The Scott rank of Polish metric spaces

TL;DR

This paper investigates the Scott rank in Polish metric spaces, showing compact spaces have rank at most omega, while some ultrametric spaces can have arbitrarily high countable Scott rank.

Contribution

It extends the notion of Scott rank to Polish metric spaces and characterizes the rank bounds for compact and ultrametric spaces.

Findings

Compact spaces have Scott rank at most ω.

Existence of discrete ultrametric spaces with arbitrarily high countable Scott rank.

Abstract

We study the usual notion of Scott rank but in the setting of Polish metric spaces. The signature consists of distance relations: for each rational $q > 0$, there is a relation $R_{<q}(x,y)$ stating that the distance of $x$ and $y $ is less than $q$. We show that compact spaces have Scott rank at most $\omega$, and that there are discrete ultrametric spaces of arbitrarily high countable Scott rank.

Equations

$$D_n(\overline{x})=d(x_i,x_j{)}_{i,j<n}.$$

$${\left|\left|D(\tilde{a},\overline{x})-D(\tilde{b},\overline{y})\right|\right|}_{max}<ϵ.$$

$${\left|\left|D(\tilde{a},\overline{x})-A\right|\right|}_{\text{max}}<p=ϵ/2.$$

$$\underset{q\in {\mathbb{Q}}^{+}}{\bigwedge }\underset{n\in \mathbb{N}}{\bigvee }\exists x_0\dots x_{n-1}\forall y\underset{i<n}{\bigvee }d(x_i,y)<q.$$

Full text

This research was carried out in 2012 and has been available on Nies’ publications web page. We are putting it up on arXiv as a reference because of later research on the open question posed in here whether the Scott rank of a Polish metric space is always countable. We include updates on the question.

1. Introduction

Definition 1.1**.**

We view a metric space $(X,d)$ as a structure for the signature

$\{R_{q}\colon\,q\in{\mathbb{Q}}^{+}\}$,

where the $R_{q}$ are binary relation symbols. The intended meaning of $R_{q}xy$ is that $d(x,y)<q$.

Clearly, isomorphism of such structures is the same as isometry.

We ask to which extent Polish metric spaces behave like countable structures. For instance, just like for countable structures, if $X_{0}\cong_{p}X_{1}$ (partial isomorphism, given by system of back and forth relations as in [1]) then $X_{0}\cong X_{1}$.

For the reader’s convenience, we recall the definitions of $\alpha$-equivalence and Scott rank of a structure.

Definition 1.2**.**

Let $M$ be an $\mathcal{L}$-structure. We define inductively what it means for finite tuples of same length $\bar{a},\bar{b}$ from $M$ to be $\alpha$-equivalent, denoted by $\bar{a}\equiv_{\alpha}\bar{b}$.

• •

$\bar{a}\equiv_{0}\bar{b}$ if and only if the quantifier-free types of the tuples are the same.

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For a limit ordinal $\alpha$, $\bar{a}\equiv_{\alpha}\bar{b}$ if and only if $\bar{a}\equiv_{\beta}\bar{b}$ for all $\beta<\alpha$.

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$\bar{a}\equiv_{\alpha+1}\bar{b}$ if and only if both of the following hold:

• –

For all $x\in M$, there is some $y\in M$ such that $\bar{a}x\equiv_{\alpha}\bar{b}y$

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For all $y\in M$, there is some $x\in M$ such that $\bar{a}x\equiv_{\alpha}\bar{b}y$

The Scott rank $\mathrm{sr}(M)$ of a structure $M$ is defined as the smallest $\alpha$ such that $\equiv_{\alpha}$ implies $\equiv_{\alpha+1}$ for all tuples of that structure. We remark that always $\mathrm{sr}(M)<|M|^{+}$.

A metric space $X$ is called ultrahomogeneous if any isometry between finite subsets of $X$ can be extended to an auto-isometry of $X$. Using a back and forth argument, one shows that a countable structure has Scott rank [math] iff it is ultrahomogeneous.

Fact 1.3**.**

A Polish space has Scott rank [math] iff it is ultrahomogeneous.

