What would the rational Urysohn space and the random graph look like if they were uncountable?
aa r X i v : . [ m a t h . L O ] F e b What would the rational Urysohn space and therandom graph look like if they were uncountable?
Ziemowit Kostana * Institute of Mathematics Czech Academy of SciencesŽitná 25, 115 67 Prague, Czech Republic;University of Warsaw,Banacha 2, 02-097 Warsaw, [email protected] 11, 2021
Abstract
We apply the technology developed in the 80s by Avraham, Rubin, and Shelah, toprove that the following is consistent with ZFC: there exists an uncountable, separablemetric space X with rational distances, such that every uncountable partial 1-1 functionfrom X to X is an isometry on an uncountable subset. This space contains a dense copyof the rational Urysohn space, and is homogeneous with respect to finite subspaces. Weprove similar results for some other classes of structures, and in some cases show that themodels we obtain are, in some sense, canonical. Keywords:
Fraisse limit, Martin’s Axiom, homogeneous structure, generic structure
MSC classification:
The original theory of universal, homogeneous structures, developed by R. Fraïssé in the 50s([4]), evolved in many directions (see for example [6]). One of them, known as the Fraïssé-Jónsson Theory, allows to construct models of cardinality ω , which are homogeneous withrespect to countable submodels. This construction, relying on the Continuum Hypothesis,essentially just replaces "finite" with "countable" and "countable" with "of size ω " in theoriginal framework. The usage of countable models in the construction gives us a strongerversion of homogeneity, but since we have plenty of countable submodels, the final structureis "very non-separable". The question arises, if we can build uncountable models using finitesubmodels? One indication that such theory should be possible to formulate is the famoustheorem of Baumgartner, concerning separable ω -dense linear orders. A linear order is ω -dense if each non-empty open interval has cardinality ω . Theorem ([3]) . It is consistent with
ZF C that there exists a unique up to isomorphismseparable ω -dense linear order. If the conclusion of this theorem holds, we can take any subfield of the reals of size ω as a model of this ordering. This shows that it is homogeneous with respect to finite subsets.Ideas of Baumgartner were extended by Avraham, Rubin, and Shelah in [2] and [1]. Amongother things, they show that Baumgartner’s Theorem does not follow from M A + ¬ CH ,but the latter axiom is a significant step in the direction of Baumgartner’s Theorem. Theyintroduce another axiom, known as OCA
ARS , and show that
M A ω + OCA
ARS impliesthat either conclusion of the Baumgartner’s Theorem holds, or there are up to isomorphismexactly three homogeneous separable ω -dense linear orders ([1], Sec. 6). This gives someinsight that M A ω might be used in place of induction, in some uncountable variant of theFraïssé theory. * Research of Z. Kostana was supported by the GAˇCR project GAˇCR project EXPRO 20-31529X and RVO:67985840.
1e take ideas from papers [2] and [1], and apply them to other structures beyond linearorders – mostly metric spaces and graphs. Specifically, we take generic models described in[5], and show that, although initially rigid, they become highly homogeneous in suitable forc-ing extensions satisfying
M A ω . Inductive arguments from the classical theory are replacedby Martin’s Axiom, and in some cases it is enough to imply certain uniqueness. Whether itentitles the theory we develop to be called "the uncountable Fraïssé theory with finite sup-ports", is left for the reader to decide.Even if aiming towards a new variant of the Fraïssé theory is perhaps too ambitious, westill find new applications for some techniques developed in [2]. As an example, let us lookat one Ramsey-like result from [2]. Theorem (Thm. 2, [2]) . It is consistent with
ZF C + M A + ”2 ω = ω ” that there exists anuncountable set A ⊆ R with the property that each uncountable 1-1 function f ⊆ A × A isstrictly increasing on an uncountable set. In this spirit, we prove in Section 2
Theorem.
It is consistent with
ZF C + M A + ”2 ω = ω ” that there exists an uncountable,separable rational metric space ( X, d ) such that each uncountable 1-1 function f ⊆ X × X is an isometry on an uncountable set. Metric spaces are not exclusive to this, apparently somewhat general, phenomenon. InSection 3 we prove similar results for other classes of structures, although they are moretechnical, and require some introduction (what is a "separable" graph?). In Section 4 weprove some results on the number of isomorphism types of models we built.
By an embedding we always mean an isomorphism into its image. A subset of a model is a substructure if the identity inclusion is an embedding. We will say that the language is purelyrelational if it doesn’t contain any function or constant symbols. We write dom f and rg f todenote respectively the domain and the image of a function f .We extend ideas from [5], where we have been looking at forcing notions of the followingtype. K stands for a class of structures in some purely relational language. All unspecifiedforcing terminology follows [7]. Definition 1.
Let λ be an infinite cardinal number, and S an infinite set. For any model A ∈ K we denote by F ( A ) the underlying set. We introduce the forcing Fn( S, K , λ ) as theset { A ∈ K| F ( A ) ∈ [ S ] <λ } , ordered by the reversed substructure relation.We proved in [5] that models added generically by Fn( ω , K , ω ) tend to be rigid, while those added by Fn( ω, K , ω ) are isomorphic to the usual Fraïssé limits of K . Of course we must impose some restrictionson the class K in order to Fn( ω, K , ω ) be an interesting forcing notion. The following issufficient to ensure the c.c.c. property in all cases in the range of our interest. The name Splitting Property was coined by W. Kubi´s. Two embeddings f : A ֒ → B and g : A ֒ → C are isomorphic , if there exists an isomorphism h : B ֒ → C , such that h ◦ f = g . Definition 2. K has the Splitting Property (SP) if for all R ∈ K , all pairs of isomorphicextensions R ⊆ X , R ⊆ Y , there exists Z ∈ K , containing X ∪ Y . Proposition 1.
