Pseudo-finite sets, pseudo-o-minimality

TL;DR

The paper constructs two ordered structures with identical one-variable definable sets, illustrating limitations in axiomatizing o-minimal theories and implications for definable completeness and the pigeonhole principle.

Contribution

It provides a counterexample showing that definable completeness does not imply the pigeonhole principle, challenging previous assumptions in o-minimal theory.

Findings

Counterexample to axiomatization of o-minimal theories with one-variable sets

Definable completeness does not imply the pigeonhole principle

Partial answer to questions by Schoutens and Fornasiero

Abstract

We give an example of two ordered structures M, N in the same language L with the same universe, the same order and admitting the same one-variable definable subsets such that M is a model of the common theory of o-minimal L-structures and N admits a definable, closed, bounded, and discrete subset and a definable injective self-mapping of that subset which is not surjective. This answers negatively two questions by Schoutens; the first being whether there is an axiomatization of the common theory of o-minimal structures in a given language by conditions on one-variable definable sets alone. The second being whether definable completeness and type completeness imply the pigeonhole principle. It also partially answers a question by Fornasiero asking whether definable completeness of an expansion of a real closed field implies the pigeonhole principle.

Equations

$$\exists v\forall w\left(U(w)\to w<v\right)\to$$

$$\exists v\left(\forall w\left(U(w)\to w<v\right)\wedge \forall v^{\prime }\left(\forall w\left(U(w)\to w<v\right)\right)\to v\le v^{\prime }\right).$$

$$T^{omin}\supseteq T^{\prime }\supseteq \cup \left\{T_{\tau }\right\}\tau \text{ is an {<,U}-sentence}.$$

$$C_{<}(a,b,c)⟺(a<b<c)\vee (b<c<a)\vee (c<a<b)$$

$$C(a,-,b):=\left\{x\in A\right\}C(a,x,b).$$

$$C(a,-,c)=C(a,-,b)\cup \left\{b\right\}\cup C(b,-,c).$$

$$x\notin \left(C(a,-,b)\cup \left\{b\right\}\cup C(b,-,c)\right),$$

$$X\cap C(a,-,c)=\left(X\cap C(a,-,b)\right)\cup \left(X\cap \left\{b\right\}\right)\cup \left(X\cap C(b,-,c)\right)$$

$$X\cap C(c,-,b)=\left(X\cap C(c,-,a)\right)\cup \left(X\cap \left\{a\right\}\right)\cup \left(X\cap C(a,-,b)\right)$$

$$X\cap C(b,-,a)=\left(X\cap C(b,-,c)\right)\cup \left(X\cap \left\{c\right\}\right)\cup \left(X\cap C(c,-,a)\right)$$

$$a{\sim }_X{a}^{\prime },b{\sim }_X{b}^{\prime },c{\sim }_X{c}^{\prime },$$

$$a/{\sim }_{X}b,a/{\sim }_{X}c,b/{\sim }_{X}c.$$

$$C(a,b,c)$$

$$C(c^{\prime },b,a)$$

$$C(a,c^{\prime },b)$$

$$C(c,a,c^{\prime })$$

$$C(c^{\prime },b,c)$$

$$\left(\forall x\right)(\pi (x)\in Z)\wedge (\forall y\in Z\left(\neg C_{<}(x,y,\pi (x))\right))$$

$$S(x)=y\leftrightarrow \left(x\notin Z\wedge x=y\right)\vee \left(x\in Z\wedge y\in Z\wedge \neg \exists z\in Z\left(C_{<}\left(x,z,y\right)\right)\right)$$

$$Z\cap C_{<}\left(z,-,f^n(z)\right)\text{ and }Z\cap C_{<}\left(f^n(z),-,z\right)$$

$$f((a,b)):=\left\{\begin{array}[]{ll}(a-1\mod(N+1),b)&\mbox{if }x=(a,b)\in Z\\ x&\mbox{if }x\notin Z\end{array}\right.$$

$$f^m((x,y))=(x-m\mathrm{mod}(N+1),y)$$

$$f^n((x,y))=(x-n\mathrm{mod}(N+1),y)$$

$$\mathcal{M}\models \left[\exists x\in Z(a<x<b)\right]\leftrightarrow \left[\pi \circ S(a)<b\right]$$

$$T_0\models \forall \bar{y}\left[\left(\exists x\underset{i\in I}{\bigwedge }{\theta }_i\left(\bar{y},x\right)\right)\leftrightarrow \phi (\bar{y})\right]$$

$$⊢\left[\exists x\underset{i\in I}{\bigwedge }{\theta }_i\left(\bar{y},x\right)\right]\leftrightarrow \left[\exists x\left(\underset{i\in I}{\bigwedge }{\theta }_i\left(\bar{y},x\right)\wedge \left(x\in Z\vee x\notin Z\right)\right)\right]\leftrightarrow$$

$$\left[\left(\exists x\left(\underset{i\in I}{\bigwedge }{\theta }_i\left(\bar{y},x\right)\wedge x\in Z\right)\right)\vee \left(\exists x\left(\underset{i\in I}{\bigwedge }{\theta }_i\left(\bar{y},x\right)\wedge x\notin Z\right)\right)\right].$$

$$\varphi (\bar{y})=\exists x\left(\underset{i=1}{\overset{k}{\bigwedge }}{t}_i{\square }_{i}x\wedge x\notin Z\right)$$

$$\varphi (\bar{y})=\exists x\left(\underset{i=1}{\overset{m}{\bigwedge }}{l}_i<x\wedge \underset{j=1}{\overset{n}{\bigwedge }}{u}_i>x\wedge x\in Z\right)$$

$$\varphi (\bar{y})=\exists x\left(\underset{i=1}{\overset{m}{\bigwedge }}{l}_i<x\wedge \underset{j=1}{\overset{n}{\bigwedge }}{u}_i>x\wedge x\notin Z\right)$$

$$\underset{i=1}{\overset{n}{\bigwedge }}\underset{j=1}{\overset{m}{\bigwedge }}\left(\pi \circ S(l_i)<u_j\right).$$

$$\underset{i=1}{\overset{n}{\bigwedge }}\underset{j=1}{\overset{m}{\bigwedge }}\left(l_i<u_j\right).$$

$$\displaystyle f^{\mathcal{M}_{2}}(x):=\left\{\begin{array}[]{ll}P\circ f^{\mathcal{M}_{1}}(x)&\mbox{if }S^{n}(x)=c_{4}\text{ for some }n\in\mathbb{N}\\ f^{\mathcal{M}_{1}}(x)&\mbox{if }S^{n}(x)\neq c_{4}\text{ for all }n\in\mathbb{N}\end{array}\right..$$

Full text

The work in this paper is part of the author’s Ph.D. studies at the Department of Mathematics, Ben-Gurion University of the Negev under the supervision of Assaf Hasson.

The author was partially supported by ISF Grant 181/60 and the Hillel Gauchman scholarship.

1. Introduction

o-minimality is not preserved under ultraproducts, as shown in the following example:

Example \theexample.

Let $\mathcal{L}=\Set{<,U}$ where $<$ is a binary relation symbol and $U$ is a unary predicate. For every $n\in\mathbb{N}$, let $\mathcal{M}_{n}$ be a structure interpreting $<$ as a dense linear order without end points and $U$ as a set of points of size $n$. Then each $\mathcal{M}_{n}$ is o-minimal. But for any non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$, in the ultraproduct $\prod_{\mathbb{N}}\mathcal{M}_{n}/\mathcal{U}$, the definable set $U$ is infinite and discrete, thus the ultraproduct of o-minimal structures need not be o-minimal.

Section 1 can be generalized to any first-order language $\mathcal{L}\supsetneq\{<\}$. So By Łos’ Theorem, given a first-order language $\mathcal{L}\supsetneq\{<\}$, there is no first-order theory $T$, such that $\mathcal{M}\models T\iff\mathcal{M}$ is o-minimal for every $\mathcal{L}$-structure $\mathcal{M}$.

Here we focus our attention on some properties implied by o-minimality which are first-order, i.e., those properties which both hold in all o-minimal structures, and, given a language $\mathcal{L}=\Set{<,\dots}$, can be axiomatized by a set of $\mathcal{L}$-sentences. Rigorously, we follow the conventions from [Sch14], defined below:

Definition \thedefinition.

Given a language $\mathcal{L}=\Set{<,\dots}$, let $T^{omin}_{\mathcal{L}}$ be the set of all $\mathcal{L}$-sentences satisfied in every o-minimal $\mathcal{L}$-structure.

An $\mathcal{L}$-structure $\mathcal{M}$ for $\mathcal{L}=\Set{<,\dots}$ is pseudo-o-minimal if $\mathcal{M}\models T_{\mathcal{L}}^{omin}$.

Fact \thefact ([Sch14, Corollary 10.2]).

An $\mathcal{L}$-structure $\mathcal{M}$ for $\mathcal{L}=\Set{<,\dots}$ is pseudo-o-minimal if and only if $\mathcal{M}$ is elementarily equivalent to an ultraproduct of o-minimal structures.

The following two definitions are examples of first-order weakenings of o-minimality.

Definition \thedefinition.

An expansion of a dense linear order without endpoints $\mathcal{M}=\Braket{M;<,\dots}$ is definably complete if every definable subset of $M$ has a least upper bound.

