Existentially Closed Brouwerian Semilattices

TL;DR

This paper provides a finite, simple axiomatization for the model completion of the variety of Brouwerian semilattices, which are known to be amalgamable and locally finite.

Contribution

It introduces a concise axiomatization for the existentially closed Brouwerian semilattices' model completion, advancing understanding of their logical structure.

Findings

Model completion exists for Brouwerian semilattices.

A finite axiomatization is provided for this model completion.

The axiomatization simplifies the understanding of existentially closed structures.

Abstract

The variety of Brouwerian semilattices is amalgamable and locally finite, hence by well-known results due to W. H. Wheeler, it has a model completion (whose models are the existentially closed structures). In this paper, we supply for such a model completion a finite and rather simple axiomatization.

Equations

$$c\le a\to b\text{iff}c\wedge a\le b.$$

$$\begin{array}[]{r c l}a\wedge a=a&&a\wedge(a\rightarrow b)=a\wedge b\\ a\wedge b=b\wedge a&&b\wedge(a\rightarrow b)=b\\ a\wedge(b\wedge c)=(a\wedge b)\wedge c&&a\rightarrow(b\wedge c)=(a\rightarrow b)\wedge(a\rightarrow c)\\ a\wedge 1=a&&a\rightarrow a=1\end{array}$$

$$\begin{array}{r}{a}_1\ll b\\ a_2\ll b\\ b\ll c\\ d\to b=b\end{array}$$

$$\begin{array}{rl}& b_1\ge a_1,b_2\ge a_2\\ & a_2\to a=a_1\\ & a_1\to a=a_2\\ & a_2\to b_1=b_2\to b_1\\ & a_1\to b_2=b_1\to b_2\end{array}$$

$$a-b\le c\text{ iff }a\le b\vee c.$$

$$g\le b_1\vee \dots \vee b_n\text{implies}g\le b_i\text{ for some }i=1,\dots ,n.$$

$$\begin{array}{rl}0-a=0& a-0=a\\ (a-b)\vee b=a\vee b& (a-b)\vee a=a\\ (a-b)\vee (a-(a-b))=a& a-(a-(a-b))=a-b\\ (a_1\vee \cdots \vee a_n)-b=& (a_1-b)\vee \cdots \vee (a_n-b)\\ a-(b_1\vee \cdots \vee b_n)=& ((a-b_1)-\cdots )-b_n\end{array}$$

$$(a-b)-c=(a-c)-b$$

$$a\le b$$

$$\text{if }b\le c\text{ then }b-a$$

$$\begin{array}{rl}& a=\bigvee \{\text{join-irreducible components of }a\}\\ a-b=\bigvee \{g|g& \text{ is a join-irreducible component of }a\text{ such that }g\nleqq b\}\end{array}$$

$$S$$

$$\{f(a)|a\in \text{dom }f\text{ and }a\ge p\}\subseteq Q$$

$$S_p=\{(\{p\},\{r_1,\dots ,r_n\})|r_i\in \text{dom }g\text{ and }g(r_i)=q_i\text{ for }i=1,\dots ,n\}$$

$$\{f(a)|a\in \text{dom }g\text{ and }a\ge r\}\subseteq Q$$

$$S_r=\{(\{p_1,\dots ,p_n\},\{r\})|p_i\in \text{dom }f\text{ and }f(p_i)=q_i\text{ for }i=1,\dots ,n\}$$

$$S_P=\bigcup _{p\in P}{S}_p{S}_R=\bigcup _{r\in R}{S}_r$$

$$S_p=\{(\{p\},\{r\})|r\in \text{dom }g\text{ and }f(p)=g(r)\}.$$

$$S_P\cap S_R=\{(\{p\},\{r\})|p\in \text{dom }f,r\in \text{dom }g\text{ and }f(p)=g(r)\}.$$

$$\begin{array}{rl}(A_1,A_2)\le (B_1,B_2)\text{iff}& \forall y\in B_1\exists x\in A_1\text{ such that }x\le y\\ & \text{and}\\ & \forall y\in B_2\exists x\in A_2\text{ such that }x\le y\end{array}$$

$$\begin{array}{rl}\text{dom }{g}^{\prime }=S_P& \text{dom }{f}^{\prime }=S_R\\ g^{\prime }(\{p\},A_2)=p& f^{\prime }(A_1,\{r\})=r\end{array}$$

$$\begin{array}{rl}\text{dom }f\circ g^{\prime }& =(g^{\prime }{)}^{-1}(\text{dom }f)=\{(\{p\},A_2)\in S_P|p\in \text{dom }f\}\\ & =\{(\{p\},\{r\})|p\in \text{dom }f,r\in \text{dom }g\text{ and }f(p)=g(r)\}\\ & =(f^{\prime }{)}^{-1}(\text{dom }g)=\{(A_1,\{r\})\in S_R|r\in \text{dom }g\}\\ & =\text{dom }g\circ f^{\prime }\end{array}$$

$$(f\circ g^{\prime })((\{p\},\{r\}))=f(p)=g(r)=(g\circ f^{\prime })((\{p\},\{r\}))$$

$$L$$

$$a-x=(a_1-x)\vee \cdots \vee (a_n-x)$$

$$x-a=x-(a_1\vee \cdots \vee a_n)=((x-a_1)-\cdots )-a_n$$

$$x-a=((x-a_i)-\cdots )-a_n=(0-\cdots )-a_n=0.$$

$$a-x_i=(g-x_i)\vee \cdots \vee (a_n-x_i)=x_j\vee (a_2-x_i)\vee \cdots \vee (a_n-x_i)$$

$$a-x=(a_1-x)\vee \cdots \vee (a_n-x)$$

$$x_i-a=(g-x_j)-a=(g-a)-x_j=g-x_j=x_i$$

$$a<x\text{ iff }a\le h\text{ and }x<a\text{ iff }g\le a\text{ for some }g\in G$$

Full text

March 9, 2024

0.1. Introduction

In algebraic logic some attention has been paid to the class of existentially closed structures in varieties coming from the algebraization of common propositional logics. In fact, there are relevant cases where such classes are elementary: this includes, besides the easy case of Boolean algebras, also Heyting algebras [GZ97, GZ02], diagonalizable algebras [Sha93, GZ02] and some universal classes related to temporal logics [GvG01],[GvG16]. However, very little is known about the related axiomatizations, with the remarkable exception of the case of the locally finite amalgamable varieties of Heyting algebras recently investigated in [DJ10] and of the simpler cases of posets and semilattices studied in [AB86]. In this paper, we use a methodology similar to [DJ10] (relying on classifications of minimal extensions) in order to investigate the case of Brouwerian semilattices, i.e. the algebraic structures corresponding to the implication-conjunction fragment of intuitionistic logic. We obtain the finite axiomatization reported below, which is similar in spirit to the axiomatizations from [DJ10] (in the sense that we also have kinds of ‘density’ and ‘splitting’ conditions). The main technical problem we must face for this result (making axioms formulation slightly more complex and proofs much more involved) is the lack of joins in the language of Brouwerian semilattices.

0.1.1. Statement of the main result

The first researcher to consider the Brouwerian semilattices as algebraic objects in their own right was W. C. Nemitz in [Nem65]. A Brouwerian semilattice is a poset $(P,\leq)$ having a greatest element (which we denote with $1$), inf’s of pairs (the inf of $\{a,b\}$ is called ‘meet’ of $a$ and $b$ and denoted with $a\wedge b$) and relative pseudo-complements (the relative pseudo-complement of $a$ and $b$ is denoted with $a\rightarrow b$). We recall that $a\rightarrow b$ is characterized by the the following property: for every $c\in P$ we have

[TABLE]

Brouwerian semilattices can also be defined in an alternative way as algebras over the signature $1,\wedge,\rightarrow$, subject to the following equations

[TABLE]

In case this equational axiomatization is adopted, the partial order $\leq$ is recovered via the definition $a\leq b$ iff $a\wedge b=a$.

By a result due to Diego and McKay [Die66, McK68], Brouwerian semilattices are locally finite (meaning that all finitely generated Brouwerian semilattices are finite); since they are also amalgamable, it follows [Whe76] that the theory of Brouwerian semilattices has a model completion. We prove that such a model completion is given by the above set of axioms for the theory of Brouwerian semilattices together with the three additional axioms (Density1, Density2, Splitting) below.

We use the shorthand $a\ll b$ to mean that $a\leq b$ and $b\rightarrow a=a$.

[Density 1] For every $c$ there exists an element $b$ different from $1$ such that $b\ll c$.

[Density 2] For every $c,a_{1},a_{2},d$ such that $a_{1},a_{2}\neq 1$, $a_{1}\ll c$, $a_{2}\ll c$ and $d\rightarrow a_{1}=a_{1}$, $d\rightarrow a_{2}=a_{2}$ there exists an element $b$ different from $1$ such that:

[TABLE]

[Splitting] For every $a,b_{1},b_{2}$ such that $1\neq a\ll b_{1}\land b_{2}$ there exist elements $a_{1}$ and $a_{2}$ different from $1$ such that:

[TABLE]

As testimony of the usefulness of this result, the following proposition shows some properties of the existentially closed Brouwerian semilattices that can be deduced from our investigation as an easy exercise.

Proposition 0.1.1**.**

Let $L$ be an existentially closed Brouwerian semilattice. Then:

1. (1)

$L$* has no bottom element.* 2. (2)

If $a,b\in L$ are incomparable, i.e. $a\nleq b$ and $b\nleq a$, then the join of $a$ and $b$ in $L$ does not exist. 3. (3)

There are no meet-irreducible elements in $L$.

The paper is structured as follows: Section 0.2 gives the basic notions and definitions, in particular it describes the finite duality and characterizes the existentially closed structures by means of embeddings of finite extensions of finite sub-structures. In Section 0.3 we investigate the minimal extensions and use them to give an intermediate characterization of the existentially closed structures. Section 0.4 focuses on the axiomatization, it is split into two subsections: the first about the Splitting axiom and the second about the Density axioms. Finally, in Section 0.5 we present and prove some properties of the existentially closed structures whose validity follows from this investigation.

0.2. Preliminary Background

A co-Brouwerian semilattice, CBS for short, is a structure obtained by reversing the order of a Brouwerian semilattice.

We will work with CBSes instead of Brouwerian semilattices.

Definition 0.2.1**.**

A poset $(P,\leq)$ is said to be a co-Brouwerian semilattice if it has a least element, which we denote with [math], and for every $a,b\in P$ there exists the sup of $\{a,b\}$, which we call join of $a$ and $b$ and denote with $a\vee b$, and the difference $a-b$ satisfying for every $c\in P$

[TABLE]

$a\ll b$ will mean that $a\leq b$ and $b-a=b$.

Clearly, there is also an alternative equational definition for co-Brouwerian semilattices (which we leave to the reader, because it is dual to the equational definition for Brouwerian semilattices given above).

Moreover, we will call co-Heyting algebras the structures obtained reversing the order of Heyting algebras. Obviously any co-Heyting algebra is a CBS.

Definition 0.2.2**.**

Let $A,B$ be co-Brouwerian semilattices. A map $f:A\to B$ is a morphism of co-Brouwerian semilattices if it preserves [math], the join and difference of any two elements of $A$.

Notice that such a morphism $f$ is an order preserving map because, for any $a,b$ elements of a co-Brouwerian semilattice, we have $a\leq b$ iff $a\vee b=b$.

Definition 0.2.3**.**

Let $L$ be a CBS.

We say that $g\in L$ is join-irreducible iff for every $n\geq 0$ and $b_{1},\ldots,b_{n}\in L$, we have that

[TABLE]

Notice that taking $n=0$ we obtain that join-irreducibles are different from [math].

Remark 0.2.4**.**

Let $L$ be a CBS and $g\in L$. Then the following conditions are equivalent:

1. (1)

$g$ is join-irreducible. 2. (2)

$g\neq 0$ and for any $b_{1},b_{2}\in L$ we have that $g\leq b_{1}\vee b_{2}\;$ implies $\;g\leq b_{1}$ or $g\leq b_{2}$. 3. (3)

For every $n\geq 0$ and $b_{1},\ldots,b_{n}\in L$ we have that

$g=b_{1}\vee\ldots\vee b_{n}\;$ implies $\;g=b_{i}$ for some $i=1,\ldots,n$. 4. (4)

$g\neq 0$ and for any $b_{1},b_{2}\in L$ we have that $g=b_{1}\vee b_{2}\;$ implies $\;g=b_{1}$ or $g=b_{2}$. 5. (5)

$g\neq 0$ and for any $a\in L$ we have that $g-a=0\;$ or $\;g-a=g$.

Proof.

The implications 1 $\Leftrightarrow$ 2, 3 $\Leftrightarrow$ 4 and 1 $\Rightarrow$ 3 are straightforward. For the remaining ones see Lemma 2.1 in [Köh81]. ∎

Definition 0.2.5**.**

Let $L$ be a CBS and $a\in L$.

A join-irreducible component of $a$ is a maximal element among the join-irreducibles of $L$ that are smaller than or equal to $a$.

Remark 0.2.6**.**

The following is a list of facts that might be used without explicit mention.

These identities hold in any CBS:

[TABLE]

In particular

[TABLE]

Furthermore in any CBS:

[TABLE]

The following facts are true in any finite CBS:

[TABLE]

Moreover, in a finite CBS, $g$ is join-irreducible iff it has a unique predecessor, i.e. a maximal element among the elements strictly smaller than $g$, and in that case we denote it by $g^{-}$ and it is equal to $\bigvee_{a<g}a$.

Recall that $a\ll b$ means $a\leq b$ and $b-a=b$. Thus, in any finite CBS, $a\ll b$ if and only if $a\leq b$ and there are no join-irreducible components of $b$ that are less than or equal to $a$. Finally, if $g$ is join-irreducible then $g^{-}\ll g$.

Example 0.2.7**.**

Let $(P,\leq)$ be a poset. For any $a\in P$ we define $\downarrow\!a=\{p\in P\>|\>p\leq a\}$ and for any $A\subseteq P$ we define $\downarrow\!A=\bigcup_{a\in A}\downarrow\!a$. A subset $D\subseteq P$ such that $D=\downarrow\!D$ is called a downset, i.e. a downward closed subset, of $P$. The downsets $\downarrow\!a$ and $\downarrow\!A$ are called the downsets generated by $a$ and $A$.

Given a poset $P$, the set of downsets of $P$ denoted by $\mathcal{D}(P)$ has naturally a structure of CBS given by the usual inclusion of subsets. Joins coincide with the union of subsets and the zero element with the empty subset. It turns out that the difference of two downsets $A,B\in\mathcal{D}(P)$ is $A-B=\downarrow\!(A\setminus B)$.

Note that if $P$ is finite then also $\mathcal{D}(P)$ is. In that case any downset $A\in\mathcal{D}(P)$ is generated by the set of its maximal elements and for any $A,B\in\mathcal{D}(P)$ we have that $A-B$ is the downset generated by the maximal elements of $A$ that are not in $B$. Moreover the join-irreducibles of $\mathcal{D}(P)$ are the downsets of the form $\downarrow\!p$ for $p\in P$ and the downsets generated by the maximal elements of a given downset are its join-irreducible components. Notice that this is not always the case when $P$ is infinite.

Finally, when $P$ is finite, for $A,B\in\mathcal{D}(P)$ satisfying $A\ll B$ means that $A\subseteq B$ and $A$ does not contain any maximal element of $B$.

0.2.1. Locally finiteness

Theorem 0.2.8**.**

The variety of CBSes is locally finite.

Proof.

We just sketch the proof first presented in [McK68]. A CBS $L$ is subdirectly irreducible iff $L\setminus\{0\}$ has a least element, or equivalently $L$ has a single atom, i.e. a minimal element different from [math].

Let $L$ be subdirectly irreducible and $u$ the least element of $L\setminus\{0\}$. Then $L\setminus\{u\}$ is a sub-CBS of $L$. This implies that any generating set of $L$ must contain $u$.

Moreover if $L$ is generated by $n$ elements then $L\setminus\{u\}$ can be generated by $n-1$ elements. It follows that the cardinality of subdirectly irreducible CBSes generated by $n$ elements is bounded by $\#F_{n-1}+1$ where $F_{m}$ is the free CBS on $m$ generators. Since $\#F_{0}=1$ by induction we obtain that $F_{m}$ is finite for any $m$ because it is a subdirect product of a finite family of subdirectly irreducibles which are generated by $m$ elements. ∎

Computing the cardinality of $F_{m}$ is a hard task. It is known that $\#F_{0}=1,\#F_{1}=2,\#F_{2}=18$ and $\#F_{3}=623,662,965,552,330$. The size of $F_{4}$ is still unknown. In [Köh81] it is proved that the number of join-irreducible elements of $F_{4}$ is $2,494,651,862,209,437$. This shows that although the cardinality of the free CBS on a finite number of generators is always finite, it grows very rapidly.

0.2.2. Finite duality

Proposition 0.2.9**.**

Any finite CBS is a distributive lattice.

Proof.

A finite CBS is complete, hence also co-complete, so it is a lattice. The map $a\vee(-)$ preserves infima because it has a left adjoint given by $(-)-a$. Thus the distributive laws hold. ∎

Remark 0.2.10**.**

Every finite Brouwerian semilattice is a Heyting algebra but it is not true that every Brouwerian semilattices morphism among finite Brouwerian semilattices is a Heyting algebra morphism.

The following theorem presents the finite duality result due to Köhler:

Theorem 0.2.11**.**

The category $\mathbf{CBS}_{fin}$ of finite CBSes is dual to the category $\mathbf{P}$ whose objects are finite posets and whose morphisms are partial mappings $\alpha:P\rightarrow Q$ satisfying:

1. (i)

$\forall p,q\in\text{dom }\alpha\text{ if }p<q\text{ then }\alpha(p)<\alpha(q)$. 2. (ii)

$\forall p\in\text{dom }\alpha\text{ and }\forall q\in Q\text{ if }\alpha(p)<q\text{ then }\exists r\in\text{dom }\alpha\text{ such that }p<r\text{ and }\alpha(r)=q$.