Let K be a class of structures in a purely relational language. We assume thefollowing • K has countably many isomorphism types of finite models, K is hereditary (if A ∈ K , B ⊆ A, then B ∈ K ), • K has SP.Then the forcing Fn( λ, K , ω ) has c.c.c. for any cardinal λ . The proof is essentially the same as the proof of the c.c.c. property for the Cohen forcing(see [7]), so we omit it.
We adapt the technology from [2] and [1] to prove that it is relatively consistent with
ZF C that there exists a separable rational metric space ( X, d ) of size ω , such that any uncountable1-1 function from X to itself is an isometry on an uncountable subset. The original resultstates that it is relatively consistent with ZF C that there exists an uncountable set A ⊆ R ,such that every function f : A → A is non-decreasing on an uncountable set. Unlike thelatter statement, our result does not seem to follow from Semiopen Coloring Axiom, OpenColoring Axiom or any other combinatorial principle described in [1].We first introduce a metric analog of a k -increasing linear order, introduced in [2]. Definition 3.
Let ( X, d ) be a metric space.• We call a pair of tuples x = ( x , . . . , x n ) , y = ( y , . . . , y n ) ∈ X n alike if they satisfythe following axioms:A1 ∀ i, j = 1 , . . . , n ( d ( x i , y i ) = d ( x j , y j )) A2 ∀ i, j = 1 , . . . , n ( d ( x i , x j ) = d ( y i , y j )) A3 ∀ i, j = 1 , . . . , n ( x i = x j = ⇒ d ( x i , x j ) = d ( x i , y j )) We then write x ⊛ y .• We call ( X, d ) rectangular if it is uncountable, and for any sequence of pairwisedisjoint tuples { ( x ξ , . . . , x ξn ) | ξ < ω } ⊆ X n , there are ξ = η < ω , such that ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) .Denote by M etr the class of rational metric spaces. Keeping up with the general notation,
Fn( ω , M etr, ω ) is the partial order { ( Y, d ) | Y ∈ [ ω ] <ω , and ( Y, d ) is a rational metric space } , with the ordering relation being the reversed inclusion preserving the metric.We begin with a technical Lemma to ensure that we can amalgamate metric spaces in aspecific way. Lemma 1.
Let ( R, d R ) , ( X, d X ) , ( Y, d Y ) be finite metric spaces, such that ( X, d X ) ∩ ( Y, d Y ) =( R, d R ) , and suppose h : ( X, d X ) → ( Y, d Y ) is an isometric bijection, which is identity on R . ( X, d X )( R, d R ) ( X ∪ Y, d ∗ )( Y, d Y ) h Then there exists a metric d ∗ on X ∪ Y extending both d X and d Y , such that if ( x , . . . , x n ) is a bijective enumeration of X \ R , then ( x , . . . , x n ) ⊛ ( h ( x ) , . . . , h ( x n )) .Proof. Let s = min { d X ( x, x ′ ) | x = x ′ ∈ X } . Given that d ∗ must extend the metrics of X and Y , we must set the distances between elements from X \ R and Y \ R . Therefore we set d ∗ ( x, h ( x )) = s and d ∗ ( x, h ( x ′ )) = d X ( x, x ′ ) , for all x = x ′ ∈ X \ R . A standard com-putation shows that this definition gives a well-defined metric structure on X ∪ Y , satisfyingthe required conditions. 3n the light of Proposition 1, we have proved the c.c.c. property for Fn( ω , M etr, ω ) .Even more generally, for any countable set K ⊆ [0 , ∞ ) the proof of Lemma 1 shows that theclass of finite metric spaces with distances in K has SP. Proposition 2.
Fn( ω , M etr, ω ) (cid:13) " ( ω , ˙ d ) is rectangular".Proof. Let { ( ˙ x ξ , . . . , ˙ x ξn ) | ξ < ω } be a sequence of Fn( ω , M etr, ω ) -names for pairwisedisjoint n -tuples from ( ω , ˙ d ) . Fix a condition p . For every ξ < ω , we find a condition p ξ = ( p ξ , d ξ ) ≤ p , deciding values of ˙ x ξi and ˙ d ↾ { x ξ , . . . , x ξn } × { x ξ , . . . , x ξn } . We choosean uncountable set S ⊆ ω , satisfying the following conditions:• { p ξ | ξ ∈ S } is a ∆ -system with the root R = ( R, d R ) ,• ∀ ξ, η ∈ S there exists an isometry h : p ξ → p η , which is identity on R , and h ( x ξi ) = x ηi , for i = 1 , . . . , n , as shown in the diagram. ( p ξ , d ξ )( R, d R ) ( p η , d η ) h This can be easily done, since given any point α ∈ ω outside of R , there are onlycountably many possible configurations of distances between this point and R . Now wechoose ξ = η ∈ S , and apply Lemma 1 to obtain q ≤ p ξ , p η , which forces that ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) .It is easy to verify that any infinite subset of ω from the ground model is forced by Fn( ω , M etr, ω ) to be dense. It has even a much stronger property, which we isolate, sinceit will be important later. Definition 4.
Let ( E, d E ) be any metric space with distances in some countable set K ⊆ [0 , ∞ ) . A subset D ⊆ E is a saturated subset of E if for any finite subset E ⊆ E , for anysingle-point extension F = ( E ∪ { f } , d F ) , with distances in K , there exists d ∈ D suchthat for all e ∈ E , we have d E ( e, d ) = d F ( e, f ) .To put it shortly, every possible configuration of distances from a tuple of points in E canbe realized by a point in D . The rational Urysohn space is a saturated subspace of itself, inthe class of rational metric spaces. Definition 5.