Definition \thedefinition.

An expansion of a dense linear order without endpoints $\mathcal{M}=\Braket{M;<,\dots}$ is locally o-minimal if for any definable subset $A\subseteq M$ and any $a\in M$ there are $b_{1},b_{2}\in M$ such that $b_{1}<a<b_{2}$ and if $I=(b_{1},a)$ or $(a,b_{2})$ then either $I\subset A$ or $I\cap A=\emptyset$.

Notice that both definable completeness and local o-minimality, in a given language $\mathcal{L}$, are axiomatized by first-order schemes which hold in any o-minimal structure. Thus, any pseudo-o-minimal $\mathcal{L}$-structure is definably complete and locally o-minimal.

Fornasiero, Hieronymi, Miller, Schoutens, Servi and others proved many tameness properties for definably complete and for locally o-minimal structures. (See, e.g, [Mil01, Hie11, For10, FS11, For13, Hie13, Sch14, FH15].) Citing all tameness properties proved in this area will be longer than this paper, so we give two elementary examples by Miller:

Fact \thefact ([Mil01, Corollary 1.5]).

Let $\mathcal{M}=\Braket{M;<,\dots}$ be an expansion of a dense linear order without endpoints. Then the following are equivalent:

1. (1)

$\mathcal{M}$* is definably complete.* 2. (2)

$\mathcal{M}$* has the intermediate value property, i.e., the image of an interval under a definable continuous map is an interval.* 3. (3)

Intervals in $M$ are definably connected, i.e. for every interval $A\subseteq M$ and every disjoint open definable subsets $U,V\subseteq M$, if $A=(A\cap U)\cup(A\cap V)$, then either $A\cap U=\emptyset$ or $A\cap V=\emptyset$. 4. (4)

$M$* is definably connected.*

Fact \thefact ([Mil01, Proposition 1.10]).

Let $\mathcal{M}=\Braket{M;<,\dots}$ be definably complete. Let $f:A\to M^{n}$ be definable and continuous with $A$ closed and bounded. Then $f(A)$ is closed and bounded. In particular, If $f:A\to M$ is definable and continuous with $A$ closed and bounded, then $f$ achieves a maximum and a minimum on $A$.

In [Sch14], Schoutens presented a strengthening of local o-minimality by the name of type completeness, as defined below. In a sense this strengthening extends the locality to $\pm\infty$:

Definition \thedefinition.

An expansion of a dense linear order without endpoints $\mathcal{M}=\Braket{M;<,\dots}$ is type complete if it is locally o-minimal and, in addition, for any definable subset $A\subseteq M$ there are $c_{1},c_{2}\in M$ such that if $I=(-\infty,c_{1})$ or $(c_{2},+\infty)$, then either $I\subset A$ or $I\cap A=\emptyset$.

Type completeness is a first-order scheme, and therefore satisfied by any pseudo-o-minimal structure.

Several tameness results were proved for definably complete type complete structures in [Sch14]. For example, a version of o-minimal cell decomposition called quasi-cell decomposition ([Sch14, Theorem 8.10]) and the following monotonicity theorem:

Fact \thefact ([Sch14, Theorem 3.2]).

Let $\mathcal{M}=\Braket{M;<,\dots}$ be a definably complete type complete structure. The set of discontinuities of a one-variable definable map $f:Y\to M$ is discrete, closed, and bounded, and consists entirely of jump discontinuities. Moreover, there is a definable discrete, closed, bounded subset $D\subseteq Y$ so that in between any two consecutive points of $D\cup\Set{\pm\infty}$, the map is monotone, that is to say, either strictly increasing, strictly decreasing, or constant.

Of particular importance in the study of definably complete structures are the definable pseudo-finite sets, as defined below.

Definition \thedefinition.

Let $\mathcal{M}=\Braket{M;<,\dots}$ be a definably complete structure. A definable subset $A\subset M^{n}$ is pseudo-finite if it is closed, bounded, and discrete.

These definable sets play a role in each of the papers cited above. We follow the convention in [For10, For13], where there is an extensive study of pseudo-finite sets and their tameness properties. In [For13], the wording was justified in the definably complete context by saying that pseudo-finite sets are first-order analogue of finite subsets of $\mathbb{R}^{n}$, with evidence given by numerous tameness properties of such sets.

One must not confuse pseudo-finite sets defined above with pseudo-$o$-finite sets, as we define below, coined in [Sch14]. Though, as we will see in Section 1 the two definitions coincide if $\mathcal{M}$ is assumed to be pseudo-o-minimal.

Definition \thedefinition.

Let $\mathcal{M}=\Braket{M;<,\dots}$ be a pseudo-o-minimal structure. A definable set $X\subseteq M^{n}$ is pseudo-$o$-finite if $(\mathcal{M},X)$ satisfies the common theory of o-minimal structures expanded by a unary predicate for a distinguished finite subset.

The following section can be immediately extracted from [Sch14, Corollary 12.6] together with [Sch14, Theorem 12.7].

Fact \thefact.

Let $\mathcal{M}=\Braket{M;<,\dots}$ be a pseudo-o-minimal structure. A definable set $A\subset M^{n}$ is pseudo-finite if and only if it is pseudo-$o$-finite.

A tameness property of pseudo-finite sets occurring naturally is “the discrete pigeonhole principle” [Sch14]. (Or just “the pigeonhole principle” in [For10, For13].)

Definition \thedefinition.

An expansion of a dense linear order without endpoints $\mathcal{M}=\Braket{M;<,\dots}$ has the pigeonhole principle if for any pseudo-finite $X\subset K^{n}$ and definable $f:X\to X$, if f is injective, then it is surjective.

We remark that the pigeonhole principle can be formulated as “every pseudo-finite set is definably Dedekind finite”, and as this is a first-order scheme, every pseudo-o-minimal structure has the pigeonhole principle.

In [For10] and [For13], Fornasiero conjectured the following:

Conjecture \theconjecture.

If $\mathcal{K}=$ $\Braket{K,+,\cdot,<,\dots}$ is a definably complete expansion of a real closed field, then $\mathcal{K}$ has the pigeonhole principle.

This conjecture remained open even for $\mathcal{K}$ a definably complete expansion of a dense linear order. Clearly, the conjecture holds for $\mathcal{K}$ pseudo-o-minimal. Consequently, it is connected to two other questions asked by Schoutens in [Sch14]:

Question \thequestion.

Does every definably complete type complete structure have the pigeonhole principle?

Question \thequestion.

Is there an axiomatization of pseudo-o-minimality by first-order conditions on one-variable formulae only?

To clarify the meaning of a first-order conditions on one-variable formulae only, this does not mean a first-order sentence conditioned on a specific one-variable formula, as the following example demonstrates how any first-order theory is axiomatized by such sentences, in particular $T^{omin}$.

Example \theexample.

Let $\mathcal{L}$ be any language and $T$ be any $\mathcal{L}$-theory (not necessarily complete). For every sentence $\sigma\in T$, let $\psi_{\sigma}(x):=x=x\land\sigma$ and let $\phi_{\sigma}:=\exists x\,\psi_{\sigma}(x)$. Then $\phi_{\sigma}$ is a first order condition on $\psi_{\sigma}$, however $\vdash\psi_{\sigma}\iff\sigma$, so $\Set{\psi_{\sigma}}{\sigma\in T}$ is an axiomatization of $T$.

Clearly, this is not the intended meaning in the question. Rather, following the terminology of [Sch14], we interpret a first-order condition on one-variable formulae as first-order scheme ranging over all one-variable formulae. Rigorously, a first-order condition on one-variable formulae is obtained as follows:

• •

Let $\tau$ be a first-order sentence in the language $\Set{<,U}$ where $U$ is a unary predicate.

• •

Let $\Phi$ be the set of partitioned $\mathcal{L}$-formulae $\varphi(x;\bar{y})$ where $x$ is a single variable and $\bar{y}$ is a finite tuple of variables not appearing in $\tau$.

• •

For every $\varphi(x;\bar{y})\in\Phi$, let $\tau_{\varphi}(x;\bar{y})$ be the $\mathcal{L}$-formula obtained by replacing any instance of $U(x)$ by $\varphi(x;\bar{y})$.

• •

$T_{\tau}:=\Set{\forall\bar{y}\,\tau_{\varphi}(x;\bar{y})}{\varphi(x;\bar{y})\in\Phi}$.

For example, definable completeness is axiomatized in the above fashion by setting $\tau$ to be

[TABLE]

Namely, $\tau$ is the $\Set{<,U}$-sentence stating if $U$ is bounded, then it has a least upper bound. Following the same terminology, an axiomatization of pseudo-o-minimality by first-order conditions on one-variable formulae only is an $\mathcal{L}$-theory $T^{\prime}$ such that

[TABLE]

In [Ren14], Rennet showed that there is no recursive first-order axiomatization of pseudo-o-minimality in the language of rings $\Set{+,-,\cdot,0,1}$. In particular, as definable completeness and type completeness are both recursive first-order schemes, given a recursive language, they cannot axiomatize pseudo-o-minimality.