Proof.

The proof can be found in [Köh81]. We just recall how the equivalence works. To a finite poset $P$ it is associated the CBS $\mathcal{D}(P)$ of downsets of $P$.

To a $\mathbf{P}$-morphism among finite posets it is associated the morphism of CBSes that maps a downset to the downset generated by its preimage. More explicitly, to a $\mathbf{P}$-morphism $f:P\to Q$ is associated the morphism that maps a downset $D$ of $Q$ to $\downarrow\!f^{-1}(D)=\{p\in P\mid\exists p^{\prime}\geq p\;(p^{\prime}\in\text{dom}f\leavevmode\nobreak\ \&\leavevmode\nobreak\ f(p^{\prime})\in D)\}$.

On the other hand, to a finite CBS $L$ it is associated the poset of its join-irreducible elements. ∎

The following proposition is easily checked:

Proposition 0.2.12**.**

Let $P,Q$ be finite posets and $f:P\rightarrow Q$ a $\mathbf{P}$-morphism. Let $\alpha$ be the associated morphism of CBSes. Then

1. (i)

$\alpha$* is injective if and only if $f$ is surjective.* 2. (ii)

$\alpha$* is surjective if and only if $\text{dom }f=P$ and $f$ is injective.*

Duality results involving all Brouwerian semilattices can be found in the recent paper [BJ13] due to G. Bezhanishvili and R. Jansana. Other dualities are described in [VM86] and [Cel03].

Using finite duality we can show that the variety of CBSes has the amalgamation property.

The amalgamation property for Brouwerian semilattices is the algebraic counterpart of a syntactic fact about the implication-conjunction fragment of intuitionistic propositional logic: the interpolation property. The proof that such fragment satisfies this property can be found in [RdL89].

Theorem 0.2.13**.**

The theory of CBSes has the amalgamation property.

Proof.

First, we show that the pushout of given monomorphisms (= injective maps) $m:L_{0}\to L_{1}$ and $n:L_{0}\to L_{2}$ among finite CBSes is still formed by monomorphisms. Then we extend the result to the general case.

To do this, by finite duality, it is sufficient to show that the category $\mathbf{P}$ has the coamalgamation property. This means that, given two surjective $\mathbf{P}$-morphisms among finite posets $f:P\to Q$ and $g:R\to Q$ there exist a finite poset $S$ and two surjective $\mathbf{P}$-morphisms $f^{\prime}:S\to R$ and $g^{\prime}:S\to P$ making the following diagram commute.

[TABLE]

For any $p\in P$ let $q_{1},\ldots,q_{n}$ be the minimal elements of

[TABLE]

it could be that $n=0$ when such set is empty. Define:

[TABLE]

Analogously for any $r\in R$ let $q_{1},\ldots,q_{n}$ be the minimal elements of

[TABLE]

and define

[TABLE]

Let

[TABLE]

And take $S=S_{P}\cup S_{R}$.

We can immediately observe that if $p\in\text{dom }f$ then

[TABLE]

And that

[TABLE]

And finally that if $(\{p\},\{r_{1},\ldots,r_{n}\})\in S_{p}$ then the $r_{i}$’s are two-by-two incomparable, indeed $g$ is order preserving and the $g(r_{i})$’s are incomparable since they are the minimal elements of a subset of $Q$. Thus the elements of the two components of any element of $S$ are two-by-two incomparable.

We define an order on $S$ in the following way:

let $(A_{1},A_{2}),(B_{1},B_{2})\in S$ where $A_{1},B_{1}\subseteq P$ and $A_{2},B_{2}\subseteq R$, we define

[TABLE]

This order relation is clearly reflexive.

It is antisymmetric, indeed let $(A_{1},A_{2})\leq(B_{1},B_{2})$ and $(B_{1},B_{2})\leq(A_{1},A_{2})$ then for any $y\in B_{1}$ there exists $x\in A_{1}$ such that $x\leq y$ and there exists $z\in B_{1}$ such that $z\leq x$. Since the elements of $B_{1}$ are incomparable we get $z=y$ and thus $x=y$. Therefore $B_{1}\subseteq A_{1}$. Symmetrically we get $A_{1}\subseteq B_{1}$ and then $A_{1}=B_{1}$. Reasoning similarly we get $A_{2}=B_{2}$ and then $(A_{1},A_{2})=(B_{1},B_{2})$.

It is transitive, indeed let $(A_{1},A_{2})\leq(B_{1},B_{2})$ and $(B_{1},B_{2})\leq(C_{1},C_{2})$ then for any $z\in C_{1}$ there exists $y\in B_{1}$ such that $y\leq z$ and there exists $x\in A_{1}$ such that $x\leq y$ and hence also $x\leq z$. Analogously for the second components. Therefore $(A_{1},A_{2})\leq(C_{1},C_{2})$.

Thus we have defined a partial order on $S$.

Take $g^{\prime}:S\to P$ and $f^{\prime}:S\to R$ as:

[TABLE]

Then

[TABLE]

and if $p\in\text{dom }f,\,r\in\text{dom }g\text{ and }f(p)=g(r)$ then

[TABLE]

$g^{\prime}$ is surjective: indeed let $p\in P$ and $q_{1},\ldots,q_{n}$ be the minimal elements of $\{f(a)\>|\>a\in\text{dom }f\text{ and }a\geq p\}$, by surjectivity of $g$ there exist $r_{1},\ldots,r_{n}\in\text{dom }g$ such that $g(r_{i})=q_{i}$, then $(\{p\},\{r_{1},\ldots,r_{n}\})\in S_{p}\subseteq\text{dom }g^{\prime}$ and $g^{\prime}((\{p\},\{r_{1},\ldots,r_{n}\}))=p$. Analogously for the surjectivity of $f^{\prime}$.

It remains to show that $g^{\prime},f^{\prime}$ are $\mathbf{P}$-morphisms.

Let $(\{p\},A),(\{p^{\prime}\},B)\in S_{P}=\text{dom }g^{\prime}$ such that $(\{p\},A)<(\{p^{\prime}\},B)$, we show that $p<p^{\prime}$.

Clearly $p\leq p^{\prime}$ by the definition of the order on $S$.

Suppose that $p=p^{\prime}$, let $q_{1},\ldots,q_{n}$ be the minimal elements of $\{f(a)\>|\>a\in\text{dom }f\text{ and }a\geq p\}$. Let $A=\{r_{1},\ldots,r_{n}\}$ and $B=\{r_{1}^{\prime},\ldots,r_{n}^{\prime}\}$ be such that $g(r_{i})=g(r_{i}^{\prime})=q_{i}$ for $i=1,\ldots,n$. Then for any $r_{i}^{\prime}\in B$ there exists $r_{j}\in A$ such that $r_{j}\leq r_{i}^{\prime}$. If $r_{j}\leq r_{i}^{\prime}$ with $j\neq i$ then $q_{j}=g(r_{j})\leq g(r_{i}^{\prime})=q_{i}$ and this is absurd because the $q_{i}$’s are incomparable. Therefore $r_{i}\leq r_{i}^{\prime}$ for any $i=1,\ldots,n$, if $r_{i}<r_{i}^{\prime}$ then $q_{i}=g(r_{i})<g(r_{i}^{\prime})=q_{i}$ which is absurd. Thus $r_{i}=r_{i}^{\prime}$ and $A=B$, we have obtained a contradiction. Analogous for $f^{\prime}$.

Therefore $g^{\prime},f^{\prime}$ preserve the strict order.

Let $(\{p\},A)\in S_{P}$ and $p<p^{\prime}$.

Let $q_{1},\ldots,q_{n}$ be the minimal elements of $\{f(a)\>|\>a\in\text{dom }f\text{ and }a\geq p\}$ and $q_{1}^{\prime},\ldots,q_{m}^{\prime}$ be the minimal elements of $\{f(a)\>|\>a\in\text{dom }f\text{ and }a\geq p^{\prime}\}$; since the latter set is included in the former and they are both finite we have that for any $q_{j}^{\prime}$ there exist $q_{i}$ such that $q_{i}\leq q_{j}^{\prime}$.

Let $A=\{r_{1},\ldots,r_{n}\}$ with $g(r_{i})=q_{i}$. Since $g$ is a $\mathbf{P}$-morphism and for any $q_{j}^{\prime}$ there exists $i$ such that $g(r_{i})=q_{i}\leq q_{j}^{\prime}$, there exists $r_{j}^{\prime}\in\text{dom }g$ such that $r_{i}\leq r_{j}^{\prime}$ and $g(r_{j}^{\prime})=q_{j}^{\prime}$. Take $B=\{r_{1}^{\prime},\ldots,r_{m}^{\prime}\}$ then for any $r_{j}^{\prime}$ there exists $r_{i}$ such that $r_{i}\leq r_{j}^{\prime}$, therefore $(\{p\},A)<(\{p^{\prime}\},B)\in S_{P}$. Analogous for $f^{\prime}$.

Thus $f^{\prime},g^{\prime}$ are surjective $\mathbf{P}$-morphisms and they coamalgamate $f,g$.

We now want to prove the general case: pushouts of monos along monos in the category of CBSes are monos.

Suppose $m:L_{0}\to L_{1}$ and $n:L_{0}\to L_{2}$ are monos and $L_{0},L_{1},L_{2}$ are CBSes. Since the variety is locally finite by Theorem 0.2.8, we can consider $L_{0},L_{1},L_{2}$ as filtered colimits of families of finite CBSes. Assume without loss of generality that $L_{1}\cap L_{2}=L_{0}$ and $m,n$ are inclusions, then we can consider the families indexed by $\mathcal{P}_{fin}(L_{1}\cup L_{2})$ given for any finite subset $S\subseteq L_{1}\cup L_{2}$ by the sub-CBSes respectively of $L_{1},L_{2}$ and $L_{0}$ generated respectively by $S\cap L_{1}$, $S\cap L_{2}$ and $m^{-1}(S\cap L_{1})\cap n^{-1}(S\cap L_{2})=S\cap L_{0}$. Then we can compute the pushouts of the restrictions of the monos for any index, the colimit of all these pushouts is a mono because each of them is a mono. Thus we have obtained that the pushout of $m$ along $n$ and the pushout of $n$ along $m$ are monomorphisms. ∎

0.2.3. Existentially closed CBSes

In this subsection we want to characterize the existentially closed CBSes using the finite extensions of their finite sub-CBSes.

Definition 0.2.14**.**

Let $T$ be a first order theory and $\mathcal{A}$ a model of $T$. $\mathcal{A}$ is said to be existentially closed for $T$ if for every model $\mathcal{B}$ of $T$ such that $\mathcal{A}\subseteq\mathcal{B}$ every existential sentence in the language extended with names of elements of $\mathcal{A}$ which holds in $\mathcal{B}$ also holds in $\mathcal{A}$

The following proposition is well-known from textbooks [CK90]:

Proposition 0.2.15**.**

Let $T$ be a universal theory. If $T$ has a model completion $T^{*}$, then the class of models of $T^{*}$ is the class of models of $T$ which are existentially closed for $T$.

Thanks to the locally finiteness and the amalgamability, by an easy model-theoretic reasoning we obtain the following characterization of the existentially closed CBSes:

Theorem 0.2.16**.**

Let $L$ be a CBS. $L$ is existentially closed iff for any finite sub-CBS $L_{0}\subseteq L$ and for any finite extension $C\supseteq L_{0}$ there exists an embedding $C\to L$ fixing $L_{0}$ pointwise.

Proof.

First, we prove that if for any finite sub-CBS $L_{0}\subseteq L$ and for any finite extension $C\supseteq L_{0}$ there exists an embedding $C\to L$ fixing $L_{0}$ pointwise, then $L$ is existentially closed. Let $D$ be an extension of $L$ and $\exists x_{1},\ldots,x_{m}\varphi(x_{1},\ldots,x_{m},a_{1},\ldots,a_{n})$ an existential $\mathcal{L}_{L}$-sentence, where $\varphi(x_{1},\ldots,x_{m},a_{1},\ldots,a_{n})$ is quantifier free and $a_{1},\ldots,a_{n}\in L$.

Suppose $D\vDash\exists x_{1},\ldots,x_{m}\varphi(x_{1},\ldots,x_{m},a_{1},\ldots,a_{n})$.

Let $d_{1},\ldots,d_{m}$ be elements of $D$ such that $D\vDash\varphi(d_{1},\ldots,d_{m},a_{1},\ldots,a_{n})$.

Consider the sub-CBS $L_{0}\subseteq L$ generated by $a_{1},\ldots,a_{n}$ and the sub-CBS $C\subseteq D$ generated by $d_{1},\ldots,d_{m},a_{1},\ldots,a_{n}$. They are both finite because they are finitely generated and the CBSes form a locally finite variety.

By hypothesis there exists an embedding $C\to L$ fixing $L_{0}$ pointwise.

Let $d_{1}^{\prime},\ldots,d_{m}^{\prime}$ be the images of $d_{1},\ldots,d_{m}$ by this embedding. Thus $L\vDash\varphi(d_{1}^{\prime},\ldots,d_{m}^{\prime},a_{1},\ldots,a_{n})$ because $\varphi$ is quantifier free.

Therefore $L\vDash\exists x_{1},\ldots,x_{m}\varphi(x_{1},\ldots,x_{m},a_{1},\ldots,a_{n})$. It follows that $L$ is existentially closed.

To prove the other implication, suppose $L$ is existentially closed.

By amalgamation property there exists a CBS $D$ amalgamating $L$ and $C$ over $L_{0}$.

[TABLE]

Let $\Sigma$ be the set of quantifier free $\mathcal{L}_{C}$-sentences of the form $c*c^{\prime}=c^{\prime\prime}$ true in $C$ where $c,c^{\prime},c^{\prime\prime}\in C$ and $*$ is either $\vee$ or $-$. Hence $(C,\Sigma)$ is a finite presentation of $C$.

Now let $c_{1},\ldots,c_{r},a_{1},\ldots,a_{n}$ be an enumeration of the elements in $C$ where the $a_{i}$’s are the elements in $L$. We obtain the quantifier free $\mathcal{L}_{C}$-sentence $\sigma(c_{1},\ldots,c_{r},a_{1},\ldots,a_{n})$ by taking the conjunction of all the sentences in $\Sigma$ and all the sentences of the form $\neg(c=c^{\prime})$ for every $c,c^{\prime}\in C$ such that $c\neq c^{\prime}$.

Clearly $\exists x_{1},\ldots,x_{r}\sigma(x_{1},\ldots,x_{r},a_{1},\ldots,a_{n})$ is an existential $\mathcal{L}_{L}$-sentence true in $D$. Since $L$ is existentially closed, $L\vDash\exists x_{1},\ldots,x_{r}\sigma(x_{1},\ldots,x_{r},a_{1},\ldots,a_{n})$. Let $c_{1}^{\prime},\ldots,c_{r}^{\prime}\in L$ be such that $L\vDash\sigma(c_{1}^{\prime},\ldots,c_{r}^{\prime},a_{1},\ldots,a_{n})$. The map $C\to L$ fixing $L_{0}$ pointwise and mapping $c_{i}$ to $c_{i}^{\prime}$ is an embedding. Indeed it is injective and an homomorphism by definition of the sentence $\sigma$. ∎

0.3. Minimal finite extensions

In this section we focus on the finite extensions of CBSes. We are interested in particular to the minimal ones since any finite extension can be decomposed in a finite chain of minimal extensions. We will study minimal finite extensions by describing the properties of some elements which generate them. This investigation will lead us to another characterization of the existentially closed CBSes.

Definition 0.3.1**.**

Let $P$ be a poset, $P_{0}\subseteq P$ and $\mathcal{F}$ a partition of $P_{0}$, let $A,B\in\mathcal{F}$.

We say that $A\leq B$ iff there exist $a\in A,b\in B$ such that $a\leq b$.

Proposition 0.3.2**.**

*Let $P$ be a finite poset.

To give a surjective $\mathbf{P}$-morphism $f$ from $P$ to any finite poset is equivalent, up to isomorphism, to give a partition $\mathcal{F}$ of a subset of $P$ such that:*

1. (1)

for all $A,B\in\mathcal{F}$ we have that if $A\leq B$ and $B\leq A$ then $A=B$, 2. (2)

for all $A,B\in\mathcal{F}$ and $a\in A$ if $A\leq B$ then there exists $b\in B$ such that $a\leq b$, 3. (3)

for all $A\in\mathcal{F}$ we have that all the elements of $P$ in $A$ are two-by-two incomparable.

Proof.

Given a surjective $\mathbf{P}$-morphism $f:P\rightarrow Q$, the partition $\mathcal{F}$ of $\text{dom }f\subseteq P$ is obtained by taking the collection of the fibers of $f$. $\mathcal{F}$ satisfies 1 because $f$ is order preserving and the order on $Q$ is antisymmetric. Furthermore $\mathcal{F}$ satisfies 2 as a consequence of condition (ii) in the definition of $\mathbf{P}$-morphism. Finally, $\mathcal{F}$ satisfies 3 because $\mathbf{P}$-morphisms are strict order preserving.

On the other hand, given a partition $\mathcal{F}$ of a subset $P_{0}$ of $P$ satisfying the conditions 1, 2 and 3, we obtain a poset $Q$ by taking the quotient set of $P_{0}$ given by $\mathcal{F}$ with the order defined in Definition 0.3.1. The partial map $f:P\rightarrow Q$ is just the projection onto the quotient.

$Q$ is a poset: the order of $Q$ is clearly reflexive, it is antisymmetric because $\mathcal{F}$ satisfies 1. It is also transitive because if $A\leq B$ e $B\leq C$ then there exist $a\in A,\ b,b^{\prime}\in B,\ c\in C$ such that $a\leq b,\ b^{\prime}\leq c$; since 2 holds, there exist $c^{\prime}\in C$ such that $b\leq c^{\prime}$, hence $a\leq c^{\prime}$ and $A\leq C$. The projection $f$ is order preserving, it is a $\mathbf{P}$-morphism because 2 holds and it is obviously surjective.

