When is a partially ordered set really a fragment of a Boolean algebra, and why is this an
interesting question?
Rob Egrot
Ordered structures
Definition (Poset)
A poset is a set equipped with a binary relation≤that is reflexive, transitive and antisymmetric.
Definition (Semilattice)
A (meet) semilattice is a poset where every pair of elements has an infimum. This is enough to guarantee that every non-empty finite set of elements has an infimum.
Definition (Lattice)
A lattice is a poset in which every pair of elements has both a supremum and an infimum (expressed using binary operations∨ and∧respectively). This is enough to guarantee that every non-empty finite set has both a supremum and an infimum. We call these meets and joins respectively.
Boolean algebras
Definition (Bounded distributive lattices)
I A lattice is bounded if it has greatest and least elements 1 and0.
I A lattice is distributive if
a∧(b∨c) = (a∧b)∨(a∧c)for all a,b,c. Or, equivalently, if
a∨(b∧c) = (a∨b)∧(a∨c)for all a,b,c.
Definition (Boolean algebra)
A Boolean algebra is a bounded distributive lattice where every element has a (necessarily unique) complement (i.e. a∨ ¬a= 1, and a∧ ¬a= 0).
Prime filters
Definition (Prime filter)
In a lattice, a prime filter is a setΓthat is closed upwards and under finite meets, and is prime (i.e., if T is finite andW
T ∈Γ then T∩Γ6=∅).
In Boolean algebras prime filters are calledultrafilters.
Definition ((m,n)-filter)
In a poset P, an(m,n)-filter is an upclosed set Γsuch that:
1. If S ⊆Γ, and|S|<m andV
S is defined, then V S ∈Γ.
2. If T ⊆P, and|T|<n andW
T is defined and in Γ, then T ∩Γ6=∅.
(m,n)-filters generalize prime filters to posets.
Motivating examples
Definition (Field of sets)
If X is a set then℘(X) defines a Boolean algebra with operations union, intersection and taking complements. 1 and 0 are X and∅ respectively. A field of sets is a sub-Boolean algebra of a powerset algebra.
Definition (Ring of sets)
A ring of sets is the lattice reduct of a field of sets.
Representations for Boolean algebras and distributive lattices
Theorem (Stone 1936 [10])
Every Boolean algebra is isomorphic to a field of sets.
Theorem (Birkhoff 1933 [2])
A lattice is isomorphic to a ring of sets if and only if it is distributive.
Proof idea: given an algebraA, use the set of all prime filters ofA and the maph:a7→ {Γ :a∈Γ}. We need the ‘prime ideal theorem’ to guarantee enough prime filters. This is a non-constructive choice principle weaker than AC.
Representations for posets and semilattices
Definition
Let2≤m,n≤ω. Then:
I A meet semilattice A is n-representable if there is a powerset algebra℘(X) and a 1-1 semilattice homomorphism
h :A→℘(X) such that if, T ⊆A with |T|<n andW T defined in A, then h(W
T) =S h[T].
I A poset P is(m,n)-representable if there is a powerset algebra
℘(X) and an order embedding h:P →℘(X) such that:
1. If S ⊆P withVS defined and|S|<m, then h(VS) =Th[S].
2. If T ⊆P andWT is defined and |T|<n, then h(WT) =Sh[T].
When are semilattices representable?
Theorem (Balbes 1969 [1])
A meet semilattice A is n-representable iff for all k<n, for all y∈A, whenever y∧(x1∨. . .∨xk) is defined,
(y∧x1)∨. . .∨(y∧xk) is also defined and the two are equal.
Like the representation theorems for distributive lattices and Boolean algebras, this comes down to the semilattice having enough prime filter-like subsets. This follows from the above property by Zorn’s lemma, or the implication can be adopted as a weaker choice axiom.
Clearly if a semilattice isn-representable then it is k-representable for allk ≤n. However, given any 2≤n< ω we can construct a semilattice that isn-representable but not (n+ 1)-representable (Kearnes 1997 [6]).
When are posets representable?
Theorem
A poset is(m,n)-representable iff for all p6≤q ∈P there is an (m,n)-filter of P containing p but not q.
Again, a representation of a posetP amounts to using the set of all (m,n)-filters of P as a base, and the embedding
p7→ {Γ :p∈Γ}. The condition of the theorem just guarantees thatP has enough (m,n)-filters.
Corollary
Every poset is(ω,2)-representable and (2, ω)-representable.
Proof.
Givenp 6≤q just use{x:x ≥p} or {x:x6≤q}.
