aa r X i v : . [ m a t h . L O ] D ec DIFFERENT SIMILARITIES
Miloˇs S. Kurili´c Abstract
We establish the hierarchy among twelve equivalence relations (similari-ties) on the class of relational structures: the equality, the isomorphism,the equimorphism, the full relation, four similarities of structures inducedby similarities of their self-embedding monoids and intersections of theseequivalence relations. In particular, fixing a language L and a cardinal κ ,we consider the interplay between the restrictions of these similarities to theclass Mod L ( κ ) of all L -structures of size κ . It turns out that, concerningthe number of different similarities and the shape of the corresponding Hassediagram, the class of all structures naturally splits into three parts: finitestructures, infinite structures of unary languages, and infinite structures ofnon-unary languages (where all these similarities are different).2010MathematicsSubjectClassification: 03C07, 20M20, 06A06, 03E40.Keywords: relational structure, isomorphic substructure, partial order, self-embedding monoid, isomorphism, equimorphism, forcing-equivalence. If X is a relational structure, Emb( X ) the monoid of its self-embeddings and P ( X ) = { f [ X ] : f ∈ Emb( X ) } the set of copies of X inside X , then the poset h P ( X ) , ⊂i (isomorphic to the inverse of the right Green’s order on Emb( X ) ) con-tains a certain information about X and the equality P ( X ) = P ( Y ) defines anequivalence relation on the class of all relational structures. Writing P ( X ) insteadof h P ( X ) , ⊂i , some coarser classifications of structures are obtained if the equal-ity is replaced by the following weaker conditions: P ( X ) ∼ = P ( Y ) (implied by Emb( X ) ∼ = Emb( Y ) ), sq P ( X ) ∼ = sq P ( Y ) (where sq P denotes the separative quo-tient of a poset P ), and P ( X ) ≡ P ( Y ) (the forcing equivalence of posets of copies).Concerning the last (and the coarsest non-trivial) similarity relation we note thatthe forcing related properties of posets of copies was investigated for countablestructures in general in [6], for equivalence relations and similar structures in [7],for ordinals in [8], for scattered and non-scattered linear orders in [9] and [11], andfor several ultrahomogeneous structures in [10],[11],[12], and [13].In this paper we investigate the interplay between the four similarity relationsmentioned above and the similarities defined by the conditions X = Y , X ∼ = Y ,and X ⇄ Y (equimorphism, bi-embeddability). Department of Mathematics and Informatics, University of Novi Sad, Trg Dositeja Obradovi´ca4, 21000 Novi Sad, Serbia. e-mail: [email protected] pairs h X , L i , where L is a language and X an L -structure. (The language must be in-cluded in the game because, otherwise, since the structure X = h ω, h∅ii can beregarded as an L -structure for each language L of size 1, it is not clear what X ∼ = Y means). So, the conditions displayed in the diagram define when the pairs h X , L i and h Y , L i are similar (clearly, the equality L = L follows from X ∼ = Y and X ⇄ Y and we omit it). Thus, for example, line n denotes the statement thatequimorphic structures have forcing-equivalent posets of copies.In Section 3 we fix a language L and a set X and restrict our analysis to theclass Mod L ( X ) of L -structures with the domain X . It turns out that for a non-unary language L and infinite set X in the diagram from Figure 1 restricted to theclass Mod L ( κ ) all the implications a - o are proper and there are no new impli-cations (except the ones following from transitivity). On the other hand, for finitestructures or unary languages the diagram collapses significantly.A few words on notation. Let L = h R i : i ∈ I i be a relational language, where ar L ( R i ) = n i ∈ N , i ∈ I and let X be a non empty set. If X = h X, h ρ i : i ∈ I ii is an L -structure and ∅ 6 = A ⊂ X , then h A, h ρ i ↾ A : i ∈ I ii is a substructure of X , where ρ i ↾ A = ρ i ∩ A n i , for i ∈ I . If Y = h Y, h σ i : i ∈ I ii is an L -structuretoo, a mapping f : X → Y is an embedding (we write f : X ֒ → Y ) iff f is aninjection and for all i ∈ I and x , . . . x n i ∈ X we have h x , . . . , x n i i ∈ ρ i ⇔h f ( x ) , . . . , f ( x n i ) i ∈ σ i . Let Emb( X , Y ) denote the set of such embeddings and P ( X , Y ) = { B ⊂ Y : h B, h σ i ↾ B : i ∈ I ii ∼ = X } = { f [ X ] : f ∈ Emb( X , Y ) } . In particular,
Emb( X ) := Emb( X , X ) and P ( X ) := { f [ X ] : f ∈ Emb( X ) } = { A ⊂ X : h A, h ρ i ↾ A : i ∈ I ii ∼ = X } . If f ∈ Emb( X , Y ) is a surjection, it is an isomorphism , we write f ∈ Iso( X , Y ) , and the structures X and Y are isomorphic ,in notation X ∼ = Y . If, in particular, Y = X , then f is called an automorphism ofthe structure X and Aut( X ) denotes the set of all such mappings. Structures X and Y are called equimorphic , in notation X ⇄ Y , iff X ֒ → Y and Y ֒ → X .The right Green’s pre-order (cid:22) R on the monoid h Emb X , ◦ , id X i is defined by: f (cid:22) R g iff f ◦ h = g , for some h ∈ Emb X . The right Green’s equivalencerelation ≈ R on Emb X , given by: f ≈ R g iff f (cid:22) R g and g (cid:22) R f , determinesthe antisymmetric quotient h Emb X / ≈ R , (cid:22) R i , the right Green’s order . It is easyto check that h Emb X / ≈ R , (cid:22) R i ∼ = h P ( X ) , ⊃i so the results of this paper can beregarded as statements about transformation semigroups.A partial order P = h P, ≤i is called separative iff for each p, q ∈ P satisfying p q there is r ∈ P such that r ≤ p and r ⊥ q . The separative modification of P is the pre-order sm P = h P, ≤ ∗ i , where p ≤ ∗ q iff ∀ r ≤ p ∃ s ≤ r s ≤ q . The separative quotient of P is the separative partial order sq P = h P/ = ∗ , E i , where p = ∗ q ⇔ p ≤ ∗ q ∧ q ≤ ∗ p and [ p ] E [ q ] ⇔ p ≤ ∗ q . If P is a separative partialifferent similarities 3 ✟✟✟✟✟✟✟✟❍❍❍❍❍❍❍❍ ✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟ ❍❍❍❍❍❍❍❍ ❍❍❍❍❍❍❍❍ ✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟ ❍❍❍❍❍❍❍❍ rrr rr rr rr rrr X = Y ∧ L = L P ( X ) = P ( Y ) X ∼ = YP ( X ) ∼ = P ( Y ) X ⇆ Y sq P ( X ) ∼ = sq P ( Y )ro sq P ( X ) ∼ = ro sq P ( Y ) ⇔ P ( X ) ≡ P ( Y ) P ( X ) = P ( Y ) ∧ X ∼ = YP ( X ) = P ( Y ) ∧ X ⇆ Y P ( X ) ∼ = P ( Y ) ∧ X ⇆ Y sq P ( X ) ∼ = sq P ( Y ) ∧ X ⇆ Y the full relation ab cd e fg h ij k lm no Figure 1: The hierarchy of similarities between relational structuresorder, by ro P we will denote the Boolean completion of P . For a pre-order P let ge( P ) = { V P [ G ] : G is a P -generic filter over V } . Two pre-orders P and Q are saidto be forcing equivalent , in notation P ≡ Q , iff ge( P ) = ge( Q ) . Fact 1.1
Let P , Q and P i , i ∈ I , be partial orderings. Then(a) P ∼ = Q ⇒ sm P ∼ = sm Q ⇒ sq P ∼ = sq Q ⇒ ro sq P ∼ = ro sq Q ⇒ P ≡ Q ;(b) P ≡ sm P ≡ sq P ≡ (ro sq P ) + ;(c) sq( Q i ∈ I P i ) ∼ = Q i ∈ I sq P i . In this section we establish the implications a - o from Figure 1. In Section 3 wewill show that, regarding the class of all relational structures, there are no new im-plications in Figure 1 (except the ones which follow from the transitivity). First, the Miloˇs S.Kurili´cimplications a , b , c , d , e , g , h , k , l , and o are evident, while i , j and m follow fromFact 1.1(a). In the sequel we prove the equivalence ro sq P ( X ) ∼ = ro sq P ( Y ) ⇔ P ( X ) ≡ P ( Y ) and the implications f and n (see Theorems 2.7 and 2.10). Here we prove that the Boolean completion of the poset of copies of a relationalstructure is a homogeneous Boolean algebra. We recall that a partial order P = h P, ≤i is called homogeneous iff it has a largest element and P ∼ = p ↓ , for each p ∈ P and that a Boolean algebra B is called a homogeneous Boolean algebra iff B ∼ = b ↓ , for each b ∈ B + . It is known that the Boolean completion of a separativehomogeneous partial order P is a homogeneous Boolean algebra (see [4], p. 181)and, by Theorem 2.2 of [6], the posets of the form P ( X ) are homogeneous but it iseasy to see that they are not separative in most of the cases. So, in order to provethat the Boolean completions ro sq P ( X ) are homogeneous algebras, we show thatin the theorem mentioned above the separativity of P can be omitted and that theassumption of homogeneity can be relaxed. Namely, defining a partial order P tobe quasi homogeneous iff for each p ∈ P there is a dense subset of P isomorphicto a dense subset of p ↓ , we have the following generalization. Theorem 2.1
The Boolean completion of a quasi homogeneous partial order P isa homogeneous Boolean algebra. Proof.