It remains to show that a surjective $\mathbf{P}$-morphism $f:P\rightarrow Q$ differs by an isomorphism to the projection onto the quotient defined by the partition given by the fibers of $f$. This follows from the fact that for any $a,b\in\text{dom }f$ it is $f(a)\leq f(b)$ iff $f^{-1}(f(a))\leq f^{-1}(f(b))$ (notice that $f^{-1}(f(a))$ is the element of $\mathcal{F}$ containing $a$). Indeed if $f(a)\leq f(b)$, since $f$ is a $\mathbf{P}$-morphism, there exists $b^{\prime}$ such that $a\leq b^{\prime}$ and $f(b^{\prime})=f(b)$, therefore since $a\leq b^{\prime}$ it is $f^{-1}(f(a))\leq f^{-1}(f(b))$. The other direction of the implication holds because $f$ is order preserving. ∎

Definition 0.3.3**.**

Let $P,Q$ be finite posets and $f:P\rightarrow Q$ a surjective $\mathbf{P}$-morphism (or equivalently: let $\mathcal{F}$ satisfy conditions 1, 2 and 3 of Proposition 0.3.2). We say that $f$ (or $\mathcal{F}$) is minimal if $\#P=\#Q+1$.

Remark 0.3.4**.**

If $\mathcal{F}$ is minimal, then at most one element of $\mathcal{F}$ is not a singleton.

Theorem 0.3.5**.**

Let $f:P\rightarrow Q$ be a surjective $\mathbf{P}$-morphism between finite posets. Let $n=\#P-\#Q$. Then there exist $Q_{0},\ldots,Q_{n}$ with $Q_{0}=P$, $Q_{n}=Q$ and $f_{i}:Q_{i-1}\rightarrow Q_{i}$ which are minimal surjective $\mathbf{P}$-morphisms for $i=1,\ldots,n$ such that $f=f_{n}\circ\cdots\circ f_{1}$.

Proof.

Let $R=\text{dom }f$, we can decompose $f=f^{\prime\prime}\circ f^{\prime}$ where $f^{\prime\prime}:R\rightarrow Q$ is just the restriction of $f$ on its domain and $f^{\prime}:P\rightarrow R$ is the partial morphism with domain $R$ that acts as the identity on $R$.

The morphism $f^{\prime\prime}:R\rightarrow Q$ is a total morphism111Since it is a total map, its dual preserves the maximum downset and intersections of downsets. Therefore it is dual to a co-Heyting algebras morphism., we prove by induction on $\#R-\#Q$ that it can be decomposed in a chain of minimal surjective $\mathbf{P}$-morphisms.

Suppose $\#R-\#Q>1$ and let us consider the partition $\mathcal{F}$ of $R$ given by the fibers of $f^{\prime\prime}$. Let $x\in P$ be minimal among the elements of $R$ that are not in a singleton of $\mathcal{F}$. Let $G$ be the element of $\mathcal{F}$ containing $x$, then $\#G>1$ and all the elements of $R$ inside $G$ are incomparable to each other.

Let $Q_{n-1}$ be the quotient of $R$ defined by the refining of $\mathcal{F}$ in which $G$ is substituted by $\{x\}$ and $G\backslash\{x\}$, we name this new partition $\mathcal{F}^{\prime}$.

The projection onto the quotient $\pi:R\rightarrow Q_{n-1}$ is a $\mathbf{P}$-morphism because $\mathcal{F}^{\prime}$ satisfies the conditions 1, 2 and 3 of Proposition 0.3.2. Indeed, it satisfies 1 and 3 because $\mathcal{F}$ satisfies them and the elements in $G$ are incomparable. To show that 2 holds it is sufficient to show that for the pairs of sets in $\mathcal{F}^{\prime}$ in which exactly one of the two is $\{x\}$ or $G\backslash\{x\}$ because $\mathcal{F}$ satisfies 2 and $\{x\}$ and $G\backslash\{x\}$ are incomparable.

Let $A\in\mathcal{F}$ be different from $\{x\}$ and $G\backslash\{x\}$.

If $\{x\}\leq A$ then 2 holds because $\{x\}$ is a singleton.

If $A\leq\{x\}$ then there exists $a\in A$ such that $a\leq x$, hence $A$ is a singleton by minimality of $x$, therefore 2 holds.

If $G\backslash\{x\}\leq A$ then we have that $G\leq A$, thus for any $y\in G\backslash\{x\}$ there exists $y^{\prime}\geq y$ such that $y^{\prime}\in A$.

If $A\leq G\backslash\{x\}$ it is $A\leq G$ thus for any $y\in A$ there exists $y^{\prime}\geq y$ such that $y^{\prime}=x$ or $y^{\prime}\in G\backslash\{x\}$. Suppose there exists $y\in A$ such that there is no $y^{\prime}\geq y$ such that $y^{\prime}\in G\backslash\{x\}$: then $x\geq y$, by minimality of $x$ it has to be $A=\{y\}$ then $A\nleq G\backslash\{x\}$, this is absurd.

Therefore $\pi:R\rightarrow Q_{n-1}$ is a surjective total $\mathbf{P}$-morphism and we can apply the inductive hypothesis on $\pi$.

Then it suffices to show that the order-preserving map $f_{n}:Q_{n-1}\rightarrow Q$ induced by $f^{\prime\prime}$ is a $\mathbf{P}$-morphism, because in that case it is obviously surjective and minimal. But this is easy to show because the fibers of $f_{n}$ are all singletons except one and because $f^{\prime\prime}$ is a $\mathbf{P}$-morphism.

It remains to decompose $f^{\prime}$, to do that just enumerate the elements of $P\setminus R=\{p_{1},\ldots,p_{k}\}$ and let $f_{1}^{\prime}:R\cup\{p_{1}\}\rightarrow R$ be the partial morphism with domain $R$ that acts as the identity on $R$. Then construct $f_{2}^{\prime}:R\cup\{p_{1},p_{2}\}\rightarrow R\cup\{p_{1}\}$ in the same way and so on until $p_{k}$. ∎

Definition 0.3.6**.**

We say that a proper extension $L_{0}\subseteq L$ of finite CBSes is minimal if there is no intermediate proper extension $L_{0}\subsetneq L_{1}\subsetneq L$.

Proposition 0.3.7**.**

An extension $L_{0}\subseteq L$ of finite CBSes is minimal iff the surjective $\mathbf{P}$-morphism that is dual to the inclusion is minimal.

Proof.

Let $f:P\rightarrow Q$ be a surjective $\mathbf{P}$-morphism with $\#P=\#Q+1$. And suppose there exist two surjective $\mathbf{P}$-morphisms $g_{1}:P\rightarrow R$ and $g_{2}:R\rightarrow Q$ such that $f=g_{2}\circ g_{1}$, being $g_{1}$ and $g_{2}$ surjective $\#R$ must be equal to $\#P$ or $\#Q$. In the former case the domain of $g_{1}$ must be all $P$ and the relative fiber partition could only be the one formed exclusively by singletons because of cardinality, in the latter case the same holds for $g_{2}$. So either $g_{1}$ or $g_{2}$ has to be an isomorphism of posets.

Hence if we have two consecutive extensions that form an inclusion whose dual is minimal, then the dual of one of the two extensions is an isomorphism and so the relative extension is the identity.

The other implication follows easily from Theorem 0.3.5. ∎

Remark 0.3.8**.**

By Definition 0.3.3 it follows immediately that there are two different kinds of minimal surjective $\mathbf{P}$-morphisms between finite posets.

We call a minimal surjective $\mathbf{P}$-morphism of the first kind when there is exactly one element outside its domain and thus the restriction of such map on its domain is bijective and therefore an isomorphism of posets (any bijective $\mathbf{P}$-morphism is an isomorphism of posets). Some of these maps are dual to co-Heyting algebras embeddings but some are not.

We call a minimal surjective $\mathbf{P}$-morphism of the second kind when it is total, i.e. there are no elements outside its domain, and thus there is exactly a single fiber which is not a singleton and it contains exactly two elements. The maps of the second kind are dual to co-Heyting algebras embeddings.

Figures 1 and 2 show some examples of minimal surjective $\mathbf{P}$-morphisms and relative extensions of CBSes.

We call a finite minimal extension of CBSes either of the first or of the second kind if the corresponding minimal surjective $\mathbf{P}$-morphism is respectively of the first or of the second kind.

Therefore, a finite minimal extension of CBSes of the first kind preserves the join-irreducibility of all the join-irreducibles in the domain. Indeed, since the corresponding $\mathbf{P}$-morphism is an isomorphism when restricted on its domain, we have that the downset generated by the preimage of a principal downset is still principal.

A finite minimal extension of CBSes of the second kind preserves the join-irreducibility of all the join-irreducibles in the domain except one which becomes the join of the two new join-irreducible elements in the codomain. Indeed, the corresponding $\mathbf{P}$-morphism is total and all its fibers are singletons except one, this implies that the preimage of any principal downset is principal except for one whose preimage is a downset generated by two elements.

It turns out that we can characterize the finite minimal extensions of CBSes by means of their generators.

Definition 0.3.9**.**

Let $L_{0}$ be a finite CBS and $L$ an extension of $L_{0}$. We call an element $x\in L$ primitive of the first kind over $L_{0}$ if the following conditions are satisfied:

1. (1)

$x\notin L_{0}$

and for any $a$ join-irreducible of $L_{0}$:

1. (2)

$a-x\in L_{0}$, 2. (3)

$x-a=x\text{ or }x-a=0$.

Theorem 0.3.10**.**

Let $L_{0}$ be a finite CBS and $L$ an extension of $L_{0}$.222Notice that we do not require $L$ to be a finite CBS. If $x\in L$ is primitive of the first kind over $L_{0}$ then the sub-CBS $L_{0}\langle x\rangle$ of $L$ generated by $x$ over $L_{0}$ is a finite minimal extension of $L_{0}$ of the first kind.

Before proving Theorem 0.3.10 we need the following lemma:

Lemma 0.3.11**.**

Let $L_{0}$ be a finite CBS, $L$ an extension of $L_{0}$ and $x\in L$ primitive of the first kind over $L_{0}$, then the two following properties hold:

1. (i)

$\forall a\in L_{0}\quad a-x\in L_{0}$, 2. (ii)

$\forall a\in L_{0}\quad x-a=x\text{ or }x-a=0$.

Proof.

Let $a\in L_{0}$ and $a_{1},\ldots,a_{n}$ be its join-irreducible components in $L_{0}$, since $L_{0}$ is finite we have $a=a_{1}\vee\cdots\vee a_{n}$. To prove (i) observe that

[TABLE]

which is an element of $L_{0}$ because it is join of elements of $L_{0}$ as a consequence of 2 of Definition 0.3.9.

Furthermore to prove (ii) notice that

[TABLE]

and that 3 of Definition 0.3.9 implies that there are two possibilities: $x-a_{i}=x$ for any $i=1,\ldots,n$ or $x-a_{i}=0$ for some $i$. In the former case we have $x-a=x$, in the latter suppose that $i$ is the smallest index such that $x-a_{i}=0$ then

[TABLE]

∎

Proof of Theorem 0.3.10.

Let $L^{\prime}$ be the sub $\vee$-semilattice of $L$ generated by $x$ over $L_{0}$, we show that $L^{\prime}$ actually coincides with $L_{0}\langle x\rangle$.

$L^{\prime}$ is clearly finite, its elements are the elements of $L_{0}$ and the elements of the form $a\vee x$ with $a\in L_{0}$. It follows from (i) and (ii) of Lemma 0.3.11 that if $a,b,c,d\in L_{0}\cup\{x\}$ then $(a\vee b)-(c\vee d)=(a-(c\vee d))\vee(b-(c\vee d))=((a-c)-d)\vee((b-c)-d)$ belong to $L^{\prime}$. Therefore $L^{\prime}=L_{0}\langle x\rangle$.

We want to show that the join-irreducibles of $L_{0}\langle x\rangle$ are exactly the join-irreducibles of $L_{0}$ and $x$.

$x$ is a join-irreducible element of $L_{0}\langle x\rangle$, indeed $x\neq 0$ since by hypothesis $x\notin L_{0}$ and suppose that $x\leq a\vee b$ with $a,b\in L_{0}\langle x\rangle$ and $a,b\ngeq x$; therefore $a$ and $b$ must be elements of $L_{0}$ because they cannot be of the form $c\vee x$ with $c\in L_{0}$. It follows from (ii) of Lemma 0.3.11 and $a,b\ngeq x$ that $x-a=x-b=x$ and so $0=x-(a\vee b)=(x-b)-a=x-a=x$, this is absurd because $x\neq 0$.

The join-irreducible elements of $L_{0}$ are still join-irreducible in $L_{0}\langle x\rangle$. It is sufficient to show that for any $g$ join-irreducible in $L_{0}$ if $g\leq a\vee x$ with $a\in L_{0}$ then $g\leq a$ or $g\leq x$. Notice that being $L$ a CBS it is $g=(g-x)\vee(g-(g-x))$ (see Remark 0.2.6), we also have by 2 of Definition 0.3.9 that $g-x$ and $g-(g-x)$ are in $L_{0}$. Then being $g$ join-irreducible in $L_{0}$ we get $g=g-x$ or $g=g-(g-x)$. In the latter case $g-x=g-(g-(g-x))=g-g=0$ so $g\leq x$. In the former case $0=g-(a\vee x)=(g-x)-a=g-a$ so $g\leq a$.

Clearly if an element of the form $x\vee a$ with $a\in L_{0}$ is different from $a$ and $x$ it cannot be join-irreducible in $L_{0}\langle x\rangle$. Also if an element of $L_{0}$ is not join-irreducible in $L_{0}$ it cannot be join-irreducible in $L_{0}\langle x\rangle$. Hence the join-irreducible elements of $L_{0}\langle x\rangle$ are exactly the join-irreducible elements of $L_{0}$ and $x$.

Therefore the extension $L_{0}\hookrightarrow L_{0}\langle x\rangle$ is minimal since $L_{0}\langle x\rangle$ contains exactly one join-irreducible element more than $L_{0}$. Notice that $L_{0}\langle x\rangle$ is a minimal extension of $L_{0}$ of the first kind because the join-irreducibility of all the join-irreducibles of $L_{0}$ is preserved. ∎

Definition 0.3.12**.**

Let $L_{0}$ be a finite CBS and $L$ an extension of $L_{0}$.333Again we do not require $L$ to be a finite CBS. We call a couple of elements $(x_{1},x_{2})\in L^{2}$ primitive of the second kind over $L_{0}$ if the following conditions are satisfied:

1. (1)

$x_{1},x_{2}\notin L_{0}$ and $x_{1}\neq x_{2}$

and there exists $g$ join-irreducible element of $L_{0}$ such that:

1. (2)

$g-x_{1}=x_{2}$ and $g-x_{2}=x_{1}$, 2. (3)

for any join-irreducible element $a$ of $L_{0}$ such that $a<g$ we have $a-x_{i}\in L_{0}$ for $i=1,2$.

Remark 0.3.13**.**

$g$ in Definition 0.3.12 is univocally determined by $(x_{1},x_{2})$ since $g=x_{1}\vee x_{2}$.

Indeed, by property 2 of Definition 0.3.12 we have $x_{1}\leq g$, $x_{2}\leq g$ and also $g-(x_{1}\vee x_{2})=(g-x_{1})-x_{2}=x_{2}-x_{2}=0$ that implies $g\leq x_{1}\vee x_{2}$.

Theorem 0.3.14**.**

Let $L_{0}$ be a finite CBS and $L$ an extension of $L_{0}$. If $(x_{1},x_{2})\in L^{2}$ is primitive of the second kind over $L_{0}$ then the sub-CBS $L_{0}\langle x_{1},x_{2}\rangle$ of $L$ generated by $\{x_{1},x_{2}\}$ over $L_{0}$ is a finite minimal extension of $L_{0}$ of the second kind.

Before proving Theorem 0.3.14 we need the following lemma:

Lemma 0.3.15**.**

Let $L_{0}$ be a finite CBS, $L$ an extension of $L_{0}$ and $(x_{1},x_{2})\in L^{2}$ primitive of the second kind over $L_{0}$, then the two following properties hold:

1. (i)

$\forall a\in L_{0}\quad a-x_{i}\in L_{0}\text{ or }a-x_{i}=b\vee x_{j}$* with $b\in L_{0}$ for $\{i,j\}=\{1,2\}$.* 2. (ii)

$\forall a\in L_{0}\quad x_{i}-a=x_{i}\text{ or }x_{i}-a=0$* for $i=1,2$.*

Proof.

To show (i) we first prove that if $a\neq g$ is join-irreducible in $L_{0}$, then $a-x_{i}\in L_{0}$. If $a<g$ this is covered by the hypothesis 3 of Definition 0.3.12. Now suppose that $a$ is a join-irreducible element of $L_{0}$ such that $a\nleq g$ then $a-g=a$ because $a$ is join-irreducible. Thus $a=a-g\leq a-x_{i}\leq a$ since $x_{i}\leq g$ (because $x_{i}=g-x_{j}\leq g$ with $i\neq j$) and thus $a-x_{i}=a\in L_{0}$ for $i=1,2$. We now prove (i) for all $a\in L_{0}$.

Let $a\in L_{0}$ and $a_{1},\ldots,a_{n}$ be its join-irreducible components in $L_{0}$, since $L_{0}$ is finite we have $a=a_{1}\vee\cdots\vee a_{n}$. To prove (i) we consider two cases: $g$ is a join-irreducible component of $a$ or $g$ is not a join-irreducible component of $a$. In the former case, when $g$ is a join-irreducible component of $a$, suppose $a_{1}=g$, then

[TABLE]

with $\{i,j\}=\{1,2\}$, notice that $(a_{2}-x_{i})\vee\cdots\vee(a_{n}-x_{i})\in L_{0}$ because it is join of elements of $L_{0}$ by what we have just proved. In the latter case, $g$ is not a join-irreducible component of $a$, we have

[TABLE]

which is an element of $L_{0}$ because it is join of elements of $L_{0}$ as a consequence of what we have just proved.