Let ( X, d ) be any metric space, with distances in a given countable set.1. ( X, d ) is separably saturated if it has a countable saturated subset.2. ( X, d ) is hereditarily separably saturated (HSS) if for any countable subset A ⊆ X ,the space X \ A is separably saturated. Proposition 3.
Fn( ω , M etr, ω ) (cid:13) " ( ω , ˙ d ) is HSS (with distances in Q )".Proof. Each infinite subset of ω which belongs to the ground model is saturated.Let ( M , d ) be the metric space we added to our model by Fn( ω , M etr, ω ) . Our nexttask is to force M A ω , while preserving ( M , d ) being rectangular. Following the ideas from[2], we introduce the special class of partial orders. Definition 6.
A partial order P is appropriate if given any natural number n > , for eachdisjoint family { ( p ξ , x ξ , . . . , x ξn ) | ξ < ω } ⊆ P × ( M , d ) n , there exist ξ = η < ω , suchthat p ξ and p η are comparable, and ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) . Proposition 4. If P is appropriate then P (cid:13) " ( M , d ) is rectangular".Proof. Fix a sequence of P -names { ( ˙ x ξ , . . . , ˙ x ξn ) | ξ < ω } ⊆ ( M , d ) n for pairwise disjoint n -tuples. For a given condition p ∈ P , and ξ < ω , we fix a condition p ξ ≤ p deciding ( ˙ x ξ , . . . , ˙ x ξn ) . Then we apply appropriateness for the family { ( p ξ , x ξ , . . . , x ξn ) | ξ < ω } .This way we obtain ξ = η < ω , and q ≤ p ξ , p η , such that q (cid:13) ( ˙ x ξ , . . . , ˙ x ξn ) ⊛ ( ˙ x η , . . . , ˙ x ηn ) .4he argument showing that appropriateness is preserved under iterations is really notdifferent than the one showing that the c.c.c. is preserved, applied for example in [7]. Weinclude it for completeness. Proposition 5.
Finite support iterations of appropriate posets are appropriate.Proof.
Assume that P is appropriate, and P (cid:13) " ˙ Q is appropriate". Take a sequence { ( p ξ , ˙ q ξ , x ξ , . . . , x ξn ) | ξ < ω } , and towards contradiction assume that it witnesses P ∗ ˙ Q not being appropriate. Let ˙ σ be a P -name defined ˙ σ = { ( ξ, p ξ ) | ξ < ω } . If G ⊆ P is an M -generic filter, then M [ G ] | = ( ξ ∈ σ ⇐⇒ p ξ ∈ G ) . We claim that in M [ G ] , for any two conditions η, ξ ∈ σ , if ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) ,then q ξ and q η are inconsistent. For otherwise, there exist q ≤ q η , q ξ , and p ∈ G , whichforces it. Since p ξ and p η are in G , which is a filter, we may choose p ≤ p ξ , p η . Then ( p, ˙ q ) ≤ ( p ξ , ˙ q ξ ) , ( p η , ˙ q η ) , and ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) contrary to the choice of thesequence { ( p ξ , ˙ q ξ , x ξ , . . . , x ξn ) | ξ < ω } . G was any generic filter, and remember that P (cid:13) " ˙ Q is appropriate". The conclusion ofthis is that P (cid:13) | ˙ σ | < ω . Then there exists a P -name for a countable ordinal ˙ a , for which P (cid:13) ˙ σ ⊆ ˙ a . Since P is c.c.c. there are only countably many possible values of ˙ a , and bytaking supremum of them, we can replace ˙ a by a canonical name a . But note that p a (cid:13) a ∈ ˙ σ .This is a contradiction.Consider now a finite support iteration of an infinite length κ of appropriate forcings P = { P α ∗ ˙ Q α | α < κ } . We prove by induction on κ , that P is appropriate. The successorstep has just been taken care of, so suppose that the conclusion holds for any ordinal less than κ , and κ is limit. Take any disjoint sequence { ( p α , x α ) | α < ω } ⊆ P × ( M , d ) n . We canassume that the supports of conditions p α form a ∆ -system with the root R , and that for any α , supp p α \ R is above R . Let δ = max R . There exist two different α, β < ω , for which x α ⊛ x β and p α ↾ δ is comparable with p β ↾ δ . From the general theory of finite supportiterations it follows that p α and p β are comparable.It may look suspicious that in the proof above we never actually used P being appropriate,only c.c.c. However one can verify that if some c.c.c. forcing forces a poset to be appropriate,then it must be appropriate itself. So equally good we could have assumed that P is c.c.c. Theimmediate consequence of this proposition is Lemma 2.
It is consistent with
ZF C + M A ( appropriate ) + ”2 ω = ω ” that ( M , d ) isrectangular. Conveniently, the full Martin’s Axiom will hold in such model. Recall that if P is aforcing notion, a set D ⊆ P is predense if each element of P is comparable with an elementof D . Lemma 3.
Let P be a c.c.c. forcing notion of size ω . There exists p ∈ P and a family of ω many subsets of P , predense below p , such that any filter G ⊆ P containing p and intersectingall of them is uncountable.Proof. Enumerate bijectively P as { p γ | γ < ω } . Let D α = { p γ | α < γ < ω } . We aim tofind p ∈ P such that uncountably many of sets D α are predense below p – clearly this willfinish the proof. If p with this property doesn’t exist, then the following assertion holds: ∀ γ < ω ∃ γ ′ > γ " D γ ′ is not predense below p γ " , and consequently ∀ γ < ω ∃ p ′ γ ≤ p γ ∃ γ ′ > γ ∀ η > γ ′ p ′ γ ⊥ p η . Using this, we can easily define an uncountable antichain { p γ | γ ∈ E } , ensuring at eachstep of induction that ∀ γ ∈ E ∃ γ ′ > γ ∀ η > γ ′ p γ ⊥ p η . Lemma 4.