The poset representation decision problem
Theorem (Van Alten 2016 [11])
For3≤m,n≤ω with at least one of m,n strictly greater than 3, the problem of deciding whether an arbitrary finite poset is (m,n)-representable isNP-complete.
Theorem (Vardi 1995 [12])
There is a p-time algorithm for checking whether an arbitrary first-order sentence holds in a finite structure.
Corollary
If a suitable class of(m,n)-representable posets is finitely axiomatizable thenP=NP
Needless to say, this is very strong evidence that these classes are not finitely axiomatizable, but what can we say for sure?
Ultraproducts
LetL be a first-order signature, letI be an indexing set, and for eachi ∈I letAi be an L-structure.
Definition (Ultraproduct)
Let U be an ultrafilter of℘(I) (considered as a powerset algebra).
Then the ultraproduct of{Ai}I with respect to U is defined to be the set of equivalence classes ofQ
IAi with respect to the equivalence relation∼defined by
x ∼y ⇐⇒ {i ∈I :x(i) =y(i)} ∈U. We write
Y
U
Ai,
and we use e.g. [x]to denote elements of the ultraproduct.
IfAi =A for alli ∈I, thenQ
UAi is an ultrapower, andAis an ultraroot.
Ultraproducts as L -structures
Theorem
The ultraproductQ
UAi is an L-structure if we interpret the non-logical symbols ofL as follows:
I If R is an n-ary relation symbol define Y
U
Ai |=R([x1], . . . ,[xn])
⇐⇒ {i ∈I :Ai |=R(x1(i), . . . ,xn(i))} ∈U.
I If f is an n-ary function symbol define
f([x1], . . . ,[xn]) = [f(x1, . . . ,xn)]
where f(x1, . . . ,xn)(i) =f(x1(i), . . . ,xn(i)).
I If c is a constant symbol define
c = [¯c]wherec¯(i) =c for all i.
Ultraproducts as models
Theorem ( Lo´s 1955 [8])
Letφ(x1, . . . ,xn) be anL-formula. Then Y
U
Ai |=φ([x1], . . . ,[xn]) ⇐⇒ {i ∈I :Ai |=φ(x1(i), . . . ,xn(i))} ∈U.
Corollary
If T is anL-theory and Ai is a model of T for all i ∈I , then Q
UAi is a model of T .
The Keisler-Shelah theorem
Theorem
A class ofL-structures is elementary iff it is closed under taking isomorphisms, ultraproducts and ultraroots.
One direction of this follows directly from Lo´s’ theorem. The other follows, after some effort, from the fact that elementarily
equivalent structures have isomorphic ultrapowers. This difficult result was proved by Keisler in 1961 (assuming the generalized continuum hypothesis) [7], and in 1971 by Shelah (without the assumption of GCH) [9].
The class of (m, n)-representable posets is finitely axiomatizable in first-order logic!
But we have to cheat.
Definition (Pseudoelementary class)
A classAof L-structures is pseudoelementary if there is an extensionL+ of L, and anL+ theory T such thatAis the class ofL-reducts of models of T .
Theorem (Egrot 2016 [3])
For m,n≤ω the class of(m,n)-representable posets is
pseudoelementary. When m,n< ω the class can be axiomatized with a finite theory in the extended language.
Proof sketch
1. Use the extended language to define a theory with two sorts,
‘Set’ and ‘Element’.
2. Formalize the concept of a ‘Set’ being an (m,n)-filter, with respect to other ‘Sets’.
3. Write down axioms demanding that ifp andq are ‘Elements’
andp 6≤q there is a ‘Set’ sorted ‘(m,n)-filter’ containingp and notq.
4. Every (m,n)-representable poset will be a model of this theory.
5. For the converse we need to guarantee that being an
‘(m,n)-filter’ with respect to ‘Sets’ in a model is enough to guarantee being an actual (m,n)-filter with respect to the actual sets that exist in a model. With a little cunning we can write down axioms that do this.
Axiomatizing the class of (m, n)-representable posets
Theorem (Egrot 2016 [3])
For2≤m,n≤ω the class of(m,n)-representable posets is elementary.
Proof sketch:
1. Show that the class of (m,n)-representable posets is closed under ultraproducts.
1.1 It follows from Lo´s theorem that pseudoelementary classes are closed under ultraproducts.
2. Notice that a poset P embeds elementarily intoQ
UP, so (m,n)-filters of Q
UP restrict to (m,n)-filters ofP, thus the class is closed under ultraroots.
3. Note that the class is obviously closed under taking isomorphisms.
4. Appeal to Keisler-Shelah.
Finite axiomatizations for (m, n)-representable posets?