The statement is a consequence of the following two claims. Namely, if P isa quasi homogeneous partial ordering, then, by Claim 2.2, sq P is a separative quasihomogeneous partial order and, by Claim 2.3, the algebra ro sq P is homogeneous. Claim 2.2
The separative quotient of a quasi homogeneous partial order is quasihomogeneous.
Proof.
Let P = h P ≤i be a quasi homogeneous partial order, sq P = h P/ = ∗ , E i and p ∈ P . Let D be a dense subset of P and f : h D, ≤i → h p ↓ , ≤i an embeddingsuch that f [ D ] is a dense subset of p ↓ . First we prove that ∀ q, r ∈ D ( q ≤ ∗ r ⇔ f ( q ) ≤ ∗ f ( r )) . (1)Let q, r ∈ D . If q ≤ ∗ r , then each s ≤ q is compatible with r and we prove thateach u ≤ f ( q ) is compatible with f ( r ) . If u ≤ f ( q ) , then u ≤ p and, since f [ D ] is dense in p ↓ , there is s ∈ D such that f ( s ) ≤ u . Since f is an embedding and f ( s ) ≤ f ( q ) we have s ≤ q and, since q ≤ ∗ r , there is t ≤ s, r , and, moreoverthere is t ′ ∈ D such that t ′ ≤ t which implies f ( t ′ ) ≤ f ( s ) ≤ u and f ( t ′ ) ≤ f ( r ) .Thus u f ( r ) .ifferent similarities 5Assuming that f ( q ) ≤ ∗ f ( r ) and s ≤ q we show that s r . If s ≤ q and s ′ ∈ D , where s ′ ≤ s , then f ( s ′ ) ≤ f ( q ) and, since f ( q ) ≤ ∗ f ( r ) , there is v ≤ f ( s ′ ) , f ( r ) ≤ p . Since f [ D ] is dense in p ↓ , there is t ∈ D such that f ( t ) ≤ v .Since f is an embedding we have t ≤ s ′ , r and, hence, s r . So (1) is true.It is evident that the set D := { [ q ] : q ∈ D } is a dense suborder of the partialorder h P/ = ∗ , E i and we prove that the mapping F : hD , E i → h [ p ] ↓ , E i , given by F ([ q ]) = [ f ( q )] , is an embedding. First, for q, r ∈ D by (1) we have [ q ] = [ r ] iff q = ∗ r iff q ≤ ∗ r ∧ r ≤ ∗ q iff f ( q ) ≤ ∗ f ( r ) ∧ f ( r ) ≤ ∗ f ( q ) iff f ( q ) = ∗ f ( r ) iff [ f ( q )] = [ f ( r )] iff F ([ q ]) = F ([ r ]) and, thus, F is a well definedinjection. Second, for q ∈ D we have f ( q ) ≤ p , which implies f ( q ) ≤ ∗ p and,hence, [ f ( q )] E [ p ] , that is F ([ q ]) ∈ [ p ] ↓ . Thus F [ D ] ⊂ [ p ] ↓ . Finally, by (1),for q, r ∈ D we have [ q ] E [ r ] iff q ≤ ∗ r iff f ( q ) ≤ ∗ f ( r ) iff [ f ( q )] E [ f ( r )] iff F ([ q ]) E F ([ r ]) and, thus, F is a strong homomorphism.Now we prove that F [ D ] is a dense set in the poset h [ p ] ↓ , E i . If [ q ] E [ p ] , thenthere is s ≤ p, q and, since f [ D ] is dense in p ↓ , there is u ∈ D such that f ( u ) ≤ s .Now, f ( u ) ≤ q implies f ( u ) ≤ ∗ q thus F ([ u ]) = [ f ( u )] E [ q ] and F ([ u ]) ∈ F [ D ] .Thus the partial order sq P is quasi homogeneous indeed. ✷ Claim 2.3
The Boolean completion of a separative quasi homogeneous partialordering is a homogeneous complete Boolean algebra.
Proof.
Let P = h P ≤i be a separative quasi homogeneous partial order. First weshow that ∀ p ∈ P ro P ∼ = ro( p ↓ ) . (2)If p ∈ P , then there is a dense subset D of P and an embedding f : D ֒ → p ↓ such that f [ D ] is a dense subset of p ↓ . Thus D and f [ D ] are isomorphic separativeposets, which implies that ro D ∼ = ro f [ D ] . In addition, D is a dense suborderof the separative order P , which, by the uniqueness of the Boolean completion,implies ro P ∼ = ro D and, similarly, ro f [ D ] ∼ = ro( p ↓ ) and (2) is true.Let B = ro P , b ∈ B + and w.l.o.g. suppose that P is a dense suborder of B + .Then there is p ∈ P such that p ≤ B b . Clearly the set ( p ↓ ) B ∩ P = ( p ↓ ) P is adense suborder of the relative algebra ( p ↓ ) B , which implies ( p ↓ ) B ∼ = ro(( p ↓ ) P ) so, by (2), ( p ↓ ) B ∼ = ro P ∼ = B . It is well known that, if B is a σ -complete Booleanalgebra, a, b ∈ B , a ≤ b and B ∼ = a ↓ , then B ∼ = b ↓ (see [4], p. 180). So we have b ↓∼ = B . ✷ Miloˇs S.Kurili´c
Example 2.4
Clearly homogeneous partial orders are quasi homogeneous, but theconverse is not true. Let R be the real line and P = D { ( a, b ] : a, b ∈ R ∧ a < b } ∪ { R } , ⊂ E . Then for p = ( a, b ] we have p ↓6∼ = P , since the largest element of P is not thesupremum of two smaller elements. Thus P is not a homogeneous partial order.On the other hand, if f : R → ( a, b ) is an isomorphism, then it is easy to show thatthe mapping F : P → p ↓ defined by F ( R ) = p and F (( c, d ]) = ( f ( c ) , f ( d )] is anembedding and that F [ P ] is a dense subset of p ↓ . Thus the partial order P is quasihomogeneous. We note that P is, in addition, separative. Theorem 2.5
For each relational structure X the Boolean completion ro sq P ( X ) of the poset P ( X ) is a homogeneous complete Boolean algebra, forcing equivalentto P ( X ) . All generic extensions by P ( X ) are elementarily equivalent. Proof.
By Theorem 2.2 of [6] the poset P ( X ) is homogeneous and, by Theorem2.1, ro sq P ( X ) is a homogeneousthe algebra. By Fact 1.1(b) the posets P ( X ) and ro sq P ( X ) are forcing equivalent. By Theorem 4.3 of [6] either | sq P ( X ) | = 1 ,and then all generic extensions are trivial, or sq P ( X ) is an atomless poset, andthen B := ro sq P ( X ) is an infinite homogeneous complete Boolean algebra. Thisimplies that for each a, b ∈ B \ { , } there is f ∈ Aut( B ) such that f ( a ) = b (see[4], Proposition 9.13) and, hence B + is a weakly homogeneous partial order (werecall that a partial order P = h P, ≤i is called weakly homogeneous iff for each p, q ∈ P there is f ∈ Aut( P ) such that f ( p ) q ). By a known fact concerningweakly homogeneous partial orders (see [5], p. 245), for each sentence ϕ of thelanguage of set theory we have (cid:13) ϕ or (cid:13) ¬ ϕ . Thus all generic extensions by P ( X ) satisfy the same set of sentences. ✷ Here we show that the posets of copies of two structures are forcing equivalent ifftheir Boolean completions are isomorphic.
Fact 2.6 If B and C are complete Boolean algebras such that some B -genericextension is equal to some C -generic extension, then(a) There are b ∈ B and c ∈ C such that b ↓∼ = c ↓ (see [3], p. 267);(b) If B and C are homogeneous algebras, then B ∼ = C . So, B ≡ C ⇔ B ∼ = C . Proof. (b) If b ∈ B and c ∈ C are the elements from (a), by the homogeneity wehave B ∼ = b ↓ and C ∼ = c ↓ and, hence, B ∼ = C . ✷ ifferent similarities 7 Theorem 2.7
Let X and Y be arbitrary relational structures. Then(a) P ( X ) ≡ P ( Y ) iff ro sq P ( X ) ∼ = ro sq P ( Y ) ;(b) The collections ge P ( X ) and ge P ( Y ) are either disjoint or equal. Proof. (a) By Fact 1.1(b), Fact 2.6(b) and Theorem 2.5, P ( X ) ≡ P ( Y ) iff ro sq P ( X ) ≡ ro sq P ( Y ) iff ro sq P ( X ) ∼ = ro sq P ( Y ) .(b) If ge P ( X ) ∩ ge P ( Y ) = ∅ , then by Fact 1.1(b) and Fact 2.6(b) we have ro sq P ( X ) ∼ = ro sq P ( Y ) , which implies ge(ro sq P ( X )) = ge(ro sq P ( Y )) , that is ge P ( X ) = ge P ( Y ) . ✷ In this section we prove that the posets of copies of isomorphic (resp. equimorphic)structures are isomorphic (resp. have isomorphic Boolean completions). We willuse the following elementary fact.