Furthermore, to prove (ii) notice that since $g$ is join-irreducible in $L_{0}$ we have that for any $a\in L_{0}$ there are two cases to consider: $g\leq a$ or $g-a=g$. In the former case we have, since $x_{i}\leq g$ by 2 of Definition 0.3.12, that $x_{i}-a=0$ for $i=1,2$ because $x_{i}\leq g\leq a$. In the latter case, since $g-a=g$, we have

[TABLE]

for $\{i,j\}=\{1,2\}$.

∎

Proof of Theorem 0.3.14.

Let $L^{\prime}$ be the sub $\vee$-semilattice of $L$ generated by $\{x_{1},x_{2}\}$ over $L_{0}$.

As shown in Remark 0.3.13 we have $g=x_{1}\vee x_{2}$. Also $x_{2}-x_{1}=x_{2}$ and $x_{1}-x_{2}=x_{1}$. Indeed $x_{2}-x_{1}=(g-x_{1})-x_{1}=g-x_{1}=x_{2}$, the other case is symmetrical.

Hence by reasoning in a similar way as in the proof of Theorem 0.3.10, using properties (i) and (ii) of Lemma 0.3.15, we get that $L^{\prime}=L_{0}\langle x_{1},x_{2}\rangle$.

We now want to show that the join-irreducibles of $L_{0}\langle x_{1},x_{2}\rangle$ are exactly $x_{1},x_{2}$ and the join-irreducibles of $L_{0}$ different from $g$.

First, notice that if an element of $L_{0}$ is not join-irreducible in $L_{0}$ it cannot be join-irreducible in $L_{0}\langle x_{1},x_{2}\rangle$. Furthermore, the only elements of $L_{0}\langle x_{1},x_{2}\rangle$ not in $L_{0}$ that could be join-irreducible in $L_{0}\langle x_{1},x_{2}\rangle$ are $x_{1},x_{2}$ because $L_{0}\langle x_{1},x_{2}\rangle$ is the $\vee$-semilattice generated by $\{x_{1},x_{2}\}$ over $L_{0}$.

We now show that $x_{1},x_{2}$ are join-irreducible in $L_{0}\langle x_{1},x_{2}\rangle$.

Suppose $x_{1}$ is not join-irreducible in $L_{0}\langle x_{1},x_{2}\rangle$ and let $y_{1},\ldots,y_{r}$ be its join-irreducible components. One of them must be $x_{2}$ because $x_{1}\notin L_{0}$ and we observed that all the join-irreducible elements of $L_{0}\langle x_{1},x_{2}\rangle$ are in $L_{0}\cup\{x_{1},x_{2}\}$. But then $x_{2}\leq x_{1}$ and therefore, by what was shown above, $0=x_{2}-x_{1}=x_{2}$ which is absurd because $x_{2}\notin L_{0}$. The same reasoning holds for the join-irreducibility of $x_{1}$.

It remains to show that the only element join-irreducible of $L_{0}$ which is not join-irreducible in $L_{0}\langle x_{1},x_{2}\rangle$ is $g$.

Observe that $g$ is not join-irreducible in $L_{0}\langle x_{1},x_{2}\rangle$ because $g=x_{1}\vee x_{2}$ and $x_{1},x_{2}\neq g$ since $x_{1},x_{2}\notin L_{0}$.

Let $b\in L_{0}$ be join-irreducible in $L_{0}$ but not in $L_{0}\langle x_{1},x_{2}\rangle$, let $y_{1},\ldots,y_{r}$ be the join-irreducible components of $b$ in $L_{0}\langle x_{1},x_{2}\rangle$. From what we observed above it follows that the $y_{i}$’s are in $L_{0}\cup\{x_{1},x_{2}\}$ and since $b$ is join-irreducible in $L_{0}$ at least one of them is not in $L_{0}$. We can suppose $y_{1}=x_{1}$, so $x_{1}\leq b$. This implies that $g\leq b$, indeed one among $y_{2},\ldots,y_{r}$ has to be $x_{2}$ because otherwise $y_{2}\vee\cdots\vee y_{r}\in L_{0}$ and being the $y_{i}$’s the join-irreducible components of $b$ we have that $x_{1}=b-(y_{2}\vee\cdots\vee y_{r})$ must be in $L_{0}$, this is absurd. If $g<b$ then $b-g=b$ because $b$ is join-irreducible in $L_{0}$ but in this case $x_{1}=y_{1}\leq b=b-g\leq b-x_{1}=y_{2}\vee\cdots\vee y_{r}$ and this is not possible because the $y_{i}$’s are the join-irreducible components of $b$. This implies $b=g$.

Therefore the extension $L_{0}\hookrightarrow L_{0}\langle x_{1},x_{2}\rangle$ is minimal since the number of join-irreducibles of $L_{0}\langle x_{1},x_{2}\rangle$ is greater by one than the number of the join-irreducibles of $L_{0}$.

Notice that $L_{0}\langle x_{1},x_{2}\rangle$ is a minimal extension of $L_{0}$ of the second kind because the join-irreducibility of all but one of the join-irreducibles of $L_{0}$ is preserved. ∎

Theorem 0.3.16**.**

Let $L_{0}$ be a finite CBS and $L$ a finite minimal extension of $L_{0}$, then $L$ is generated over $L_{0}$ either by a primitive element $x\in L$ of the first kind over $L_{0}$ or by $x_{1},x_{2}\in L$ forming a primitive couple $(x_{1},x_{2})$ of the second kind over $L_{0}$.

Proof.

Let $f:P\rightarrow Q$ be the surjective minimal $\mathbf{P}$-morphism dual to the inclusion of $L_{0}$ into $L$. Recall that $P$ and $Q$ are respectively the posets of the join-irreducible elements of $L$ and $L_{0}$.

We consider two cases:

The first case is when $f$ is of the first kind, i.e. $\text{dom }f\neq P$ and there exists only one element $p\in P\setminus\text{dom }f$. In this case, by minimality of $f$, the restriction of $f$ on its domain is an isomorphism of posets. We want to prove that $x=p$ is a primitive element of $L$ of the first kind over $L_{0}$.

We observe that the downset $\downarrow\!p$ cannot be generated by the preimage of any downset in $Q$ because $p$ is not in the domain of $f$, therefore $x\notin L_{0}$.

For any $q\in Q$ let $q^{\prime}$ be the unique element of $P$ in the preimage of $q$ by $f$, then $\downarrow\!f^{-1}(\downarrow\!q)=\downarrow\!q^{\prime}$ because $f$ is a $\mathbf{P}$-morphism. Hence if $q^{\prime}\leq p$ then $\downarrow\!q^{\prime}-\downarrow\!p=\emptyset$ and if $q^{\prime}\nleq p$ then $\downarrow\!q^{\prime}-\downarrow\!p=\downarrow\!q^{\prime}$. This translates to the fact that for any $a$ join-irreducible of $L_{0}$ we have $a-x\in L_{0}$ because both $\emptyset$ and $\downarrow\!q^{\prime}$ are generated by the preimage of a downset of $Q$. Furthermore, for any $q\in Q$ if $p\leq q^{\prime}$ then $\downarrow\!p-\downarrow\!q^{\prime}=\emptyset$ and if $p\nleq q^{\prime}$ then $\downarrow\!p-\downarrow\!q^{\prime}=\downarrow\!p$. Thus for any $a$ join-irreducible of $L_{0}$ either $x-a=0$ or $x-a=x$.

The second case is when $f$ is of the second kind, i.e. $\text{dom }f=P$ and only two elements $p_{1},p_{2}$ have the same image by $f$, recall that $p_{1},p_{2}$ are incomparable. We want to prove that $x_{1}=p_{1}$ and $x_{2}=p_{2}$ form a primitive couple of elements of $L$ of the second kind over $L_{0}$.

$x_{1}\neq x_{2}$ and $x_{1},x_{2}\in L_{0}$ because the downsets $\downarrow\!p_{1}$ and $\downarrow\!p_{2}$ are distinct and neither of them is generated by the preimage of a downset in $Q$. Indeed, since $f$ is a total map the preimages of downsets of $Q$ are already downsets of $P$ and any preimage contains $p_{1}$ iff it contains $p_{2}$.

Let $f(p_{1})=f(p_{2})=g\in Q$ then $f^{-1}(g)=\{p_{1},p_{2}\}$, since $f$ is total $f^{-1}(\downarrow\!g)$ is a downset and thus $f^{-1}(\downarrow\!g)=\downarrow\!f^{-1}(\downarrow\!g)$. We have that $f^{-1}(\downarrow\!g)=\downarrow\!p_{1}\cup\downarrow\!p_{2}$ because $f$ is a $\mathbf{P}$-morphism and therefore $f^{-1}(\downarrow\!g)-\downarrow\!p_{1}=\downarrow\!p_{2}$ and $f^{-1}(\downarrow\!g)-\downarrow\!p_{2}=\downarrow\!p_{1}$ because $p_{1}$ and $p_{2}$ are incomparable. Therefore $g-x_{1}=x_{2}$ and $g-x_{2}=x_{1}$.

Let $q\in Q$ such that $q<g$ and $q^{\prime}$ be the unique element of $P$ in the preimage of $q$ by $f$, then $\downarrow\!f^{-1}(\downarrow\!q)=\downarrow\!q^{\prime}$ because $f$ is a $\mathbf{P}$-morphism. Hence if $q^{\prime}\leq p$ then $\downarrow\!q^{\prime}-\downarrow\!p=\emptyset$ and if $q^{\prime}\nleq p$ then $\downarrow\!q^{\prime}-\downarrow\!p=\downarrow\!q^{\prime}$. Both $\emptyset$ and $\downarrow\!q^{\prime}$ are generated by the preimage of a downset of $Q$. This means that for any $a$ join-irreducible of $L_{0}$ such that $a<g$ we have $a-x_{i}\in L_{0}$ for $i=1,2$. ∎

Definition 0.3.17**.**

Let $L_{0}$ be a finite CBS.

We call signature of the first kind in $L_{0}$ a couple $(h,G)$ where $h\in L_{0}$ and $G$ is a set of two-by-two incomparable join-irreducible elements of $L_{0}$ such that $h<g$ for all $g\in G$. We allow $G$ to be empty.

We call signature of the second kind in $L_{0}$ a triple $(h_{1},h_{2},g)$ where $h_{1},h_{2}\in L_{0}$, $g$ is a join-irreducible element of $L_{0}$ such that $h_{1}\vee h_{2}=g^{-}$ the unique predecessor of $g$ in $L_{0}$.

Theorem 0.3.18**.**

Let $L_{0}$ be a finite CBS. To give a minimal finite extension either of the first or of the second kind of $L_{0}$ (up to isomorphism over $L_{0}$) is equivalent to give respectively:

1. (1)

A signature $(h,G)$ of the first kind in $L_{0}$. 2. (2)

A signature $(h_{1},h_{2},g)$ of the second kind in $L_{0}$.

Once again finite duality shows its usefulness. Indeed, we have the following lemma:

Lemma 0.3.19**.**

Let $Q$ be a finite poset. To give a minimal surjective $\mathbf{P}$-morphism $f$ with codomain $Q$ either of the first or of the second kind (up to isomorphism) is equivalent to give respectively:

1. (1)

$D,U$* respectively a downset and an upset444The definitions of upset and of the upset $\uparrow\!a$ generated by an element $a$ are analogous to the definitions for the downsets replacing $\leq$ with $\geq$.*of $Q$ such that $D\cap U=\emptyset$ and for any $d\in D,u\in U$ we have $d\leq u$. 2. (2)

$g\in Q$* and $D_{1},D_{2}$ downsets of $Q$ such that $D_{1}\cup D_{2}=\downarrow\!g\setminus\{g\}$.*

Proof.

Let $f:P\to Q$ be a minimal surjective $\mathbf{P}$-morphism.

If $f$ is of the first kind and $\text{dom }f=P\setminus\{x\}$ take $D=f(\downarrow\!x\setminus\{x\})$ and $U=f(\uparrow\!x\setminus\{x\})$.

If $f$ is of the second kind, i.e. $\text{dom }f=P$, then there is exactly one $g\in Q$ such that $f^{-1}(g)=\{x_{1},x_{2}\}$ consisting of two elements of $P$. Take $D_{i}=f(\downarrow\!x_{i}\setminus\{x_{i}\})$ for $i=1,2$.

On the other hand, given $D,U$ as in 4 ,we obtain a minimal surjective $\mathbf{P}$-morphism $f:P\to Q$ by taking $P=Q\sqcup\{x\}$ and extending the order of $Q$ setting $q<x$ iff $q\in D$ and $x<q$ iff $q\in U$ for any $q\in Q$. Take $\text{dom }f=Q\subset P$ and $f$ as the identity on its domain.

Given $g\in Q$, $D_{1},D_{2}$ as in 2 obtain a minimal surjective $\mathbf{P}$-morphism $f:P\to Q$ taking $P=Q\setminus\{g\}\sqcup\{x_{1},x_{2}\}$ and extending the order of $Q\setminus\{g\}$ setting $q<x_{i}$ iff $q\in D_{i}$ and $x_{i}<q$ iff $g<q$ for any $q\in Q$. Take $\text{dom }f=P$ and $f$ maps $x_{1},x_{2}$ into $g$ and acts as the identity on $Q\setminus\{g\}$.

Now let $f_{1}:P_{1}\rightarrow Q$ and $f_{2}:P_{2}\rightarrow Q$ be two surjective $\mathbf{P}$-morphisms to which are associated the same $(D,U)$ or $(D_{1},D_{2},g)$, we show that there exists an isomorphism of posets $\varphi:P_{1}\rightarrow P_{2}$ such that $f_{2}\circ\varphi=f_{1}$.

Suppose $f_{1},f_{2}$ are of the first kind and the same $(D,U)$ is associated to both of them. Then $\text{dom }f_{1}=P_{1}\setminus\{p_{1}\}$ and $\text{dom }f_{2}=P_{2}\setminus\{p_{2}\}$. Being $f_{1},f_{2}$ two $\mathbf{P}$-morphisms which are isomorphisms when restricted on their domains, we can invert the restriction of $f_{2}$ and compose it with the restriction of $f_{1}$ to obtain an isomorphism of posets $\varphi^{\prime}:\text{dom }f_{1}\rightarrow\text{dom }f_{2}$.

It remains to extend $\varphi^{\prime}$ to an isomorphism $\varphi:P_{1}\rightarrow P_{2}$, just set $\varphi(p_{1})=p_{2}$; $\varphi$ so defined is an isomorphism of posets, we need to show that it reflects and preserves the order of $P_{1}$. $f_{1}$ and $f_{2}$ map respectively the elements smaller than $p_{1}$ and $p_{2}$ into the same elements of $Q$ and the elements greater than $p_{1}$ and $p_{2}$ into the same elements of $Q$ by hypothesis. Hence $\varphi^{\prime}$ maps the elements smaller than $p_{1}$ into the elements smaller than $p_{2}$ and the elements greater than $p_{1}$ into the elements greater than $p_{2}$ and so does its inverse. It follows that $\varphi$ is an isomorphism of posets.

Suppose $f_{1},f_{2}$ are of the second kind and the same $(D_{1},D_{2},g)$ is associated to both of them. Then $f_{1},f_{2}$ are total, i.e. $\text{dom }f_{1}=P_{1}$ and $\text{dom }f_{2}=P_{2}$. The orders restricted on $P_{1}\backslash f_{1}^{-1}(g)$ and $P_{2}\backslash f_{2}^{-1}(g)$ are both isomorphic to $Q\backslash\{g\}$ with isomorphisms given by the restrictions of $f_{1},f_{2}$, indeed given two elements $a,b\in P_{1}\backslash f_{i}^{-1}(g)$ it is $f_{i}(a)\leq f_{i}(b)$ iff $a\leq b$ because $f_{i}$ is a $\mathbf{P}$-morphism for $i=1,2$. Composing these two isomorphisms we obtain an isomorphism $\varphi^{\prime}:P_{1}\backslash f_{1}^{-1}(g)\rightarrow P_{2}\backslash f_{2}^{-1}(g)$.

We now extend it to $\varphi:P_{1}\to P_{2}$.

Let $f_{i}^{-1}(g)=\{x_{1,i},x_{2,i}\}$ for $i=1,2$, we can suppose to have ordered the indices in such a way that $f_{i}(\downarrow\!x_{j,i}\backslash\{x_{j,i}\})=D_{j}$ for $i,j=1,2$. Clearly we extend $\varphi^{\prime}$ to $\varphi$ defining $\varphi(x_{j,1})=x_{j,2}$. It remains to show that $\varphi$ is order preserving and reflecting.

Let $p\in P_{1}$ be such that $p\notin\{x_{1,1},x_{2,1}\}$.

Since $f_{1},f_{2}$ are $\mathbf{P}$-morphisms and $f_{i}(x_{j,i})=g$ we get $x_{j,2}=\varphi(x_{j,1})\leq\varphi(p)$ iff $g\leq f_{2}(\varphi(p))=f_{1}(p)$ iff $x_{j,1}\leq p$ for $j=1,2$.

Furthermore it is $p\leq x_{1,1}$ iff $f_{1}(p)\in D_{1}$ and $p\leq x_{2,1}$ iff $f_{1}(p)\in D_{2}$, similarly it is $\varphi(p)\leq x_{1,2}$ iff $f_{1}(p)=f_{2}(\varphi(p))\in D_{1}$ and $\varphi(p)\leq x_{2,2}$ iff $f_{1}(p)=f_{2}(\varphi(p))\in D_{2}$.