The nontrivial, left-to right direction is proved via a back and forth argument where the $\forall$ player takes points from the countable dense set. The resulting partial isometry can be extended to a full isometry by completeness.

Question 1.4**.**

• (a)

Does every Polish metric space $M$ have countable Scott rank?

• (b)

Can every Polish metric space be described within the class of Polish spaces by an $L_{\omega_{1},\omega}$ sentence?

By analogy with countable models, some evidence for an affirmative answer to (b) can be obtained as follows. By Gao and Kechris [6], isometry of Polish spaces (suitably encoded by a real) is $\leq_{B}$ the orbit equivalence relation $E_{I}$ obtained by the action of $Iso(\mathbb{U})$ on the Effros Space $F(\mathbb{U})$. Each orbit is Borel (Luzin-Nardzewsky). So each isometry class of Polish spaces is Borel.

Remark 1.5** (May 2019).**

Doucha [3, 4] proved that the Scott rank of $M$ is at most $\omega_{1}$. Whether it is always countable remains open.

Ben Yaacov et al [2, Section 10] answer part (b) in the affirmative. They produced a Scott sentence in infinitary continuous logic, and observed that it can be translated to classical infinitary logic.

Can we show directly that each isometry class is Borel?

Remark 1.6** (May 2019).**

Ben Yaacov et al [2] have given a different argument that each isometry class not relying on the descriptive set theory approach of [6]. Rather, they construct a Scott sentence in infinitary continuous logic, and the models of such a sentence always form a Borel set. This is not the case for $L_{\omega_{1},\omega}$ sentences in classical logic.

One can verify that the Scott rank of a metric structure is less than the least hyperprojective $\alpha$; that is, the least $\alpha$ such that $L_{\alpha}({\mathbb{R}})$ is a model of Kripke-Platek set theory. This is in fact true for Borel structures.

There is a Borel linear order $S$ with Scott rank $\omega_{1}$: reals $b\in[0,1]$ encode binary relations $R_{b}$. Let $L=\{b\in[0,1]\colon R_{b}\text{ is linear}\}$. Let $S$ be the sum of all $R_{b}$ for $b\in L$.

2. Compact metric spaces are $\exists$-homogenous and hence have Scott rank at most $\omega$.

Let $X$ be a metric space. For a tuple $\overline{x}\in X^{n}$ consider the $n\times n$ distance matrix

[TABLE]

We often view this matrix as a tuple in ${\mathbb{R}}^{n^{2}}$ with the max norm ${\left|\!\left|.\right|\!\right|}_{\text{max}}$. Note that for any matrix $A\in{\mathbb{Q}}^{n^{2}}$ and any positive rational $p$, there is a quantifier-free positive first-order formula $\phi_{A,n,p}(\overline{x})$ in the metric signature (Definition 1.1) expressing that ${\left|\!\left|D_{n}(\overline{x})-A\right|\!\right|}_{\text{max}}<p$.

Let $D_{n}(X)\subseteq{\mathbb{R}}^{n^{2}}$ denote the set of $n\times n$ distance matrices. Let now $X$ be a compact metric space. Then for each $n$, the set $D_{n}(X)$ is compact. Gromov [7] showed that the space $X$ is described up to isometry by this sequence of compact sets $D_{n}(X)_{n\in{\mathbb{N}}}$ (see [5, proof of 14.2.1], but note that our $D_{n}$ is denoted $M_{n-1}$ there). This shows that isometry of compact spaces is smooth, i.e., Borel reducible to the identity on ${\mathbb{R}}$.

An existential positive formula is a first-order formula not using $\forall,\lnot$.

Theorem 2.1**.**

Let $X,Y$ be compact metric spaces. Suppose that tuples $\widetilde{a}\in X^{p},\widetilde{b}\in Y^{p}$ satisfy the same existential positive formulas. Then there is an isometry from $X$ to $Y$ mapping $\widetilde{a}$ to $\widetilde{b}$. In particular, each compact metric space $(X,d)$ is $\exists$-homogeneous.