M A ω ( appropriate ) implies that any family of pairwise disjoint tuples { ( x ξ , . . . , x ξn ) | ξ < ω } ⊆ ( M , d ) n contains an uncountable subfamily of pairwise aliketuples. roof. Let P = { F ∈ [ ω ] <ω | ∀ ξ = η ∈ F x ξ ⊛ x η } , where x ξ = ( x ξ , . . . , x ξn ) . The or-dering is given by the reversed inclusion. We claim that P is appropriate. Fix an uncountablefamily { F α | α < ω } ⊆ P , and a family of tuples { ( v α , . . . , v αs ) | α < ω } ⊆ ( M , d ) s .Without loss of generality we may assume that F α = { e , . . . , e k , e αk +1 , . . . , e αk + m } , where sets { e αk +1 , . . . , e αk + m } are pairwise disjoint for different α . For each α , let y α ∈ ( M , d ) n · m + s be a concatenation of all tuples x e αk +1 , . . . , x e αk + m , and ( v α , . . . , v αs ) . Thereexists α = β < ω , such that y α ⊛ y β , and we claim that they witness the fact that P isappropriate. Let us write: y α = ( x , . . . , x n , x , . . . , x n , . . . , x m , . . . , x mn , v , . . . , v s ) ,y β = ( y , . . . , y n , y , . . . , y n , . . . , y m , . . . , y mn , u , . . . , u s ) . What is clear, is that ( v , . . . , v s ) ⊛ ( u , . . . , u s ) , and for all k ≤ m , ( x k , . . . , x kn ) ⊛ ( y k , . . . , y kn ) . What is not clear, is that in this case the ⊛ relation is "shift-invariant", i.e. forall ≤ p = r ≤ m we have ( x p , . . . , x pn ) ⊛ ( y r , . . . , y rn ) . We will check that this is the case.A1 For all ≤ r = p ≤ m , and ≤ i, j ≤ n , we have d ( x pi , y ri ) = d ( x pi , x ri ) by A3 for y α and y β , d ( x pi , x ri ) = d ( x pj , x rj ) by A1 for ( x p , . . . , x pn ) and ( x r , . . . , x rn ) , d ( x pj , x rj ) = d ( x pj , y rj ) by A3 for y α and y β . A2 For all ≤ r = p ≤ m , and ≤ i, j ≤ n , we have d ( x pi , x pj ) = d ( x ri , x rj ) by A2 for ( x p , . . . , x pn ) , ( x r , . . . , x rn ) , d ( x ri , x rj ) = d ( y ri , y rj ) by A2 for y α , y β .A3 If x ri = x rj then d ( x ri , x rj ) = d ( x ri , x pj ) by A3 for ( x p , . . . , x pn ) , ( x r , . . . , x rn ) , d ( x ri , x pj ) = d ( x ri , y pj ) by A3 for y α , y β .Therefore F α ∪ F β ∈ P , and this concludes the proof that P is appropriate. The conclusionfollows from Martin’s Axiom together with Lemma 3. Proposition 6.
M A ω ( appropriate ) implies that any c.c.c. partial order of size ω is appro-priate.Proof. Suppose that P is a c.c.c. partial order of cardinality ω , and fix some disjoint family { ( p ξ , x ξ , . . . , x ξn ) | ξ < ω } ⊆ P × ( M , d ) n . If M A ω ( appropriate ) holds, we can assumethat all tuples ( p ξ , x ξ , . . . , x ξn ) are pairwise alike, and since P is c.c.c. we will find two p ξ and p η , which are comparable.The immediate consequence is Theorem 1.
It is consistent with
ZF C + M A + ”2 ω = ω ” that ( M , d ) is rectangular. We stress that the notion of an appropriate poset did rely on the specific model ( M , d ) .But since we now have access to full Martin’s Axiom, we don’t have to think of ( M , d ) asany distinguished structure. It just witnesses the fact that Martin’s Axiom is consistent withthe existence of a metric space, which is rectangular and HSS. For this reason, further resultsapply to any space with these properties, not necessarily ( M , d ) . Theorem 2.
Let ( X, d ) be a rectangular rational metric space of size ω . M A ω impliesthat any uncountable 1-1 function f ⊆ X × X is an isometry on an uncountable set. roof. Let f ⊆ X × X be an uncountable 1-1 function. Consider the partial order P f = ( { E ∈ [dom f ] <ω | f ↾ E is an isometry } , ⊆ ) . We will check that P f is c.c.c. Take a sequence { e ξ | ξ < ω } ⊆ P f . Applying ∆ -systemLemma we may assume that• ∀ ξ < ω | e ξ | = m ,• ∀ ξ = η < ω e ξ ∩ e η = r ,• e ξ = ( e ξ , . . . , e ξl , e ξl +1 , . . . , e ξm ) , where { e ξ , . . . , e ξl } = r .Look at the family of tuples { ( e ξl +1 , . . . , e ξm , f ( e ξl +1 ) , . . . , f ( e ξm )) | ξ < ω } . Using for example ∆ -system Lemma one can easily trim this sequence so that all tuples arepairwise disjoint. Now, given that ( X, d ) is rectangular, we find ξ = η < ω , such that ( e ξl +1 , . . . , e ξm , f ( e ξl +1 ) , . . . , f ( e ξm )) ⊛ ( e ηl +1 , . . . , e ηm , f ( e ηl +1 ) , . . . , f ( e ηm )) . We must check that e ξ and e η are comparable, that is f ↾ ( e ξ ∪ e η ) is an isometry. Butnotice that for i = j = l + 1 , . . . , m , d ( e ξi , e ηj ) = d ( e ξi , e ξj ) = d ( f ( e ξi ) , f ( e ξj )) = d ( f ( e ξi ) , f ( e ηj )) . This proves that e ξ ∪ e η ∈ P f . Conclusion of the theorem follows from Martin’s Axiom andLemma 3. Corollary 1.