Therefore $\varphi$ is order preserving and reflecting. ∎

Proof of Theorem 0.3.18.

We just need to translate Lemma 0.3.19 in the language of CBSes using the finite duality:

A signature of the first kind $(h,G)$ in $L_{0}$ corresponds to a couple $(D,U)$ in $P$ as in 4 of Lemma 0.3.19. Indeed, by Köhler duality, downsets of $P$ correspond to elements of $L_{0}$ and upsets of $P$ correspond to the sets of their minimal elements, i.e. sets of two-by-two incomparable join-irreducible elements of $L_{0}$. The conditions $D\cap U=\emptyset$ and $\forall d\in D,u\in U$ $d\leq u$ translate in the condition $h<g$ for any $g\in G$.

A signature of the second kind $(h_{1},h_{2},g)$ in $L_{0}$ corresponds to a triple $(D_{1},D_{2},g)$ in $P$ as in 2 of Lemma 0.3.19. Indeed, $h_{1},h_{2}\in L_{0}$ correspond to the downsets $D_{1},D_{2}$ and $g$ join-irreducible of $L_{0}$ is an element of $P$ (recall that $P$ is the poset of the join-irreducibles of $L_{0}$). The condition that $D_{1}\cup D_{2}=\downarrow\!g\setminus\{g\}$ translates into $h_{1}\vee h_{2}=g^{-}$ since the predecessor $g^{-}$ of $g$ in $L_{0}$ corresponds to the downset $\downarrow\!g\setminus\{g\}$ of $P$. ∎

*Therefore signatures inside a finite CBS $L_{0}$ are like ‘footprints’ left by the minimal finite extensions of $L_{0}$: any minimal finite extension of $L_{0}$ leaves a ‘footprint’ inside $L_{0}$ given by the corresponding signature. On the other hand, given a signature inside $L_{0}$ we can reconstruct a unique (up to isomorphism over $L_{0}$) minimal extension of $L_{0}$ corresponding to that signature.

Since, by Theorems 0.3.10, 0.3.14 and 0.3.16, minimal finite extension of a finite CBS $L_{0}$ are exactly the ones generated over $L_{0}$ either by a primitive element or by a primitive couple, to any element or couple primitive over $L_{0}$ it is associated a unique signature in $L_{0}$. This is exactly what the next definition and theorem talk about.

Definition 0.3.20**.**

Let $L_{0}$ be a finite CBS and $L$ an extension of $L_{0}$.

We say that a primitive element $x\in L$ of the first kind over $L_{0}$ induces a signature of the first kind $(h,G)$ in $L_{0}$ if for any $a$ join-irreducible of $L_{0}$ we have that

[TABLE]

We say that a primitive couple $(x_{1},x_{2})\in L^{2}$ of the second kind over $L_{0}$ induces a signature of the second kind $(h_{1},h_{2},g)$ in $L_{0}$ if $g=x_{1}\vee x_{2}$ and for any $a$ join-irreducible of $L_{0}$ we have that

[TABLE]

Theorem 0.3.21**.**

*Let $L_{0}$ be a finite CBS and $L$ an extension of $L_{0}$.

A primitive element $x\in L$ induces a signature $(h,G)$ iff the extension $L_{0}\subseteq L_{0}\langle x\rangle$ corresponds to that signature.

A primitive couple $(x_{1},x_{2})\in L^{2}$ induces a signature $(h_{1},h_{2},g)$ iff the extension $L_{0}\subseteq L_{0}\langle x_{1},x_{2}\rangle$ corresponds to that signature.*

Proof.

For a primitive element $x$ of the first kind over $L_{0}$ to induce a signature $(h,G)$ means that $h$ is the predecessor of $x$ in $L_{0}\langle x\rangle$ and $G$ is the set of the join-irreducibles of $L_{0}$ which are minimal among the ones that are strictly greater than $x$ in $L_{0}\langle x\rangle$. This is the same as saying that the signature $(h,G)$ is associated to the extension $L_{0}\subseteq L_{0}\langle x\rangle$.

For a primitive couple $(x_{1},x_{2})$ of the second kind over $L_{0}$ to induce a signature $(h_{1},h_{2},g)$ means that $h_{i}$ is the predecessor of $x_{i}$ in $L_{0}\langle x_{1},x_{2}\rangle$ for $i=1,2$. This is the same as saying that the signature $(h_{1},h_{2},g)$ is associated to the extension $L_{0}\subseteq L_{0}\langle x_{1},x_{2}\rangle$. ∎

We have thus finally obtained an intermediate characterization of existentially closed CBSes:

Theorem 0.3.22**.**

A CBS $L$ is existentially closed iff for any finite sub-CBS $L_{0}\subseteq L$ we have:

1. (1)

Any signature of the first kind in $L_{0}$ is induced by a primitive element $x\in L$ of the first kind over $L_{0}$. 2. (2)

Any signature of the second kind in $L_{0}$ is induced by a primitive couple $(x_{1},x_{2})\in L^{2}$ of the second kind over $L_{0}$.

Proof.

By the characterization of the existentially closed CBSes given in Theorem 0.2.16 we have that a CBS $L$ is existentially closed iff for any finite sub-CBS $L_{0}$ and for any finite extension $L_{0}^{\prime}$ of $L_{0}$ we have that $L_{0}^{\prime}$ embeds into $L$ fixing $L_{0}$ pointwise. Since any finite extension of $L_{0}$ can be decomposed into a chain of minimal extensions, we can restrict to the case in which $L_{0}^{\prime}$ is a minimal finite extension of $L_{0}$. Then the claim follows from Theorem 0.3.18 and Theorem 0.3.21. ∎

Thanks to Theorem 0.3.22 we already get an axiomatization for the class of the existentially closed CBSes, indeed the quantification over the finite sub-CBS $L_{0}$ can be expressed elementarily using an infinite number of axioms. But this axiomatization is clearly unsatisfactory: other than being infinite, it is not conceptually clear.

0.4. Axioms

In this section we will prove that the existentially closed CBSes are exactly the ones satisfying the Splitting, Density 1 and Density 2 axioms. Each subsection focuses on one axiom. We will use extensively the characterization of existentially closed CBSes given by Theorem 0.3.22. To show the validity of the axioms in any existentially closed CBS we will use the following lemma.

Lemma 0.4.1**.**

*Let $\theta(\underline{x})$ and $\phi(\underline{x},\underline{y})$ be quantifier-free formulas in the language of CBSes. Assume that for every finite CBS $L_{0}$ and every tuple $\underline{a}$ of elements of $L_{0}$ such that $L_{0}\vDash\theta(\underline{a})$, there exists an extension $L_{1}$ of $L_{0}$ which satisfies $\exists\underline{y}\phi(\underline{a},\underline{y})$.

Then every existentially closed CBS satisfies the following sentence:*

[TABLE]

Proof.

Let $L$ be an existentially closed CBS.

Let $\underline{a}=(a_{1},\ldots,a_{n})\in L^{n}$ be such that $L\vDash\theta(\underline{a})$. Let $L_{0}$ be the sub-CBS of $L$ generated by $a_{1},\ldots,a_{n}$, by local finiteness $L_{0}$ is finite. By hypothesis there exists an extension $L_{1}$ of $L_{0}$ and $\underline{b}=(b_{1},\ldots,b_{m})\in L_{1}^{m}$ such that $L_{1}\vDash\phi(\underline{a},\underline{b})$.

Denote by $L_{0}^{\prime}$ the sub-CBS of $L_{1}$ generated by $b_{1},\ldots,b_{m}$ over $L_{0}$, it is a finite extension of $L_{0}$. By Theorem 0.2.16 $L_{0}^{\prime}$ embeds into $L$ fixing $L_{0}$ pointwise.

We thus get $L\vDash\phi(\underline{a},\underline{b^{\prime}})$ where $\underline{b^{\prime}}=(b_{1}^{\prime},\ldots,b_{m}^{\prime})\in L^{m}$ are the images of $b_{1},\ldots,b_{m}$ by the embedding.

Therefore we have proved that:

[TABLE]

∎

0.4.1. Splitting axiom

[Splitting Axiom] For every $a,b_{1},b_{2}$ such that $b_{1}\vee b_{2}\ll a\neq 0$ there exist elements $a_{1}$ and $a_{2}$ different from [math] such that:

[TABLE]

Theorem 0.4.2**.**

Any existentially closed CBS satisfies the Splitting Axiom.

Proof.

It is sufficient to show, by Lemma 0.4.1, that for any finite CBS $L_{0}$ and $a,b_{1},b_{2}\in L_{0}$ such that $b_{1}\vee b_{2}\ll a\neq 0$ there exists a finite extension $L_{0}\subseteq L$ with $a_{1},a_{2}\in L$ different from [math] such that:

[TABLE]

Let $Q$ be the poset dual to $L_{0}$ and $A,B_{1},B_{2}$ its downsets corresponding to $a,b_{1},b_{2}$.

We obtain a surjective $\mathbf{P}$-morphism $\pi:P\to Q$ in the following way:

For any $x\in Q$ such that $x\notin B_{2}$ (respectively $x\notin B_{1}$) let $\xi_{x,1}$ (respectively $\xi_{x,2}$) be a new symbol.

For any $x\in Q$ such that $x\in B_{1}\cap B_{2}$ let $\xi_{x,0}$ be a new symbol.

Let $P$ be the set of all these symbols, we define an order on $P$ setting:

[TABLE]

Intuitively $P$ is composed by a copy of $B_{1}\cup B_{2}$ and two copies of $Q\backslash(B_{1}\cup B_{2})$, one of the two copies is placed over $B_{1}$ and the other over $B_{2}$.

We define $\pi:P\rightarrow Q$ setting $\text{dom }\pi=P$ and $\pi(\xi_{x,i})=x$.

Let $\downarrow\!a_{1},\ldots,\downarrow\!a_{r}$ be the join-irreducible components of $A$, for any $i$ we have $a_{i}\notin B_{1}\cup B_{2}$ because by hypothesis $B_{1}\cup B_{2}\ll A$. Therefore $\pi^{-1}(\downarrow\!a_{i})=\downarrow\!\xi_{a_{i},1}\cup\downarrow\!\xi_{a_{i},2}$

We take:

[TABLE]

We obtain $\pi^{-1}(A)-A_{1}=A_{2}$ and $\pi^{-1}(A)-A_{2}=A_{1}$, they are both not empty because $r\geq 1$ and $A$ is not empty.

Furthermore for any $x\in B_{1}\cup B_{2}$ we have that $x\leq a_{i}$ for some $i$. Therefore if $x\in B_{1}\backslash B_{2}$ it is $\xi_{x,1}\leq\xi_{a_{i},1}$, if $x\in B_{2}\backslash B_{1}$ it is $\xi_{x,2}\leq\xi_{a_{i},2}$, finally if $x\in B_{1}\cap B_{2}$ then $\xi_{x,0}\leq\xi_{a_{i},1}$ and $\xi_{x,0}\leq\xi_{a_{i},2}$. This implies that $\pi^{-1}(B_{1})\subseteq A_{1}$ and $\pi^{-1}(B_{2})\subseteq A_{2}$.

We now show that $A_{1}\cap A_{2}=\pi^{-1}(B_{1})\cap\pi^{-1}(B_{2})$.

Let $\xi\in P$, we show that $\xi\in A_{1}\cap A_{2}$ iff $\xi\in\pi^{-1}(B_{1})\cap\pi^{-1}(B_{2})$.

If $\xi\in\pi^{-1}(B_{1})\cap\pi^{-1}(B_{2})$ then $\pi(\xi)\in B_{1}\cap B_{2}$, therefore $\xi=\xi_{x,0}$ and $x\leq a_{i}$ for some $i$. It implies that $\xi_{x,0}\leq\xi_{a_{i},1}$, thus $\xi_{x,0}\in A_{1}$ and $\xi_{x,0}\leq\xi_{a_{i},2}$, therefore $\xi_{x,0}\in A_{2}$ and $\xi\in A_{1}\cap A_{2}$.

On the other hand if $\xi\in A_{1}\cap A_{2}$ then there exist $i,j$ such that $\xi\leq\xi_{a_{i},1}$ and $\xi\leq\xi_{a_{j},2}$. By definition of the order on $P$ it has to be $\xi=\xi_{x,0}$ with $x\in B_{1}\cap B_{2}$, therefore $\xi\in\pi^{-1}(B_{1})\cap\pi^{-1}(B_{2})$.

Then

[TABLE]

Therefore

[TABLE]

Analogously we can show

[TABLE]

Thus taking the embedding $L_{0}\hookrightarrow L$ dual to $\pi$ and $a_{1},a_{2}\in L$ corresponding to $A_{1},A_{2}$ we have obtained what we were looking for. ∎

Lemma 0.4.3**.**

If $L$ is a CBS generated by a finite subset $X$ then any join-irreducible element of $L$ is a join-irreducible component in $L$ of some element of $X$.

Proof.

In any CBS the following identities hold:

[TABLE]

It follows by an easy induction that any term in the language of CBS is equivalent to a term of the form $x_{1}\vee\cdots\vee x_{n}$ with $x_{1},\ldots x_{n}$ containing only the difference symbol and variables. Notice that if an element $x_{1}\vee\cdots\vee x_{m}$ with $x_{1},\ldots x_{m}\in L$ is join-irreducible then it coincides with $x_{i}$ for some $i=1,\ldots,m$; thus any join-irreducible element $g$ of $L$ is the interpretation of a term $t$ over the variables $X$ containing only the difference symbol. This implies that $g$ is the join of some join-irreducible components of the leftmost variable in $t$. Indeed, this can be proved by induction on the complexity of the term observing that if $c_{1},\ldots,c_{m}$ are the join-irreducible components of an element $c\in L$ then for any $b\in L$:

[TABLE]

because $c_{i}-b=0$ or $c_{i}-b=c_{i}$, respectively, when $c_{i}\leq b$ or $c_{i}\nleq b$ since the $c_{i}$’s are join-irreducibles.

Thus $g$ is the join of the join-irreducible components of some $x\in X$, so, since it is join-irreducible, it is a join-irreducible component of $x$. ∎

Remark**.**

Lemma 0.4.3 is not true for co-Heyting algebras.

Indeed, consider the inclusion $L_{0}\hookrightarrow L_{1}$ of co-Heyting algebras described by Figure 3. $L_{1}$ is generated by $L_{0}$ and $a$ but $b=a\wedge(1-a)$ is join-irreducible in $L_{1}$ and it is not a join-irreducible component of any element of $L_{0}$ or $a$.

Lemma 0.4.4**.**

*Let $L_{0}$ be a finite sub-CBS of $L$ and let $L$ be generated by $L_{0}$ and $a_{1},\ldots,a_{n}\in L$.

If $a_{1},\ldots,a_{n}$ are joins of join-irreducible components in $L$ of elements of $L_{0}$555For instance, this happens when we have that all the $a_{i}$’s are of the kind $a_{i}=b_{i}-c_{i}$ with $b_{i}\in L_{0}$ and $c_{i}\in L$. Indeed, in such case, the join-irreducible components in $L$ of the $a_{i}$’s are among those of $b_{i}$.then the surjective $\mathbf{P}$-morphism $\varphi:P\to Q$ dual to the inclusion $L_{0}\hookrightarrow L$ is such that $\text{dom }\varphi=P$. In particular, the inclusion is also a co-Heyting algebras morphism, i.e. it preserves meets and $1$. *

Proof.

By Lemma 0.4.3 all the join-irreducible elements of $L$ are join-irreducible components in $L$ of elements of $L_{0}$ or of $a_{1},\ldots,a_{n}$. Since, by hypothesis, $a_{1},\ldots,a_{n}$ are joins of join-irreducible components of elements of $L_{0}$, any join-irreducible components of $a_{i}$ is a join-irreducible component of a join-irreducible component of an element of $L_{0}$ and thus it is a join-irreducible component of an element of $L_{0}$. Therefore any join-irreducible element of $L$ is a join-irreducible component in $L$ of an element of $L_{0}$.

Suppose that there is $x\in P$ such that $x\notin\text{dom }\varphi$, then $x$ corresponds to a join-irreducible element of $L$ which is not a join-irreducible component of any element of $L_{0}$. Indeed, if it is a join-irreducible component of $a\in L_{0}$ then $x\in P$ would be a maximal element of the downset $\downarrow\!\varphi^{-1}(A)$ where $A\subseteq Q$ is the downset relative to $a$, but this is not possible since if $x\in\downarrow\!\varphi^{-1}(A)$ then $x$ would be less than or equal to an element in $\varphi^{-1}(A)\subseteq\text{dom }\varphi$ which would be different from $x$, this is absurd because $x$ is maximal in $\downarrow\!\varphi^{-1}(A)$. Therefore, $\text{dom }\varphi=P$ because the existence of an element $x\notin\text{dom }\varphi$ would imply the existence of a join-irreducible element of $L$ which is not a join-irreducible component in $L$ of any element of $L_{0}$, but this contradicts what we have proven in the first part of this proof. ∎

Lemma 0.4.5**.**

Let $L$ be a CBS and $L_{0}$ a finite sub-CBS of $L$, $g$ be join-irreducible in $L_{0}$ and $y_{1},y_{2}\in L$ be nonzero elements such that

[TABLE]

Let also $L_{0}\langle y_{1},y_{2}\rangle$ be the sub-CBS of $L$ generated by $L_{0}$ and $\{y_{1},y_{2}\}$. We have that:

1. (1)

$g=y_{1}\vee y_{2}$, 2. (2)

any join-irreducible $a$ of $L_{0}$ such that $a\nleq g$ is still join-irreducible in $L_{0}\langle y_{1},y_{2}\rangle$, 3. (3)

$y_{1},y_{2}$* are distinct, not in $L_{0}$ and they are the join-irreducible components of $g$ in $L_{0}\langle y_{1},y_{2}\rangle$.*

Proof.

