aa r X i v : . [ m a t h . L O ] S e p Cohen-like first order structures
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] 22, 2020
Abstract
We study forcing notions similar to the Cohen forcing, which add some struc-tures in given first-order language. These structures can be seen as versions ofuncountable Fraïssé limits with finite conditions. Among them, we are primarilyinterested in linear orders.
Keywords:
Cohen forcing, Fraisse limit, homogeneous structure, generic structure
MSC classification:
As one looks at the classical construction of Fraïssé limit, described for instance in [7]or [8], one might notice that it is much in the spirit of the Baire Theorem. Namely,we show the existence of a universal homogeneous structure by proving that almostany , in a suitable sense, countable structure is universal and homogeneous. In fact,universal homogeneous structures form a residual set in certain Polish space. Havingthat in mind, one might try to construct specific instance of a universal homogeneousstructure, mimicking the definition of a Cohen real from the forcing theory. Roughlyspeaking, a real number is Cohen over some model if it belongs to each residual setfrom that model. So it is very generic , in a sense that for any typical property a realmight have, the Cohen real has this property. Of course the same can be said aboutrandom numbers, but with different notion of typicality. This is the idea behind thiswork. From one side, we want to look at the model theoretic notion of saturation asstemming from the forcing language. From the other, we reach to model theory fortools to produce Cohen-like forcing notions (which might often be just different incar-nations of the Cohen forcing). ∗ Research of Z. Kostana was supported by the GAˇCR project EXPRO 20-31529X and RVO: 67985840. random structure, in addition to being a generic one?This is of course a very vague question, and it is not even clear what the measure spaceunder consideration should be. This idea was undertaken by Petrov and Vershik forgraphs [11], and extended to other structures by Ackerman, Freer, and Pater [1]. Theyobtain an elegant internal characterization of Fraïssé classes for which the Fraïssé limitis a structure appearing with probability one in certain probability measure space. Thishappens precisely in the case of Fraïssé classes in purely relational language with theStrong Amalgamation Property (Definition 1 here). The reader is encouraged to con-sult [1] for the precise formulation.We assume the reader is familiar with the basics of forcing theory, the level of [9]will just do, and the very basics of model theory. In the first section we develop thelanguage, and prove, or just state, some general properties of the forcings under consid-eration. In the second section, we prove that unlike ordinary Fraïssé limits, uncountablestructures of this kind tend to be rigid. The third section is devoted to the constructionof an uncountable real order type, with ( Z , +) as the group of automorphisms. In thelast, fourth section, we collect some open questions, which look relevant for this line ofresearch. Finally, it should be mentioned that this topic used to be informally discussedfrom time to time already, as kind of folklore idea known to the community. However,up to the author’s knowledge, no systematic study of this idea was ever carried out. Theclosest to it was perhaps a note by M. Golshani [6], unfortunately not free of significantmistakes.As an initial example, look at the following poset. P = { ( A, ≤ ) | A ∈ [ κ ] <ω , A is a linear order } , where κ is any cardinal, and the ordering is the reversed inclusion. The followingsubsets are dense, for α = β ∈ κ . • D α = { ( A, ≤ ) | α ∈ A } , • D α,β = { ( A, ≤ ) | ∃ n < ω n is between α and β } ,Therefore, for κ = ω , the generic filter produces an isomorphic copy of rationals,and for any κ it gives some separable κ -dense order type. We say that a linear orderis κ -dense, if every open interval has cardinality κ .It is a general phenomenon that for κ = ω this forcing gives the Fraïssé limit of the given class. It is an interesting remark,made by M. Golshani in [6], that every infinite subset of ω from the ground model isdense in the obtained structure.For this section I adopt the convention that boldface letters A , B denote first-orderstructures, while the corresponding capital letters A , B denote underlying sets. Infurther sections I will denote structures and underlying sets with the same letters, ascommon in mathematics.In the whole paper K is a class of structures in some countable, relational, first-order language. By K κ we denote the class of structures from K of cardinality less2han κ . Relational means in particular that we do not allow constants in our language.We make the following assumptions on K : • K has Joint Embedding Property (JEP), • K Amalgamation Property (AP), • K is hereditary, so if A ∈ K , and B ⊆ A , then B ∈ K , • K has infinitely many isomorphism types, • K κ is closed under increasing unions of length < κ .For all undefined notions, we refer reader to [7] or [8]. Note, that K ω is a Fraïsséclass, and if κ <κ = κ , for some uncountable cardinal κ , then K κ is an uncountableFraïssé class in the sense of [8]. In the subsequent part of the paper we are going tomake more assumptions on K , like the Strong Amalgamation Property (SAP), recalledbelow.