In this paper, we show a stronger result (with respect to one-variable definable sets) by constructing two ordered structures $\mathcal{M},\mathcal{N}$ on the same universe, in the same language, with the same definable subsets in one variable, where $\mathcal{M}$ is pseudo-o-minimal and $\mathcal{N}$ does not have the pigeonhole principle. This gives a negative answer to both Sections 1 and 1, as well as a partial answer to Section 1. Furthermore, this gives a stronger result then a negative answer to Section 1. It shows that not only is there no first order axiomatization $T^{\prime}$ as above, but also there is no second order theory in the language $\mathcal{L}_{\operatorname{Def}}:=\Set{<,\operatorname{Def}}$ where $\operatorname{Def}$ is a unary predicate on subsets interpreted as the definable subsets. This result is strictly stronger as any axiomatization $T^{\prime}$ as above is equivalent to a second order theory in $\mathcal{L}_{\operatorname{Def}}$, but not vice-versa.

This also implies that there is no result analogous to Section 1 in the theory of definably complete type complete structures, namely there is a definably complete type complete structure $\mathcal{M}$ and a pseudo-finite subset $X\subset M$ such that $(\mathcal{M},X)$ does not satisfy the common theory of definably complete type complete structures expanded by a unary predicate for a finite set.

It is still open whether we can extend this result to the case where $\mathcal{M}_{0}$ is an expansion of a real closed field and fully answer Section 1.

Acknowledgements

The author is grateful to Phillip Hieronymi for presenting the question which motivated this paper, as well as for the fruitful discussions. The author is grateful to Assaf Hasson for the fruitful discussions and the warm support along the way. The author thanks Itay Kaplan for his helpful comments on a preliminary version of this paper and for the discussion which helped clarify and strengthen the main result.

Outline

The construction is done as follows: In Section 3, the theory $T_{0}$ is constructed as an expansion of a dense linear without endpoints by a predicate for a discrete, closed, and bounded set $Z$ and some extra structure in the language $\mathcal{L}_{0}$ such that $T_{0}\supseteq T_{\mathcal{L}_{0}}^{omin}$. We then introduce an expansion $\mathcal{L}_{1}\supset\mathcal{L}_{0}$ and $T_{1}\supset T_{0}$ an $\mathcal{L}_{1}$-theory containing a function symbol $f$ which is bijective on $Z$. We show $T_{0}$ and $T_{1}$ are consistent. In Section 4 we prove quantifier elimination for $T_{0}$.

In Section 5, we give the construction of $\mathcal{M}_{2}$ which will be an expansion of some model $\mathcal{M}_{0}$ of $T_{0}$ to $\mathcal{L}_{1}$ with the same one-variable definable sets as $\mathcal{M}_{0}$ such that $\mathcal{M}_{2}$ does not have the pigeonhole principle. This is done by tweaking a given model $\mathcal{M}_{1}$ of $T_{1}$ expanding $\mathcal{M}_{0}$ so that $f$ is now injective but not surjective. It is done carefully enough, so that any definable set in $\mathcal{M}_{2}$ differs from a set definable in $\mathcal{M}_{1}$ by finitely many constant terms. In Section 6 we show quantifier elimination in $\mathcal{M}_{2}$ and deduce that any definable subset of $\mathcal{M}_{2}$ is definable in $\mathcal{M}_{0}$. We then define $\mathcal{M}$ to be a trivial expansion of $\mathcal{M}_{0}$ to $\mathcal{L}_{0}$ and $\mathcal{N}$ to be $\mathcal{M}_{2}$ and show that $\mathcal{M},\mathcal{N}$ possess the properties proclaimed in the introduction.

2. Preliminaries - cyclic orders

In this section, we present the standard definition of a cyclic order, as defined below, and present some of its properties needed for the construction following.

Definition \thedefinition.

A cyclic order on a set $A$ is a ternary relation $C$ satisfying the following axioms:

1. (1)

Cyclicity: If $C(a,b,c)$, then $C(b,c,a)$. 2. (2)

Asymmetry: If $C(a,b,c)$, then not $C(c,b,a)$. 3. (3)

Transitivity: If $C(a,b,c)$ and $C(a,c,d)$, then $C(a,b,d)$. 4. (4)

Totality: If $a,b,c$ are distinct, then either $C(a,b,c)$ or $C(c,b,a)$.

The following fact is folklore (e.g., [Hun35] and [Č66, Part I, §4]) and can be easily verified:

Fact \thefact.

If $\Braket{A,<}$ is a linearly ordered set, then the relation defined by

[TABLE]

is a cyclic order on $A$.

We call $C_{<}$ the cyclic order induced by $<$.**

Definition \thedefinition.

Let $(X,<)$ be a linearly ordered set. A $<$-cut in $X$ is a pair of subsets $(A,B)$ of $X$ such that $X=A\mathbin{\mathaccent 0{\cdot}\cup}B$ and $a<b$ for every $a\in A,b\in B$.

Fact \thefact ([Nov84, Lemma 3.8]).

Let $X$ be a set with two linear orders, $<_{1},<_{2}$, on $X$. Let $(A_{1},B_{1})$ be a $<_{1}$-cut in $X$ and $(A_{2},B_{2})$ be a $<_{2}$-cut in $X$. If $(A_{1},<_{1})\cong(B_{2},<_{2})$ and $(B_{1},<_{1})\cong(A_{2},<_{2})$, then $C_{<_{1}}=C_{<_{2}}$.

Definition \thedefinition.

Let $C$ be a cyclic order on a set $A$. For any $a,b\in A$, denote

[TABLE]

Lemma \thelemma.

Let $C$ be a cyclic order on a set $A$ and let $a,b,c\in A$. If $C(a,b,c)$ then

[TABLE]

Proof.

• •

To prove $C(a,-,c)\supseteq C(a,-,b)\cup\Set{b}\cup C(b,-,c)$:

• –

By definition, $b\in C(a,-,c)$.

• –

If $C(a,x,b)$, then together with $C(a,b,c)$, and transitivity, we get $C(a,x,c)$.

• –

If $C(b,x,c)$, then by cyclicity, $C(c,b,x)$. By cyclicity again, $C(c,a,b)$. Now by transitivity, $C(c,a,x)$, which is equivalent by cyclicity to $C(a,x,c)$.

• •

To prove $C(a,-,c)\subseteq C(a,-,b)\cup\Set{b}\cup C(b,-,c)$, if

[TABLE]

then $x\notin\Set{a,b,c}$ and by totality, $C(b,x,a)$ and $C(c,x,b)$. By cyclicity, we get that $C(x,a,b)$ and $C(x,b,c)$, which in turn, by transitivity, implies $C(x,a,c)$ which by cyclicity is equivalent to $C(c,x,a)$ which by asymmetry, implies that $x\notin C(a,-,c)$.

∎

Definition \thedefinition.

Let $C$ be a cyclic order on a set $A$ and let $X\subseteq A$. Two elements $a,b\in A$ are $X$-close if either $X\cap C(a,-,b)$ or $X\cap C(b,-,a)$ is finite.

Denote $a\sim_{X}b$ if $a,b\in A$ are $X$-close.

Lemma \thelemma.

Let $C$ be a cyclic order on a set $A$ and let $X\subseteq A$. Then $\sim_{X}$ is an equivalence relation on $A$.

Proof.

• •

$X\cap C(a,-,a)=\emptyset$ for all $a\in A$, so reflexivity holds.

• •

Symmetry is obvious by definition.

• •

To prove transitivity, let $a,b,c\in A$ such that $a\sim_{X}b$ and $b\sim_{X}c$. Assume towards a contradiction that $X\cap$ $C(a,-,c)$ and $X\cap C(c,-,a)$ are both infinite. We may further assume, without loss of generality, that $X\cap C(a,b,c)$. So by cyclicity, also $X\cap C(c,a,b)$ and $X\cap C(b,c,a)$. By Section 2,

[TABLE]

By Equation (2), $X\cap C(c,-,b)$ is infinite and by Equation (3), $X\cap C(b,-,a)$ is infinite. But by Equation (1), either $X\cap C(a,-,b)$ or $X\cap C(b,-,c)$ is infinite, so either $a\not\sim_{X}b$ or $b\not\sim_{X}c$.

∎

Lemma \thelemma.

Let $C$ be a cyclic order on a set $A$. Let $a,a^{\prime},b,b^{\prime},c,c^{\prime}\in A$ and let $X\subseteq A$ such that

[TABLE]

Then $C(a,b,c)\iff C(a^{\prime},b^{\prime},c^{\prime})$.

Proof.

By symmetry of $\sim_{X}$ and cyclicity of $C$, it suffices to show that $C(a,b,c)\implies C(a,b,c^{\prime})$. So assume towards a contradiction

[TABLE]

By cyclicity on (5), we get

[TABLE]

By transitivity applied to (4) and (6) we get $C(a,c^{\prime},c)$, which in turn by cyclicity is equivalent to (7) below. By cyclicity and transitivity applied to (4) and (5), we get (8) below.

[TABLE]

By the assumption of the section, either $C(c^{\prime},-,c)$ or $C(c,-,c^{\prime})$ is finite. By Section 2 and by (7) and (8), this implies that at least one of the following is finite: $C(c^{\prime},-,b),C(b,-,c),C(c,-,a),C(a,-,c^{\prime})$, so $a\sim_{X}c$ or $b\sim_{X}c$. Contradiction. ∎

3. Definitions of $T_{0}$ and $T_{1}$

Definition \thedefinition.