Notice that $y_{1}\vee y_{2}=g$ because $y_{1}\leq g$ and $y_{2}\leq g$ and

[TABLE]

Furthermore $y_{1},y_{2}\notin L_{0}$. Indeed, suppose that $y_{1}\in L_{0}$, then $y_{2}=g-y_{1}\in L_{0}$, since $g$ is join-irreducible in $L_{0}$ and $g=y_{1}\vee y_{2}$, we have that $g=y_{1}$ or $g=y_{2}$, by hypothesis it follows respectively that $y_{2}=0$ or $y_{1}=0$, in both cases we have a contradiction because $y_{1},y_{2}\neq 0$. Similarly, we obtain that $y_{2}\notin L_{0}$.

We also have that $y_{1}\neq y_{2}$. Indeed, suppose $y_{1}=y_{2}$, then $g-y_{1}=y_{1}$ implies that $g=y_{1}=0$ and this is absurd.

We now show that any join-irreducible $a$ of $L_{0}$ such that $a\nleq g$ is still join-irreducible in $L_{0}\langle y_{1},y_{2}\rangle$.

Any element of $L_{0}\langle y_{1},y_{2}\rangle$ is the join of repeated differences of $y_{1},y_{2}$ and of join-irreducibles of $L_{0}$ different from $g$, this is implied by the identities (+ ‣ 0.4.1) as noted in the proof of Lemma 0.4.3.

It is sufficient to show that for any $x$ obtained as repeated differences of $y_{1},y_{2}$ and join-irreducibles of $L_{0}$ different from $g$ we have $a-x=a$ or $a-x=0$. This will ensures that $a$ is join-irreducible in $L\langle y_{1},y_{2}\rangle$. Since $x$ is obtained as repeated differences of $y_{1},y_{2}$ and join-irreducibles $a_{1},\ldots,a_{n}$ of $L_{0}$ different from $g$, there is a term $t$ in the language of CBSes containing only $-$ and variables expressing $x$ as $t(y_{1},y_{2},a_{1},\ldots,a_{n})$. We prove that $a-x=a$ or $a-x=0$ by induction on the length of $t$. If the length is $1$, then $x\in\{y_{1},y_{2},a_{1},\ldots,a_{n}\}$

If $x\in L_{0}$ then $a-x=a$ or $a-x=0$ since $a$ is join-irreducible of $L_{0}$. Moreover $a-y_{i}=a$ for $i=1,2$. Indeed, $a\geq a-y_{i}\geq a-g=a$ because $a$ is a join-irreducible of $L_{0}$ such that $a\nleq g$.

Suppose the length of $t$ is greater than $1$. If the leftmost element among the ones whose differences give $x$ is $y_{i}$ for $i=1,2$ then $a-x=a$ because $a=a-y_{i}\leq a-x\leq a$.666Notice the following fact: if a term $t$ in the language of CBSes only contains $-$ (and variables) and $z$ is the leftmost variable in $t$, then the inequality $t\leq z$ is valid in every CBS (this is established by an easy induction on the length of $t$).

If the leftmost element among the ones whose differences give $x$ is $b$ join-irreducible of $L_{0}$ different from $g$. If $b<g$, since $a\nleq g$ yields $a\nleq b$ and thus $a=a-b$ by join-irriducibility of $a$, we obtain that $a=a-b\leq a-x\leq a$. If $b\nleq g$ we can obtain $x$ with a smaller number of differences because we can apply the induction hypothesis to replace its subterm of the kind $b-c$ with $b$ or [math] (with $b$ and $c$ playing respectively the role previously played by $a$ and $x$) and apply again the inductive hypothesis because $x$ can be expressed by a term shorter than $t$.

Now we prove that $y_{1},y_{2}$* are join-irreducibles of $L_{0}\langle y_{1},y_{2}\rangle$.* We show that for $y_{1}$, for $y_{2}$ is analogous.

Again, we show that for all $x\in L_{0}\langle y_{1},y_{2}\rangle$ we have $y_{1}-x=y_{1}$ or $y_{1}-x=0$. Let $x$ be obtained as repeated differences of $y_{1},y_{2}$ and join-irreducibles of $L_{0}$ different from $g$ as above and let us proceed by induction on the number of such differences.

First of all

[TABLE]

Let $x\in L_{0}$, if $g\leq x$ then $y_{1}-x\leq y_{1}-g=0$, if $g\neq x$, namely $g-x=g$ (recall that $g$ is join-irreducible in $L_{0}$) then

[TABLE]

If the leftmost element among the ones whose differences give $x$ is $y_{i}$ for $i=1,2$ then applying the inductive hypothesis, possibly many times, we obtain that $x=0$ or $x=y_{i}$ and in either case $y_{1}-x=0$ or $y_{1}-x=y_{1}$.

Suppose the leftmost element among the ones whose differences give $x$ is $b$ join-irreducible of $L_{0}$ different from $g$. If $b<g$ then

[TABLE]

and $y_{1}=y_{1}-b\leq y_{1}-x\leq y_{1}$.777Recall footnote 6.

If $b\nleq g$ we cannot have $b=g$ because $g$ is not join-irreducible, then using what we have proved above (that any join-irreducible $b$ of $L_{0}$ such that $b\nleq g$ is still join-irreducible in $L_{0}\langle y_{1},y_{2}\rangle$) we obtain $x=b$ or $x=0$ and in either case $y_{1}-x=0$ or $y_{1}-x=y_{1}$.

Finally, to prove that $y_{1},y_{2}$* are the join-irreducible components of $g$ in $L_{0}\langle y_{1},y_{2}\rangle$*, we simply have to notice that $y_{1}\nleq y_{2}$ and $y_{2}\nleq y_{1}$. Just observe that if $y_{1}\leq y_{2}$ then $g=y_{1}\vee y_{2}=y_{2}\notin L_{0}$ which is absurd. Analogously it cannot be $y_{2}\leq y_{1}$. ∎

Theorem 0.4.6**.**

*Let $L$ be a CBS satisfying the Splitting Axiom.

Then for any finite sub-CBS $L_{0}\subseteq L$ and for any signature $(h_{1},h_{2},g)$ of the second kind in $L_{0}$ there exists a primitive couple $(x_{1},x_{2})\in L^{2}$ of the second kind over $L_{0}$ inducing such signature.*

Proof.

We follow this strategy: we use the Splitting Axiom to ‘split’ $g$ obtaining two elements, one over $h_{1}$ and another over $h_{2}$. When $h_{1}=h_{2}$ these two elements form a primitive couple that induces the signature $(h_{1},h_{2},g)$, but unfortunately this is not true in general because these two elements may be too big. So we may have to ‘split’ these two new elements too in order to obtain other elements which we may have to ‘split’ again and again. This process has to stop after a finite number of steps, intuitively the more the element $h_{1}\wedge h_{2}$ (the meet is taken inside $L_{0}$) is smaller that $h_{1}$ and $h_{2}$, the more the process lasts. Then we accurately partition the set of all these ‘shards’ into two disjoint subsets and we take the joins of these two subsets. In this way we obtain two elements that form a primitive couple that induces the signature $(h_{1},h_{2},g)$ and we are done.

The statement of the Theorem require that, according to the definition of primitive couple inducing a given signature, we need to do the following. Given $h_{1},h_{2}\in L_{0}$ and $g$ join-irreducible of $L_{0}$ such that $h_{1}\vee h_{2}=g^{-}$, we have to find $x_{1},x_{2}\in L$ such that:

1. (1)

$x_{1}\neq x_{2}$ and $x_{1},x_{2}\notin L_{0}$, 2. (2)

$g-x_{1}=x_{2}$ and $g-x_{2}=x_{1}$

and for any $a$ join-irreducible of $L_{0}$:

1. (3)

if $a<g$ then $a-x_{i}\in L_{0}$ for $i=1,2$, 2. (4)

$a<x_{i}$ iff $a\leq h_{i}\;$ for $i=1,2$.

We recall that $L_{0}$ is a co-Heyting algebra because it is finite. In particular we can consider meets inside $L_{0}$ and they distribute with the joins.

Let $n_{i}$ for $i=1,2$ the maximum length of chains of join-irreducible elements of $L_{0}$

[TABLE]

such that $k_{n_{i}}\leq h_{i}$ and $k_{1}\nleq h_{j}$ with $i\neq j$, or equivalently $k_{1}\nleq h_{1}\wedge h_{2}$ where the meet is taken inside $L_{0}$.

Let $n=n_{1}+n_{2}$.

Intuitively, the natural number $n$ measures how much $h_{1}\wedge h_{2}$ is smaller than $h_{1}$ and $h_{2}$.

We prove the claim by induction on $n$.

Case 1: $n=0$.

Then $h_{1}\wedge h_{2}=h_{1}=h_{2}=g^{-}$. We denote $h_{1}=h_{2}$ by $h$.

Since $h\ll g$, we can apply the splitting axiom to $g,h,h$, hence there exist elements $x_{1},x_{2}\in L$ different from [math] such that:

[TABLE]

We now show that $(x_{1},x_{2})$ is a primitive couple of the second kind and induces the signature $(h_{1},h_{2},g)$:

1. (1)

As shown in Lemma 0.4.5 we have that $x_{1}\neq x_{2}$ and $x_{1},x_{2}\notin L_{0}$. 2. (2)

$g-x_{1}=x_{2}$ and $g-x_{2}=x_{1}$ follow directly from the splitting axiom, see (1).

Let $a$ be a join-irreducible element of $L_{0}$, then for $i=1,2$:

1. (3)

If $a<g$ then $a\leq g^{-}=h$ thus $a-x_{i}=0$ because $h\leq x_{i}$ as a consequence of the splitting axiom, see (1). 2. (4)

If $a<x_{i}$ then $a<g$ because $x_{i}<g$ by (1) and therefore $a\leq g^{-}=h$.

If $a\leq h$ then, since $x_{i}\notin L_{0}$ and $h\leq x_{i}$, we have $a<x_{i}$.

Case 2: $n>0$.

Suppose that the claim is true for any $m<n$.

Since $h_{1}\vee h_{2}=g^{-}\ll g$, we can apply the splitting axiom to $g,h_{1},h_{2}$, hence there exist elements $y_{1},y_{2}\in L$ different from [math] such that:

[TABLE]

Let $L_{0}\langle y_{1},y_{2}\rangle$ be the sub-CBS of $L$ generated by $L_{0}$ and $\{y_{1},y_{2}\}$. By local finiteness $L_{0}\langle y_{1},y_{2}\rangle$ is finite and thus a co-Heyting algebra.

Before continuing the proof we show a series of claims.

Claim 1*.*

The two following triples are signatures of the second kind in $L_{0}\langle y_{1},y_{2}\rangle$:

[TABLE]

where the meets are taken inside $L_{0}\langle y_{1},y_{2}\rangle$.

Proof of Claim 0: By Lemma 0.4.5 $y_{1},y_{2}\notin L_{0}$ are join-irreducibles in $L_{0}\langle y_{1},y_{2}\rangle$.

Moreover, in $L_{0}\langle y_{1},y_{2}\rangle$ we have that:

[TABLE]

Indeed

[TABLE]

and we have that this coincides with $y_{1}^{-}$, the predecessor of $y_{1}$ in $L_{0}\langle y_{1},y_{2}\rangle$. To show this, observe that, as a consequence of Lemma 0.4.4, the inclusion $L_{0}\hookrightarrow L_{0}\langle y_{1},y_{2}\rangle$ is dual to a surjective $\mathbf{P}$-morphism $\varphi:P\to Q$ with $\text{dom }\varphi=P$. Recall that $Q$ and $P$ are the posets of the join-irreducibles respectively of $L_{0}$ and $L_{0}\langle y_{1},y_{2}\rangle$. Notice that the preimage of an element $q$ of $Q$, i.e. a join-irreducible element of $L_{0}$ consists of the join-irreducible components of such element inside $L_{0}\langle y_{1},y_{2}\rangle$. This is a consequence of the fact that $\downarrow\!\varphi^{-1}(q)\subseteq P$ corresponds to $q$ as element of $L_{0}\langle y_{1},y_{2}\rangle$ and the set of maximal elements of $\downarrow\!\varphi^{-1}(q)$ is exactly the preimage of $q$ because $\varphi$ preserves the strict order.

Then $\varphi^{-1}(g)=\{y_{1},y_{2}\}$ because $y_{1},y_{2}$ are the join-irreducible components of $g$ in $L_{0}\langle y_{1},y_{2}\rangle$ by Lemma 0.4.5.

Notice that for any downset $D\subseteq Q$ we have $\downarrow\!\varphi^{-1}(D)=\varphi^{-1}(D)$ because $\text{dom }\varphi=P$ and $\varphi^{-1}(D)$ is a downset in the domain of $\varphi$.

Since $\downarrow\!\varphi^{-1}(\downarrow\!g)=\varphi^{-1}(\downarrow\!g)=\downarrow\!y_{1}\cup\downarrow\!y_{2}$ we have:

[TABLE]

Thus $(h_{1}\vee h_{2})\wedge y_{1}=g^{-}\wedge y_{1}$ is equal to $y_{1}^{-}$ because of the following equations

[TABLE]

To prove that $(h_{1}\wedge y_{2})\vee h_{2}=y_{2}^{-}$ the reasoning is analogous. $\blacksquare$

Notice that $h_{1}\wedge h_{2}$ is the same taken in $L_{0}$ and in $L_{0}\langle y_{1},y_{2}\rangle$ because by Lemma 0.4.4 the inclusion $L_{0}\hookrightarrow L_{0}\langle y_{1},y_{2}\rangle$ preserves meets.

Claim 2*.*

For $i=1,2$ the maximum length of chains of join-irreducibles of $L_{0}\langle y_{1},y_{2}\rangle$ less than or equal to $h_{i}$ that are not less than or equal to $h_{1}\wedge h_{2}$ is the same as $n_{i}$ defined above (recall that $n_{i}$ is defined taking the join-irreducibles of $L_{0}$).

Proof of Claim 0: Suppose there exists a chain of join-irreducibles in $L_{0}\langle y_{1},y_{2}\rangle$

[TABLE]

such that $k_{r}\leq h_{i}$ and $k_{1}\nleq h_{1}\wedge h_{2}$. Let, as above, $\varphi:P\to Q$ the surjective total $\mathbf{P}$-morphism dual to the inclusion $L_{0}\hookrightarrow L_{0}\langle y_{1},y_{2}\rangle$, then

[TABLE]

is a chain of join-irreducibles in $L_{0}$ such that $\varphi(k_{r})\leq h_{i}$ and $\varphi(k_{1})\nleq h_{1}\wedge h_{2}$. 888We remind that $\mathbf{P}$-morphisms preserve the strict order.

On the other hand a chain of join-irreducibles in $L_{0}$

[TABLE]

such that $b_{r}\leq h_{i}$ and $b_{1}\nleq h_{1}\wedge h_{2}$ can be lifted to a chain of join-irreducibles of $L_{0}\langle y_{1},y_{2}\rangle$

[TABLE]

such that $\varphi(k_{s})=b_{s}$ for $s=1,\ldots,r$ using the fact that $\varphi$ is a surjective $\mathbf{P}$-morphism, we obtain that $k_{r}\leq h_{i}$ and $k_{1}\nleq h_{1}\wedge h_{2}$. $\blacksquare$

Claim 3*.*

If $h_{2}\nleq h_{1}$ then the maximum length of chains of join-irreducibles of $L_{0}\langle y_{1},y_{2}\rangle$ less than or equal to $h_{2}\wedge y_{1}$ that are not less than or equal to $h_{1}\wedge h_{2}=h_{1}\wedge(h_{2}\wedge y_{1})$ is strictly smaller than $n_{2}$ (notice that $n_{2}\neq 0$ because $h_{2}\nleq h_{1}$).

When $h_{1}\nleq h_{2}$ in the same way we obtain the analogous result switching $y_{1}$ with $y_{2}$ and $h_{1}$ with $h_{2}$.

Proof of Claim 0: Suppose there exists a chain of join-irreducibles in $L_{0}\langle y_{1},y_{2}\rangle$

[TABLE]

such that $k_{n_{2}}\leq h_{2}\wedge y_{1}$ and $k_{1}\nleq h_{1}\wedge h_{2}$. Notice that this chain is not empty because $n_{2}\neq 0$. We have that $k_{n_{2}}$ is not a join-irreducible component of $h_{2}$ in $L_{0}\langle y_{1},y_{2}\rangle$. Indeed, $k_{n_{2}}\leq y_{1}$ and $y_{1}$ is not greater than or equal to any join-irreducible component of $h_{2}$ which is not less than or equal to $h_{1}\wedge h_{2}$ because $h_{2}-y_{1}=h_{2}-h_{1}=h_{2}-(h_{1}\wedge h_{2})$. Thus there would exist a continuation of such chain given by $k_{n_{2}+1}$ join-irreducible component of $h_{2}$ in $L_{0}\langle y_{1},y_{2}\rangle$, but this is absurd because we have proved in Claim 2 that $n_{2}$ is the maximum length of such chains. $\blacksquare$

We can now apply the inductive hypothesis, to do so we shall consider different cases.

Subcase 2.1: $h_{1},h_{2}$ incomparables. First, we consider the case in which $h_{1}\nleq h_{2}$ and $h_{2}\nleq h_{1}$, i.e. $h_{1},h_{2}$ are incomparable.