Fact 2.8
Let h P , ≤i be a pre-order and p ∈ P . Then(a) If G is a P -generic filter over V and p ∈ G , then G ∩ p ↓ is a p ↓ -genericfilter over V and V P [ G ] = V p ↓ [ G ∩ p ↓ ] ;(b) If H is a p ↓ -generic filter over V , then H ↑ is a P -generic filter over V and V p ↓ [ H ] = V P [ H ↑ ] . Lemma 2.9 If X and Y are structures of the same language, h : X ֒ → Y , and C = h [ X ] , then the mapping F : P ( X ) → ( C ↓ ) P ( X , Y ) defined by F ( A ) = h [ A ] ,for A ∈ P ( X ) , is an isomorphism of the posets h P ( X ) , ⊂i and h ( C ↓ ) P ( X , Y ) , ⊂i . Proof.
For A ∈ P ( X ) there is ϕ : X ֒ → X such that ϕ [ X ] = A and, clearly, h ◦ ϕ : X ֒ → Y , thus h [ ϕ [ X ]] = h [ A ] ∈ P ( X , Y ) and h [ A ] ⊂ h [ X ] = C , whichimplies that h [ A ] ∈ ( C ↓ ) P ( X , Y ) . So F [ P ( X )] ⊂ ( C ↓ ) P ( X , Y ) .Since h is an injection, for each A, B ∈ P ( X ) we have F ( A ) ⊂ F ( B ) iff h [ A ] ⊂ h [ B ] iff h − [ h [ A ]] ⊂ h − [ h [ B ]] iff A ⊂ B , thus F is an embedding.If D ∈ P ( X , Y ) and D ⊂ C , then h [ h − [ D ]] = D and the surjective restriction h | h − [ D ] : h − [ D ] → D is an isomorphism, which implies h − [ D ] ∈ P ( X ) . Inaddition F ( h − [ D ]) = h [ h − [ D ]] = D thus F is onto. ✷ Theorem 2.10 If X and Y are structures of the same relational language, then(a) X ∼ = Y ⇒ P ( X ) ∼ = P ( Y ) ;(b) X ⇄ Y ⇒ ro sq P ( X ) ∼ = ro sq P ( Y ) . Miloˇs S.Kurili´c
Proof. (a) If h : X → Y is an isomorphism, then, by Lemma 2.9, h P ( X ) , ⊂i ∼ = h P ( X , Y ) , ⊂i and, clearly, h P ( X , Y ) , ⊂i = h P ( Y ) , ⊂i .(b) Let f : X ֒ → Y , g : Y ֒ → X and P ( Y ) ↑ = { S ⊂ Y : ∃ B ∈ P ( Y ) B ⊂ S } .First we show that P ( X , Y ) := { h [ X ] | h : X ֒ → Y } is a dense suborder of h P ( Y ) ↑ , ⊂i . If C ∈ P ( X , Y ) and h : X ֒ → Y , where C = h [ X ] , then, clearly, h ◦ g : Y ֒ → Y and, hence, h [ g [ Y ]] ∈ P ( Y ) and h [ g [ Y ]] ⊂ h [ X ] = C , whichimplies C ∈ P ( Y ) ↑ . Thus P ( X , Y ) ⊂ P ( Y ) ↑ . Let S ∈ P ( Y ) ↑ , B ∈ P ( Y ) , where B ⊂ S and ψ : Y ֒ → Y , where B = ψ [ Y ] . Now ψ ◦ f : X ֒ → Y and, hence, ψ [ f [ X ]] ∈ P ( X , Y ) and ψ [ f [ X ]] ⊂ ψ [ Y ] = B ⊂ S . Thus P ( X , Y ) is dense in h P ( Y ) ↑ , ⊂i . Since P ( Y ) is dense in h P ( Y ) ↑ , ⊂i as well, we have h P ( X , Y ) , ⊂i ≡ h P ( Y ) , ⊂i . (3)Now let W be a generic extension by P ( Y ) . By (3) W = V P ( X , Y ) [ G ] , where G isa P ( X , Y ) -generic filter over V . Let C ∈ G . By Fact 2.8(a) we have V P ( X , Y ) [ G ] = V C ↓ [ G ∩ C ↓ ] and, if F : h P ( X ) , ⊂i → h ( C ↓ ) P ( X , Y ) , ⊂i is the isomorphismdefined in Lemma 2.9, then H := F − [ G ∩ C ↓ ] is a P ( X ) -generic filter over V and V C ↓ [ G ∩ C ↓ ] = V P ( X ) [ H ] . Thus W = V P ( X ) [ H ] and, by Theorem 2.7(b), P ( X ) ≡ P ( Y ) . Now, by Theorem 2.7(a), ro sq P ( X ) ∼ = ro sq P ( Y ) . ✷ L (X) Now we restrict our consideration to some smaller classes of structures. If L = h R i : i ∈ I i is a language, X a fixed set and ρ = h ρ i : i ∈ I i ∈ Int L ( X ) , we willabuse notation writing P ( ρ ) instead of P ( h X, ρ i ) and h P ( h X, ρ i ) , ⊂i whenever thecontext admits it. So, restricting our similarity relations to the set Mod L ( X ) or,equivalently, to the corresponding set of interpretations, Int L ( X ) , we obtain thefollowing equivalence relations: for ρ = h ρ i : i ∈ I i , σ = h σ i : i ∈ I i ∈ Int L ( X ) (writing ρ ∼ = σ instead of h X, ρ i ∼ = h X, σ i and similarly for ρ ⇄ σ ) we define ρ ∼ σ ⇔ ρ = σ ρ ∼ σ ⇔ P ( ρ ) ∼ = P ( σ ) ρ ∼ σ ⇔ P ( ρ ) = P ( σ ) ∧ ρ ∼ = σ ρ ∼ σ ⇔ sq P ( ρ ) ∼ = sq P ( σ ) ∧ ρ ⇄ σρ ∼ σ ⇔ P ( ρ ) = P ( σ ) ∧ ρ ⇄ σ ρ ∼ σ ⇔ sq P ( ρ ) ∼ = sq P ( σ ) ρ ∼ σ ⇔ ρ ∼ = σ ρ ∼ σ ⇔ ρ ⇄ σρ ∼ σ ⇔ P ( ρ ) = P ( σ ) ρ ∼ σ ⇔ P ( ρ ) ≡ P ( σ ) ρ ∼ σ ⇔ P ( ρ ) ∼ = P ( σ ) ∧ ρ ⇄ σ ρ ∼ σ ⇔ .Then some implications between the similarities on the set Mod L ( X ) are displayedin Figure 2.It is natural to ask are there more implications in it (except the ones whichfollow from the transitivity), that is, are some of the implications a - o , in fact,ifferent similarities 9 ✟✟✟✟✟✟✟✟❍❍❍❍❍❍❍❍ ✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟ ❍❍❍❍❍❍❍❍ ❍❍❍❍❍❍❍❍ ✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟ ❍❍❍❍❍❍❍❍ rrr rr rr rr rrr ρ ∼ σρ = σ ρ ∼ σ P ( ρ ) = P ( σ ) ∧ ρ ∼ = σρ ∼ σ P ( ρ ) = P ( σ ) ∧ ρ ⇄ σ ρ ∼ σρ ∼ = σρ ∼ σ P ( ρ ) = P ( σ ) ρ ∼ σ P ( ρ ) ∼ = P ( σ ) ∧ ρ ⇄ σρ ∼ σ P ( ρ ) ∼ = P ( σ ) ρ ∼ σ sq P ( ρ ) ∼ = sq P ( σ ) ∧ ρ ⇄ σρ ∼ σ sq P ( ρ ) ∼ = sq P ( σ ) ρ ∼ σρ ⇄ σρ ∼ σ ro sq P ( ρ ) ∼ = ro sq P ( σ ) ⇔ P ( ρ ) ≡ P ( σ ) the full relation ρ ∼ σ ab cd e fg h ij k lm no Figure 2: Some implications between the similarities on
Mod L ( X ) equivalences. Concerning this question we will show that the class of all relationalstructures splits into the following three parts: finite structures, infinite structuresof unary languages, and infinite structures of non-unary languages. (A language L = h R i : i ∈ I i is called unary iff ar( R i ) = 1 , for all i ∈ I . Structures ofunary languages will be called unary structures ). Let us call a class C of structuresa Cantor-Schr¨oder-Bernstein (CSB) class iff ∀ X , Y ∈ C ( X ⇄ Y ⇒ X ∼ = Y ) . For finite structures the diagram from Figure 2 collapses significantly.