Theorem (Egrot 2018 [4])
Let3≤m,n≤ω, then the class of(m,n)-representable posets is not finitely axiomatizable in the language of posets.
Proof sketch:
1. Since the class of (m,n)-representable posets is elementary, it will be finitely axiomatizable iff its complement is elementary.
2. For each 0≤k < ω construct a poset that is not (m,n)-representable, in such a way that Q
UPk is (m,n)-representable for some ultrafilterU of ℘(ω).
3. By Lo´s’ theorem the complement is not elementary, thus the class is not finitely axiomatizable.
Describing the class of posets that are not (m, n)-representable
Lemma
If anyNP-complete problem is in coNP thenNP=coNP.
Theorem (Fagin 1973 [5])
A class isNP iff it has a finite axiomatization in existential 2nd order logic.
Corollary
Let3≤m,n≤ω with either m or n strictly greater than 3. If the class of posets that are not(m,n)-representable has a finite axiomatization in existential second order logic thenNP=coNP.
Constructing explicit axiomatizations for (m, n)-representable posets
I We have (hopefully) seen that classes of (m,n)-representable posets are worth investigating.
I We know they cannot be finitely axiomatized except in trivial cases.
I We do, however, know that all these classes are, in principle, axiomatizable in first-order logic using the standard language of posets (i.e. just the ≤relation).
I We will construct an explicit axiomatization.
A plan
Our plan has two parts:
1. Define a two player game, played between playersA andE using a posetP, such that E has a winning strategy if and only if P is (m,n)-representable.
2. Write down first-order sentences formalizing the property ofE having a winning strategy for a given poset.
Games and strategies
I Games are played between two players,A andE, over a poset P.
I A game is played in rounds indexed by natural numbers (starting at 0).
I In each round, A plays first, thenE responds.
I Each move E makes adds an element of P to a set U.
I The initial state ofU, along with the state of another setV, is the starting position of the game.
I E wins ann-round game ifA does not win till at least the (n+ 1)th round.
I E wins anω-round game ifA does not win at any point.
I E has an n-strategy if she can guarantee a win in ann-round game however Aplays.
I E has an ω-strategy is she can guarantee a win in anω-round game.
A game when m = n = 3
Moves forA:
1. Ifb ≥afor somea∈U, then Acan play (b).
2. Ifa,b ∈U, anda∧b is defined in P, then Acan play (a∧b).
3. Ifa,b ∈U, anda∨b is defined in P, then Acan play (a∨b).
Responses forE:
1. E must add b to U.
2. E must add a∧b to U.
3. E must add either aor b to U.
Awins in round n ifU∩V 6=∅ at the start of that round. By choosingU and V correctly, a game won byE constructs an (m,n)-filter containing p but not containingq, for any given p6≤q ∈P.
The game models representability
Proposition
If P is(3,3)-representable then, for all p6≤q∈P, E has an ω-strategy in the game with starting position ({p},{q}). If P is countable, then the converse is also true.
Proof idea (forward direction):
1. IfP is (3,3)-representable then, givenp 6≤q ∈P,E chooses an (3,3)-filter containing p and not q.
2. IfE responds to moves by Aby playing only elements that are in Γ she will never lose.
The game models representability: part 2
Proof idea (backward direction):
1. Given p6≤q∈P, sinceP is countable we can assume thatA plays every useful move eventually. I.e. if a move is available for Ain round n, then either he plays that move in roundk (for some k≥n), or the move becomes useless during the course of play (a move is useless ifE can respond without adding any new element to U).
2. Since Amakes every move, the setU constructed by E during a winning ω-game must be a (3,3)-filter.
3. Since E has an ω-strategy, there must be a (3,3)-filter for every p 6≤q ∈P.
Axioms for winning games
For allm,n ∈ω we use ¯xm to denote them-tuple (x1, . . . ,xm), and we define formulasφmn, with m+ 1 free variables, by recursion as follows.
φm0(¯xm,y) =Dm(¯xm,y)
φm(n+1)(¯xm,y) =
∀ab
(∃c(Cm(¯xm,c)∧(c≤a))→φ(m+1)n(¯xm,a,y)
∧ (Cm(¯xm,a)∧Cm(¯xm,b)∧ ∃cM(a,b,c))→φ(m+1)n(¯xm,c,y)
∧ ∃c(Cm(¯xm,c)∧J(a,b,c))→(φ(m+1)n(¯xm,a,y)∨φ(m+1)n(¯xm,b,y))
HereJ(x,y,z) and M(x,y,z) should be interpreted as sayingz is the join (respectively, meet) ofx andy.
Cm(¯xm,y) saysy ∈ {x1, . . . ,xm}, andDm(¯xm,y) saysy∈ {x/ 1, . . . ,xm}.