Let $\mathcal{L}_{0}:=\Braket{<,Z;S,P,\pi;c_{1},c_{2},c_{3},c_{4}}$ where $<$ is a binary relation symbol, $Z$ is a unary predicate, $S,P,\pi$ are function symbols and $c_{1},c_{2},c_{3},c_{4}$ are constant symbols. Let $T_{0}$ be the $\mathcal{L}_{0}$-theory consisting of the following axioms:

1. (1)

$T^{omin}_{\mathcal{L}_{{0}}}$. 2. (2)

$<$ is a dense linear order without end points. 3. (3)

$Z$ is discretely ordered, i.e., every non-maximal (respectively, non-minimal) element in $Z$ has an immediate successor (respectively, predecessor) in $Z$. 4. (4)

$Z$ is closed, i.e., for all $x\notin Z$, there is an interval disjoint from $Z$ containing $x$. 5. (5)

$\min(Z)=c_{1},\max(Z)=c_{4}$. 6. (6)

$c_{2},c_{3}\in Z$ are such that $c_{1}<c_{2}<c_{3}<c_{4}$ and there are infinitely many elements in $Z$ between any two of them. 7. (7)

$\pi$ is the cyclic forward projection on $Z$:

[TABLE] 8. (8)

$S$ is defined as the cyclic successor function on $Z$, and as the identity outside of $Z$:

[TABLE]

$P$ is defined as $S^{-1}$.

The consistency of $T_{0}$ will be proven together with the consistency of $T_{1}$ defined in Section 3 below.

Definition \thedefinition.

Let $\mathcal{L}_{1}:=\mathcal{L}_{0}\cup\{f,g\}$ where $f,g$ are unary function symbols.

Let $T_{1}$ be $T_{0}$ together with the following axioms:

1. (9)

$f$ is bijective and $g=f^{-1}$. 2. (10)

$f(Z\cap[c_{1},c_{2}])=Z\cap[c_{3},c_{4}]$ and $f\upharpoonright\left(Z\cap[c_{1},c_{2}]\right)$ is a partial order isomorphism. 3. (11)

$f(Z\cap(c_{2},c_{4}])=Z\cap[c_{1},c_{3})$ and $f\upharpoonright\left(Z\cap(c_{2},c_{4}]\right)$ is a partial order isomorphisms. 4. (12)

For all $n>1$ and for every $z\in Z$

[TABLE]

are infinite, i.e. $z\not\sim_{Z}f^{n}(z)$.

Notice that this is a first-order scheme. 5. (13)

$f(x)=x$ for every $x\notin Z$ 6. (14)

$C_{<}\left(f^{m}(z),f^{n}(z),z\right)$ for all $m>n>0$ and for every $z\in Z$.

Proposition \theproposition.

$T_{1}$* is consistent.*

Proof.

We prove finite satisfiability of $T_{1}$ take some sufficiently large natural number $N$. Take $Z=\Set{0,\dots,N}\times\Set{0,\dots,N}$ with the lexicographic order and consider a structure $\mathcal{M}$ which is a DLO containing $Z$ as an ordered subset.

Let $c_{1}:=(0,0),c_{2}:=(0,N),c_{3}:=(N,0),c_{4}:=(N,N)$.

Let

[TABLE]

and let $g:=f^{-1}$

Let $\pi$ the circular projection, as defined in Axiom 7.

Let $S$ be the circular successor function, as defined in Axiom 8 and let $P:=S^{-1}$

Then $\mathcal{M}$ satisfies Axioms 1, 2, 3, 5, 4, 8, 7, 9, 10, 11 and 13 by definition. As for Axioms 6, 12 and 14:

Any finite segment of Axiom 6 is contained in the following axiomatization, for a fixed $k\in\mathbb{N}$:

1. LABEL:axiomInfBetweenk*.

$c_{2},c_{3}\in Z$ are such that $c_{1}<c_{2}<c_{3}<c_{4}$ and there are at least $k$ elements in $Z$ between any two of them. 2. LABEL:finfk*.

For all $k>n>0$ and for every $z\in Z$.

1. (a)

There are at least $k$ elements in $Z\cap C_{<}\left(z,-,f^{n}(z)\right)$. 2. (b)

There are at least $k$ elements in $Z\cap C_{<}\left(f^{n}(z),-,z\right)$. 3. LABEL:circularAxiomk*.

$C_{<}\left(f^{m}(z),f^{n}(z),z\right)$ for all $k>m>n>0$.

If $N>k$ then $\mathcal{M}$ satisfies Axioms LABEL:axiomInfBetweenk* and LABEL:finfk*, by definition.

Under the assumption $N>k$, we prove that $\mathcal{M}$ satisfies Axiom LABEL:circularAxiomk*, thus $T_{1}$ is finitely satisfiable.

For all $(x,y)\in Z$:

[TABLE]

So proving Axiom 14k reduces to proving that for any $x\in\Set{0,\dots,N}$ and $0<n<m<N$ one of the following holds:

1. (a)

$x\ominus m<x\ominus n<x$ 2. (b)

$x\ominus n<x<x\ominus m$ 3. (c)

$x<x\ominus m<x\ominus n$

where $\ominus$ is subtraction modulo $N+1$.

If $m\leq x$         then (a) holds.

If $n\leq x<m$  then (b) holds.

If $x<n$         then (c) holds. ∎

4. Quantifier Elimination in $T_{0}$

We now show that $T_{0}$ eliminates quantifiers:

Remark \theremark.

Let $\mathcal{M}\models T_{0}$ and $\tau,a\in\mathcal{M}$. Then the following hold:

1. (1)

$S(\tau)\in Z\iff\tau\in Z$** 2. (2)

$P(\tau)\in Z\iff\tau\in Z$** 3. (3)

$S(\tau)=a\iff\tau=P(a)$** 4. (4)

$S(\tau)<a\iff\tau<P(a)$** 5. (5)

$S(\tau)>a\iff\tau>P(a)$** 6. (6)

$\pi(\tau)\in Z\iff c_{1}\in Z$** 7. (7)

$\pi(\tau)=a\iff\left[\tau\leq c_{4}\land a\in Z\land P(a)<\tau\leq a\right]\lor\left[\tau>c_{4}\land c_{1}=a\right]$** 8. (8)

$\pi(\tau)<a\iff\left[\tau\leq c_{4}\land\tau\leq P\circ\pi(a)\right]\lor\left[\tau>c_{4}\land c_{1}<a\right]$** 9. (9)

$\pi(\tau)>a\iff\left[\tau\leq c_{4}\land\tau\geq P\circ\pi\circ S(a)\right]\lor\left[\tau>c_{4}\land c_{1}>a\right]$**

Remark \theremark.

If $x\in Z$ then:

1. (1)

$S^{m_{1}}\circ\pi^{n_{1}}\circ\dots\circ S^{m_{k}}\circ\pi^{n_{k}}\circ S^{l}(x)=S^{m_{1}+\dots+m_{k}+l}(x)$. 2. (2)

$P^{m}(x)=S^{-m}(x)$* and $P^{-m}(x)=S^{m}(x)$ for all $m\in\mathbb{N}$.* 3. (3)

$S^{m}(x)\square x\iff S^{m}(c_{2})\square c_{2}$* for all $m\in\mathbb{N},\square\in\Set{<,>,=}$, $x\notin\Set{c_{4},P(c_{4}),\dots,P^{m}(c_{4}),}$. $P^{m}(x)\square x\iff S^{m}(c_{2})\square c_{2}$ for all $m\in\mathbb{N},\square\in\Set{<,>,=}$, $x\notin\Set{c_{1},S(c_{1}),\dots,S^{m}(c_{1}),}$.*

If $x\notin Z$:

1. (1)

$S^{m_{1}}\circ\pi^{n_{1}}\circ\dots\circ S^{m_{k}}\circ\pi^{n_{k}}\circ S^{l}(x)=S^{m_{1}+\dots+m_{k}}\circ\pi(x)$. 2. (2)

$S^{m}(x)\square x\iff c_{1}\square c_{1}$* for all $m\in\mathbb{Z},\square\in\Set{<,>,=}$.* 3. (3)

$S^{m}\circ\pi(x)=x\iff c_{1}\neq c_{1}$* for $m\neq 0$.* 4. (4)

$S^{m}\circ\pi(x)>x\iff S^{m+1}\circ\pi(x)>\pi(x)$. 5. (5)

$S^{m}\circ\pi(x)<x\iff S^{m}\circ\pi(x)<\pi(x)$.

Lemma \thelemma.

For any $\mathcal{M}\models T_{0}$ and $a,b\in\mathcal{M}$,

[TABLE]

Proof.

If $a\in Z$ then $\mathcal{M}\models\left[\exists x\in Z(a<x<b)\right]\leftrightarrow\left[S(a)<b\right]$ and $S(a)=\pi\circ S(a)$.

If $a\notin Z$ then $\mathcal{M}\models\left[\exists x\in Z(a<x<b)\right]\leftrightarrow\left[\pi(a)<b\right]$ and $\pi(a)=\pi\circ S(a)$. ∎

Proposition \theproposition.

$T_{0}$* admits quantifier elimination.*

Proof.