Example 3.1 If L is an arbitrary relational language and X a finite set, then foreach ρ ∈ Int L ( X ) we have P ( ρ ) = { X } , because X ∈ P ( X, ρ ) ⊂ [ X ] | X | = { X } . Thus, ∼ is the full relation, which implies that ∼ = ∼ = ∼ = ∼ = ∼ .0 Miloˇs S.Kurili´cIn addition, Mod L ( X ) is a CSB class. Namely, if ρ, σ ∈ Int L ( X ) and ρ ⇄ σ , then there is an embedding f : h X, ρ i → h
X, σ i , and, since X is a finiteset, f is an isomorphism, thus ρ ∼ = σ . So we have ∼ ⊂∼ , which implies ∼ = ∼ = ∼ = ∼ = ∼ = ∼ . Since h X, h∅ , ∅ , . . . ii 6∼ = h X, h X n i : i ∈ I ii , wehave ∼ = ∼ . If | X | > , let a and b be different elements of X , i ∈ I , and let ρ, σ ∈ Int L ( X ) , where ρ i = {h a, a, . . . , a i} , σ i = {h b, b, . . . , b i} ⊂ X n i and ρ i = σ i = ∅ , for i = i . Then ρ = σ , but ρ ∼ = σ and, hence, ∼ = ∼ . Thus Figure3 describes the hierarchy of the similarities ∼ k on the set Mod L ( X ) , if | X | > .We prove that ∼ = ∼ ⇔ | X | = 1 . Let X = { x } and ρ, σ ∈ Int L ( X ) , where ρ ∼ σ . Then there is an isomorphism f : h{ x } , ρ i → h{ x } , σ i and, consequently,for each i ∈ I we have h x, x, . . . , x i ∈ ρ i ⇔ h x, x, . . . , x i ∈ σ i and, hence, ρ i = σ i . So ρ = σ , that is ρ ∼ σ and the inclusion ∼ ⊂∼ is proved. rrr ∼ = ∼ = ∼ = ∼ = ∼ = the full relation ∼ = ∼ = ∼ = ∼ = ∼ = ∼ = the isomorphism ∼ = the equality Figure 3: The similarities on the class
Mod L ( X ) , if < | X | < ω In this subsection we assume that L = h R i : i ∈ I i is a unary relational language.If X = h X, h ρ i : i ∈ I ii is an L -structure, it is easy to check that the binary relation ≈ on the set X defined by: x ≈ y ⇔ ∀ i ∈ I ( x ∈ ρ i ⇔ y ∈ ρ i ) is an equivalencerelation. Then [ x ] := { y ∈ X : y ≈ x } is the equivalence class of x ∈ X , andif X/ ≈ = { X j : j ∈ J } is the corresponding partition we define κ j := | X j | , for j ∈ J , and J := { j ∈ J : | X j | < ω } . Theorem 3.2
Let X = h X, h ρ i : i ∈ I ii be a unary structure. Then(a) If f : X → X is an injection, then f ∈ Emb( X ) ⇔ ∀ x ∈ X f [[ x ]] ⊂ [ x ] ;(b) If J = J , then P ( X ) = { X } ;(c) If J = J , then the poset P ( X ) is atomless and we have P ( X ) ∼ = Q j ∈ J \ J h [ κ j ] κ j , ⊂i and sq P ( X ) ∼ = Q j ∈ J \ J ( P ( κ j ) / [ κ j ] <κ j ) + . ifferent similarities 11 Proof. (a) If f ∈ Emb( X ) and x ∈ X , then x ∈ ρ i ⇔ f ( x ) ∈ ρ i , for each i ∈ I ,thus x ≈ f ( x ) . So for y ∈ [ x ] we have f ( y ) ≈ y ≈ x and, hence, f ( y ) ∈ [ x ] .Let f [[ x ]] ⊂ [ x ] , for all x ∈ X . Then, for x ∈ X , x ∈ [ x ] implies f ( x ) ∈ [ x ] ,that is f ( x ) ≈ x . Hence ∀ i ∈ I ∀ x ∈ X ( x ∈ ρ i ⇔ f ( x ) ∈ ρ i ) , so f ∈ Emb( X ) .(b) Let J = J . By (a), for f ∈ Emb( X ) and x ∈ X we have f [[ x ]] ⊂ [ x ] and,since | [ x ] | < ω , f [[ x ]] = [ x ] , which implies f [ X ] = X .(c) If f ∈ Emb( X ) , then, by (a), f [ X j ] = X j , for all j ∈ J , and C j := f [ X j ] ∈ [ X j ] κ j , for all j ∈ J \ J . Thus the inclusion “ ⊂ ” in the equality P ( X ) = n S j ∈ J X j ∪ S j ∈ J \ J C j : h C j : j ∈ J \ J i ∈ Q j ∈ J \ J [ X j ] κ j o (4)is proved. On the other hand, if h C j : j ∈ J \ J i ∈ Q j ∈ J \ J [ X j ] κ j and ifwe choose bijections ϕ j : X j → C j , for all j ∈ J \ J , then by (a) we have f = S j ∈ J id X j ∪ S j ∈ J \ J ϕ j ∈ Emb( X ) and, hence, S j ∈ J X j ∪ S j ∈ J \ J C j ∈ P ( X ) , so (4) is true. Thus the mapping F : Q j ∈ J \ J h [ X j ] κ j , ⊂i → h P ( X ) , ⊂i given by F ( h C j : j ∈ J \ J i ) = S j ∈ J X j ∪ S j ∈ J \ J C j is a well-defined surjection and, since { X j : j ∈ J } is a partition of X , it is aninjection. It is easy to see that F is an order isomorphism. By Fact 1.1(c) we have sq h P ( X ) , ⊂i ∼ = Q j ∈ J \ J sq h [ κ j ] κ j , ⊂i = Q j ∈ J \ J ( P ( κ j ) / [ κ j ] <κ j ) + . ✷ Lemma 3.3
Let κ ≥ ω be a cardinal, U ⊂ κ and λ := min {| U | , | κ \ U |} . Then ρ = h U, ∅ , ∅ , . . . i ∈ Int L ( κ ) and we have(a) P ( ρ ) = { C ∪ C : C ∈ [ U ] | U | ∧ C ∈ [ κ \ U ] | κ \ U | } ;(b) P ( ρ ) ∼ = h [ κ ] κ , ⊂i × h [ λ ] λ , ⊂i ;(c) sq P ( ρ ) ∼ = ( P ( κ ) / [ κ ] <κ ) + × ( P ( λ ) / [ λ ] <λ ) + , where, by convention, for λ ∈ ω , by ( P ( λ ) / [ λ ] <λ ) + we denote the one-element poset. Proof.
For x, y ∈ κ we have: x ≈ y iff x ∈ ρ i ⇔ y ∈ ρ i , for all i ∈ I , iff x ∈ U ⇔ y ∈ U . Thus κ/ ≈ = { U, κ \ U } and we apply Theorem 3.2. ✷ Fact 3.4 (a) If κ > ω is a regular cardinal and κ = κ + , then ro( P ( κ ) / [ κ ] <κ ) ∼ =Col( ω, κ ) (Balcar, Vopˇenka [1]; see also [2], p. 380).(b) Under CH, all separative atomless ω -closed posets of size ω are forcingequivalent (for example to Col( ω , ω ) ) (folklore).(c) If λ > ω is a cardinal and P a poset of size λ such that P (cid:13) | ˇ λ | = ˇ ω , then ro P ∼ = Col( ω, λ ) (see [3], p. 277).(d) If B is a Boolean algebra of size > , then B + = B + × B + . Proof. (d) The sentence ∀ x = 1 ∃ y ( x ⊥ y & x ∨ y = 1) is true in the poset B + , but it is not true in B + × B + . Namely, since | B | > , there is a ∈ B + \ { } andwe have x := h , a i ∈ ( B + × B + ) \ {h , i} and a ′ ∈ B + , but for each b ∈ B + wehave h , a i ⊥ h b, a ′ i and h , a i ∨ h b, a ′ i = h , i . ✷ Theorem 3.5
For any unary language L and infinite cardinal κ we have(a) Mod L ( κ ) is a CSB class;(b) Figure 4 describes the hierarchy of the similarities ∼ k , for k = 8 , , onthe set Mod L ( κ ) . In addition we have ∼ = ∼ .(c) If κ is a regular cardinal and κ = κ + , then ∼ = ∼ . ❅❅❅ (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) ❅❅❅ rrr rrr ∼ = the full relation ∼ = ∼ ∼ = the equality of P ( X ) ∼ = the isomorphism of P ( X ) ∼ = ∼ = ∼ = ∼ = the isomorphism = the equimorphism ∼ = the equality Figure 4: The similarities on
Mod L ( κ ) , for unary L and infinite κ Proof.