Definition 1.
A Fraïssé class F has the Strong Amalgamation Property if for anystructures a, b, c ∈ F and embeddings f : a ֒ → b , g : a ֒ → c , there exists d ∈ F ,together with embeddings f ′ : b ֒ → d , g ′ : c ֒ → d , satisfying f ′ ◦ f = g ′ ◦ g , andmoreover rg f ′ ∩ rg g ′ = rg f ′ ◦ f . The Strong Amalgamation Property for a Fraïssé class F , with the Fraïssé limit F , corresponds to a certain property of F . The structure F has no algebraicity if foreach subset A ⊆ F , and each f ∈ F \ A , f has infinite orbit under the action of thepointwise stabilizer of A in Aut F . Let us recall Theorem 7.1.8 from [7]. Theorem 1.
Let F be a Fraïssé class with the Fraïssé limit F . The following areequivalent.1. F has SAP2. F has no algebraicity It will prove convenient to introduce a notation paraphrasing the notation for theCohen forcing in [9].
Definition 2.
Let λ be an infinite cardinal number, and S be any infinite set. Denoteby Fn( S, K , λ ) the set { A ∈ K| A ∈ [ S ] <λ } , ordered by the reversed inclusion. The following claims hold true, and proofs are straightforward modifications ofanalogous results for the Cohen forcing [9].
Proposition 1. If K satisfies SAP, and K ω has at most countably many isomorphismtypes, then Fn( S, K , ω ) satisfies c.c.c., and even the Knaster’s condition, for any set S . K is a class of structures in a finite language. When the language is countable, itmay or may not be true. Finite metric spaces can be viewed as structures in countablelanguage, and still there are continuum many pairwise non-isomorphic (non-isometric)2-element structures. If we restrict to finite metric spaces with rational distances, thereare clearly only countably many isomorphism types. The relevance of SAP is visible inthe example discovered by Wiesław Kubi´s. Let F be the class of all finite linear graphs,i.e. connected, acyclic, and with degree of every vertex at most 2. It can be easilychecked that F has AP, but not SAP. If S is any infinite set, then Fn( S, F , ω ) forcesthat S is a linear graph, and each two points of S are in a finite distance. Therefore itcollapses | S | to ω . Proposition 2.
Let S be any set, and assume K λ satisfies SAP. We assume moreover,that for any δ < λ there are at most λ many structures from K , with the universe δ .Then Fn( S, K , λ ) is λ -closed, and if λ <λ = λ , then Fn( S, K , λ ) is λ + -c.c. It should be stressed out that in this proposition we don’t count isomorphic types of K -structures of cardinality less than λ . We take into account the number of different,not only non-isomorphic, ways the ordinal δ can be endowed with a first-order struc-ture, so that it becomes a member of K . In all but one relevant examples, a bound onthis number will be guaranteed by finite language. Corollary 1. If K is a class of structures in a finite language, and CH holds, then Fn( S, K , ω ) is ω -c.c. For start we describe structures added by
Fn( S, K , ω ) . Proposition 3.