Let $\phi=\exists x\bigwedge_{i\in I}\theta_{i}\left(\bar{y},x\right)$ such that $\{\theta_{i}\}_{i\in I}$ are atomic and negated atomic formulas. We need to find a quantifier-free $\mathcal{L}_{0}$-formula $\varphi$ such

[TABLE]

Firstly, since \vdash\exists x\big{(}\chi\left(\bar{y},x\right)\land\theta\left(\bar{y}\right)\big{)}\leftrightarrow\exists x\big{(}\chi\left(\bar{y},x\right)\big{)}\land\theta\left(\bar{y}\right) we may assume that $x$ occurs in $\theta_{i}$ for all $i\in I$. Secondly,

[TABLE]

[TABLE]

So we may assume $\phi$ is either of the form $\exists x\left(\bigwedge_{i\in I}\theta_{i}\left(\bar{y},x\right)\land x\in Z\right)$ or of the form $\exists x\left(\bigwedge_{i\in I}\theta_{i}\left(\bar{y},x\right)\land x\notin Z\right)$ where $\theta_{i}$ are atomic and negated atomic formulas such that $x$ occurs in each $\theta_{i}$. We may assume that $\theta_{i}$ is neither ‘$x\in Z$’ nor ‘$x\notin Z$’ for any $i\in I$, as such occurrence would be either superfluous or inconsistent. So each $\theta_{i}$ is of the form $t_{1}\square t_{2}$ where $t_{1},t_{2}$ are terms with variables in $x,\bar{y}$.

By Section 4, we may assume either

[TABLE]

or

[TABLE]

where $t_{i}$ are with variables from $\Set{x,\bar{y}}$, $\square\in\Set{<,>,=,\leq,\geq,\neq}$. By Section 4, we may assume that $x$ does not occur in any $t_{i}$. Next, notice that $\geq,\leq,\neq$ are positive Boolean combinations of $<,>,=$ and if $\square_{i}$ is $``="$ for some $i$ we can just replace $x$ with $t_{i}$. So we may assume $\square_{i}\in\Set{<,>}$, i.e. either

[TABLE]

or

[TABLE]

where $l_{i},u_{i}$ are terms not containing $x$.

If $\phi$ is as in (9), then by Section 4, $\phi(\bar{y})$ is equivalent to

[TABLE]

If $\phi$ is as in (10), then since $Z$ is co-dense, $\phi(\bar{y})$ is equivalent to

[TABLE]

∎

5. Definition of $T_{2}$ and the relation to $T_{1}$

Definition \thedefinition.

Let $\mathcal{M}_{1}\models T_{1}$ be arbitrary, with universe $M$.

Let $\mathcal{M}_{0}$ be the restriction of $\mathcal{M}$ to $\mathcal{L}_{0}$, i.e., $\mathcal{M}_{0}=\mathcal{M}_{1}\upharpoonright\mathcal{L}_{0}$. Consequently, $\mathcal{M}_{0}\models T_{0}$.

Let $\mathcal{M}_{2}$ be the same $\mathcal{L}_{1}$-structure as $\mathcal{M}$ with a slight modification on $f$ and $g$, as follows.

[TABLE]

In words, there is some convex set $X$ with maximum $c_{4}$ such that the order type of $X\cap Z$ is $\omega^{*}$. $f^{\mathcal{M}_{1}}$ maps $X\cap Z$ to a convex subset $f^{\mathcal{M}_{1}}(X\cap Z)$ of $Z$ of order type $\omega^{*}$ with maximum $P(c_{3})$, by Axiom 11 in Section 3.

Then $f^{\mathcal{M}_{2}},g^{\mathcal{M}_{2}}$ are obtained from $f^{\mathcal{M}_{1}},g^{\mathcal{M}_{1}}$ by applying a shift by one element in $X\cap Z$, $f(X\cap Z)$ respectively.

Lemma \thelemma.

$f^{\mathcal{M}_{1}}$* preserves the cyclic order on $Z$, i.e.*

[TABLE]

Proof.

Define a new ordering $<^{\prime}$ on $Z$ by

[TABLE]

By Axioms 10 and 11 in Section 3, $\left([c_{1},c_{2}],<\right)\cong([c_{3},c_{4}],<^{\prime})$ and $\left((c_{2},c_{4}],<\right)\cong([c_{1},c_{3}),<^{\prime})$ and

[TABLE]

Additionally, by definition of $<^{\prime}$, it follows that $\left([c_{3},c_{4}],[c_{1},c_{3})\right)$ is a $<^{\prime}$-cut in $Z$ and $\left([c_{1},c_{2}],(c_{2},c_{4}]\right)$ is a $<$-cut in $Z$. So by Section 2, $C_{<^{\prime}}=C_{<}$. In conclusion

[TABLE]

∎

Lemma \thelemma.

Let $f,g,S,P$ be as in Section 3. Then $\Braket{f,g,S,P}_{cl}$, the closure of $\Set{f,g,S,P}$ under composition is an Abelian group.

Proof.

By definition, $g\circ f=I=P\circ S$, so $f,S$ are invertible and $\Braket{f,g,S,P}_{cl}=\Braket{f,S}_{grp}$ where $\Braket{f,S}_{grp}$ is the group generated by $\Set{f,S}$.

Since $S$ is definable by the cyclic order on $Z$ (Axiom 8 in Section 3) and $f$ preserves the cyclic order on $Z$ (Section 5), it follows that $f\circ S(x)=S\circ f(x)$. Now $\Braket{f,S}_{grp}$ is Abelian, as the group defined by $\Braket{a,b}{ab=ba}$ is Abelian. ∎

Corollary \thecorollary.

Let $n\geq 1$ and $x\in Z$.

[TABLE]

Proof.

Since $x\in Z$, so is $\left(g^{\mathcal{M}_{1}}\right)^{n}(x)$. Therefore, by Axiom 12,

[TABLE]

By Section 5, $\left(f^{\mathcal{M}_{1}}\right)^{n}\circ\left(g^{\mathcal{M}_{1}}\right)^{n}(x)=x$. ∎

Lemma \thelemma.

Let $x\in M$ and $n\in\mathbb{N}$. There are $k_{1},k_{2}\in\mathbb{N}$ such that

[TABLE]

Proof.

By definition of $f^{\mathcal{M}_{2}},g^{\mathcal{M}_{2}}$ (Section 5), there are $\varepsilon_{1},\dots,\varepsilon_{n},\nu_{1},\dots,\nu_{k}\in\Set{0,1}$ such that

[TABLE]

By Section 5, the right hand side in Equation (11) is equal to

[TABLE]

and the right hand side in Equation (12) is equal to

[TABLE]

∎

Corollary \thecorollary.

For all $n\in\mathbb{N}$ and every $x\in M$:

[TABLE]

6. Quantifier elimination in $T_{2}$

In this section, unless otherwise specified, we work inside $\mathcal{M}_{2}$, so $f$ is $f^{\mathcal{M}_{2}}$ and $g$ is $g^{\mathcal{M}_{2}}$.

Lemma \thelemma.

$\mathcal{M}_{2}$* satisfies the following:*

1. (1)

$f(Z\cap[c_{1},c_{2}])=Z\cap[c_{3},c_{4}]$* and $f\upharpoonright\left(Z\cap[c_{1},c_{2}]\right)$ is a partial order isomorphism, and its inverse is $g\upharpoonright Z\cap[c_{3},c_{4}]$.* 2. (2)

$f(Z\cap(c_{2},c_{4}])=Z\cap[c_{1},P(c_{3}))$* and $f\upharpoonright\left(Z\cap(c_{2},c_{4}]\right)$ is a partial order isomorphisms, and its inverse is $g\upharpoonright Z\cap[c_{1},P(c_{3}))$.* 3. (3)

$g(x)=f(x)=x$* for every $x\notin Z$* 4. (4)

$f$* is injective and not surjective on $Z$. Moreover, $f(Z)=Z\setminus\Set{P(c_{3})}$.* 5. (5)

$g\circ f(x)=x$* for all $x\in M$.* 6. (6)

$f\circ g(x)=x$* for all $x\in M\setminus\Set{P(c_{3})}$.* 7. (7)

For all $n\geq 1$ and for every $z\in Z$

[TABLE]

are infinite, i.e. $z\not\sim_{Z}f^{n}(z)$. 8. (8)

For all $n\geq 1$ and for every $z\in Z$

[TABLE]

are infinite, i.e. $z\not\sim_{Z}g^{n}(z)$.

Proof.

• •

Items 1, 2 and 3 follow by definition of $f^{\mathcal{M}_{2}}$ and by Axioms 10, 11 and 13 in Section 3.

• •

Item 4 follows from Items 1 and 2, as

[TABLE]

• •

To prove Item 5, we separate into two cases:

• –

if $S^{n}(x)=c_{4}$ for some $n\in\mathbb{N}$, then $S^{n}\circ f^{\mathcal{M}_{1}}(x)=P(c_{3})$, so

[TABLE]

So by definition of $g^{\mathcal{M}_{2}}$,

[TABLE]

• –

if $S^{n}(x)\neq c_{4}$ for all $n\in\mathbb{N}$, then $S^{n}\circ f^{\mathcal{M}_{1}}(x)\neq P(c_{3})$ for all $n\in\mathbb{N}$. So

[TABLE]

• •

To prove Item 6, by Items 4 and 3, for all $x\in M\setminus\Set{P(c_{3})}$, $x=f(y)$ for some $y\in M$, therefore by Item 5

[TABLE]

• •

Item 7 follows from Axiom 12 in Section 3 and Section 5.