Let L = { R i : i ∈ I } .(a) Assuming that ρ = h ρ i : i ∈ I i , σ = h σ i : i ∈ I i ∈ Int L ( κ ) and ρ ⇄ σ weshow that ρ ∼ = σ . By the assumption, there are embeddings f : h κ, ρ i ֒ → h κ, σ i and g : h κ, σ i ֒ → h κ, ρ i . (5)Let ≈ ρ and ≈ σ be the equivalence relations determined by the interpretations ρ and σ respectively (see Theorem 3.2) and, for x ∈ κ , let [ x ] ρ and [ x ] σ be thecorresponding equivalence classes. First we prove that ∀ x ∈ κ f [[ x ] ρ ] ⊂ [ f ( x )] σ and ∀ x ∈ κ g [[ x ] σ ] ⊂ [ g ( x )] ρ . (6)For a proof of the first statement we take x ∈ κ and y ∈ [ x ] ρ . Then y ≈ ρ x , that is ∀ i ∈ I ( x ∈ ρ i ⇔ y ∈ ρ i ) , (7)ifferent similarities 13and, since f is an embedding, we have ∀ i ∈ I ∀ x ∈ κ ( x ∈ ρ i ⇔ f ( x ) ∈ σ i ) . (8)We prove that f ( y ) ∈ [ f ( x )] σ , which means that f ( y ) ∈ σ i ⇔ f ( x ) ∈ σ i , for all i ∈ I . So, f ( y ) ∈ σ i iff (by (8)) y ∈ ρ i iff (by (7)) x ∈ ρ i iff (by (8)) f ( x ) ∈ σ i .Thus the first statement of (6) is proved and the second has a symmetric proof.Let κ = S j ∈ J X j and κ = S k ∈ K Y k be the partitions determined by therelations ≈ ρ and ≈ σ respectively. By (6), if j ∈ J and X j = [ x ] ρ , then f [ X j ] ⊂ [ f ( x )] σ = Y k , for (a unique) k ∈ K . Similarly, for each k ∈ K there is a unique j ∈ J satisfying g [ Y k ] ⊂ X j so we define the functions F : J → K by: F ( j ) = k iff f [ X j ] ⊂ Y k , (9) G : K → J by: G ( k ) = j iff g [ Y k ] ⊂ X j , (10)and prove that G ◦ F = id J and F ◦ G = id K . (11)By (5) we have g ◦ f : h κ, ρ i ֒ → h κ, ρ i and, by Theorem 3.2(a), ∀ x ∈ X g [ f [[ x ] ρ ]] ⊂ [ x ] ρ . (12)For j ∈ J we prove that G ( F ( j )) = j . Let F ( j ) = k and x ∈ X j . Then X j = [ x ] ρ , by (6) f [ X j ] = f [[ x ] ρ ] ⊂ [ f ( x )] σ = Y k ′ , for some k ′ ∈ K , and, by (9) f [ X j ] ⊂ Y k , which implies k ′ = k . Thus f [[ x ] ρ ] ⊂ Y k and, hence, g [ f [[ x ] ρ ]] ⊂ g [ Y k ] . (13)Let G ( k ) = j ′ . Then by (10) and (13) we have g [ f [[ x ] ρ ]] ⊂ g [ Y k ] ⊂ X j ′ and, by(12), g [ f [[ x ] ρ ]] ⊂ X j , which implies j ′ = j . Thus G ( F ( j )) = G ( k ) = j and thefirst equality in (11) is proved. The second equality has a similar proof.Now we prove that ∀ j ∈ J | X j | = | Y F ( j ) | . (14)By (9) we have | X j | = | f [ X j ] | ≤ | Y F ( j ) | and, by (10) and (11), | Y F ( j ) | = | g [ Y F ( j ) ] | ≤ | X G ( F ( j )) | = | X j | . So (14) is true.By (14) there are bijections ϕ j : X j → Y F ( j ) ; let ϕ = S j ∈ J ϕ j : κ → κ . Since { X j : j ∈ J } is a partition of κ the mapping ϕ is well defined. By (11) F : J → K is a bijection and, since the mappings ϕ j are surjections, ϕ is a surjection as well.Since { Y k : k ∈ K } is a partition of κ and the mappings ϕ j are injections, ϕ is ainjection too. Thus ϕ is a bijection from κ onto κ .4 Miloˇs S.Kurili´cIn order to show that ϕ : h κ, ρ i → h κ, σ i is an isomorphism, that is ∀ i ∈ I ∀ x ∈ κ ( x ∈ ρ i ⇔ ϕ ( x ) ∈ σ i ) , (15)we take i ∈ I and x ∈ κ . Let j ∈ J , where x ∈ X j . Then X j = [ x ] ρ and ϕ ( x ) = ϕ j ( x ) ∈ Y F ( j ) and, by (6) and (9), f ( x ) ∈ [ f ( x )] σ = Y F ( j ) . Thus ϕ ( x ) ≈ σ f ( x ) , that is ∀ i ∈ I ( ϕ ( x ) ∈ σ i ⇔ f ( x ) ∈ σ i ) . (16)Now x ∈ ρ i iff (by (8)) f ( x ) ∈ σ i iff (by (16)) ϕ ( x ) ∈ σ i and (15) is proved.Thus ϕ : h κ, ρ i → h κ, σ i is an isomorphism and, hence, ρ ∼ = σ .(b) By (a) we have ∼ ⊂ ∼ which, according to Figure 2, implies that ∼ = ∼ = ∼ = ∼ and ∼ = ∼ .Let us prove that ∼ ∼ . If κ = A ∪ B , where A ∩ B = ∅ and A, B ∈ [ κ ] κ ,then ρ := h A, ∅ , ∅ , . . . i 6 = σ := h B, ∅ , ∅ , . . . i . By Lemma 3.3(a) we have P ( ρ ) = { C ∪ C : C ∈ [ A ] κ ∧ C ∈ [ B ] κ } = P ( σ ) . If f : κ → κ is a bijection satisfying f [ A ] = B , then f : h κ, ρ i → h κ, σ i is an isomorphism and, hence, ρ ∼ σ , but ρ σ .Now we prove that ∼
6⊂ ∼ and, hence, ∼ ∼ and ∼ ∼ . Let x, y ∈ κ , x = y and let ρ := h{ x } , ∅ , ∅ , . . . i and σ := h{ y } , ∅ , ∅ , . . . i . If f : κ → κ is abijection satisfying f ( x ) = y , then f : h κ, ρ i → h κ, σ i is an isomorphism and,hence, ρ ∼ σ . By Lemma 3.3(a) we have P ( ρ ) = { C ∪ C : C ∈ [ { x } ] ∧ C ∈ [ κ \ { x } ] κ } = { C ∈ [ κ ] κ : x ∈ C } and, similarly, P ( σ ) = { C ∈ [ κ ] κ : y ∈ C } , which implies that κ \ { y } ∈ P ( ρ ) \ P ( σ ) . Thus ρ σ .Further we prove that ∼
6⊂ ∼ and, hence, ∼ ∼ and ∼ ∼ . Let x ∈ κ and ρ := h{ x } , ∅ , ∅ , . . . i and σ := h κ \ { x } , ∅ , ∅ , . . . i . Then, clearly, ρ = σ , that is ρ σ . As above we have P ( ρ ) = { C ∈ [ κ ] κ : x ∈ C } and, by Lemma 3.3(a), P ( σ ) = { C ∪ C : C ∈ [ κ \{ x } ] κ ∧ C ∈ [ { x } ] } = { C ∈ [ κ ] κ : x ∈ C } = P ( ρ ) .Thus ρ ∼ σ .Finally we prove that ∼ = ∼ , which implies ∼ = ∼ . Let U ⊂ κ , where | U | = | κ \ U | = κ and let ρ := h∅ , ∅ , ∅ , . . . i and σ := h U, ∅ , ∅ , . . . i . Then, byLemma 3.3 (b) and (c), P ( ρ ) = h [ κ ] κ , ⊂i , P ( σ ) = h [ κ ] κ , ⊂i × h [ κ ] κ , ⊂i , and sq P ( ρ ) ∼ = ( P ( κ ) / [ κ ] <κ ) + , (17) sq P ( σ ) ∼ = ( P ( κ ) / [ κ ] <κ ) + × ( P ( κ ) / [ κ ] <κ ) + . (18)By Fact 3.4(d), the poset ( P ( κ ) / [ κ ] <κ ) + is not isomorphic to its square. So, by(17) and (18) we have ρ σ .ifferent similarities 15(c) For ρ and σ defined in the previous paragraph we have ρ σ and weprove that ρ ∼ σ . First we consider the case when κ > ω . By (17) and Fact 3.4, P ( ρ ) ≡ (Col( ω, κ )) + . By (18), forcing by the poset sq P ( σ ) collapses κ to ω and, since the poset is of size κ , by Fact 3.4(c) we have ro sq P ( σ ) ∼ = Col( ω, κ ) .Thus the posets P ( ρ ) and P ( σ ) are forcing equivalent, that is ρ ∼ σ . If κ = ω we use Fact 3.4(b). ✷ The following theorem shows that the equivalence of the similarities ∼ (theisomorphism of sq P ( X ) ) and ∼ (the isomorphism of ro sq P ( X ) ) is independentof ZFC even for the simplest unary language. Theorem 3.6 If L is the language containing only one unary relational symbol,then on Mod L ( ω ) we have ∼ = ∼ and ∼ = (cid:26) ∼ if the poset ( P ( ω ) / Fin) + is forcing equivalent to its square, ∼ otherwise.So, the equality ∼ = ∼ is independent of ZFC. Proof.