Let P = Fn( ω, K , ω ) . Let G ⊆ P be a generic filter. Then S G is astructure with the universe ω , isomorphic to the Fraïssé limit K of the class K ω .Proof. In order to ensure that S G is defined on all ω , we must verify density of thesets D n = { A ∈ P | n ∈ A } , for n < ω , which is straightforward. To see that we obtain the Fraïssé limit we mustcheck that each finite extension of a finite substructure is realized. For this purpose, set E i,f B = { A | i : B ֒ → A is an embedding = ⇒ ∃ g : B ′ ֒ → A g is an embedding, and i = g ◦ f } , where B , B ′ ∈ K , f : B ֒ → B ′ is an embedding, and i : B ֒ → ω is any − function.We also make a technical assumtion that both B and B ′ are disjoint from ω . One coulddirectly apply AP to show that the sets E i,f B are dense, however it may be easier tomake use of a simple trick. This trick is due to Kubi´s.Fix a structure A ∈ P , and assume that i , B , B ′ , f are as above. Since A ⊂ ω , wemay extend A to a structure Ω , isomorphic to K , with the universe ω . Then, since thisstructure is injective, there exists g : B ′ → Ω , such that i = g ◦ f . Clearly if we define A ′ = A ∪ g [ B ′ ] ⊆ Ω , then A ′ ∈ E i,f B .The proof that the generic structure is universal is left to the reader.Note that we used only countably many dense subsets of P , so the Propositionworks under Rasiowa-Sikorski Lemma, without requiring G being "generic" in thesense of the forcing theory. 4 Results about rigidity
The generic structure added by
Fn( ω, K , ω ) is homogeneous, so it can be of somesurprise, that forcing on uncountable set gives rise to a rigid structure, at least in most ofthe cases. This is obviously not true for example if K is the class of all finite sets, but itseems to be true in all sufficiently nontrivial cases. This is proved in the first subsection.In the second, we study linear orders added by forcing with countable support, andshow that they are not only rigid, but also remain so in any generic extension via ac.c.c. forcing. Note that this is in contrast with the "finite-support-generic" linear orderssince, as proved by Baumgartner [4], under CH we can add a nontrivial automorphismto any ω -dense separable linear order, using a c.c.c. partial order. Recall that a linearorder is ω -dense, if every open interval has cardinality ω . Fn( ω , K , ω ) We prove that the uncountable partial order and the uncountable undirected graphadded by the forcing
Fn( ω , K , ω ) are rigid. Proofs for linear orders, directed graphs,tournaments or finite rational metric spaces are all easy modifications of either of these. Theorem 2.
Let F be the class of (undirected) graphs, and S be an uncountable set.Then the generic graph added by Fn( S, F , ω ) is rigid. Assume that p (cid:13) ” ˙ h : ( S, ˙ E ( S )) → ( S, ˙ E ( S )) is a non-identity isomorphism ” .It is easy to check that for every infinite set F ⊆ S from the ground model, andevery two different s, t ∈ S , there exists a vertex e ∈ F , with { s, e } ∈ E ( S ) , and { t, e } / ∈ E ( S ) . There are clearly uncountably many pairwise disjoint, infinite subsetsof S in the ground model, so h must be non-identity on each of them. Therefore thereexists an uncountable set { p s | s ∈ S ′ ⊆ S } of conditions stronger than p , with p s (cid:13) ˙ h ( s ) = s = s. Without loss of generality we can assume that { p s | s ∈ S ′ } form a ∆ -system with aroot R , disjoint with S ′ , and the graph structures of all p s agree on the root.Fix two different s, t ∈ S ′ . We can amalgamate p s , and p t over R in such a way,that { s, t } ∈ E ( S ) , and { s, t } / ∈ E ( S ) , obtaining some stronger condition q ∈ Fn( S, F , ω ) . But then q forces, that ˙ h is not a graph homomorphism. Theorem 3.
