class of metric spaces $S$ (consisting of one or two elements) can be described using the 1st order language: $$ \exists x\in M \wedge \forall x, y, z\in M, x=y \vee y=z \vee z=x. $$

With more work, one can show that for each $n\ge 1$ there exists a Lipschitz topological embedding $$ f_n: \Delta^n\to {\mathcal G}{\mathcal H} $$ whose image consists of finite metric spaces with explicitly given metrics. Here $\Delta^n$ is the standard $n$-dimensional simplex. (The map $f_n$ is defined using a geodesics cone over previously defined $f_{n-1}$.) The Hausdorff dimension of the image of $f_n$ then will be exactly $n$. I am quite sure that for each $n$ one can describe the class of the corresponding metric spaces using the 1st order language, just the description will be quite unpleasant.

What I do not see is any reasonable class whose Hausdorff dimension will be non-integral. One can use images of fractals in $\Delta^n$ under $f_n$, I am just not sure how to define these metrics using 1st order sentences.

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  • $\begingroup$ Neat! I'm going to hold off on accepting for a bit to see if any fractional-dimension examples come along, but this is great, thanks! $\endgroup$ Nov 21, 2021 at 5:00
  • $\begingroup$ @NoahSchweber: Sure, not a problem. $\endgroup$ Nov 21, 2021 at 20:39

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