• •

Item 8 follows from 5 and Section 5.

∎

Corollary \thecorollary.

Let $a,b\in Z$, $\square\in\Set{<,>,=}$

1. (1)

If $a\in[c_{1},c_{2}]$ and $b\in[c_{3},c_{4}]$, then $\mathcal{M}_{2}\models f(a)\square b\iff a\square g(b)$. 2. (2)

If $a\in[c_{1},c_{2}]$ and $b\notin[c_{3},c_{4}]$, then $\mathcal{M}_{2}\models f(a)\square b\iff c_{3}\square b$. 3. (3)

If $a\in(c_{2},c_{4}]$ and $b\in[c_{1},P(c_{3}))$, then $\mathcal{M}_{2}\models f(a)\square b\iff a\square g(b)$. 4. (4)

If $a\in(c_{2},c_{4}]$ and $b\notin[c_{1},P(c_{3}))$, then $\mathcal{M}_{2}\models f(a)\square b\iff c_{1}\square b$. 5. (5)

If $a\in[c_{1},P(c_{3}))$ and $b\notin(c_{2},c_{4}]$, then $\mathcal{M}_{2}\models g(a)\square b\iff c_{4}\square b$. 6. (6)

If $a\in[c_{3},c_{4}]$ and $b\notin[c_{1},c_{2}]$, then $\mathcal{M}_{2}\models g(a)\square b\iff c_{1}\square b$.

Proof.

$(1)$ and $(2)$ follow from Section 6, Items 1 and 6.

$(3)$ and $(4)$ follow from Section 6, Items 2 and 6.

$(5)$ follows from Section 6, Items 2 and 5.

$(6)$ follows from Section 6, Items 1 and 5. ∎

Corollary \thecorollary.

Let $x\in M,y\in Z$, $\square\in\Set{<,>,=}$.

[TABLE]

Remark \theremark.

If $x\notin Z$ then $\mathcal{M}_{2}\models f(x)=g(x)=x$. In particular,

• •

$\mathcal{M}_{2}\models f(x)\in Z\leftrightarrow x\in Z$* for all $x\in M$.*

• •

$\mathcal{M}_{2}\models g(x)\in Z\leftrightarrow x\in Z$* for all $x\in M$.*

• •

$\mathcal{M}_{2}\models f(x)\square y\leftrightarrow g(x)\square y\leftrightarrow x\square y$* for any $x\in M\setminus Z,y\in M$, $\square\in\Set{<,>,=}$.*

Remark \theremark.

If $x\in Z,y\notin Z$ then:

• •

$\mathcal{M}_{2}\models x>y\leftrightarrow x\geq\pi(y)$.

• •

$\mathcal{M}_{2}\models x<y\leftrightarrow x\leq P\circ\pi(y)$.

Corollary \thecorollary.

[TABLE]

Definition \thedefinition.

Following standard terminology, a constant term is a term with no free variables.

Definition \thedefinition.

Given two $\mathcal{L}_{1}$-definable maps $F,G:M\to M$, denote $F\approx G$ if there are finitely many constant terms $\tau_{1},\dots,\tau_{k}$, such that

[TABLE]

$\approx$ is an equivalence relation. For any $\mathcal{L}_{1}$-definable map $F:M\to M$, let $[F]$ be its equivalence class.

Lemma \thelemma.

$f\circ S\approx S\circ f$.

Proof.

• •

If $x\notin Z$ then both $S$ and $f$ are the identity on $x$, so the equality $f\circ S(x)=S\circ f(x)$ is trivial.

• •

If $x\in Z$ and $c_{1}<x<c_{2}$ or $c_{2}<x<c_{4}$ then the equality $f\circ S(x)=S\circ f(x)$ follows by Items 1 and 2 in Section 6.

In conclusion, the equality $f\circ S(x)=S\circ f(x)$ holds for all $x\neq c_{1},c_{2},c_{4}$.

∎

For any finite-to-one map $F,F^{\prime},G,G^{\prime}:M\to M$, if $F\approx F^{\prime}$ and $G\approx G^{\prime}$ then $F\circ G\approx F\circ G$. Since $f,S,P$ are injective and $g$ is injective outside $\{P(c_{3})\}$, the composition $[F]\circ[G]:=[F\circ G]$ is well defined, for any composition of $f,g,S,P$.

Proposition \theproposition.

$\Braket{[f],[g],[S],[P]}_{cl}$, the closure of $\Set{[f],[g],[S],[P]}$ under composition is an Abelian group.

Proof.

[TABLE]

So $[g][f]=[f][g]=[P][S]=[S][P]=1$. In particular $[f],[S]$ are invertible and $\Braket{[f],[g],[S],[P]}=\Braket{[f],[S]}_{grp}$ where $\Braket{[f],[S]}_{grp}$ is the group generated by $\Set{[f],[g]}$. By Section 6, $[f][S]=[S][f]$. The claim now follows from the fact the group defined by $\Braket{a,b}{ab=ba}$ is Abelian. ∎

Remark \theremark.

Let $x\in M$ and $F\in\Set{S,P,\pi}$. If there are infinitely many elements in $Z$ between $x$ and $F(x)$, then $F(x)\in\Set{c_{1},c_{4}}$.

By infinitely many elements in $Z$ between $x$ and $F(x)$, we mean with respect to the order $<$ and not the cyclic order $C_{<}$, i.e., either $x<F(x)$ and $Z\cap[x,F(x)]\geq\aleph_{0}$, or $F(x)<x$ and $Z\cap[F(x),x]\geq\aleph_{0}$.

This is weaker than $x\not\sim_{Z}F(x)$; for example, $c_{1}\sim_{Z}c_{4}$ but there are infinitely many elements in $Z$ between $c_{1}$ and $c_{4}$.

Lemma \thelemma.

Let $F,G\in\Braket{S,P,\pi}_{cl}$. Then there are finitely many constant terms $\tau_{1},\dots,\tau_{k}$, such that if $F(x)\notin\Set{\tau_{1},\dots,\tau_{k}}$, then there are only finitely many elements in $Z$ between $x$ and $F(x)$.

Proof.

Let $F=G_{k}\circ\dots\circ G_{1}$ where $G_{1},\dots,G_{k}\in\Set{S,P,\pi}$. Let $F_{i}:=Id$, $F_{i}:=G_{i}\circ\dots\circ G_{1}$ for any $1\leq i\leq k$, so $F=F_{k}$. If there are infinitely many elements in $Z$ between $x$ and $F(x)$, then there is some $1\leq i\leq k$ with infinitely many elements in $Z$ between $F_{i}(x)$ and $F_{i-1}(x)$, so by Section 6, $F_{i}(x)\in\Set{c_{1},c_{4}}$ and thus $F(x)=F_{k}(x)=F_{k-i}\circ F_{i}(x)\in\Set{F_{k-i}(c_{1}),F_{k-i}(c_{4})}$. So if

[TABLE]

then there are finitely many elements in $Z$ between $x$ and $F(x)$. ∎

Lemma \thelemma.

$C_{<}\left(f^{m}(z),f^{n}(z),z\right)$* for all $m>n>0$ and for every $z\in Z$.*

Proof.

Let $m>n>0$ and $z\in Z$. By Axiom 14 in Section 3,

[TABLE]

By Section 5,

[TABLE]

and the section follows from Section 2. ∎

Lemma \thelemma.

for any $n\in\mathbb{N}$ and $z\in Z$:

[TABLE]

Proof.

We prove the section by induction on $n$. For $n=0$ the claim holds by definition of $f$. For $n\geq 1$, By Section 6, $C_{<}(f^{n+1}(z),f^{n}(z),z)$. So

[TABLE]

By the induction hypothesis, $f^{n}(z)<z$ is equivalent to $\bigwedge_{i=0}^{n-1}(f^{i}(z)>c_{2})$ and $f^{n+1}(z)<f^{n}(z)$ is equivalent to $f^{n}(z)>c_{2}$. ∎

Definition \thedefinition.

1. (1)

$\Phi:=\Set{\phi^{n}}{\phi\in\Set{f,g},\ n\in\mathbb{N}}$. 2. (2)

$\Sigma:=\Set{\sigma^{m}}{\sigma\in\set{S,P},m\in\mathbb{N}}$. 3. (3)

$\Pi:=\Set{\pi^{\epsilon}}{\epsilon\in\set{0,1}}$. 4. (4)

For any functions $h_{1},\dots,h_{n}$ and $A,B\subseteq\Braket{h_{1},\dots,h_{n}}_{cl}$, let $AB:=\Set{a\circ b}{a\in A,b\in B}$.

Lemma \thelemma.

Let $n\geq 1$, $\psi_{1},\psi_{2}\in\Sigma\Pi$, and $\square\in\Set{<,>,=}$. Then

1. (1)

There are constant terms $\tau_{1},\dots,\tau_{k}$ such that

[TABLE] 2. (2)

There are constant terms $\sigma_{1},\dots,\sigma_{l}$ such that

[TABLE]

Proof.