Let F be the class of partial orders, and S be an uncountable set. Thenthe generic partial order added by Fn( S, F , ω ) is rigid.Proof. Assume that p (cid:13) ” ˙ h : ( S, ˙ ≤ ) → ( S, ˙ ≤ ) is a non-identity isomorphism ” . It iseasy to check that for every infinite set E ⊆ S from the ground model, Fn( S, F , ω ) (cid:13) ” E is strongly dense ” . Strongly dense means that for every s < t ∈ S , there exists e ∈ E , such that s < e < t , and for every s, t ∈ S incomparable, there exists e i ∈ E , i = 0 , , , , , with e > s , e incomparable with t , e < s, t ; e < s ,incomparable with t ; e > s, t , and e incomparable with both s and t . Long storyshort, each type with parameters (not necessarily from E ) is realized in E . There areclearly uncountably many pairwise disjoint, infinite subsets of S in the ground model,5nd h must be non-identity on each of them. Therefore there exists an uncountable set { p s | s ∈ S ′ ⊆ S } of conditions stronger than p , and p s (cid:13) ˙ h ( s ) = s = s. Without loss of generality we can assume that { p s | s ∈ S ′ } form a ∆ -system with aroot R , disjoint with S ′ , and the order structures of all p s agree on the root. Supposealso, that for each s ∈ S ′ , s > s (the other cases are handled similarily). Since S ′ isuncountable, we can further thin it out, so that all embeddings of the form R ⊂ R ∪{ s } are pairwise isomorphic, and similarly for s . Recall that two extensions of a givenstructure R are isomorphic if there is an isomorphism between them, which is identityon R .Fix two different s, t ∈ S ′ . There exists an extension R ⊂ R ∪ { s, t, s, t } , with { s < t < t < s } . We can amalgamate p s ∪ { t < t } and p t ∪ { s < s } over R ∪ { s < t < t < s } , to obtain some condtion q ∈ Fn( S, F , ω ) . But then q (cid:13) s < t , and q (cid:13) ˙ h ( s ) > ˙ h ( t ) , exhibiting the contradiction.It is worth to remark that uncountable linear ordering added this way satisfies somestrong variant of rigidity. Following [2] and [3], we say that an uncountable separablelinear order ( L, ≤ ) is k -entangled, for some k ∈ N , if for every tuple t ∈ { T, F } k , andany family { ( a ξ , . . . , a ξk − ) | ξ < ω } of pairwise disjoint k -tuples from L , one canfind ξ = η < ω , such that for i = 0 , . . . , k − a ξi ≤ a ηi iff t ( i ) = T . This in particularimplies that no two uncountable, disjoint subsets of L are isomorphic. Property ofbeing k -entangled for all natural k is featured for example by uncountable set of Cohenreals, added over some model. Martin’s Axiom with negation of CH implies that nouncountable set of reals is k -entangled for all k [2]. Fn( ω , LO , ω ) We will prove that under CH forcing with countable supports on the set of biggercardinality gives rise to a rigid linear order, for which we cannot add an automorphismusing a c.c.c. forcing. This result holds under CH, however the c.c.c.-absolute rigidityis clearly preserved by any c.c.c. forcing. In effect, existence of a rigid ω -dense linearorder is consistent with any possible value of ω , and for example M A + 2 ω = κ ,for any κ = κ <κ . Also we can’t replace ω with ω in results of this section. UnderCH there exists a unique ω -saturated linear order of cardinality ω and as such, it issurely not rigid. However Fn( ω , LO , ω ) forces that the generic order is ω -saturatedof cardinality ω , for the same reasons that Fn( ω, LO , ω ) forces the generic order tobe ω -saturated (i.e. dense, without endpoints).6 heorem 4. Let P = Fn( ω , LO , ω ) , where LO denotes the class of all linear or-ders. Let ( ω , ≤ ) be a generic order added by P over a countable, transitive model V ,satisfying CH. Denote by V [ ≤ ] the corresponding generic extension. Let Q ∈ V [ ≤ ] be any forcing notion, such that V [ ≤ ] | = " Q is c.c.c." , and H be a Q -generic filter in V [ ≤ ] . Then the linear order ( ω , ≤ ) is rigid in V [ ≤ ][ H ] . We will use a simple Lemma assuring that we can amalgamate linear orders in asuitable way.
Lemma 1.
Let ( L , ≤ ) , ( L , ≤ ) be any linear orders, R = L ∩ L ,and ( R, ≤ ) = ( R, ≤ ) . There exists a linear order ≤ on L ∪ L , extending both ≤ and ≤ and satisfying ∀ l ∈ L \ R ∀ l ∈ L \ R l < l ⇐⇒ ∃ r ∈ R l < r < l . Proof.
We take the above formula as the definition.
Lemma 2.