By Lemma 3.3, for U ⊂ ω , writing P ( U ) instead of h P ( ω, U ) , ⊂i , we have P ( U ) ∼ = (cid:26) h [ ω ] ω , ⊂i if | U | < ω or | ω \ U | < ω, h [ ω ] ω , ⊂i otherwise ; (19) sq P ( U ) ∼ = (cid:26) ( P ( ω ) / Fin) + if | U | < ω or | ω \ U | < ω, (( P ( ω ) / Fin) + ) otherwise . (20)If U , U ⊂ ω and U U , that is P ( U ) = P ( U ) , then, by (19) and (20),for example, sq P ( U ) ∼ = ( P ( ω ) / Fin) + and sq P ( U ) ∼ = (( P ( ω ) / Fin) + ) and,by Fact 3.4(d), sq P ( U ) = sq P ( U ) , that is U U . Thus ∼ ⊂ ∼ , whichimplies ∼ = ∼ .If ( P ( ω ) / Fin) + ≡ (( P ( ω ) / Fin) + ) , then by (20) for each U ⊂ ω we have P ( U ) ≡ ( P ( ω ) / Fin) + and, hence ∼ = ∼ . Otherwise, if ( P ( ω ) / Fin) + (( P ( ω ) / Fin) + ) , then for U , U ⊂ ω satisfying U ∼ U by Fact 1.1(b) wehave sq P ( U ) ≡ sq P ( U ) so, by the assumption and (20), sq P ( U ) ∼ = sq P ( U ) .Thus ∼ ⊂ ∼ and, hence ∼ = ∼ = ∼ .By Fact 3.4(b), CH implies that ( P ( ω ) / Fin) + ≡ (( P ( ω ) / Fin) + ) . But, bya result of Shelah and Spinas [14], in the Mathias model these two posets havedifferent distributivity numbers and, hence, they are not forcing equivalent. ✷ ❅❅❅ (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) ❅❅❅ rrr rrr ∼ = ∼ ∼ ∼ = ∼ ∼ = ∼ = ∼ = ∼ ∼ ∼ = ∼ = the full relation Figure 5: The similarities on
Mod h R i ( ω ) if ( P ( ω ) / Fin) + ≡ (( P ( ω ) / Fin) + ) For infinite structures of non-unary languages the diagram from Figure 2 does notcollapse at all. Namely the main result of this subsection is the following theorem.
Theorem 3.7 If L is a non-unary relational language and κ an infinite cardi-nal, then in the diagram from Figure 2 describing the similarities ∼ k on the set Mod L ( κ ) all the implications a - o are proper and there are no new implications(except the ones following from transitivity). Consequently, the same holds for thediagram from Figure 1 related to the class of all relational structures. Theorem 3.7 will be proved in two steps. First we will prove the statement for theclass
Mod L b ( ω ) of countable binary structures (where L b = h R i and ar ( R ) = 2 )and then, roughly speaking, make a correspondence between the classes Mod L b ( ω ) and Mod L ( κ ) preserving all the similarities ∼ k and their negations. First, giving examples (i.e. constructing pairs of structures), we show that for L = L b and | X | = ω , in the diagram from Figure 2 all the implications a - o are proper.We will use the following auxiliary claim. Lemma 3.8 If P = h P, ≤ P i and Q = h Q, ≤ Q i are partial orders and f : P → Q a surjection such that for each p , p ∈ P we have(i) p ≤ P p ⇒ f ( p ) ≤ ∗ Q f ( p ) ,(ii) p ⊥ P p ⇒ f ( p ) ⊥ Q f ( p ) ,then sq P ∼ = sq Q . ifferent similarities 17 Proof.
First we prove that for each p , p ∈ P we have p ≤ ∗ P p ⇔ f ( p ) ≤ ∗ Q f ( p ) . (21)( ⇒ ) Assuming p ≤ ∗ P p we have to prove that ∀ q ≤ Q f ( p ) q Q f ( p ) . (22)Let q ≤ Q f ( p ) . Since f is onto, there is p ∈ P such that f ( p ) = q . Thus f ( p ) ≤ Q f ( p ) and, by (ii), there is p ≤ P p , p . So, since p ≤ ∗ P p wehave p P p namely there is p ≤ P p , p . By (i) we have f ( p ) ≤ ∗ Q f ( p ) ,which implies f ( p ) Q f ( p ) and, hence, there is q ≤ Q f ( p ) , f ( p ) . Since p ≤ P p ≤ P p , by (i) we have f ( p ) ≤ ∗ Q f ( p ) = q and, hence, q ≤ ∗ Q q , whichimplies q Q q , so there is q ′ ≤ Q q , q . Now q ′ ≤ Q q, f ( p ) and (22) is proved.( ⇐ ) Assuming (22) we prove that p ≤ ∗ P p . So, taking p ≤ P p we showthat p P p . By (i) we have f ( p ) ≤ ∗ Q f ( p ) which implies that there is q ≤ Q f ( p ) , f ( p ) . By (22) we have q Q f ( p ) and, hence, there is q ′ ≤ Q q, f ( p ) .Now q ′ ≤ Q f ( p ) , f ( p ) and, by (ii), p P p . Thus (21) is proved.Now we show that h P/ = ∗ P , E P i ∼ = F h Q/ = ∗ Q , E Q i , where F ([ p ]) = [ f ( p )] .By (21), for p , p ∈ P we have [ p ] = [ p ] iff p = ∗ P p iff p ≤ ∗ P p ∧ p ≤ ∗ P p iff f ( p ) ≤ ∗ Q f ( p ) ∧ f ( p ) ≤ ∗ Q f ( p ) iff f ( p ) = ∗ Q f ( p ) iff [ f ( p )] = [ f ( p )] iff F ([ p ]) = F ([ p ]) and F is a well defined injection. Since f is onto, for q ∈ Q there is p ∈ P such that q = f ( p ) . Thus F ([ p ]) = [ f ( p )] = [ q ] and F is onto.By (21) again, [ p ] E P [ p ] iff p ≤ ∗ P p iff f ( p ) ≤ ∗ Q f ( p ) iff [ f ( p )] E Q [ f ( p )] iff F ([ p ]) E Q F ([ p ]) . Thus F is an isomorphism. ✷ Example 3.9
The implication a can not be reversed. Let X = h ω, ≤i and Y = h ω, ≤ f i , where f : ω → ω is a bijection different from the identity and ≤ f = {h f ( m ) , f ( n ) i : m ≤ n } . Then X ∼ = Y and P ( X ) = P ( Y ) = [ ω ] ω , but X = Y . Example 3.10
The implications b and f can not be reversed. Let X = h ω, {h n, n + 1 i : n ∈ ω } ∪ {h n, n i : n ∈ ω }i and Y = h ω, {h n, n + 1 i : n ∈ ω } ∪ {h n + 1 , n + 1 i : n ∈ ω }i .Then P ( X ) = P ( Y ) = { [2 n, ∞ ) : n ∈ ω } and X ⇄ Y but X = Y . Example 3.11
The implications c , e and g can not be reversed. Let us define X = h ω, ω \ {h , i}i and Y = h ω, ω \ {h , i}i . Then X ∼ = Y and P ( X ) = { A ∈ [ ω ] ω : 0 ∈ A } ∼ = P ( Y ) = { A ∈ [ ω ] ω : 1 ∈ A } , but P ( X ) = P ( Y ) . Example 3.12
The implications d , h , k and n can not be reversed. Let X = h ω, ≤i and Y = h ω, ω × ω i . Then P ( X ) = P ( Y ) = [ ω ] ω and, hence, P ( X ) ∼ = P ( Y ) , sq P ( X ) ∼ = sq P ( Y ) and P ( X ) ≡ P ( Y ) , but X ⇄ Y .8 Miloˇs S.Kurili´c Example 3.13
The implications i and j can not be reversed. Let X = h (0 , Q , ≤i and Y = h (0 , Q , ≤i be suborders of the rational line, Q . Then, clearly, X ⇄ Y .Since the elements of P ( X ) are dense linear orders without end points, eachchain Ł in the poset h P ( X ) , ⊂i has a supremum: S Ł. On the other hand, Ł = { (0 , − n ] Q : n ≥ } is a chain in the poset h P ( Y ) , ⊂i , S Ł = (0 , ) P ( Y ) andthe sets (0 , ) Q ∪ { q } , q ∈ [ , Q , are upper bounds for Ł, but Ł does not havea least upper bound. Thus the poset h P ( Y ) , ⊂i is not chain complete and, hence, P ( X ) = P ( Y ) .Using Lemma 3.8 we show that sq P ( X ) ∼ = sq P ( Y ) . We remind the reader thata linear order L is called scattered iff Q ֒ → L . Let Scatt denote the set of scatteredsuborders of Q . It is easy to see that for A, B ∈ P ( X ) we have A ≤ ∗ B ⇔ A \ B ∈ Scatt and A ⊥ B ⇔ A ∩ B ∈ Scatt (where ≤ ∗ is the corresponding separativemodification) and that the same holds for A, B ∈ P ( Y ) . Clearly, if A ∈ P ( Y ) , then A \ { max A } ⊂ (0 , Q and it is a copy of X , so, the function f : P ( Y ) → P ( X ) ,given by f ( A ) = A \ { max A } , is well defined and we show that it satisfies theassumptions of Lemma 3.8. First, if C ∈ P ( X ) , then C ⊂ (0 , Q and, clearly, C ∪ { } ∈ P ( Y ) and f ( C ∪ { } ) = C . Thus f is a surjection. Let A, B ∈ P ( Y ) . If A ⊂ B , then f ( A ) \ f ( B ) = ( A \{ max A } ) \ ( B \{ max B } ) ⊂ { max B } ∈ Scatt and, hence, f ( A ) ≤ ∗ f ( B ) so (i) is true. If A ⊥ B , that is A ∩ B ∈ Scatt , then,clearly, f ( A ) ∩ f ( B ) ∈ Scatt , thus f ( A ) ⊥ f ( B ) and (ii) is true as well. ByLemma 3.8 we have sq P ( X ) ∼ = sq P ( Y ) . Example 3.14