1. (1)

By Section 6 applied twice, there are constant terms $\tau_{1},\dots,\tau_{k}$ such that whenever $\psi(x)_{1},\psi_{2}(x)\notin\Set{\tau_{1},\dots,\tau_{k}}$, there are finitely many elements in $Z$ between $\psi(x)_{1}$ and $\psi_{2}(x)$.

• •

If $\psi_{1}(x)\notin Z$, then by Item 3 of Section 6, $f^{n}\circ\psi_{1}(x)=\psi_{1}(x)$. In particular,

[TABLE]

• •

If $\psi_{1}(x)\in Z$, $\psi_{1}(x),\psi_{2}(x)\notin\Set{\tau_{1},\dots,\tau_{k}}$, then by Section 6, Item 7 there are infinitely many elements in $Z$ between $\psi_{1}(x)$ and $f^{n}\circ\psi_{1}(x)$. As there are only finitely many elements in $Z$ between $\psi_{1}(x)$ and $\psi_{2}(x)$, it follows that

[TABLE] 2. (2)

The proof is similar.

∎

Definition \thedefinition.

1. (1)

We define $\deg(F)$ for $F\in\Braket{f,g,S,P,\pi}_{cl}$ inductively, as follows:

• •

$\deg(Id)=\deg(S)=\deg(P)=\deg(\pi)=0$

• •

$\deg(f)=\deg(g)=1$.

• •

$\deg(F\circ G)=\deg(F)+\deg(G)$ for all $F,G\in\Braket{f,g,S,P,\pi}_{cl}$.

Notice that this is a syntactic definition, e.g., $\deg(F\circ G)=2$. 2. (2)

For any quantifier free $\mathcal{L}_{1}$-formula $\theta(x,\bar{y})$ and variable $x$ we define $\operatorname{rank}(\theta,x)\in\left(\Set{-\infty}\cup\mathbb{N}\right)^{2}$ by induction on the complexity of $\theta$:

• •

If $x$ does not occur in $\theta$, then $\operatorname{rank}(\theta,x)=(-\infty,-\infty)$.

• •

If $\theta$ is atomic of the form $F(x)\in Z$ then $\operatorname{rank}(\theta,x)=(-\infty,\deg(F))$.

• •

If $\theta$ is atomic of the form $F(x)\square\tau$ where $F\in\Braket{f,g,S,P}_{cl}$, $\square\in\Set{<,>,=}$, and $\tau$ is an $\mathcal{L}_{1}$-term such that $x$ does not occur in $\tau$, then $\operatorname{rank}(\theta,x)=(-\infty,\deg(F))$.

• •

If $\theta$ is atomic of the form $F(x)\square G(x)$ where $F,G\in\Braket{f,g,S,P}_{cl}$, $\square\in\Set{<,>,=}$, and $\deg(F)\leq\deg(G)$, then $\operatorname{rank}(\theta,x)=(\deg(F),\deg(G))$.

• •

If $\theta$ is a Boolean combination of atomic formulas $\theta_{1},\dots,\theta_{k}$, then $\operatorname{rank}(\theta,x)$ is the lexicographic maximum of $\Set{\operatorname{rank}(\theta_{i},x)}_{i=1}^{k}$.

Definition \thedefinition.

A quantifier free $\mathcal{L}_{1}$-formula $\theta(x,\bar{y})$ is $x$-corrected if any term $F(x)$ appearing in $\theta$ belongs to $\Phi\Sigma\Pi$.

Lemma \thelemma.

For any quantifier free $\mathcal{L}_{1}$-formula $\varphi$ and variable $x$, there is some $x$-corrected formula $\phi$ such that $\operatorname{rank}(\varphi,x)\leq\operatorname{rank}(\phi,x)$ and $T_{2}\models\varphi\leftrightarrow\phi$.

Proof.

A Boolean combination of $x$-corrected formulas is $x$-corrected, so we may assume $\varphi$ is atomic.

• •

If $\varphi$ is of the form $F(x)\in Z$ for some $G\in\Braket{f,g,S,P,\pi}_{cl}\setminus\Braket{f,g,S,P}_{cl}$ then $T_{2}\models F(x)\in Z\leftrightarrow\pi(x)\in Z$.

• •

If $\varphi$ is of the form $F(x)\in Z$ for some $G\in\Braket{f,g,S,P}$ then $T_{2}\models G(x)\in Z\leftrightarrow x\in Z$.

• •

If $\varphi$ of the form $F(x)\square\tau$ for some term $\tau$ and $\square\in\Set{<,>,=}$:

• –

If $F\in\Braket{f,g,S,P}_{cl}$, then by Section 6, there is some $F^{\prime}\in\Phi\Sigma$ with $\deg(F^{\prime})=\deg(F)$, and constant terms $\tau_{1},\dots,\tau_{k}$ such that $F(x)=F^{\prime}(x)$f for all $x\notin\Set{\tau_{1},\dots,\tau_{k}}$.

So

[TABLE]

• –

If $F\in\Braket{f,g,S,P,\pi}_{cl}\setminus\Braket{f,g,S,P}_{cl}$, then there are $F_{1},F_{2}\in\Braket{f,g,S,P}_{cl}$ such that $\mathcal{M}_{2}\models F(x)=F_{1}\circ\pi\circ F_{2}(x)$ for all $x\in M$ and $\deg(F_{1})\leq\deg(F_{1}\circ F_{2})=\deg(F)$. So $\mathcal{M}_{2}\models F(x)=F_{1}\circ F_{2}(x)$ for all $x\in Z$ and $\mathcal{M}_{2}\models F(x)=F_{1}\circ\pi(x)$ for all $x\notin Z$. So

[TABLE]

and $F_{1},F_{1}\circ F_{2}\in\Braket{f,g,S,P}_{cl}$, so we can apply the previous case to get a formula where every term $F(x)$ to the left of $\square$ belongs to $\Phi\Sigma\Pi$.

If $x$ does not appear in $\tau$ we are done. Otherwise, if $\tau=G(x)$ for some term $G$, a symmetric argument applied to $G$ will an $x$-corrected formula $\phi$ equivalent to $\varphi$ as needed.

∎

Lemma \thelemma.

1. (1)

Let $\varphi$ be an $x$-corrected atomic formula of rank $(-\infty,n+1)$ or of rank $(n+1,k)$ for some $n,k\in\mathbb{N}$. Then there is some quantifier free formula $\phi$ such that $\operatorname{rank}(\phi,x)<\operatorname{rank}(\varphi,x)$ and $T_{2}\models\varphi\leftrightarrow\phi$.

Proof.

1. (1)

Assume $\operatorname{rank}(\varphi,x)=(-\infty,n+1)$.

If $\varphi$ is of the form $F\circ H(x)\in Z$ where $F\in\Phi,H\in\Sigma\Pi$, by Section 6, $\mathcal{M}_{2}\models F\circ H(x)\in Z\iff H(x)\in Z$ and $\operatorname{rank}(H(x)\in Z,x)=(-\infty,0)$.

If $\varphi$ is of the form $F\circ H(x)\square\tau$ where $F\in\Phi,H\in\Phi\Sigma\Pi$, $\deg(F)=1,\deg(H)=n$, and $\tau$ is some $\mathcal{L}_{1}$-term not containing $x$

In which case, $F\in\Set{f,g}$ and

[TABLE]

1. (a)

Applying Section 6 to $H(x),\tau$, there is some $\phi^{\prime}(x,\tau)$ $x$-corrected formula $\phi^{\prime}(x,\tau)$ of rank $(-\infty,n)$ such that $\mathcal{M}_{2}\models\phi^{\prime}(x,\tau)\leftrightarrow F\circ H(x)\square\tau$ for all $\tau\in Z$. So

[TABLE] 2. (b)

Applying Section 6 to $F\circ H(x)$ and $\tau$, we obtain that

[TABLE]

is equivalent to one of the following:

[TABLE]

As in (1)a, there are $\phi^{\prime}_{1}(x,\pi(\tau))$ and $\phi^{\prime}_{2}(x,P\circ\pi(\tau))$ of rank $(-\infty,n)$ equivalent to $F\circ H(x)\geq\pi(\tau)\land\pi(\tau)\in Z$ and $F\circ H(x)\leq P\circ\pi(\tau)\land P\circ\pi(\tau)\in Z$, respectively. So (13) is equivalent to one of the following:

[TABLE]

an each is $x$-corrected of rank $(-\infty,n)$.

By Section 6, $\left(F\circ H(x)\square\tau\land\tau\notin Z\land H(x)\notin Z\right)$ is equivalent to

[TABLE]

and the latter is an $x$-corrected formula of rank $(-\infty,n)$. 2. (2)

Assume $\operatorname{rank}(\varphi,x)=(n+1,k)$. Then $\varphi$ is of the form $F\circ H(x)\square G(x)$ where $F\in\Phi,H,G\in\Phi\Sigma\Pi$, $\deg(F)=1,\deg(H)=n,deg(G)=k$ and $n<k$. Replace $G(x)$ with $G(y)$ on the left of $\square$ to get $F\circ H(x)\square G(y)$. Apply Item (1) of this section to $F\circ H(x)\square G(y)$ and get some quantifier-free formula $\phi^{\prime}(x,y)$ with $\operatorname{rank}(\phi^{\prime}(x,y),x)\leq(-\infty,n)$. Replacing back, we get $\operatorname{rank}(\phi^{\prime}(x,x),x)<(n+1,k)$ for any $k$ and

[TABLE]

∎

Lemma \thelemma.

