aa r X i v : . [ m a t h . L O ] O c t GALVIN’S QUESTION ON NON- σ -WELL ORDEREDLINEAR ORDERS HOSSEIN LAMEI RAMANDI
Abstract.
Assume C is the class of all linear orders L such that L is not a countable union of well ordered sets, and every uncountablesubset of L contains a copy of ω . We show it is consistent that C has minimal elements. This answers an old question due to Galvinin [3]. Introduction
A linear order L is said to be σ -well ordered if it is a countableunion of well ordered subsets. Galvin asked whether or not every non- σ -well ordered linear order has to contain a real type, Aronszajn type,or ω ∗ . Baumgartner answered Galvin’s question negatively by provingthe following theorem. Theorem 1.1 ([3]) . There are non- σ -well ordered linear orders L suchthat every uncountable suborder of L contains a copy of ω . Recall that a linear order L is said to be a real type, if it is isomorphicto an uncountable set of real numbers. An uncountable linear order L is said to be an Aronszajn type, if it does not contain any real type orcopies of ω , ω ∗ . Here ω ∗ is ω with the reverse ordering.Let C be the class of all non- σ -well ordered linear orders L such thatevery uncountable suborder of L contains a copy of ω . Note that theelements in C together with real types, Aronszajn types, and ω ∗ form abasis for the class of non- σ -well ordered linear orders. Buamgartner’stheorem asserts it is essential to include C in this basis.In the final section of [3], Baumgartner mentions the following ques-tion which is due to Galvin. Question 1.2 ([3], Problem 4) . L ∈ C is said to be minimal providedthat whenever L ′ ⊂ L , | L ′ | = | L | and L ′ ∈ C then L embeds into L ′ .Does C have minimal elements? Key words and phrases. trees, linear orders, σ -well ordered, σ -scattered . Before we answer Question 1.2, we discuss the motivation behind thisquestion. The following two deep theorems are about the minimalityof non- σ -well ordered order types. Theorem 1.3 ([1]) . Assume MA ω . Then it is consistent that there isa minimal Aronszajn line. Theorem 1.4 ([2]) . Assume
PFA . Then every two ℵ -dense subsetsof the reals are isomorphic. In particular, these theorems show it is consistent that real types andAronszajn types have minimal elements. It is trivial that ω ∗ is a min-imal non- σ -well ordered linear order as well. So it is natural to askwhether or not C can have minimal elements. A consistent negativeanswer to Question 1.2 is provided in [4]. Theorem 1.5.
Assume
PFA + . Then every minimal non- σ -scatteredlinear order is either a real type or an Aronszajn type. Recall that a linear order L is said to be scattered if it does not containa copy of ( Q , ≤ ). L is called σ -scattered if it is a countable union ofscattered suborders. Theorem 1.5 provides a consistent negative answerto Question 1.2 because of the following fact: a linear order L is σ -wellordered if and only if L is σ -scattered and ω ∗ does not embed into L . In this paper we provide a consistent positive answer to Question 1.2by proving the following theorem.
Theorem 1.6.
Assume C is the class of all non- σ -well ordered linearorders L such that every uncountable suborder of L contains a copy of ω . Then it is consistent with ZFC that C has a minimal element ofsize ℵ . This theorem should be compared to the following theorem from [6].
Theorem 1.7.
It is consistent with
ZFC that there is a minimal non- σ -scattered linear order L , which does not contain any real type orAronszajn type. Theorem 1.7 does not answer Question 1.2. The reason is that thelinear orders which witness Theorem 1.7 in [6], are dense subordersof the set of all branches of a Kurepa tree K . Note that such linearorders have to contain copies of ω ∗ . Moreover, the only way to showthat a suborder L of the set of branches of K is not σ -scattered was A linear order is said to be ℵ -dense if every non-empty interval has size ℵ . This fact probably exists in classical texts. Since we do not have a reference forit, we provide a proof in the next section.