It is consistent with
ZF C + M A +”2 ω = ω ” that there exists an uncountable,separable (even hereditarily separably saturated) rational metric space ( X, d ) such that eachuncountable function f ⊆ X × X is an isometry on an uncountable set. The reader might have noticed that in the previous section the triangle inequality for the space ( M , d ) was never applied. The crucial property we used was a variant of SP, which ensuresthat "remainders" will be alike (see Lemma 1). In this section we will define a version of SPwhich allows to proceed with the proof of Theorem 2 in case of other classes of structures.One assumption which seems hard to remove is that the language consists of binary relationsonly.Each language consisting of finitely many binary relational symbols can be identifiedwith a finite coloring of ordered pairs – given a model A , we assign to each element of A its isomorphism type. There are only finitely many symbols in the language, so this coloringis indeed finite, and moreover, it determines the model A completely. Also, any functionis a homomorphism precisely when it preserves this coloring. This observation allows togeneralize results from Section 2 to other classes, besides metric spaces. In fact, the finite-ness of language is not relevant as long as there are only countably many isomorphism typesof 2-element models. Let K be some class of structures in a countable language {R i } i<ω consisting of binary relational symbols. Assume also that K has only countably many iso-morphism types of finite models. Let c be the corresponding coloring of pairs in models from K – by the remark above we may forget about the relational symbols, and think of modelsfrom K as having only one "relation", namely c . We introduce the ⊛ relation by axioms sim-ilar to the metric case, but unlike metrics, the coloring c might not be symmetric. This is theonly significant difference. Definition 7.
Let X ∈ K , and ( x , . . . , x n ) , ( y , . . . , y n ) ∈ X n be disjoint. We will say thatthey are alike , and write ( x , . . . , x n ) ⊛ ( y , . . . , y n ) , if the following axioms are satisfiedA1a ∀ i, j = 1 , . . . , n c ( x i , y i ) = c ( x j , y j ) A1b ∀ i, j = 1 , . . . , n c ( y i , x i ) = c ( y j , x j ) ∀ i, j = 1 , . . . , n c ( x i , x j ) = c ( y i , y j ) A3a ∀ i, j = 1 , . . . , n ( x i = x j = ⇒ c ( x i , x j ) = c ( x i , y j ) = c ( y i , x j )) If all relations R k are anti-reflexive ( ∀ x ¬R k ( x, x ) ), then we can omit the clause ( x i = x j ) = ⇒ in A3a. It is standard to check that ( x , . . . , x n ) ⊛ ( y , . . . , y n ) ⇐⇒ ( y , . . . , y n ) ⊛ ( x , . . . , x n ) . Definition 8. X ∈ K is rectangular if | X | > ω , and for any family of pairwise disjointtuples { ( x ξ , . . . , x ξn ) | ξ < ω } ⊆ X n , there exist ξ = η < ω , such that ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) . Definition 9. K has the Rectangular Splitting Property (RSP) if for all R ∈ K , for all pairsof isomorphic extensions R ⊆ X , R ⊆ Y , with the corresponding isomorphism h : X → Y ,there exists Z ∈ K , with the universe X ∪ Y , such that for any sequence ( x , . . . , x n ) enumerating bijectively X \ R , Z satisfies ( x , . . . , x n ) ⊛ ( h ( x ) , . . . , h ( x n )) .RSP ensures the conclusion of Lemma 1. Moreover, since the definition of ⊛ relationis independent of the ordering of the tuples, if there exists an enumeration ( x , . . . , x n ) likeabove, it can be replaced by any other enumeration, even not 1-1. Theorem 3. If K has RSP then Fn( ω , K , ω ) forces the generic structure to be rectangular.Proof. Exactly like the proof of Proposition 2.Let us fix a rectangular model ( X , c ) . The appropriate forcings are defined in the sameway as in Definition 6: Definition 10.
A partial order P is appropriate if given any natural number n > , for eachdisjoint family { ( p ξ , x ξ , . . . , x ξn ) | ξ < ω } ⊆ P × ( X , c ) n , there exist ξ = η < ω , such that p ξ and p η are comparable, and ( x ξ , . . . , x ξn ) ⊛ ( x η , . . . , x ηn ) . Proposition 7. If P is appropriate, then P (cid:13) " ( X , c ) is rectangular".Proof. Exactly like the proof of Proposition 4.
Proposition 8.
M A ω ( appropriate ) implies that any family of pairwise disjoint tuples { ( x ξ , . . . , x ξn ) | ξ < ω } ⊆ ( X , c ) n contains an uncountable subfamily of pairwise aliketuples.Proof. Same as Lemma 4.What follows, is a generalization of Theorem 1.
Theorem 4.
It is consistent with
ZF C + M A + ”2 ω = ω ” that ( X , c ) is rectangular. The proof that appropriateness is preserved under finite support iterations is also the same. ( X , c ) was a specific model providing us with the notion of an appropriate partial order, butonce Theorem 4 is proved, we may really forget about it. It just witnesses the fact, that some rectangular model can exist in the presence of Martin’s Axiom. Proposition 9.
Let ( X, c ) ∈ K be rectangular. M A ω implies that any uncountable 1-1function f ⊆ X × X is a homomorphism on an uncountable set.Proof. Almost exactly like the proof of Theorem 2 – the only difference is that we write c instead of d , isomorphism instead of isometry , and in the end use A1a and A1b instead of A1,since c might not be symmetric. Proposition 10.
The classes of graphs, directed graphs, tournaments, linear orders, andpartial orders have RSP.Proof.