Proof.

Recall that any isometric self-embedding of a compact metric space is onto (see [5, proof of 14.2.1]). So by the symmetry it suffices to find an isometric embedding of $X$ into $Y$ mapping $\widetilde{a}$ to $\widetilde{b}$.

The following slightly extends the above-mentioned result of Gromov (see [5, Exercise 14.2.3]).

Lemma 2.2**.**

Suppose that for any $\epsilon>0$, for any $n$ and tuple $\overline{x}\in X^{n}$ there is a tuple $\overline{y}\in Y^{n}$ such that

[TABLE]

Then there is an an isometric embedding of $X$ to $Y$ mapping $\widetilde{a}$ to $\widetilde{b}$.

Proof.

For each $k$ let $\overline{y}_{k}$ be a tuple such that $\left|\!\left|D(\widetilde{a},\overline{x})-D(\widetilde{b},\overline{y})\right|\!\right|_{max}<1/k$. Let $\overline{y}$ be the limit of a convergent subsequence of $(\overline{y}_{k})_{k\in{\mathbb{N}}}$. This shows that for each $\overline{x}$ there is $\overline{y}$ such that $D(\widetilde{a},\overline{x})=D(\widetilde{b},\overline{y})$.

We can now proceed almost exactly as in [5, proof of 14.2.1]. Let $(x_{i})_{i\in{\mathbb{N}}}$ be a dense sequence in $X$. For each $n\in{\mathbb{N}}$ there are $y^{n}_{0},\ldots,y^{n}_{n}\in Y$ such that

$D(\widetilde{a},x_{0},\ldots,x_{n})=D(\widetilde{b},y^{n}_{0},\ldots,y^{n}_{n})$.

There is $A_{0}\subset{\mathbb{N}}$ with $0\not\in A_{0}$ such that $z_{0}:=\lim_{n\in A_{0}}y^{n}_{0}$ exists. There is $A_{1}\subset A_{0}$ with $\min A_{0}\not\in A_{1}$ such that $z_{1}:=\lim_{n\in A_{1}}y^{n}_{1}$ exists. Proceeding that way we obtain a sequence of points $z_{n}\in Y$ and a descending sequence of sets $A_{0}\supset A_{1}\supset A_{2}\supset\ldots$ with $\min A_{k}\not\in A_{k+1}$ such that $z_{k}=\lim_{n\in A_{k}}y^{n}_{k}$.

Let $A=\{\min A_{k}\colon\,k\in{\mathbb{N}}\}$. Then $A\setminus A_{k}$ is finite for each $k$, so $z_{k}=\lim_{n\in A}y^{n}_{k}$ for each $k$. One uses this to show that $d(x_{i},x_{j})=d(z_{i},z_{j})$ for each $i,j$, and $d(a_{r},x_{i})=d(b_{r},z_{i})$ for each $r,j$. Hence the map $x_{i}\mapsto z_{i}$ can be extended to the required isometric embedding of $X$ into $Y$ ∎

It now suffices to show that if $\widetilde{a}\in X^{p},\widetilde{b}\in Y^{p}$ satisfy the same existential positive formulas, then the hypothesis of the lemma is satisfied. Recall that the formula $\phi_{A,n,p}(\overline{x})$ expresses that ${\left|\!\left|D_{n}(\overline{x})-A\right|\!\right|}_{\text{max}}<p$. Given $\overline{x}\in\ X^{n}$ choose a rational $(k+n)\times(k+n)$ matrix $A$ such that

[TABLE]

Thus $\exists\overline{x}\,\phi_{A,n+k,p}(\widetilde{a},\overline{x})$ holds in $X$. Hence there is $\overline{y}\in Y^{n}$ such that $\phi_{A,n+k,p}(\widetilde{b},\overline{y})$ holds in $Y$. This implies $\left|\!\left|D(\widetilde{a},\overline{x})-D(\widetilde{b},\overline{y})\right|\!\right|_{max}<\epsilon$ as required. ∎

The case $\widetilde{a}=\widetilde{b}=\emptyset$ yields:

Corollary 2.3**.**

Within the class of compact metric spaces, each member is uniquely described by its existential positive first-order theory.