What we have proved in Claim 3 implies that the sum of the lengths of the chains considered above for either of the two signatures (3) is strictly smaller than $n$. Therefore we can apply the inductive hypothesis on both the two signatures (3) considered inside $L_{0}\langle y_{1},y_{2}\rangle$ to obtain two primitive couples $(y_{11},y_{12})\in L^{2}$ and $(y_{21},y_{22})\in L^{2}$ of the second kind over $L_{0}\langle y_{1},y_{2}\rangle$ such that they induce respectively the signatures $(h_{1},h_{2}\wedge y_{1},y_{1})$ and $(h_{1}\wedge y_{2},h_{2},y_{2})$. This means that:

1. (1)

$y_{11}\neq y_{12}$ and $y_{11},y_{12}\notin L_{0}\langle y_{1},y_{2}\rangle$, 2. (2)

$y_{1}-y_{11}=y_{12}$ and $y_{1}-y_{12}=y_{11}$

and for any $a$ join-irreducible of $L_{0}\langle y_{1},y_{2}\rangle$:

1. (3)

if $a<y_{1}$ then $a-y_{1i}\in L_{0}\langle y_{1},y_{2}\rangle\;$ for $i=1,2$, 2. (4)

$a<y_{11}$ iff $a\leq h_{1}\;$ and $\;a<y_{12}$ iff $a\leq(h_{2}\wedge y_{1})$.

furthermore

1. (1)

$y_{21}\neq y_{22}$ and $y_{21},y_{22}\notin L_{0}\langle y_{1},y_{2}\rangle$, 2. (2)

$y_{2}-y_{21}=y_{22}$ and $y_{2}-y_{22}=y_{21}$

and for any $a$ join-irreducible of $L_{0}\langle y_{1},y_{2}\rangle$:

1. (3)

if $a<y_{2}$ then $a-y_{2i}\in L_{0}\langle y_{1},y_{2}\rangle\;$ for $i=1,2$, 2. (4)

$a<y_{21}$ iff $a\leq(h_{1}\wedge y_{2})\;$ and $\;a<y_{22}$ iff $a\leq h_{2}$.

Notice that properties 4 of $y_{11},y_{12}$ and 4 of $y_{21},y_{22}$ actually hold for any $a\in L_{0}\langle y_{1},y_{2}\rangle$ since any element in a finite CBS is the join of join-irreducible elements. Observe also that for $a\in L_{0}$ we have $a\leq y_{ij}$ iff $a<y_{ij}$ because $y_{ij}\notin L_{0}$.

We want to prove that $x_{1}=y_{11}\vee y_{21}$ and $x_{2}=y_{12}\vee y_{22}$ are the two elements of $L$ we are looking for, i.e. $(x_{1},x_{2})$ is a primitive couple of the second kind over $L_{0}$ inducing the signature $(h_{1},h_{2},g)$.

First of all, we observe that

[TABLE]

Indeed,

[TABLE]

the second equation is shown analogously. (6) also shows that $y_{1}-y_{2}=y_{1}$ and $y_{2}-y_{1}=y_{2}$.

Moreover

[TABLE]

Indeed,

[TABLE]

and thus $y_{11}-y_{21}=y_{11}$, the remaining cases are analogous.

Notice the following fact about the extensions generated by the $y_{ij}$’s:

Claim 4*.*

The two extensions of finite CBSes given by $L_{0}\langle y_{1},y_{2}\rangle\subseteq L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle\subseteq L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$ are both minimal of the second kind. This implies that any $b$ join-irreducible of $L_{0}\langle y_{1},y_{2}\rangle$ different from $y_{1},y_{2}$ is still join-irreducible in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$.

Proof of Claim 0: It suffices to prove that $(y_{21},y_{22})$ is a primitive couple of the second kind over $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$.

First of all, as a consequence of Lemma 0.4.5, $y_{1},y_{2}$ are join-irreducible in $L_{0}\langle y_{1},y_{2}\rangle$, thus $y_{2}$ is join-irreducible in $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$.

1. (1)

$y_{21}\neq y_{22}$ by property 1 of $y_{21},y_{22}$.

$y_{21},y_{22}\in L\setminus L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$. Indeed, if $y_{21}\in L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$ then $y_{22}=y_{2}-y_{21}\in L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$ and vice versa. In that case $y_{2}=y_{21}\vee y_{22}\in L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$ with $y_{21},y_{22}\neq y_{2}$ because they are not in $L_{0}\langle y_{1},y_{2}\rangle$, but this is absurd because $y_{2}$ is join-irreducible. 2. (2)

$y_{2}-y_{21}=y_{22}$ and $y_{2}-y_{22}=y_{21}$ by property 2 of $y_{21},y_{22}$. 3. (3)

Since $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$ is a minimal finite extension of $L_{0}\langle y_{1},y_{2}\rangle$, its join-irreducibles are $y_{11},y_{12}$ and the join-irreducibles of $L_{0}\langle y_{1},y_{2}\rangle$ except $y_{1}$. If $a$ is a join-irreducible of $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$ such that $a<y_{2}$ then $a$ is join-irreducible in $L_{0}\langle y_{1},y_{2}\rangle$ because $a\neq y_{11},y_{12}$ since $y_{11},y_{12}\nless y_{2}$: indeed $y_{1i}-y_{2}=y_{1i}-(y_{21}\vee y_{22})=(y_{1i}-y_{21})-y_{22}=y_{1i}\neq 0$ by (7). Thus $a-y_{2i}\in L_{0}\langle y_{1},y_{2}\rangle$ by property 3 of of $y_{21},y_{22}$.

$\blacksquare$

Moreover, we observe that

[TABLE]

because thanks to equations (5) we have:

[TABLE]

showing the second equation of (8) is analogous.

We are now ready to show that $(x_{1},x_{2})$ is a primitive couple of the second kind over $L_{0}$ inducing the signature $(h_{1},h_{2},g)$.

1. (1)

Equations (8) imply that $g=x_{1}\vee x_{2}$, thus if $x_{1}=x_{2}$ then $x_{1}=g=0$ but this is absurd because $x_{1},x_{2}\neq 0$ since $y_{11},y_{12},y_{21},y_{22}\neq 0$ because they are not in $L_{0}$.

Furthermore $x_{1},x_{2}\notin L_{0}$; this is because $g$ is join-irreducible in $L_{0}$ and $g-x_{1}=x_{2}$ and $g-x_{2}=x_{1}$ are different from [math] and $g$. 2. (2)

See equations (8).

Let now $a$ be a join-irreducible element of $L_{0}$ and $i=1,2$:

1. (3)

If $a<g$ then $a\leq g^{-}=h_{1}\vee h_{2}$ and $a\leq h_{1}$ or $a\leq h_{2}$

• •

If $a\leq h_{1}$ then $a-x_{1}=0$ because $h_{1}\leq y_{11}\leq x_{1}$ by property 4 of $y_{11}$.

• •

If $a\leq h_{2}$ and $a\nleq h_{1}$ we want to prove that $a-x_{1}=a$.

Claim 5*.*

The join-irreducible components of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$ coincide with the join-irreducible component of $a$ in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$.

Proof of Claim 0: Since $a$ is the join of its join-irreducible components in $L_{0}\langle y_{1},y_{2}\rangle$, it is sufficient to prove that any join-irreducible component $b$ of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$ is join-irreducible in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$. We have $b\neq y_{1},y_{2}$ because $b\leq h_{2}$ and $y_{1},y_{2}\nleq h_{2}$ since if $y_{i}\leq h_{2}$ then $0=y_{i}-h_{2}=(g-y_{j})-h_{2}=(g-h_{2})-y_{j}=g-y_{j}=y_{i}$ with $i\neq j$ which is absurd. Thus by Claim 4 we have that $b$ is also join-irreducible in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$. $\blacksquare$

Since $a$ is join-irreducible of $L_{0}$ and $a\nleq h_{1}$ it is $a-h_{1}=a$. For any $b$ join-irreducible component of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$ we have $b\nleq h_{1}$ because $a-h_{1}=a$ means that $h_{1}$ is not greater than or equal to any join-irreducible component of $a$. Since $b\nleq h_{1}$ and in particular $b\nleq h_{1}\wedge y_{2}$ then property 4 of $y_{11}$ and property 4 of $y_{21}$ imply that $b\nleq y_{11},y_{21}$. Therefore $b\nleq y_{11}\vee y_{21}=x_{1}$ because $b$ is join-irreducible in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$. This implies that $a-x_{1}=a$ because $x_{1}$ is not greater than or equal to any join-irreducible component of $a$ in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$.

For $a-x_{2}$ the property is checked in an analogous way. 2. (4)

If $a\leq h_{i}$ then $a<y_{ii}\leq x_{i}$ by property 4 of $y_{11}$ and property 4 of $y_{22}$

If $a<x_{1}$ then $a<g$ by (8) and $a\leq h_{1}\vee h_{2}=g^{-}$. Let $b$ be a join-irreducible component of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$. We claim that $b\leq h_{1}$. We have that $b$ is join-irreducible in $L_{0}\langle y_{1},y_{2}\rangle$ and $b\neq y_{1},y_{2}$ because $b<x_{1}$ and $y_{1},y_{2}\nless x_{1}$. Indeed, by equations (7), we have:

[TABLE]

Suppose $b\nleq h_{1}$, then by property 4 of $y_{11}$ we would get $b\nless y_{11}$, furthermore $b\nleq h_{1}\wedge y_{2}$ and by property 4 of $y_{21}$ we would get $b\nless y_{21}$. Then $b$ would also be join-irreducible in $L_{0}\langle y_{ij}\>|\>i,j=1,2\rangle$ (see Claim 4). Therefore $b\nless y_{11}\vee y_{21}=x_{1}$ but this is absurd. Thus for any $b$ join-irreducible component of $a$ we have $b\leq h_{1}$ and hence $a\leq h_{1}$.

For $x_{2}$ the reasoning is analogous

Subcase 2.2: $h_{1},h_{2}$ comparables. The remaining cases are when $h_{1}<h_{2}$ or $h_{2}<h_{1}$ since $h_{1}=h_{2}$ only occurs when $n=0$.

We now consider the case $h_{1}<h_{2}$, for $h_{2}<h_{1}$ the reasoning is analogous.

In this case, by equations (4), we have $y_{2}^{-}=(h_{1}\wedge y_{2})\vee h_{2}=h_{2}$ in $L_{0}\langle y_{1},y_{2}\rangle$. Notice that we can apply the inductive hypothesis only on the first signature in (3) because $h_{2}\nleq h_{1}$ but it is not true that $h_{1}\nleq h_{2}$. Then we obtain the existence of $y_{11},y_{12}$ with the same properties 1, 2, 3, 4 as in the previous subcase. We define $x_{1}=y_{11}$ and $x_{2}=y_{12}\vee y_{2}$.

We want to prove that $x_{1},x_{2}$ form a primitive couple $(x_{1},x_{2})$ of the second kind over $L_{0}$ inducing the signature $(h_{1},h_{2},g)$.

We have

[TABLE]

because

[TABLE]

We have used that $y_{2}-y_{11}=y_{2}$, it is proven in the same way as (5) above.

1. (1)

Equations (10) imply $g=x_{1}\vee x_{2}$, thus if $x_{1}=x_{2}$ then $x_{1}=g=0$ but this is absurd because $x_{1},x_{2}\neq 0$ since $y_{11},y_{12},y_{2}\neq 0$ because they are not in $L_{0}$.

Furthermore $x_{1},x_{2}\notin L_{0}$ since $g$ is join-irreducible in $L_{0}$ and $g-x_{1}=x_{2}$ and $g-x_{2}=x_{1}$ are different from [math] and $g$. 2. (2)

See equations (10).

Let now $a$ be a join-irreducible element of $L_{0}$:

1. (3)

If $a<g$ then $a\leq g^{-}=h_{1}\vee h_{2}=h_{2}$.

• •

If $a\leq h_{1}$ then $a-x_{1}=0$ because $h_{1}\leq y_{11}=x_{1}$, moreover $a\leq h_{1}<h_{2}\leq y_{2}<x_{2}$ imply $a-x_{2}=0$.

• •

If $a\leq h_{2}$ and $a\nleq h_{1}$ clearly $a-x_{2}=0$ because $a\leq h_{2}\leq y_{2}\leq x_{2}$. We want to prove that $a-x_{1}=a$.

Claim 6*.*

The join-irreducible components of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$ coincide with the join-irreducible components of $a$ in $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$.

Proof of Claim 0: Since $a$ is the join of its join-irreducible components in $L_{0}\langle y_{1},y_{2}\rangle$, it is sufficient to prove that any join-irreducible component $b$ of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$ is join-irreducible in $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$. Pick such $a,b$. We have $b\neq y_{1}$ because $b\leq h_{2}$ and $y_{1}\nleq h_{2}$. Notice that $y_{1}\nleq h_{2}$ since $y_{1}\leq h_{2}$ would imply $0=y_{1}-h_{2}=(g-y_{2})-h_{2}=(g-h_{2})-y_{2}=g-y_{2}=y_{1}$ which is absurd. Then we have that $b$ is also join-irreducible in $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$, this follows from the fact that $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$ is a minimal extension of $L_{0}\langle y_{1},y_{2}\rangle$ of the second kind, which implies that the join-irreducibles in $L_{0}\langle y_{1},y_{2}\rangle$ different from $y_{1}$ are still join-irreducible in $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$. $\blacksquare$

Since $a$ is join-irreducible of $L_{0}$ and $a\nleq h_{1}$ it is $a-h_{1}=a$. For any $b$ join-irreducible component of $a$ in $L_{0}\langle y_{1},y_{2}\rangle$ we have $b\nleq h_{1}$ because $a-h_{1}=a$ means that $h_{1}$ is not greater than or equal to any join-irreducible component of $a$. Since $b\nleq h_{1}$ by property 4 of $y_{11}$ we have that $b\nless y_{11}=x_{1}$, therefore $b\nleq x_{1}$ because $b\neq y_{11}$ since $y_{11}\notin L_{0}\langle y_{1},y_{2}\rangle$ . This implies that $a-x_{1}=a$ because $x_{1}$ is not greater than or equal to any join-irreducible component of $a$ in $L_{0}\langle y_{1},y_{2},y_{11},y_{12}\rangle$. 2. (4)

By property 4 of $y_{11}$ we have that $a\leq h_{1}$ iff $a<y_{11}=x_{1}$.

If $a\leq h_{2}$ then $a\leq h_{2}<y_{2}\leq x_{2}$ since $h_{2}=y_{2}^{-}$.

If $a<x_{2}$, since $x_{2}=y_{12}\vee y_{2}\leq y_{1}\vee y_{2}=g$ then $a<g$ and $a\leq g^{-}=h_{1}\vee h_{2}=h_{2}$.

0.4.2. Density axioms

[Density 1 Axiom] For every $c$ there exists $b\neq 0$ such that $c\ll b$

Theorem 0.4.7*.*

Any existentially closed CBS satisfies the Density 1 Axiom.

Proof 0.4.8*.*

It is sufficient to show, by Lemma 0.4.1, that for any finite CBS $L_{0}$ and $c\in L_{0}$ there exists a finite extension $L_{0}\subseteq L$ with $b\in L$ different from [math] such that $c\ll b$.

Let $P_{0}$ be the finite poset dual to $L_{0}$ and $C$ its downset corresponding to $c$.

Let $P$ be the poset obtained by $P_{0}$ by adding a new maximum element $m\in P$ such that $m\geq p$ for any $p\in P_{0}$ and $\varphi:P\to P_{0}$ a surjective $\mathbf{P}$-morphism such that $\text{dom }\varphi=P_{0}$ and it is the identity on its domain. Then $C\ll\downarrow\!m$ and take as $L$ the CBS dual to $P$ and $b\in L$ corresponding to $\downarrow\!m$.

[Density 2 Axiom] For every $c,a_{1},a_{2},d$ such that $a_{1},a_{2}\neq 0$, $c\ll a_{1}$, $c\ll a_{2}$ and $a_{1}-d=a_{1}$, $a_{2}-d=a_{2}$ there exists an element $b$ different from [math] such that:

[TABLE]

Theorem 0.4.9*.*

Any existentially closed CBS satisfies the Density 2 Axiom.

Proof 0.4.10*.*

It is sufficient to show, by Lemma 0.4.1, that for any finite CBS $L_{0}$ and $c,a_{1},a_{2},d$ such that $a_{1},a_{2}\neq 0$, $c\ll a_{1}$, $c\ll a_{2}$ and $a_{1}-d=a_{1}$, $a_{2}-d=a_{2}$ there exists a finite extension $L_{0}\subseteq L$ with $b\in L$ different from [math] such that $c\ll b$, $b\ll a_{1}$, $b\ll a_{2}$ and $b-d=b$.

Let $P_{0}$ the poset dual to $L_{0}$ and $C,A_{1},A_{2},D$ its downsets corresponding to $c,a_{1},a_{2},d$.

If $C=\emptyset$ choose two maximal elements $\alpha^{1},\alpha^{2}$ respectively of $A_{1}$ and $A_{2}$ and obtain a poset $P$ by adding a new element $\beta$ to $P_{0}$ and setting for any $x\in P$:

• •

$\beta\leq x$ iff $x=\beta$ or $\alpha^{1}\leq x$ or $\alpha^{2}\leq x$.

If $\alpha^{1},\alpha^{2}$ are incomparable they become the only two successors of $\beta$ in $P$, otherwise if e.g. $\alpha^{1}\leq\alpha^{2}$ then $\alpha^{1}$ is the only successor of $\beta$.

• •

$x\leq\beta$ iff $x=\beta$, i.e. $\beta$ is minimal in $P$.

Define a surjective $\mathbf{P}$-morphism $\varphi:P\to P_{0}$ taking $\text{dom }\varphi=P_{0}$ and $\varphi$ acting as the identity on its domain. Take $B=\downarrow\!\beta$, we have:

• •

$\downarrow\!\varphi^{-1}(C)=\emptyset\ll B$,

• •

$B\ll\downarrow\!\varphi^{-1}(A_{1})=A_{1}\cup\{\beta\}$,

• •

$B\ll\downarrow\!\varphi^{-1}(A_{2})=A_{2}\cup\{\beta\}$,

• •

$B-\downarrow\!\varphi^{-1}(D)=B$.

Indeed, since $a_{1}-d=a_{1}$ and $a_{2}-d=a_{2}$, $D$ does not contain any maximal element of $A_{1}$ or $A_{2}$, in particular it does not contain $\alpha^{1}$ or $\alpha^{2}$

Take $L$ the CBS dual to $P$ and $b\in L$ corresponding to $B$.