We will prove RSP for linear and partial orders. Arguments for other classes are easyand left for the reader. For each partial order ≤ there exists a corresponding quasi-orderingrelation < , which is anti-reflexive.Suppose we have a diagram of linear quasi-orders R, X, Y , and h : X → Y , like in thedefinition of RSP. We define a quasi-ordering on Z = X ∪ Y , extending both < X and < Y ,by conditions: 8 ∀ i < ω x i < h ( x i ) ,• ∀ i = j < ω x i < h ( x j ) ⇐⇒ x i < X x j . Clearly < is an anti-reflexive relation on Z . For checking transitivity we must go throughseveral (somewhat boring) cases.1. x i < h ( x j ) , h ( x j ) < x k . Either x i = x j or x i < x j . In the first case h ( x i ) < x k , andso x i < x k . In the second, x i < x k , and x i < x k follows from transitivity of < on X .2. h ( x i ) < x j , x j < x k . In this case x i < x j , so x i < x k , and h ( x i ) < x k .3. x i < x j , x j < h ( x k ) . Either x j = x k or x j < x k . In both cases x i < x k , so x i < h ( x k ) .4. h ( x i ) < h ( x j ) , h ( x j ) < x k . In this case x i < x j , and x j < x k . By transitivity x i < x k , and h ( x i ) < x k follows.5. x i < h ( x j ) , h ( x j ) < h ( x k ) . If x i < x j , then we proceed like before. If x i = x j , h ( x i ) < h ( x k ) . It follows that x i < x k , and x i < h ( x k ) .6. h ( x i ) < x j , x j < h ( x k ) . We see that x i < x j . If x j < x k , we use transitivity of < on X . If x j = x k , then h ( x i ) < x k , and so x i < x k . It follows that h ( x i ) < h ( x k ) .The proof for partial orders is strictly simpler – we define the quasi-ordering by condi-tions:• ∀ i < ω x i is incomparable with h ( x i ) ,• ∀ i, j < ω x i < h ( x j ) ⇐⇒ x i < x j , • ∀ i, j < ω ¬ h ( x i ) < x j . Verification of transitivity is a run through the same cases, except this time we do nothave to care if x i , x j , x k are distinct.For each class K having RSP one can prove a variant of Corollary 1. We could of coursestate a general theorem, after introducing a notion of "separable model" for arbitrary binaryrelational class. We will refrain from doing so, and provide just two such variants as anillustration. Interested reader will easily formulate corresponding results for tournaments,directed graphs, etc. Theorem 5.
Each of the following is consistent with
ZF C + M A + ”2 ω = ω ” :1. There exists a graph G of size ω , with a countable subset D ⊆ G , satisfying thefollowing properties: • For all pairs of disjoint finite subsets
A, B ⊆ G , there exists a vertex d ∈ D ,connected with each point in A , and with no point in B . • Each uncountable 1-1 function f ⊆ G × G is a graph homomorphism on anuncountable set.2. ([2], Thm. 2) There exists a separable ω -dense linear order ( L, ≤ ) of size ω , suchthat each uncountable 1-1 function f ⊆ L × L is order preserving on an uncountablesubset. The first point of this theorem is also a consequence of Theorem 2 – we can turn a metricspace into a graph by connecting two vertices iff the distance between them is ≥ . If the uncountable models we are considering are to resemble Fraïssé limits, we should beable to do some kind of "back-and-forth" arguments, like in the classical theory. There is noway inductive arguments can work, but thanks to Theorem 4 we can rely on Martin’s Axiominstead of induction. 9 .1 HSS Rectangular Metric Spaces
We will be looking at metric spaces with distances in a given countable set K ⊆ [0 , ∞ ) . Theorem 6.
Assume
M A ω . Let ( X, d ) be any rectangular HSS metric space of size ω , withdistances in K . Let Y ⊆ X be any HSS uncountable subspace. Then X and Y are isometric.Proof. Since X is hereditarily separably saturated, we can decompose it into a disjoint unionof countable saturated subsets { X α } α<ω . Of course we can do the same with Y , so let uswrite Y = [ α<ω Y α . Let P consist of finite partial isometries between X and Y , which mapelements from X α to Y α for all α < ω . It is standard to check that the following sets aredense for x ∈ X , y ∈ Y : D x = { p ∈ P | x ∈ dom p } ,E y = { p ∈ P | y ∈ rg p } . We are left with the task of verifying the c.c.c. property. Fix any uncountable subset { p γ | γ < ω } ⊆ P . Using ∆ -system Lemma we can write dom p γ = ( x , . . . , x k , x γk +1 , . . . , x γm ) , rg p γ = ( y , . . . , y k , y γk +1 , . . . , y γm ) , where tuples ( x γk +1 , . . . , x γm ) are pairwise disjoint, and moreover p γ ( x i ) = y i , and p γ ( x γi ) = y γi for each γ < ω . Recall that X is rectangular, so we can find ξ = η < ω , such that ( x ξk +1 , . . . , x ξm , y ξk +1 , . . . , y ξm ) ⊛ ( x ηk +1 , . . . , x ηm , y ηk +1 , . . . , y ηm ) . If k < i = j ≤ m , then d ( x ηi , x ξj ) = d ( x ηi , x ηj ) = d ( y ηi , y ηj ) = d ( y ηi , y ξj ) Also for k < i ≤ m d ( x ηi , x ξi ) = d ( y ηi , y ξi ) Clearly p ξ ∪ p η ∈ P . Corollary 2.
Assume
M A ω . Let ( X, d ) be any rectangular HSS metric space of size ω ,with distances in K . X can be decomposed into a disjoint union of λ many its isometriccopies for any λ ∈ { , . . . , ω } .Proof. Let X = [ α<ω X α be a decomposition of X into countable saturated subsets. Let ω = [ α<λ A α be a decomposition of ω into pairwise disjoint uncountable subsets. For any γ < λ the space X γ = [ α ∈ A γ X α is isometric to ( X, d ) .After taking X = Y and obvious adjustments to the forcing used, we obtain the classicalhomogeneity. Corollary 3.