Let P = Fn( ω , K , ω ) , P (cid:13) " ˙ Q is a c.c.c. forcing notion" , and assumethat P ∗ ˙ Q (cid:13) h : ω → ω is a bijection . Then for every p ∈ P exists p c ≤ p with the property that ( p c , ˙ Q ) (cid:13) h [ p c ] = p c .Proof. Let { F n } n<ω be a partition of ω into infinite sets, such that ∀ n < ω n ≥ min F n . We define a sequence of conditions p n ∈ P by induction, startingwith p = p . Enumerate p = { r n | n ∈ F } . Suppose we have p n defined. We maytake a sequence of P -names with the property p n (cid:13) " { ˙ q kn +1 } k<ω is a maximal antichain deciding h ( r n ) " . Since P is σ -closed, we will find p ′ n ≤ p n deciding all the names ˙ q kn +1 for k < ω .Therefore the set A = { β < ω | ∃ k < ω ( p ′ n , ˙ q kn +1 ) (cid:13) h ( r n ) = β } is at mostcountable. Let p n +1 = p ′ n ∪ A (with relations defined arbitrarily), and enumerate p n +1 = { r k | k ∈ F n +1 } . The inductive step is completed.Take p c = S n<ω p n . We will show that for any ˙ q , with p c (cid:13) ˙ q ∈ ˙ Q , and any α ∈ p c , ( p c , ˙ q ) (cid:13) h ( α ) ∈ p c . Indeed, in this situation there is some n < ω such that α ∈ p n .Therefore we will find k < ω with α = r k , k ∈ F n . In the k -th indutive step we ensurethat ( p k +1 , ˙ q ) (cid:13) h ( r k ) ∈ p k +1 . It follows that ( p c , ˙ q ) (cid:13) h ( r k ) ∈ p c . Proof.
Work in V . Let ˙ ≤ be a P -name for ≤ . Suppose that P (cid:13) " ˙ Q is c.c.c." , and P ∗ ˙ Q (cid:13) " h : ( ω , ˙ ≤ ) → ( ω , ˙ ≤ ) is a non-identity isomorphism". Step 0
It can be easily verified, that if h was identity on a dense set, then it would beidentity everywhere. Therefore there exist P ∗ ˙ Q -names δ , δ , such that P ∗ ˙ Q (cid:13) δ ˙ <δ , ∀ x ∈ ( δ , δ ) h ( x ) = x. ( p, ˙ q ) ∈ P ∗ ˙ Q deciding δ and δ , i.e. ( p, ˙ q ) (cid:13) ∀ i ∈ { , } δ i = δ i , for some δ i ∈ ω . Without loss of generality, we can assume that ˙ q is the greatest element of ˙ Q ,so that ( p, ˙ Q ) (cid:13) δ ˙ <δ , ∀ x ∈ ( δ , δ ) h ( x ) = x. Step 1
For α ∈ ω \ { δ , δ } we fix a condition p α = ( p α , ≤ α ) ≤ p , with δ < α α < α δ . Take a sequence of names satisfying p α (cid:13) " { ˙ q nα } n<ω is a maximal antichain deciding h ( α ) " . Since P is σ -closed, we can assume that p α decides all the names ˙ q nα , so the set F ( α ) = { β < ω | ∃ n < ω ( p α , ˙ q nα ) (cid:13) h ( α ) = β } is countable. Note, that since p α ≤ p , α / ∈ F ( α ) . Finally, we can assume that F ( α ) ⊆ p α , and, due to Lemma 2, that ( p α , ˙ Q ) (cid:13) h [ p α ] = p α . Step 2
Using ∆ -Lemma for countable sets, we can find I ⊆ ω of cardinality ω ,with the following conditions satisfied • ∀ α ∈ I ∀ β ∈ I β = α = ⇒ p α ∩ p β = R , for some fixed countable R ⊆ ω , • ∀ α ∈ I ∀ β ∈ I ≤ α ↾ R × R = ≤ β ↾ R × R , • extensions R ⊂ R ∪ { α } , for α ∈ I , are pairwise isomorphic, • ∀ α ∈ I ( p α , ˙ Q ) (cid:13) h [ R ] = R .All these conditions, perhaps excluding the last one, are direct consequenes of CH. Tojustify the last claim, notice that P ∗ ˙ Q is ω -c.c. and so the set A = { β < ω | ∃ ( p, ˙ q ) ∈ P ∗ ˙ Q ∃ r ∈ R ( p, ˙ q ) (cid:13) h ( r ) = β } has cardinality at most ω . We choose to { p α | α ∈ I } only conditions with ( p α \ R ) ∩ A = ∅ . Take r ∈ R . ( p α , ˙ Q ) (cid:13) h ( r ) ∈ p α ∩ A ⊆ R . Step 3