If $C\neq\emptyset$ let $\gamma_{1},\ldots,\gamma_{n}$ be the maximal elements of $C$.

Choose for any $i=1,\ldots,n$ two maximal elements $\alpha_{i}^{1},\alpha_{i}^{2}$ respectively of $A_{1}$ and $A_{2}$ such that $\gamma_{i}\leq\alpha_{i}^{1}$ and $\gamma_{i}\leq\alpha_{i}^{2}$. Notice that they exist and $\gamma_{i}\neq\alpha_{i}^{1}$, $\gamma_{i}\neq\alpha_{i}^{2}$ because $C\ll A_{1}$ and $C\ll A_{2}$.

Obtain a poset $P$ by adding new elements $\beta_{1},\ldots,\beta_{n}$ to $P_{0}$ and setting for any $x\in P$:

• •

$\beta_{i}\leq x$ iff $x=\beta_{i}$ or $\alpha_{i}^{1}\leq x$ or $\alpha_{i}^{2}\leq x$.

If $\alpha_{i}^{1},\alpha_{i}^{2}$ are incomparable they become the only two successors of $\beta_{i}$ in $P$, otherwise if e.g. $\alpha_{i}^{1}\leq\alpha_{i}^{2}$ then $\alpha_{i}^{1}$ is the only successor of $\beta_{i}$.

• •

$x\leq\beta_{i}$ iff $x=\beta$ or $x\leq\gamma_{i}$,

i.e. $\gamma_{i}$ is the unique predecessor of $\beta_{i}$ in $P$.

Define a surjective $\mathbf{P}$-morphism $\varphi:P\to P_{0}$ taking $\text{dom }\varphi=P_{0}$ and $\varphi$ acting as the identity on its domain.

Take $B=\downarrow\!\beta_{1}\cup\cdots\cup\downarrow\!\beta_{n}$, we have:

• •

$\downarrow\!\varphi^{-1}(C)\ll B$,

• •

$B\ll\downarrow\!\varphi^{-1}(A_{1})=A_{1}\cup\{\beta_{1},\ldots,\beta_{n}\}$,

• •

$B\ll\downarrow\!\varphi^{-1}(A_{2})=A_{2}\cup\{\beta_{1},\ldots,\beta_{n}\}$,

• •

$B-\downarrow\!\varphi^{-1}(D)=B$.

Indeed $D$ does not contain any maximal element of $A_{1}$ or $A_{2}$, in particular it does not contain $\alpha_{i}^{1}$ or $\alpha_{i}^{2}$ for any $i=1,\ldots,n$.

Take $L$ the CBS dual to $P$ and $b\in L$ corresponding to $B$.

Theorem 0.4.11*.*

Let $L$ be a CBS satisfying the Splitting, Density 1 and Density 2 Axioms.

Then for any finite sub-CBS $L_{0}\subseteq L$ and for any signature $(h,G)$ of the first kind in $L_{0}$ there exists a primitive element $x\in L$ of the first kind over $L_{0}$ inducing such signature.

Proof 0.4.12*.*

We follow this strategy: if $G=\emptyset$ we use the Density 1 Axiom to take an element $m\in L$ greater than any element of $L_{0}$, then, thanks to the Splitting Axiom, using Theorem 0.4.6 we ‘split’ $m$ into two elements, one over $1_{L_{0}}$ the top element of $L_{0}$ and another over $h$. It turns out that this second element is primitive of the first kind and induces the signature $(h,\emptyset)$. If $G$ is nonempty and $G=\{g_{1},\ldots,g_{k}\}$ we suppose to have already found a primitive element $y$ inducing the signature $(h,\{g_{1},\ldots,g_{k-1}\})$. Then, using Theorem 0.4.6 again, we ‘split’ $g_{k}$ into two elements $g_{k}^{\prime},g_{k}^{\prime\prime}$, the first over $h$ and the second over the predecessor of $g_{k}$. Finally, applying the Density 2 Axiom, we obtain an element of $L$ in between $h$, $g_{k}^{\prime}$ and $y$. It turns out that this element is primitive of the first kind and induces the signature $(h,G)$.

The statement of the Theorem require that, according to the definition of primitive element inducing a given signature, we need to do the following. Given $h\in L_{0}$ and $G$ a set of join-irreducibles of $L_{0}$ such that $h<g$ for any $g\in G$, we have to find $x\in L$ such that:

1. (1)

$x\notin L_{0}$

and for any $a$ join-irreducible of $L_{0}$:

1. (2)

$a-x\in L_{0}$, 2. (3)

either $x-a=x\text{ or }x-a=0$, 3. (4)

$a<x$ iff $a\leq h\;$ and $\;x<a$ iff $g_{i}\leq a$ for some $i=1,\ldots,k$.

The proof is by induction on $k=\#G$.

Case $k=0$.

Let $1_{L_{0}}$ be the maximum element of $L_{0}$, by Density 1 there exists $0\neq m\in L$ such that $1_{L_{0}}\ll m$.

Then $L_{1}=L_{0}\cup\{m\}$ is a sub-CBS of $L$. Indeed it is closed under taking joins and differences since for any $a\in L_{0}$ we have $m>a$ and thus $a-m=0$ and $m=m-1_{L_{0}}\leq m-a\leq m$, therefore $m-a=m$. Hence $m$ is a join-irreducible of $L_{1}$. Furthermore it is clear that the join-irreducibles of $L_{1}$ are the join-irreducibles of $L_{0}$ and $m$.

$(h,1_{L_{0}},m)$ is a signature of the second kind in $L_{1}$, indeed $h\vee 1_{L_{0}}=1_{L_{0}}=m^{-}$.

Thanks to the Splitting Axiom we can apply Theorem 0.4.6 to the signature $(h,1_{L_{0}},m)$ in $L_{1}$ and obtain the existence of a primitive couple of the second kind $(x_{1},x_{2})\in L^{2}$ inducing such signature. Thus we have that there exist $x_{1},x_{2}\in L$ such that:

1. (1)

$x_{1}\neq x_{2}$ and $x_{1},x_{2}\notin L_{1}$, 2. (2)

$m-x_{1}=x_{2}$ and $m-x_{2}=x_{1}$

and for any $c$ join-irreducible of $L_{1}$:

1. (3)

if $c<m$ then $c-x_{i}\in L_{1}$ for $i=1,2$, 2. (4)

$c<x_{1}$ iff $c\leq h\;$ and $c<x_{2}$ iff $c\leq 1_{L_{0}}$.

Recall that Lemma 0.3.12 implies that for any $c\in L_{1}$:

1. (i)

$c-x_{i}\in L_{1}$ or $c-x_{i}=b\vee x_{j}$ for some $b\in L_{1}$ with $\{i,j\}=\{1,2\}$. 2. (ii)

$x_{i}-c=x_{i}\text{ or }x_{i}-c=0\;$ for $i=1,2$.

Let $x=x_{1}$, it is the element we were looking for. Indeed we now show that $x$ is a primitive element of the first kind over $L_{0}$ inducing the signature $(h,\emptyset)$

1. (1)

$x\notin L_{0}$ since $x=x_{1}\notin L_{1}$ by property 1 of $x_{1}$.

Let $a$ be a join-irreducible of $L_{0}$. Then

1. (2)

$a-x_{1}\in L_{0}$. Indeed, from $a\leq 1_{L_{0}}$ it follows (by property 4 of $x_{2}$) $a<x_{2}$; then by (ii) either $a-x_{1}\in L_{1}$ or $a-x_{1}=b\vee x_{2}$ with $b\in L_{0}$. The latter is absurd because (for $a<x_{2}$) we would get $x_{2}>a\geq a-x_{1}=b\vee x_{2}\geq x_{2}$. Then $a-x_{1}\in L_{1}$, i.e. $a-x_{1}\in L_{0}$ because $m>a\geq a-x_{1}$. 2. (3)

$x_{1}-a=x_{1}$ or $x_{1}-a=0$ by property (ii). 3. (4)

$a<x_{1}$ if and only if $a\leq h$ by property 4 of $x_{1}$.

$x_{1}\nless a$, because if $x_{1}<a$ then $x_{1}<1_{L_{0}}$ and so $0=x_{1}-1_{L_{0}}=(m-x_{2})-1_{L_{0}}=(m-1_{L_{0}})-x_{2}=m-x_{2}=x_{1}$ which is absurd because $x_{1}\notin L_{1}$ by property 1 of $x_{1}$.

Case $k\geq 1$.

Suppose that $G=\{g_{1},\ldots,g_{k}\}$. By inductive hypothesis there exists a primitive element $y\in L$ of the first kind over $L_{0}$ which induces the signature $(h,\{g_{1},\ldots,g_{k-1}\})$. This means that for any $a$ join-irreducible of $L_{0}$:

1. (1)

$y\notin L_{0}$, 2. (2)

$a-y\in L_{0}$, 3. (3)

either $y-a=y\text{ or }y-a=0$, 4. (4)

$a<y$ iff $a\leq h\;$ and $\;y<a$ iff $g_{i}\leq a$ for some $i=1,\ldots,k-1$.

Recall that Lemma 0.3.11 shows that the properties 2 and 3 actually hold for any $a\in L_{0}$.

Notice that $g_{k}$ is still join-irreducible in the sub-CBS $L_{0}\langle y\rangle\subseteq L$ generated by $L_{0}$ and $y$ since $L_{0}\subseteq L_{0}\langle y\rangle$ is a minimal finite extension of the first kind by Theorem 0.3.10.

Since $L$ satisfies the Splitting Axiom, we can apply Theorem 0.4.6 to the signature $(h,g_{k}^{-},g_{k})$ in $L_{0}\langle y\rangle$. Notice that it is a signature of the second kind because $h\vee g_{k}^{-}=g_{k}^{-}\ll g_{k}$. Therefore, there exists a primitive couple of the second kind $(g_{k}^{\prime},g_{k}^{\prime\prime})\in L^{2}$ inducing such signature. Thus we have that there exist $g_{k}^{\prime},g_{k}^{\prime\prime}\in L$ such that:

1. (1)

$g_{k}^{\prime},g_{k}^{\prime\prime}\notin L_{0}\langle y\rangle$ and $g_{k}^{\prime}\neq g_{k}^{\prime\prime}$, 2. (2)

$g_{k}-g_{k}^{\prime}=g_{k}^{\prime\prime}$ and $g_{k}-g_{k}^{\prime\prime}=g_{k}^{\prime}$

and for any $a$ join-irreducible of $L_{0}\langle y\rangle$:

1. (3)

if $a<g_{k}$ then $a-g_{k}^{\prime}\in L_{0}\langle y\rangle$ and $a-g_{k}^{\prime\prime}\in L_{0}\langle y\rangle$, 2. (4)

$a<g_{k}^{\prime}$ iff $a\leq h\quad$ and $\quad a<g_{k}^{\prime\prime}$ iff $a\leq g_{k}^{-}$.

Observe that property 4 actually holds for any $a\in L_{0}\langle y\rangle$ since any element in a finite CBS is the join of join-irreducible elements.

Apply the Density 2 Axiom on $h,y,g_{k}^{\prime},d$ where

[TABLE]

We can apply it because:

$h\ll y$ since by property 3 of $y$ we have $y-h=y$ because $h\in L_{0}$ and $h<y$.

$h\ll g_{k}^{\prime}$ since $h<g_{k}^{\prime}$ and $g_{k}^{\prime}-h=(g_{k}-g_{k}^{\prime\prime})-h=(g_{k}-h)-g_{k}^{\prime\prime}=g_{k}-g_{k}^{\prime\prime}=g_{k}^{\prime}$. Notice that $g_{k}-h=g_{k}$ because $g_{k}$ is join-irreducible in $L_{0}$.

$y-d=y$ since for any $b$ join-irreducible in $L_{0}$ such that $b\ngeq g_{1},\ldots,b\ngeq g_{k}$ we have $y-b=y$: otherwise, since $y$ is join-irreducible in $L_{0}\langle y\rangle$, it would be $y-b=0$ so $b>y$ and then by property 4 of $y$ we would have $b\geq g_{i}$ for some $i<k$ which is absurd.

$g_{k}^{\prime}-d=g_{k}^{\prime}$ since

[TABLE]

and $g_{k}^{\prime}-d=(g_{k}-g_{k}^{\prime\prime})-d=(g_{k}-d)-g_{k}^{\prime\prime}=g_{k}-g_{k}^{\prime\prime}=g_{k}^{\prime}$.

Then by the Density 2 Axiom there exists $0\neq x\in L$ such that

[TABLE]

$x$ is the element we were looking for. Indeed, it is primitive of the first kind over $L_{0}$ and induces the signature $(h,G)$:

1. (1)

We have $x\notin L_{0}$ because if $x\in L_{0}$ then since $x<y$ it would be $x\leq h$ by property 4 of $y$ but this is absurd because $x\neq 0$ and $h\ll x$.

Let $a$ be a join-irreducible of $L_{0}$:

1. (2)

If $a\leq h$ then $a-x=0$ since $h\leq x$ by (11).

If $a\nleq h$ then by property 4 of $g_{k}^{\prime}$ we have $a\nless g_{k}^{\prime}$.

• •

If $a\nleq h$ and $a\neq g_{k}$ then $a$ is still join-irreducible in $L_{0}\langle y,g_{k}^{\prime},g_{k}^{\prime\prime}\rangle$ (since $L_{0}\langle y\rangle\subseteq L_{0}\langle y,g_{k}^{\prime},g_{k}^{\prime\prime}\rangle$ is a minimal finite extension by Theorem 0.3.14), thus $a-g_{k}^{\prime}=a$. Therefore $a-x=a$ because $a=a-g_{k}^{\prime}\leq a-x\leq a$ since $x\leq g_{k}^{\prime}$.

• •

If $a=g_{k}$ then by $x\ll g_{k}^{\prime}$ (see (11))

[TABLE] 2. (3)

If $a\geq g_{i}$ for some $i=1,\ldots,k$ then:

• •

If $i\neq k$ then $a\geq y\geq x$ and $x-a=0$ by property 4 of $y$ and (11).

• •

If $i=k$ then $a\geq g_{k}\geq g_{k}^{\prime}\geq x$ and $x-a=0$.

If $a\ngeq g_{i}$ for any $i=1,\ldots,k$ then $a\leq d$ and $x-a=x$ since $x=x-d\leq x-a\leq x$ 3. (4)

If $a<x$ then $a<g_{k}^{\prime}$ and thus $a\leq h$ by property 4 of $g_{k}^{\prime}$.

If $a\leq h$ then $a<x$ because $h<x$ by (11).

If $x<a$ and $a\ngeq g_{1},\ldots,a\ngeq g_{k}$, then $a\leq d$ and $x=x-d\leq x-a=0$ which is absurd, thus $g_{i}\leq a$ for some $i=1,\ldots,k$.

If $g_{i}\leq a$ for some $i=1,\ldots,k$ then:

• •

If $i\neq k$ then, since $y<g_{i}$ by property 4 of $y$, we have $x<y<g_{i}\leq a$ and thus $x<a$.

• •

If $i=k$ then $x<g_{k}^{\prime}<g_{k}\leq a$ and thus $x<a$.

0.5. Properties of existentially closed CBSes

From our investigation we can easily obtain some properties of the existentially closed CBSes:

Proposition 0.5.1*.*

If $L$ is an existentially closed CBS, then $L$ does not have a maximum element.

Proof 0.5.2*.*

Since $L$ satisfy the Density 1 Axiom for any $c\in L$ there exists an element $b\neq 0$ such that $c\ll b$ and therefore $c<b$. This implies that there cannot be a maximum element of $L$.

Proposition 0.5.3*.*

Let $L$ be an existentially closed CBS and $a,b\in L$.

If $a$ and $b$ are incomparable, i.e. $a\nleq b$ and $b\nleq a$, then there does not exist the meet of $a$ and $b$ in $L$.

Notice that if $a\leq b$ then the meet exists and it is $a$.

Proof 0.5.4*.*

Denote by $c$ the meet of $a$ and $b$. Consider $L_{0}\subseteq L$ the sub-CBS generated by $a,b,c$. It is finite by local finiteness. $c$ is the meet of $a$ and $b$ also in $L_{0}$.

Since $a,b$ are incomparable there exist $g_{1},g_{2}$ join-irreducible components in $L_{0}$ respectively of $a$ and $b$ such that $g_{1}\nleq b$ and $g_{2}\nleq a$.

By Theorem 0.3.22 taking $h=0\in L_{0}$ we have that there exists $x\in L\setminus L_{0}$ such that for any $d\in L_{0}$:

• •

$d<x$ iff $d=0$,

• •

$x<d$ iff $g_{i}\leq d$ for $i=1$ or $i=2$.

We have that $x\nleq c$ since $x\notin L_{0}$, $g_{1}\nleq c$ and $g_{2}\nleq c$, therefore $c<c\vee x$. Notice that $x<g_{1}\leq a$ and $x<g_{2}\leq b$, thus $c\vee x\leq a$ and $c\vee x\leq b$. This implies that $c$ cannot be the meet of $a$ and $b$ in $L$.

Proposition 0.5.5*.*

If $L$ is an existentially closed CBS, then there are no join-irreducible elements of $L$.

Proof 0.5.6*.*

Let $g$ be a nonzero element of $L$. We can apply the splitting axiom on the triple $g,0,0$, then there exist $g_{1},g_{2}\in L$ such that

[TABLE]

Since $g_{1},g_{2}\leq g$ and $g-(g_{1}\vee g_{2})=(g-g_{1})-g_{2}=0$ we have that $g=g_{1}\vee g_{2}$. Moreover $g_{1},g_{2}\neq g$ because $g_{1},g_{2}\neq 0$. Therefore $g$ cannot be join-irreducible because $g=g_{1}\vee g_{2}$ with $g_{1},g_{2}\neq g$, recall that [math] is never join-irreducible.