Assume
M A ω . If ( X, d ) is any rectangular HSS metric space of size ω , withdistances in K , then any finite partial isometry of ( X, d ) extends to a full isometry. Still, this kind of homogeneity is not sufficient to prove uniqueness, like for the rationalUrysohn space.
Theorem 7.
Assume
M A ω . If there exists a rectangular HSS rational metric space of size ω , then there exist infinitely many pairwise non-isometric such spaces.Proof. If ( X, d ) is a rectangular, hereditarily separably saturated rational metric space of size ω , we introduce a family of metrics on Xd k ( x, y ) = k · d ( x, y ) , for positive integers k . If k < l are positive integers, the spaces ( X, d k ) and ( X, d l ) are notisometric. Indeed, if f : X ֒ → X is a bijection, then by Theorem 2, f is an isometry ofthe space ( X, d ) on some pair of distinct points x, y ∈ X . Then d k ( x, y ) = k · d ( x, y ) = k · d ( f ( x ) , f ( y )) < l · d ( f ( x ) , f ( y )) = d l ( f ( x ) , f ( y )) .10 .2 HSS Rectangular Graphs If we take K = { , , } , metric spaces with distances in K are graphs – think of two pointsin distance as connected, and two points in distance as not connected. Notions of saturatedset and separably saturated space translate to the following. Definition 11.
Let G be a graph. A subset D ⊆ G is a saturated subset of G if for any pairof disjoint finite subsets A, B ⊆ G , there exists d ∈ D \ ( A ∪ B ) which is connected witheach vertex in A and with no vertex in B . Definition 12. If G is any graph, then1. G is separably saturated if it has a countable saturated subset.2. G is hereditarily separably saturated (HSS) if for any countable subset E ⊆ G , thegraph G \ E is separably saturated. Proposition 11.
Fn( ω , Graphs, ω ) (cid:13) " ( ω , ˙ E ) is a rectangular HSS graph".Proof. Each infinite subset of ω which belongs to the ground model is saturated, what isclearly sufficient for being hereditarily separably saturated. Rectangularity is a consequenceof Theorem 3.From Theorem 6, with K = { , , } , it follows that HSS rectangular graphs are incertain sense minimal. Theorem 8.
Assume
M A ω . If G is a HSS rectangular graph of size ω , then each uncount-able HSS subgraph of G is isomorphic with G . Theorem 9.
Assume
M A ω . If G and H are two HSS rectangular graphs of size ω , then G ≃ H if and only if G and H c do not contain a common uncountable subgraph ( H c denotesthe complement of H ).Proof. " ⇒ ." We must show that G and G c do not contain a common uncountable subgraph. Fixarbitrary graph F of size ω , and suppose towards contradiction that there exists a pair of em-beddings i : F ֒ → G , j : F ֒ → G c . The partial function given by i ( α ) j ( α ) is a bijectionbetween uncountable subsets of G . By Proposition 9 it is a homomorphism on some pair ofpoints α = β . This contradicts the choice of i and j ." ⇐ ." Assume that there is no uncountable graph which embeds both into G and H . Weproceed like in the proof of Theorem 6. G and H are HSS, so we can decompose them intodisjoint unions of countable saturated subsets { G α } α<ω , and { H α } α<ω respectively. Let P consist of finite partial isomorphisms between G and H , which map elements from G α to H α for all α < ω . The following sets are dense for g ∈ G , h ∈ H : D g = { p ∈ P | g ∈ dom p } ,E h = { p ∈ P | h ∈ rg p } . We verify the c.c.c. property. Fix any uncountable subset { p γ | γ < ω } ⊆ P . Using ∆ -system Lemma we can write dom p γ = ( x , . . . , x k , x γk +1 , . . . , x γm ) , rg p γ = ( y , . . . , y k , y γk +1 , . . . , y γm ) , where tuples ( x γk +1 , . . . , x γm ) are pairwise disjoint, and moreover p γ ( x i ) = y i , and p γ ( x γi ) = y γi for each γ < ω . By rectangularity and Proposition 8 we can assume thattuples { ( x γk +1 , . . . , x γm ) | γ < ω } , as well as { ( y γk +1 , . . . , y γm ) | γ < ω } are pairwise alike. 11or all k < i = j ≤ m , and ξ = η < ω we have x ηi E x ξj ⇐⇒ x ηi E x ηj ⇐⇒ y ηi E y ηj ⇐⇒ y ηi E y ξj What remains is the case i = j . But look at the function given by x ηk +1 y ηk +1 , for η < ω .This is a bijection between an uncountable subgraph of G and an uncountable subgraph of H ,or equivalently, H c . By our assumption, it cannot be a homomorphism into H c . We concludethat there are η = ξ < ω such that x ηk +1 E x ξk +1 ⇐⇒ y ηk +1 E y ξk +1 , and by rectangularity x ηi E x ξi ⇐⇒ y ηi E y ξi , for all i = k + 1 , . . . , m . Clearly p ξ ∪ p η ∈ P . Corollary 4.
Assume
M A ω . If there exists a HSS rectangular graph of size ω then it isunique up to taking the complement.Proof. Fix two HSS rectangular graphs G and H , both of size ω . It is sufficient to prove thateither no uncountable graph F can be embedded both into G and H , or no uncountable graph F can be embedded both into G and H c . Suppose towards contradiction that F , F ⊆ G are uncountable subgraphs, and there exist embeddings i : F ֒ → H , i : F ֒ → H c .The function given by i ( f ) i ( f ) is a bijection between uncountable subsets of H . ByProposition 9, on some two points it must be a homomorphism. But this contradicts thechoice of i and i . Theorem 10.