Axioms for winning games: part 2
Lemma
Let P be a poset, and let v be an assignment of variables onto elements of P. Then, for all m,n∈ω,
P,v |=φmn(¯xm,y) ⇐⇒ E has an n-strategy for the game with starting position({v(x1), . . . ,v(xm)},{v(y)})
Axioms for winning games: part 3
Proof sketch (induction onn):
1. Definev[¯xm] ={v(x1), . . . ,v(xm)}. Then
P,v |=φm0(¯xm,y) ⇐⇒ v[¯xm]∩ {v(y)}=∅, i.e. E has a 0-strategy in the game with starting position (v[¯xm],{v(y)}).
2. P,v |=φm(n+1)(¯xm,y) if and only if, whenever Ahas a useful move from starting position (v[¯xm],{v(y)}),E can respond with a move (a) such that the original formula is satisfied iff a certain φ(m+1)n formula is satisfied.
3. With a suitable inductive hypothesis, this φ(m+1)n formula is satisfied iffE has a winning strategy for then-round game with starting position (v[¯xm]∪ {a},{v(y)}).
4. So the original formula is equivalent to E having an (n+ 1)-strategy, as required.
Axioms for winning games: part 4
Now, for eachn∈ω we define sentences as follows.
ψn=∀xy((x 6≤y)→φ1n(x,y))
Proposition
P |=ψn ⇐⇒ for all p,q ∈P, if p6≤q then E has an n-strategy for the game with starting position ({p},{q}).
Proof.
By the lemma.
Axioms for winning games: part 5
Theorem
Let P be a poset. Then P has a(3,3)-representation if and only if P |=ψn for all n∈ω.
Proof sketch (forward direction):
1. IfP is (3,3)-representable then it has a separating set of (3,3)-filters.
2. So, givenp 6≤q ∈P,E can pick a (3,3)-filter, Γ, containing p but notq, and win the ω-game with starting position ({p},{q}) just by always picking elements of Γ.
Axioms for winning games: part 6
Proof sketch (backward direction):
1. Suppose that P |=ψn for all n∈ω, and (for now) that P is countable.
2. Then whenever p6≤q ∈P,E has an-strategy for the game with starting position ({p},{q}).
3. By K¨onig’s tree lemma, it follows thatE has an ω-strategy, and soP is (3,3)-representable.
4. IfP is uncountable then it has a countable elementary substructure, P0, by the downward L¨owenheim-Skolem theorem.
5. Since P0|=ψn for all n∈ω, we have just shown that P0 must be (3,3)-representable.
6. But the class of (3,3)-representable posets is elementary, so, as P andP0 are elementarily equivalent, it follows thatP is (3,3)-representable too.
References I
[1] Balbes, R.: A representation theory for prime and implicative semilattices.
Trans. Amer. Math. Soc.136, 261–267 (1969)
[2] Birkhoff, G.: On the combination of subalgebras.
Proc. Camb. Philos. Soc.29, 441–464 (1933)
[3] Egrot, R.: Representable posets.
J. Appl. Log.16, 60–71 (2016)
[4] Egrot, R.: No finite axiomatizations for posets embeddable into distributive lattices.
Ann. Pure Appl. Logic169, 235–242 (2018)
[5] Fagin, R.: CONTRIBUTIONS TO THE MODEL-THEORY OF FINITE-STRUCTURES.
ProQuest LLC, Ann Arbor, MI (1973).
Thesis (Ph.D.)–University of California, Berkeley
[6] Kearnes, K.A.: The class of prime semilattices is not finitely axiomatizable.
Semigroup Forum55(1), 133–134 (1997)
[7] Keisler, H.: Ultraproducts and elementary models.
Indag. Math.23, 477–495 (1961)
[8] Lo´s, J.: Quelques remarques, th´eor`emes et probl`emes sur les classes d´efinissables d’alg`ebres.
In: Mathematical interpretation of formal systems, pp. 98–113. North-Holland Publishing Co., Amsterdam (1955)
References II
[9] Shelah, S.: Every two elementarily equivalent models have isomorphic ultrapowers.
Israel J. Math.10, 224–233 (1971)
[10] Stone, M.: The theory of representations for Boolean algebras.
Trans. Amer. Math. Soc.40, 37–111 (1936)
[11] Van Alten, C.J.: Embedding ordered sets into distributive lattices.
Order33(3), 419–427 (2016)
[12] Vardi, M.: On the complexity of bounded-variable queries.
In: ACM Symp. on Principles of Database Systems, pp. 266–276. ACM press (1995)