ALVIN’S QUESTION 3 to show that L is dense in a Kurepa subtree. In particular, it wasunclear how to keep the tree K non- σ -scattered, if K had only ℵ many branches. In this paper, aside from eliminating copies of ω ∗ , weprovide a different way of keeping ω -trees like K non- σ -scattered, incertain forcing extensions.2. Preliminaries
In this section we review some facts and terminology regarding ω -trees, linear orders and countable support iteration of some type offorcings. The material in this section can also be found in [4] and [5].Recall that an ω -tree is a tree which has height ω and countablelevels. If T is a tree we assume that it does not branch at limit heights.More precisely, if s, t are distinct elements in the same level of limitheight then they have different sets of predecessors. Moreover, we onlyconsider ω -trees T that are ever branching: for every element t ∈ T and α ∈ ω there are u, v of height more than α which are above t andwhich are incomparable.Assume T is a tree and U ⊂ T . We say that U is nowhere dense iffor all t ∈ T there is s > t such that U has no element above s . If T is a tree and A is a set of ordinals then T ↾ A is the tree consistingof all t ∈ T with ht( t ) ∈ A . Assume T, U are trees. The function f : T −→ U is said to be a tree embedding if f is one-to-one, it is levelpreserving and t < s if and only if f ( t ) < f ( s ). Assume T is a tree,then T t is the collection of all s ∈ T which are comparable with t . Wecall a chain b ⊂ T a cofinal branch, if it intersects all levels of T . If b ⊂ T is a branch then b ( α ) refers to the element t ∈ b which is ofheight α . If b, b ′ are two different maximal chains then ∆( b, b ′ ) is thesmallest ordinal α such that b ( α ) = b ′ ( α ). The collection of all cofinalbranches of T is denoted by B ( T ). We use the following fact which iseasy to check. Fact 2.1.
Assume T is a lexicographically ordered ω -tree such that ( T, < lex ) has a copy of ω ∗ . Then there is a branch b and a sequence ofbranches h b ξ : ξ ∈ ω i such that: • for all ξ ∈ ω , b < lex b ξ • sup { ∆( b, b ξ ) : ξ ∈ ω } = ω . Definition 2.2. [4] Assume L is a linear order. We use ˆ L in order torefer to the completion of L . In other words, we add all the Dedekindcuts to L in order to obtain ˆ L . For any set Z and x ∈ L we say Z captures x if there is z ∈ Z ∩ ˆ L such that Z ∩ L has no element whichis strictly in between z and x . H. LAMEI RAMANDI
Fact 2.3. [4]
Assume L is a linear order, M ≺ H θ where θ is a regularlarge enough cardinal, x ∈ L and M captures x . Then there is a unique z ∈ ˆ L ∩ M such that M ∩ L has no element strictly in between z, x . Inthis case we say that M captures x via z . Definition 2.4. [4] The invariant Ω( L ) is defined to be the set ofall countable Z ⊂ ˆ L such that Z captures all elements of L . We letΓ( L ) = [ ˆ L ] ω r Ω( L ).Assume T is a lexicographically ordered ω -tree such that for every t ∈ T , there is a cofinal branch b ⊂ T with t ∈ b . By Ω( T ) , Γ( T )we mean Ω( B ( T )) , Γ( B ( T )), where B ( T ) is considered with the lexi-cographic order. If M ≺ H θ is countable and θ is a regular cardinal,we abuse the notation and write M ∈ Ω( T ) instead of M captures allelements of B ( T ). Similar abuse of notation will be used for Γ. Theproof of the following fact is a definition chasing. Fact 2.5.
Assume T is a lexicographically ordered ever branching ω -tree such that for every t ∈ T , there is a cofinal branch b ⊂ T with t ∈ b . Let θ be a regular cardinal such that P ( T ) ∈ H θ , M ≺ H θ becountable such that T ∈ M and b ∈ B ( T ) . Then M captures b iff thereis c ∈ B ( T ) ∩ M such that ∆( b, c ) ≥ M ∩ ω . We will use the following lemma in order to characterize σ -scatteredlinear orders. Theorem 2.6. [4] L is σ -scattered iff Γ( L ) is not stationary in [ ˆ L ] ω . The following lemma will be used in order to determine which linearorders are σ -well ordered. Most likely an equivalent of this lemmaexists in classical texts, but since we did not find any proof for it andfor more clarity we include the proof. Our proof uses the ideas in theproof of the previous theorem from [4]. Lemma 2.7.