Note that a complete metric space is compact iff it is totally bounded, namely, satisfies the $L_{\omega_{1},\omega}$ sentence

[TABLE]

Thus using the corollary above, each compact metric space can be described by an $L_{\omega_{1},\omega}$ sentence within the class of Polish metric spaces.

Corollary 2.4**.**

The Scott rank of any compact metric space $X$ is at most $\omega$.

Proof.

In the notation of the theorem with $Y=X$, suppose $\widetilde{a}\equiv_{n}\widetilde{b}$ for each $n\in\omega$. Then $\widetilde{a}$ and $\widetilde{b}$ satisfy the same existential formulas. Hence there is an isometry of $X$ sending $\widetilde{a}$ to $\widetilde{b}$. ∎

3. Countable Polish spaces can have arbitrary countable Scott rank

Countable structures have countable Scott rank. In this section, we show that arbitrary large countable ordinal ranks are possible for countable Polish metric spaces. The examples we construct actually will be discrete.

We will inductively construct sub-trees of $\omega^{<\omega}$ with growing Scott ranks. Note that we can view such trees as model theoretic structures, for example in the language of a unary function $f$ with the semantics “$f(a)=b$ if and only if $b$ is the immediate predecessor of $a$”. Also, those trees can be regarded as metric spaces with the metric induced by the usual one on $\omega^{<\omega}$. It is easy to see that the relations $\equiv_{\alpha}$ (and hence the notion of Scott rank) is the same in both the tree structure and the metric space structure.

We denote finite sequences from $\omega^{<\omega}$ by $\langle n_{1},n_{2},\dots,n_{k}\rangle$ and the concatenation of two sequences $s,t$ by $s^{\frown}t$. Now we proceed to the definition of $T_{n}^{\alpha}\subset\omega^{<\omega}$ ($n\leq\omega$, $\alpha<\omega_{1}$) by induction over $\alpha$.

Definition 3.1**.**

For any $n\leq\omega$, let $T_{n}^{0}$ be the collection of the empty sequence and the sequences $\langle a\rangle$ for all $a<n$.

Now suppose we have defined all the $T_{n}^{\alpha}$ for some $\alpha<\omega_{1}$. Let $T_{n}^{\alpha+1}$ be the collection of

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the empty sequence

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all sequences of the form $\langle 2a\rangle^{\frown}s_{a}$ where $a<\omega$ and $s_{a}\in T_{a}^{\alpha}$

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for any $a<n$, all sequences of the form $\langle 2a+1\rangle^{\frown}s$ with $s\in T_{\omega}^{\alpha}$

Finally, suppose $\alpha$ is a countable limit ordinal and that the $T_{n}^{\beta}$ are already defined for all $n\leq\omega$ and $\beta<\alpha$. Fix any bijection $B:\alpha\times\omega\rightarrow\omega$ and define $T_{n}^{\alpha}$ as the collection of

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the empty sequence

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all sequences of the form $\langle a\rangle^{\frown}s_{a}$ where $a=B(c,d)$ for some $c<\alpha$ and $d\leq n$ and $s_{a}\in T_{\omega}^{c}$

Informally speaking, at each stage of the construction we glue infinitely many of the already defined trees into a new one. $T_{n}^{\alpha+1}$ contains all $T_{k}^{\alpha}$ for $k<\omega$ exactly once and $T_{\omega}^{\alpha}$ exactly $n$ times. At limit stages $\alpha$, we amalgamate $n+1$ copies of each $T_{\omega}^{\beta}$ ($\beta<\alpha$).

Inductively, we see that none of the constructed trees has an infinite path and that all of them, seen as metric spaces, are discrete and complete (thus they are Polish). The $T_{m}^{0}$ are homogeneous, so they have Scott rank [math], and we can verify inductively that $T_{n}^{\alpha}$ has Scott rank $\alpha\cdot\omega$ for all $\alpha<\omega_{1}$ and all $n\leq\omega$.