Take α, β ∈ I , α = β . Using the fact that the extensions R ⊆ R ∪ { α } and R ⊆ R ∪ { β } are isomorphic, we can extend ≤ α = ≤ β on R to ( R ∪ { α, β } , ≤ α,β ) insuch a way that there is no element from R between α and β . We can of course decidethat α < α,β β . We now apply Lemma 1 to the pair of isomorphic extensions p α ∪ { β } R ∪ { α, β } p β ∪ { α } where the vertical arrow maps β to α . 8xtend ≤ α,β to p α ∪ p β , ensuring that • ¬∃ r ∈ R α < α,β r < α,β β ; • ∀ γ ∈ p α \ ( R ∪ { α } ) ∀ η ∈ p β \ ( R ∪ { β } ) γ < α,β η ⇐⇒ ∃ r ∈ R γ < α,β r < α,β η. Take some condition r ≤ p α,β and ˙ q deciding the values of h ( α ) and h ( β ) . Then ( r, ˙ q ) (cid:13) h ( α ) = h ( α ) , h ( β ) = h ( β ) . Since there is no element from R between α and β , and R is h invariant, there is also no element from R between h ( α ) and h ( β ) . But since h ( α ) ∈ p α \ { α } , and h ( β ) ∈ p β \ { β } , h ( β ) < α,β h ( α ) . Therefore ( r, ˙ q ) (cid:13) h ( α ) > h ( β ) , giving rise to a contradiction. This finishes the proof. T. Ohkuma proved in [10] that there exist c pairwise non-isomorphic groups ( G, +) ≤ ( R , +) , with the property that Aut ( G, ≤ ) ≃ ( G, +) , meaning that G has no order-automorphisms other that translations. These groups all have cardinality c , however theauthors of [5] have shown that consistently there are uncountable groups of cardinalityless than c with this property. These are examples of separable, uncountable linearorders, with few, but more than one, automorphisms. We are going to provide onemore construction in this spirit. Theorem 5.
It is consistent that there exists an ω -dense real order type ( A, ≤ ) witha non-identity automorphism φ , such that Aut ( A, ≤ ) = { φ k | k ∈ Z } . Moreover, φ satisfies φ ( x ) > x for all x ∈ A . Let < ord denote the usual order on ω . The promised modification of Fn( ω , LO , ω ) is the poset P consisting of triples p = ( p, ≤ p , φ p ) satisfying1. ≤ p is a linear ordering of p ∈ [ ω ] <ω ,2. φ p is an increasing bijection between two subsets of p ,3. ∀ x ∈ dom p x < p φ p ( x ) ,4. ∀ x ∈ dom p φ ( x ) < ord x + ω , with respect to the ordinal addition on ω ,5. ∀ x ∈ rg p φ − ( x ) < ord x + ω , with respect to the ordinal addition on ω .We denote by ( ω , ≤ ) the ordering added by P , and by φ the corresponding au-tomorphism. Before proceeding with the main proof we will see that it is possible toamalgamate finite linear orders together with partial automorphisms in a desired way.It will be convenient to denote by Part ( L, ≤ ) the set of finite, partial automorphismsof a linear order ( L, ≤ ) . Lemma 3.