The implication m can not be reversed. By Example 4.4 of [6],if X is the directed graph h <ω , ρ i , where ρ = {h ϕ, ϕ a i i : ϕ ∈ <ω ∧ i ∈ } ,then h P ( X ) , ⊂i = sq h P ( X ) , ⊂i ∼ = h <ω , ⊃i . Let Y be the directed graph h <ω , σ i ,where σ = {h ϕ, ϕ a i i : ϕ ∈ <ω ∧ i ∈ } , then in a similar way we show that h P ( Y ) , ⊂i = sq h P ( Y ) , ⊂i ∼ = h <ω , ⊃i . Clearly sq h P ( X ) , ⊂i 6∼ = sq h P ( Y ) , ⊂i , but ro sq h P ( X ) , ⊂i ∼ = ro sq h P ( Y ) , ⊂i ∼ = Borel / M . Example 3.15
The implication l can not be reversed. Let X be the directed graphfrom Example 3.14 and let Y be the directed graph h Y, σ i , where Y ⊂ <ω and σ ⊂ Y × Y are defined by Y = {∅ , , } ∪ { jj a ϕ : j ∈ ∧ ϕ ∈ <ω } ,σ = {h∅ , i , h∅ , i , h , i , h , i} ∪ {h jj a ϕ, jj a ϕ a k i : j, k ∈ ∧ ϕ ∈ <ω } . It is easy to see that X ⇄ Y and P ( Y ) = { Y } ∪ { A kljj a ϕ : j, k, l ∈ ∧ ϕ ∈ <ω } , ifferent similarities 19where A kljj a ϕ = { jj a ϕ, jj a ϕ a , jj a ϕ a } ∪ { jj a ϕ a a k a ψ : ψ ∈ <ω } ∪{ jj a ϕ a a l a ψ : ψ ∈ <ω } . By Example 3.14, the poset sq h P ( X ) , ⊂i is isomor-phic to the reversed binary tree. Thus, in order to prove that sq P ( Y ) = sq P ( X ) we will show that [ A ] and [ A ] are incomparable but compatible elements of sq P ( Y ) = h P ( Y ) / = ∗ , E i . So we have A = { , , } ∪ { a ψ : ψ ∈ <ω } ∪ { a ψ : ψ ∈ <ω } ,A = { , , } ∪ { a ψ : ψ ∈ <ω } ∪ { a ψ : ψ ∈ <ω } . Clearly { a ψ : ψ ∈ <ω } is a copy of X and, hence, contains a copy of Y ,say B . Since B ⊂ A , A we have B ≤ ∗ A , A and [ B ] E [ A ] , [ A ] thus [ A ] and [ A ] are compatible elements of sq P ( Y ) .In order to prove that [ A ] E [ A ] we need C ∈ P ( Y ) such that C ⊂ A and D C ∩ A , for all D ∈ P ( Y ) . Now { a ψ : ψ ∈ <ω } ⊂ A is a copy of X and, hence, contains a copy of Y , say C . Since { a ψ : ψ ∈ <ω } ∩ A = ∅ ,we have C ∩ A = ∅ and we are done. Thus [ A ] E [ A ] and, similarly, [ A ] E [ A ] .Thus in Figure 2 for Mod L b ( ω ) all the implications a - o are proper and we showthat there are no new implications except the ones following from transitivity. Soit remains to be shown that the eight pairs which are incomparable in the Hassediagram in Figure 2 are really incomparable. We will use the following elementaryfact: if P = h P, ≤i is a partial order and p, q, r ∈ P , then r = p ∧ q and r < p and r < q ⇒ p k q. (23)In fact our poset of similarities is a suborder of the lattice h EQ (Int L b ( ω )) , ⊂i ofequivalence relations on the set Int L b ( ω ) , where for ∼ , ∼ ′ ∈ EQ (Int L b ( ω )) wehave ∼ ∧ ∼ ′ = ∼ ∩ ∼ ′ and ∼ ∨ ∼ ′ = trcl ( ∼ ∪ ∼ ′ ) and ∼⊂∼ ′ iff the ∼ -partitionis a refinement of the ∼ ′ -partition of Int L b ( ω ) . Now, since by our definition wehave ∼ = ∼ ∩ ∼ , by (23) we obtain ∼ k ∼ and similarly for the other sevenpairs. The following concepts and facts will be used in our proof. Let L b = h R i , where ar( R ) = 2 . If X = h X, ρ i is an L b -structure, then the transitive closure ρ rst of the relation ρ rs = ∆ X ∪ ρ ∪ ρ − (given by x ρ rst y iff there are n ∈ N and z = x, z , . . . , z n = y such that z i ρ rs z i +1 , for each i < n ) is the minimalequivalence relation on X containing ρ . The corresponding equivalence classes arecalled the components of X and the structure X is called connected iff | X/ρ rst | = 1 .0 Miloˇs S.Kurili´cThe complement of the structure X , h X, ( X × X ) \ ρ i will be denoted by X c ; its reflexification , h X, ρ ∪ ∆ X i , by X re ; and its irreflexification , h X, ρ \ ∆ X i , by X ir .If X i = h X i , ρ i i , i ∈ I , are connected L b -structures and X i ∩ X j = ∅ , fordifferent i, j ∈ I , then the structure S i ∈ I X i = h S i ∈ I X i , S i ∈ I ρ i i is the disjointunion of the structures X i , i ∈ I , and the structures X i , i ∈ I , are its components. Fact 3.16 ([6]) If X is an L b -structure, then at least one of the structures X and X c is connected. Fact 3.17 ([6]) Let X i = h X i , ρ i i , i ∈ I , and Y j = h Y j , σ j i , j ∈ J , be fami-lies of disjoint connected binary structures. Then F : S i ∈ I X i ֒ → S j ∈ J Y j iffthere are f : I → J and g i : X i ֒ → Y f ( i ) , i ∈ I , such that F = S i ∈ I g i and h g i ( x ) , g i ′ ( x ′ ) i 6∈ σ rs , whenever i = i ′ , x ∈ X i and x ′ ∈ X i ′ . Fact 3.18
Let X be a binary structure. Then(a) Emb( X ) = Emb( X c ) and P ( X ) = P ( X c ) ;(b) If X is irreflexive, then Emb( X ) = Emb( X re ) and P ( X ) = P ( X re ) ;(c) If X is reflexive, then Emb( X ) = Emb( X ir ) and P ( X ) = P ( X ir ) . Theorem 3.19 (Vopˇenka, Pultr, Hedrl´ın [15]) On any set X there is an irreflexivebinary relation ρ such that id X is the only endomorphism of the structure h X, ρ i . For a cardinal λ let Int ∗ L b ( λ ) = { ρ ⊂ λ : h λ, ρ i is connected ∧ ρ ∩ ∆ λ = ∅} . Then Int ∗ L b ( λ ) ⊂ Int L b ( λ ) and Mod ∗ L b ( λ ) := {h λ, ρ i : ρ ∈ Int ∗ L b ( λ ) } ⊂ Mod L b ( λ ) . Theorem 3.20
Let κ ≥ λ ≥ ω be cardinals and L = h R i : i ∈ I i a non-unaryrelational language. Then there is a mapping τ : Int ∗ L b ( λ ) → Int L ( κ ) such that(a) P ( κ, τ ρ ) ∼ = P ( λ, ρ ) , for each ρ ∈ Int ∗ L b ( λ ) ;(b) For each ρ ∈ Int L b ( λ ) there are ρ ′ ∈ Int ∗ L b ( λ ) and τ ∈ Int L ( κ ) such that P ( λ, ρ ′ ) = P ( λ, ρ ) ∼ = P ( κ, τ ) ;(c) τ preserves all the relations ∼ k from Figure 2, that is for each k ≤ ∀ ρ, σ ∈ Int ∗ L b ( λ ) ( ρ ∼ k σ ⇔ τ ρ ∼ k τ σ ) . (24) Proof.
First suppose that λ < κ . Then | κ \ λ | = κ and, by Theorem 3.19 we canfix an irreflexive binary relation θ ⊂ ( κ \ λ ) such that Emb( κ \ λ, θ ) = { id κ \ λ } .By Theorem 3.16 and Facts 3.18(a) and 3.18(c) we can assume that the relation θ is connected and irreflexive. The language L is not unary and we fix an i ∈ I suchthat n i ≥ . Now, for ρ ∈ Int ∗ L b ( λ ) let the interpretation τ ρ = h τ ρi : i ∈ I i ∈ Int L ( κ ) be defined by τ ρi = ( ρ ∪ θ ) × κ n i − if i = i and n i > ρ ∪ θ ) if i = i and n i = 2; ∅ if i = i . (25)ifferent similarities 21For convenience, for ρ, σ ∈ Int ∗ L b ( λ ) , instead of Emb( h κ, τ ρ i , h κ, τ σ i ) (respec-tively, Emb( h λ, ρ i , h λ, σ i ) ) we will write Emb( τ ρ , τ σ ) (resp. Emb( ρ, σ ) ). Claim 3.21
For each ρ, σ ∈ Int ∗ L b ( λ ) we have(i) Emb( τ ρ , τ σ ) = { f ∪ id κ \ λ : f ∈ Emb( ρ, σ ) } ;(ii) Iso( τ ρ , τ σ ) = { f ∪ id κ \ λ : f ∈ Iso( ρ, σ ) } ;(iii) P ( τ ρ , τ σ ) = { C ∪ ( κ \ λ ) : C ∈ P ( ρ, σ ) } ;(iv) Emb( τ ρ ) = { f ∪ id κ \ λ : f ∈ Emb( ρ ) } ;(v) Aut( τ ρ ) = { f ∪ id κ \ λ : f ∈ Aut( ρ ) } ;(vi) P ( τ ρ ) = { C ∪ ( κ \ λ ) : C ∈ P ( ρ ) } . Proof.