Assume
M A ω . If G is a rectangular HSS graph of size ω , then either G contains an uncountable clique or uncountable anticlique.Proof. In the light of Proposition 9, G clearly can’t contain both. Assume that G doesn’tcontain an uncountable anticlique. We can represent G as ( ω , E ) , and consider the partialorder P = { F ⊆ ω | F is a finite clique in G } , ordered by reversed inclusion. If we can show that P is c.c.c, Lemma 3 will provide us withan uncountable clique in G . Suppose that { F ξ | ξ < ω } is an uncountable subset of P . Wecan write F ξ = ( f , . . . , f k , f ξk +1 , . . . , f ξm ) , where tuples ( f ξk +1 , . . . , f ξm ) are pairwise disjoint. By the virtue of Martin’s Axiom, andProposition 8, we can also assume that they are pairwise alike. By our assumption the set { f ξk +1 | ξ < ω } is not an anticlique, so we will find ξ = η < ω , such that f ηk +1 E f ξk +1 . Itis now standard to check that F η ∪ F ξ ∈ P . ω -dense Linear Orders What do rectangular linear orders look like? After unwinding the axioms for the ⊛ relation,we see that ( x , . . . , x n ) ⊛ ( y , . . . , y n ) translates to the following three axioms:A1c ∀ i, j = 1 , . . . , n ( x i < y i ⇐⇒ x j < y j ) A2c ∀ i, j = 1 , . . . , n ( x i < x j ⇐⇒ y i < y j ) A3c ∀ i, j = 1 , . . . , n ( x i < x j ⇐⇒ x i < y j ⇐⇒ y i < x j ) If we omitted A3c, we would obtain what the authors of [1] call an i ncreasing order .An order added by
Fn( ω , LO , ω ) is a rectangular, separable, ω -dense linear order, whichunder M A ω is also homogeneous. M A ω imposes a great deal of regularity on the classof separable, homogeneous ω -dense linear orders – for example two such orderings areisomorphic precisely when they are bi-embeddable. Implications of M A ω for this class, aswell as other axioms like OCA
ARS , have been extensively studied in [1]. Not surprisingly,many properties of rectangular linear orders resemble those of graphs.12 heorem 11.
Assume
M A ω . Let L be a separable, ω -dense, rectangular linear order. If K is any other separable, ω -dense, rectangular linear order, then K ≃ L if and only if K and L ∗ do not contain a common uncountable suborder ( L ∗ denotes L with the reversedordering).Proof. " ⇒ ." We must show that there is no uncountable strictly decreasing function f ⊆ L × L .But this follows directly from Proposition 9." ⇐ ." Assume that there is no uncountable linear order which embeds both into L and L ∗ .We proceed like in the proofs of Theorems 6 and 9. We can decompose L and K into disjointunions of countable dense subsets { L α } α<ω , and { K α } α<ω respectively. Let P consist offinite partial isomorphisms between L and K which map elements from K α to L α , for all α < ω . The following sets are dense for l ∈ L , k ∈ K : D k = { p ∈ P | k ∈ dom p } ,E l = { p ∈ P | l ∈ rg p } . We verify the c.c.c. property. Fix any uncountable subset { p γ | γ < ω } ⊆ P . Using ∆ -system Lemma, we can write dom p γ = ( x , . . . , x k , x γk +1 , . . . , x γm ) , rg p γ = ( y , . . . , y k , y γk +1 , . . . , y γm ) , where tuples ( x γk +1 , . . . , x γm ) are pairwise disjoint, and moreover p γ ( x i ) = y i , and p γ ( x γi ) = y γi , for each γ < ω . By rectangularity and Proposition 8 we can assume, that tuples { ( x γk +1 , . . . , x γm ) | γ < ω } , as well as { ( y γk +1 , . . . , y γm ) | γ < ω } , are pairwise alike.For all k < i = j ≤ m , and ξ = η < ω we have x ηi < x ξj ⇐⇒ x ηi < x ηj ⇐⇒ y ηi < y ηj ⇐⇒ y ηi < y ξj What remains is the case i = j . Look at the function given by x ηk +1 y ηk +1 , for η < ω .This is a bijection between an uncountable subset of L and an uncountable subset of K . Byour assumption it cannot be decreasing, so there are η = ξ < ω such that x ηk +1 < x ξk +1 ⇐⇒ y ηk +1 < y ξk +1 , and by rectangularity x ηi < x ξi ⇐⇒ y ηi < y ξi , for all i = k + 1 , . . . , m . Clearly p ξ ∪ p η ∈ P .Just like in the case of graphs, one can easily prove Corollary 5.
Assume
M A ω . If there exists a rectangular, separable, ω -dense linear orderthen it is unique up to reversing the order. These orders are also mimimal in the same sense as metric spaces from Theorem 6. Thereader will have no difficulty in adjusting its proof to obtain
Theorem 12.
Assume
M A ω and let K , L be a pair of separable, ω -dense linear orders. If K embeds into L , and L is rectangular, then L and K are isomorphic. eferences [1] U. Avraham, M. Rubin, S. Shelah, On the consistency of some partition theorems for continuouscolorings, and the structure of ℵ -dense real order types , Annals of Pure and Applied Logic 29(1985), 123-206[2] U. Avraham, S. Shelah, Martin’s Axiom does not imply that every two ℵ -dense sets of reals areisomorphic , Israel Journal of Mathematics, vol. 38, Nos. 1-2, 1981[3] J.E. Baumgartner, All ℵ -dense sets of reals can be isomorphic , Fund. Math. 79 (1973), 101-106[4] R. Fraïssé, Sur quelques classifications des systémes de relations , Publ. Sci. Univ. Alger. Sér. A.1 (1954)[5] Z. Kostana,
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