Let ( L , ≤ ) , ( L , ≤ ) and ( R, ≤ R ) = ( L , ≤ ) ∩ ( L , ≤ ) be finite lin-ear orders. Fix partial automorphisms φ ∈ Part ( L , ≤ ) , φ ∈ Part ( L , ≤ ) . Weassume that ( L , φ ) and ( L , φ ) are isomorphic extensions of R , in a sense that theymake the diagram below commutative. L , ≤ ) ( L , ≤ )( R, ≤ R ) ( L , ≤ ) ( L , ≤ ) h φ hφ Take a, b ∈ L \ R lying in different orbits of φ . There exists a linear order ≤ c on L ∪ L extending ≤ and ≤ , and such that φ ∪ φ ∈ Part ( L ∪ L , ≤ c ) , andmoreover a < c h ( a ) , and h ( b ) < c b .Proof. We can assume that R ⊆ L ⊆ Q , and the usual ordering of ( Q , ≤ ) extends ≤ . We look for an increasing function f : ( L , ≤ ) → ( Q , ≤ ) such that f ↾ R = id R , f [ L \ R ] ∩ ( L \ R ) = ∅ , and f ◦ h ( a ) > a,f ◦ h ( b ) < b. Indeed, having f as above we will define x < c y ⇐⇒ x < f ( y ) , for x ∈ L and y ∈ L .It can be seen that the only reason why we can’t take f = h − is the disjointnessrequirement. So we should expect that f will be just a slight distortion of h − . Wemust also ensure that φ ∪ φ will be order-preserving.Let { x , . . . , x n } be a ≤ -increasing enumaration of L . For k = 1 , . . . , n choosean open interval I k around x k in such a way that all intervals obtained this way arepairwise disjoint, and for l = k = 1 , . . . , n if x l = φ m ( x k ) , then φ m [ I l ] = I k , where φ : ( Q , ≤ ) → ( Q , ≤ ) is an extension of φ .For each k we choose f ( h ( x k )) ∈ I k \{ x k } , so that φ m ( f ◦ h ( x k )) = f ◦ h ◦ φ m ( x k ) , for m ∈ Z , whenever this expression makes sense. We also ensure inequalities f ◦ h ( a ) >a and f ◦ h ( b ) < b . Proposition 4. P satisfies the Knaster’s condition.Proof. Let { p α = ( p α , ≤ α , φ α ) | α < ω } ⊆ P . We choose a ∆ -system { p α | α ∈ S } ,with some additional properties: • ∀ α ∈ S ∀ β ∈ S α = β = ⇒ ( p α , ≤ α ) ∩ ( p β , ≤ β ) = ( R, ≤ R ) , for some fixedordering ≤ R of R , • φ α [ R ] ⊆ R , • φ − α [ R ] ⊆ R . 10or ensuring the last two properties we use 4. and 5. from the definition of P . Toobtain an uncountable set of pairwise comparable conditions, we now only have to trim { p α | α ∈ S } , so that φ α ↾ R does not depend on α , and this is clearly possible. Lemma 4.
For every α ∈ ω , the orbit of α under φ is cofinal and coinitial in ( ω , ≤ ) Proof.
It is easy to see that the required family of dense sets is E β = { p = ( p, ≤ p , φ p ) ∈ P | { α , β } ⊆ p, ∃ k ≥ β < p φ kp ( α ) , φ − kp ( α ) < p β } , for β ∈ ω .In order to check, that E β is dense, fix some condition p = ( p, ≤ p , φ p ) ∈ P and β < ω . We can assume that { α , β } ⊆ p . In order to extend p so that it belongs to E β ,we embed ( p, ≤ p ) into the set of algebraic numbers A . Now we can extend φ p to anincreasing function φ : A → A , such that for some rational ǫ > ∀ a ∈ A φ ( a ) > a + ǫ .It is clear that the orbit of α under φ is both cofinal and coinitial in A . Finally we justcut out a suitable finite fragment of φ , and extend p accordingly. Lemma 5.