For convenience let π ρ := ρ ∪ θ , for ρ ∈ Int ∗ L b ( λ ) . First we prove that Emb( h κ, π ρ i , h κ, π σ i ) = { f ∪ id κ \ λ : f ∈ Emb( ρ, σ ) } . (26)By the construction, h κ, π ρ i = h λ, ρ i ∪ h κ \ λ, θ i and h κ, π σ i = h λ, σ i ∪ h κ \ λ, θ i are partitions of the binary structures h κ, π ρ i and h κ, π σ i into their connectivitycomponents. Since ρ ∩ ∆ λ = ∅ and θ is an irreflexive relation, we have h λ, ρ i 6 ֒ →h κ \ λ, θ i and the inequality κ > λ implies that h κ \ λ, θ i 6 ֒ → h λ, σ i . So, byTheorem 3.17, F ∈ Emb( h κ, π ρ i , h κ, π σ i ) iff F ↾ λ ∈ Emb( h λ, ρ i , h λ, σ i ) and F ↾ ( κ \ λ ) ∈ Emb( h κ \ λ, θ i ) = { id κ \ λ } and (26) is proved.Now we prove Emb( h κ, τ ρi i , h κ, τ σi i ) = { f ∪ id κ \ λ : f ∈ Emb( ρ, σ ) } . (27)If F : κ → κ is an injection, then F ∈ Emb( h κ, τ ρi i , h κ, τ σi i ) iff for each x , x , . . . , x n i ∈ κ h x , x , . . . , x n i i ∈ π ρ × κ n i − ⇔ h F ( x ) , F ( x ) , . . . , F ( x n i ) i ∈ π σ × κ n i − iff for each x , x ∈ κ we have: h x , x i ∈ π ρ ⇔ h F ( x ) , F ( x ) i ∈ π σ , iff F ∈ Emb( h κ, π ρ i , h κ, π σ i ) . Now (27) follows from (26).(i) Clearly, F ∈ Emb( τ ρ , τ σ ) iff F ∈ Emb( h κ, τ ρi i , h κ, τ σi i ) , for all i ∈ I . By(25) this holds iff F ∈ Emb( h κ, τ ρi i , h κ, τ σi i ) and we apply (27).(ii) If f ∈ Emb( ρ, σ ) then f ∪ id κ \ λ is a surjection iff f is a surjection iff f ∈ Iso( ρ, σ ) . Now we apply (i).(iii) A ∈ P ( τ ρ , τ σ ) iff there is F ∈ Emb( τ ρ , τ σ ) such that A = F [ κ ] so, by (i),iff A = f [ λ ] ∪ ( κ \ λ ) , for some f ∈ Emb( ρ, σ ) , iff A = C ∪ ( κ \ λ ) , for some C ∈ P ( ρ, σ ) .Statements (iv), (v) and (vi) follow from (i), (ii) and (iii) respectively. ✷ Now we prove the theorem.2 Miloˇs S.Kurili´c(a) By Claim 3.21(vi) we have P ( τ ρ ) = { C ∪ ( κ \ λ ) : C ∈ P ( ρ ) } and it is easyto check that the mapping F : P ( ρ ) → P ( τ ρ ) , defined by F ( C ) = C ∪ ( κ \ λ ) , isan isomorphism of the posets h P ( ρ ) , ⊂i and h P ( τ ρ ) , ⊂i .(b) Let ρ ∈ Int L b ( λ ) \ Int ∗ L b ( λ ) . If ρ is connected, then it is irreflexive, thus ρ re ∈ Int ∗ L b ( λ ) and, by Fact 3.18(b), P ( λ, ρ re ) = P ( λ, ρ ) . Otherwise, by Theorem3.16 the relation ρ c is connected and, by Fact 3.18(a), P ( λ, ρ c ) = P ( λ, ρ ) . Now, if ρ c ∩ ∆ λ = ∅ , we have ρ c ∈ Int ∗ L b ( λ ) ; otherwise ( ρ c ) re ∈ Int ∗ L b ( λ ) and, by Fact3.18(b), P ( λ, ( ρ c ) re ) = P ( λ, ρ c ) = P ( λ, ρ ) .If ρ ∈ Int L b ( λ ) and ρ ′ ∈ Int ∗ L b ( λ ) , where P ( λ, ρ ) = P ( λ, ρ ′ ) , then by (a) wehave P ( λ, ρ ′ ) ∼ = P ( κ, τ ρ ′ ) , where τ ρ ′ ∈ Int L ( κ ) . Thus P ( κ, τ ρ ′ ) ∼ = P ( λ, ρ ) .(c) It is sufficient to prove that the mapping τ : Int ∗ L b ( λ ) → Int L ( κ ) preservesthe relations ∼ k , for k ∈ { , , , , , , } . Let ρ, σ ∈ Int ∗ L b ( λ ) . ∼ : ρ = σ ⇔ τ ρ = τ σ . By (25) we have: τ ρ = τ σ iff τ ρi = τ ρi iff ρ ∪ θ = σ ∪ θ iff ρ = σ . ∼ : ρ ∼ = σ ⇔ τ ρ ∼ = τ σ . If ρ ∼ = σ and f ∈ Iso( ρ, σ ) , then, by Claim 3.21(ii), f ∪ id κ \ λ ∈ Iso( τ ρ , τ σ ) and, hence, τ ρ ∼ = τ σ . Conversely, if τ ρ ∼ = τ σ and F ∈ Iso( τ ρ , τ σ ) , then, by Claim 3.21(ii), F ↾ λ ∈ Iso( ρ, σ ) and, hence, ρ ∼ = σ . ∼ : ρ ⇄ σ ⇔ τ ρ ⇄ τ σ . If ρ ֒ → σ and f ∈ Emb( ρ, σ ) , then, by Claim3.21(i), f ∪ id κ \ λ ∈ Emb( τ ρ , τ σ ) and, hence, τ ρ ֒ → τ σ . Thus ρ ⇄ σ implies τ ρ ⇄ τ σ . Conversely, if τ ρ ֒ → τ σ and F ∈ Emb( τ ρ , τ σ ) , then, by Claim 3.21(i), F ↾ λ ∈ Emb( ρ, σ ) and, hence, ρ ֒ → σ . So τ ρ ⇄ τ σ implies ρ ⇄ σ . ∼ : P ( ρ ) = P ( σ ) ⇔ P ( τ ρ ) = P ( τ σ ) . This follows from Claim 3.21(vi). ∼ : P ( ρ ) ∼ = P ( σ ) ⇔ P ( τ ρ ) ∼ = P ( τ σ ) . This is true since by (a) we have P ( ρ ) ∼ = P ( τ ρ ) and P ( σ ) ∼ = P ( τ σ ) . (28) ∼ : sq P ( ρ ) ∼ = sq P ( σ ) ⇔ sq P ( τ ρ ) ∼ = sq P ( τ σ ) . This is true since by (28) andFact 1.1(a) we have sq P ( ρ ) ∼ = sq P ( τ ρ ) and sq P ( σ ) ∼ = sq P ( τ σ ) . ∼ : ro sq P ( ρ ) ∼ = ro sq P ( σ ) ⇔ ro sq P ( τ ρ ) ∼ = ro sq P ( τ σ ) . By (28) and Fact1.1(a) we have ro sq P ( ρ ) ∼ = ro sq P ( τ ρ ) and ro sq P ( σ ) ∼ = ro sq P ( τ σ ) .So, the theorem is proved for λ < κ . If λ = κ , then we define τ ρi := ρ × κ n i − and continue in the same way. ✷ Finally we prove Theorem 3.7. In Subsection 3.2.1 it is shown that all theimplications a - o in Figure 2 for the class Mod L b ( ω ) are proper. For example,concerning the implication a , in Example 3.9 we have constructed ρ, σ ∈ Int ∗ L b ( ω ) such that ρ ∼ σ but ρ σ . By Theorem 3.20(c) we have τ ρ ∼ τ σ and τ ρ τ σ ,which implies that in Figure 2 for the class Mod L ( κ ) the implication a is proper aswell. The reader will notice that the structures constructed in Examples 3.9 - 3.13belong to Int ∗ L b ( ω ) and that the structures constructed in Examples 3.14 and 3.15are irreflexive. But their refexifications are in Int ∗ L b ( ω ) . 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