Let $\varphi$ be an $x$-corrected atomic formula of rank $(0,k+1)$ for some $k\in\mathbb{N}$. Then there is some quantifier free formula $\phi$ such that $\operatorname{rank}(\phi,x)<\operatorname{rank}(\varphi,x)$ and $T_{2}\models\varphi\leftrightarrow\phi$.

Proof.

By Section 6, we may assume $\varphi$ is either of the form $f^{k+1}\circ\psi(x)\square\psi(x)$ or of the form $g^{k+1}\circ\psi(x)\square\psi(x)$ for some $\psi\in\Sigma\Pi,\square\in\Set{<,>,=}$.

1. (1)

In case $\varphi$ is $f^{k+1}\circ\psi(x)\square\psi(x)$, by Section 6,

[TABLE]

and the formula in (21) is of rank $(0,0)$. 2. (2)

In case $\varphi$ is $g^{k+1}\circ\psi(x)\square\psi(x)$, by Section 6, there are finitely many constant terms $\tau_{1},\dots,\tau_{m}$ such that $f^{k+1}\circ g^{k+1}(x)=x$ for all $x\notin\Set{\tau_{1},\dots,\tau_{m}}$. So

[TABLE]

and $\operatorname{rank}\left(\bigvee_{i=0}^{k}(c_{1}\leq f^{i}\circ\psi(x)\leq c_{2}),\ x\right)=(-\infty,k-1)$

Now, replacing $\psi(x)$ with $g^{k+1}\circ\psi(x)$ in (20), we get

[TABLE]

By noticing that $\vdash g^{k+1}\circ\psi(x)=g^{k+1}\circ\psi(x)\leftrightarrow x=x$, the formula in (23) is of rank $(0,0)$.

∎

Lemma \thelemma.

Let $\varphi$ be an $x$-corrected atomic formula of rank $(0,0)$. Then there is some quantifier free formula $\phi$ such that $\operatorname{rank}(\phi,x)<\operatorname{rank}(\varphi,x)$ and $T_{2}\models\varphi\leftrightarrow\phi$.

Proof.

By Sections 4 and 4 we may assume $\varphi$ is of the form $\psi(x)\square x$ where $\psi\in\Sigma\Pi$ and $\square\in\Set{<,>,=}$. Now

[TABLE]

By Section 4, the right hand side is equivalent to a quantifier free formula of rank $(-\infty,-\infty)$. ∎

Lemma \thelemma.

Let $\varphi$ be a quantifier free formula with free variable $x$. Then there is some $x$-corrected formula $\phi$ such that $\operatorname{rank}(\phi,x)\leq(-\infty,0)$ for some $k\in\mathbb{N}$ and $T_{2}\models\varphi\leftrightarrow\phi$.

Proof.

By Section 6 we may assume $\varphi$ is $x$-corrected. Since the lexicographic order on well-ordered sets is well-ordered, by induction it suffices to show that if $\operatorname{rank}(\varphi,x)>(-\infty,0)$, then there is some $x$-corrected $\phi$ such that $\operatorname{rank}(\phi,x)<\operatorname{rank}(\varphi,x)$ and $T_{2}\models\varphi\leftrightarrow\phi$. As a Boolean combination of formulas of rank at most $(-\infty,0)$ is of rank at most $(-\infty,0)$ as well, we may further assume that $\varphi$ is atomic.

• •

If $\operatorname{rank}(\varphi,x)=(n+1,k)$ for some $n,k\in\mathbb{N}$, then by Section 6 there is some quantifier free formula $\phi^{\prime}$ such that $\operatorname{rank}(\phi^{\prime},x)<\operatorname{rank}(\varphi,x)$.

• •

If $\operatorname{rank}(\varphi,x)=(0,k+1)$ for some $k\in\mathbb{N}$, then by Section 6 there is some quantifier free formula $\phi^{\prime}$ such that $\operatorname{rank}(\phi^{\prime},x)<\operatorname{rank}(\varphi,x)$.

• •

If $\operatorname{rank}(\varphi,x)=(0,0)$ for some $k\in\mathbb{N}$, then by Section 6 there is some quantifier free formula $\phi^{\prime}$ such that $\operatorname{rank}(\phi^{\prime},x)<\operatorname{rank}(\varphi,x)$.

• •

If $\operatorname{rank}(\varphi,x)=(-\infty,n+1)$ for some $k\in\mathbb{N}$, then by Section 6 there is some quantifier free formula $\phi^{\prime}$ such that $\operatorname{rank}(\phi^{\prime},x)<\operatorname{rank}(\varphi,x)$.

So in conclusion, whenever $\varphi$ is $x$-corrected and $\operatorname{rank}(\varphi,x)>(-\infty,0)$, there is some quantifier free formula $\phi^{\prime}$ such that $T_{2}\models\varphi\leftrightarrow\phi^{\prime}$ and $\operatorname{rank}(\phi^{\prime},x)<\operatorname{rank}(\varphi,x)$. By Section 6, there is some $x$-corrected formula $\phi$ such that $T_{2}\models\varphi\leftrightarrow\phi^{\prime}\leftrightarrow\phi$ and $\operatorname{rank}(\phi,x)=\operatorname{rank}(\phi^{\prime},x)<\operatorname{rank}(\varphi,x)$. ∎

Theorem 6.1**.**

$T_{2}$* admits quantifier elimination.*

Proof.

Let $\varphi(x,y_{1},\dots,y_{k})$ be a quantifier free $\mathcal{L}_{1}$-formula. It suffices to find a quantifier free formula $\phi(y_{1},\dots,y_{k})$ such that $T_{2}\models\exists x\varphi(x,y_{1},\dots,y_{k})\leftrightarrow\phi(y_{1},\dots,y_{k})$. By Section 6, we may assume $\varphi$ is $x$-corrected and $\operatorname{rank}(\varphi,x)=(-\infty,0)$. Since $\operatorname{rank}(\varphi,x)=(-\infty,0)$, there is some quantifier-free $\mathcal{L}_{0}$-formula $\varphi^{\prime}(x,z_{1},\dots,z_{l})$ and $\mathcal{L}_{1}$-terms $t_{1},\dots,t_{l}$ with variables in $\Set{y_{1},\dots,y_{k}}$ such that

[TABLE]

Now by Section 4, there is some quantifier-free formula $\phi(z_{1},\dots,z_{k})$ such that

[TABLE]

As $T_{2}\supset T_{0}$, in conclusion,

[TABLE]

and $\phi(t_{1},\dots,t_{l})$ is a quantifier-free $\mathcal{L}_{1}$-formula with variables from $\Set{y_{1},\dots,y_{k}}$. ∎

Corollary \thecorollary.

Every one-variable set definable in $\mathcal{M}_{2}$ is definable in $\mathcal{M}_{0}$.

Proof.

By Theorem 6.1, every definable set in $\mathcal{M}_{2}$ is quantifier-free definable. By Section 6, every quantifier-free one-variable set definable in $\mathcal{M}_{2}$ is equivalent to an $x$-corrected formula of rank $\leq(-\infty,0)$, which in turn is definable (with parameters) in $\mathcal{M}_{0}$. ∎

We conclude by articulating the answers to Questions 1 and 1.

Theorem 6.2**.**

There is a definably complete type complete structure without the pigeonhole property.

Proof.

The failure of the pigeonhole principle in $\mathcal{M}_{2}$ is witnessed by $Z$ and $f\upharpoonright Z$. But by Section 6, $\mathcal{M}_{0}$ and $\mathcal{M}_{2}$ have the same definable sets in one free variable. In particular, $\mathcal{M}_{2}$ is definably complete and type complete. ∎

Theorem 6.3**.**

There are two ordered structures in the same language $\mathcal{M},\mathcal{N}$ on the same universe, admitting the same order and the same definable subsets with $\mathcal{M}$ being pseudo-o-minimal and $\mathcal{N}$ not.

In particular, the answer to Section 1 is negative and there is no axiomatization of pseudo-o-minimality by first-order conditions on one-variable formulae only. Furthermore, there is no axiomatization of pseudo-o-minimality by any second order theory in the language $\mathcal{L}_{\operatorname{Def}}:=\Set{<,\operatorname{Def}}$ where $\operatorname{Def}$ is interpreted as the definable one-variable subsets.

Proof.

$\mathcal{M}_{0}$ is pseudo-o-minimal and $\mathcal{M}_{2}$ is not pseudo-o-minimal as the failure of the pigeonhole principle is witnessed by $Z$ and $f\upharpoonright Z$. But $\mathcal{M}_{0}\upharpoonright\Set{<}=\mathcal{M}_{2}\upharpoonright\Set{<}$ and by Section 6, $\mathcal{M}_{0}$ and $\mathcal{M}_{2}$ have the same definable sets in one free variable. We may now define $\mathcal{N}$ to be $\mathcal{M}_{2}$ and $\mathcal{M}$ to be a trivial expansion of $\mathcal{M}_{0}$ to $\mathcal{L}_{1}$ (letting every function symbol be interpreted as the identity map and any relation symbol be interpreted as the $\emptyset$).

∎