For each isomorphism h : ( ω , ≤ ) → ( ω , ≤ ) , and for every uncountableset F ⊆ ω , there exist α ∈ F and k ∈ Z , such that h ( α ) = φ k ( α ) .Proof. Fix a sequence of names for elements of F , { ˙ x α | α < ω } . Let p (cid:13) ˙ h : ( ω , ˙ ≤ ) → ( ω , ˙ ≤ ) is an isomorphism . For every α < ω we fix a condition p α = ( p α , ≤ α , φ α ) ≤ p , so that p α (cid:13) ˙ x α = x α , ˙ h ( x α ) = x α , for some ordinals x α , x α ∈ ω . We can also assume that for each α , x α = x α , for otherwise we just take k = 0 .We choose an uncountable ∆ -system { p α | α ∈ S } , and make it as uniform as possible: • ∀ α ∈ S ∀ β ∈ S α = β = ⇒ ( p α , ≤ α ) ∩ ( p β , ≤ β ) = ( R, ≤ R ) , for some fixedordering ≤ R of R , • φ α [ R ] ⊆ R , • φ − α [ R ] ⊆ R , • extensions ( R, ≤ R ) ⊆ ( R ∪ { x α } , ≤ α ) are pairwise isomorphic, • extensions ( R, ≤ R ) ⊆ ( R ∪ { x α } , ≤ α ) are pairwise isomorphic, • extensions ( R, ≤ R ) ⊆ ( p α , ≤ α ) are pairwise isomorphic, • The way φ α acts on p α is independent from the choice of α ∈ S . More precisely, ∀ α ∈ S ∀ β ∈ S the following diagram commutes p α p α p β p βφ α h hφ β where h is the unique isomorphism between ( p α , ≤ α ) and ( p β . ≤ β ) .11n particular, the unique isomorphism h maps x α to x β , and x α to x β . Fix α ∈ S . Weclaim that x α and x α are in the same orbit of φ α . For otherwise, we fix β ∈ S \ { α } ,and apply Lemma 3 for a = x α and b = x α . This way we obtain a condition q = ( p α ∪ p β , ≤ q , φ α ∪ φ β ) ≤ p α , p β , satisfying x α < q x β , and x β < q x α . But then q (cid:13) ˙ x α ˙ < ˙ x β , ˙ h ( ˙ x β ) ˙ < ˙ h ( ˙ x α ) , contrary to the choice of p . In conclusion p α (cid:13) ∃ k ∈ Z ˙ h ( ˙ x α ) = ˙ φ αk ( ˙ x α ) .Now we are in position to prove Theorem 3. Proof.
Since ( ω , ≤ ) is separable, we can replace it by an isomorphic copy A ⊆ R , A being ω -dense. Then φ : A → A is an increasing bijection, strictly above thediagonal. Let h : A → A be any increasing bijection. Both h and φ extend uniquelyto the whole real line, so we can assume that φ, h : R → R are continuous, increasingbijections.For k ∈ Z , put F k = { x ∈ R | h ( x ) = φ k ( x ) } . By continuity, sets F k are closed,and by Lemma 5, S i ∈ Z F i is dense. Fix some k ∈ Z for which the set F k is nonempty.We aim to prove that F k = R . If not, there exists x ∈ F k , and δ > satisfying at leastone of conditions ( x, x + δ ) ∩ F k = ∅ , and ( x − δ, x ) ∩ F k = ∅ . Assume the first case, the other being similar. Since the union of the sets F i isdense, we can find a decreasing sequence { x n } n<ω , converging to x , and integers k n ,for which h ( x n ) = φ k n ( x n ) .Suppose that for infinitely many n , the inequality k n > k holds. By replacing { x n } n<ω with a subsequence, we may assume that this is the case for all n < ω . Then φ k n ( x n ) ≥ φ k +1 ( x n ) −→ n →∞ φ k +1 ( x ) > φ k ( x ) = h ( x ) , which contradicts lim n →∞ φ k n ( x n ) = h ( x ) .If for infinitely many n the inequality k n < k holds, we proceed in analogous manner.The only way out is k n = k for all but finitely many n , but this in turn contradicts ( x, x + δ ) ∩ F k = ∅ . Therefore F k = R , and the theorem is proved. Question 1.
Are there some natural conditions for a class F , that ensure that Fn( S, F , ω ) is, up to completion, the same as the Cohen forcing? uestion 2. Theorem 4 gives an example of a linear order, whose rigidity is preservedin c.c.c. forcing extensions. Is it possible for a separable linear order to have thisproperty? By results of [4] this is clearly not possible when c = ω . Question 3.
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