Assume L is a linear order which does not have a copyof ω ∗ . Then L is σ -well ordered iff it is σ -scattered.Proof. Assume L is a linear order of size κ which does not have a copyof ω ∗ and which is σ -scattered. We use induction on κ . In particular,assume that every suborder L ′ ⊂ L of size less than κ is σ -well ordered.We will show that L is σ -well ordered. Let θ be a regular cardinalsuch that P ( L ) ∈ H θ . Let h M ξ : ξ ∈ κ i be a continuous ∈ -chain ofelementary submodels of H θ such that L, Ω( L ) are in M . Moreoverassume that, for each ξ ∈ κ , ξ ⊂ M ξ and | M ξ | = | ξ | + ℵ . Observethat for all x ∈ L and ξ ∈ κ there is a unique z ∈ ˆ L ∩ M ξ such that M ξ captures x via z . Moreover, if M ξ captures x via z then x ≤ z . ALVIN’S QUESTION 5
This is because ω ∗ does not embed into L . Let ˆ L ξ be the set of all z ∈ ˆ L ∩ M ξ such that for some x ∈ L , M ξ captures x via z . Inparticular, | ˆ L ξ | ≤ | M ξ | < κ for each ξ ∈ κ . We note that for each ξ ∈ κ , ˆ L ξ embeds into L . Then since the size of ˆ L ξ is less than κ , it is σ -well ordered.For each x ∈ L let g x : κ −→ ˆ L such that for all ξ ∈ κ , g x ( ξ ) ∈ M ξ ∩ ˆ L and M ξ captures x via g x ( ξ ). We note that the map x g x is orderpreserving when we consider the lexicoraphic order on all functionsfrom κ to ˆ L . The function g x is decreasing because L does not have acopy of ω ∗ . Also range( g x ) is finite, because for all ξ ∈ κ , M ξ capturesall elements of L .For each x ∈ L with | range( g x ) | = n + 1 we consider the strictlydecreasing finite sequence h z ( x ) , z ( x ) , ..., z n ( x ) i , such that for each i ≤ n, z i ( x ) ∈ range( g x ). Note that if i < j ≤ n, g x ( ξ ) = z i ( x ) , g x ( η ) = z j ( x ) then ξ < η . In other words, the cuts z i ( x ) appear in the range of g x in the order we considered in the finite sequence associated to x . Inparticular, z n ( x ) = x . For each i ≤ n , let ξ i ( x ) = min { ξ ∈ κ : g x ( ξ ) = z i ( x ) } . For each x ∈ L , let σ ( x ) = h z ( x ) , ξ ( x ) , z ( x ) , ξ ( x ) , z ( x ) , ..., ξ n ( x ) , z n ( x ) i . Let U = { σ ( x ) ↾ (2 n + 1) : x ∈ L ∧ n ∈ ω } and U m = { σ ( x ) ↾ (2 n + 1) : x ∈ L ∧ n ∈ m } for each m ∈ ω . We consider an order on U as follows. For σ, τ in U we let σ < τ iff either τ is an initial segmentof σ or σ < lex τ . Also let L m = { x ∈ L : σ ( x ) ∈ U m } . It is easy to seethat x σ ( x ) is an order preserving map from L m to U m . But each U m is an iterated sum of σ -well ordered sets. Therefore, L = S m ∈ ω L m is σ -well ordered. (cid:3) Now we review some definitions and facts about the forcings whichwe are going to use.
Definition 2.8. [5] Assume X is uncountable and S ⊂ [ X ] ω is station-ary. A poset P is said to be S -complete if every descending ( M, P )-generic sequence h p n : n ∈ ω i has a lower bound, for all M with M ∩ X ∈ S and M suitable for X, P .We note that S -complete posets preserve the stationary subsets of S .It is also easy to see that if X, S are as above and P is an S -completeforcing, then it preserves ω and adds no new countable sequences ofordinals. Lemma 2.9. [5]
Assume X is uncountable and S ⊂ [ X ] ω is stationary.Then S -completeness is preserved under countable support iterations. H. LAMEI RAMANDI
Lemma 2.10. [5]
Assume T is an ω -tree which has no Aronszajnsubtree in the ground model V . Also assume Ω( T ) is stationary and P is an Ω( T ) -complete forcing. Then T has no Aronszajn subtree in V P .Moreover, P adds no new branches to T . A Generic Element of C In this section we introduce the forcing which adds a generic lexico-graphically ordered ω -tree T . The tree T has no Aronszajn subtrees.Moreover, the set of all cofinal branches of T , which is denoted by B ,has no copy of ω ∗ . In the next section, by iterating two types of posets,we make B a minimal element of C . In particular, we will ensure that B is not σ -scattered. Definition 3.1.
Fix a set Λ of size ℵ . The forcing Q is the posetconsisting of all conditions ( T q , b q , d q ) such that the following hold.(1) T q ⊂ Λ is a lexicographically ordered countable tree of height α q + 1 with the property that for all t ∈ T q there is s ∈ ( T q ) α q such that t ≤ T q s .(2) b q is a bijective map from a countable subset of ω onto ( T q ) α q .(3) The map d q : dom( b q ) −→ ω has the property that if b q ( ξ ) = t,b q ( η ) = s and t < lex s then ∆( t, s ) < d q ( ξ ).We let q ≤ p if the following hold.(1) T p ⊂ T q and ( T p ) α p = ( T q ) α p .(2) For all s, t in T p , s < lex t in T p if and only if s < lex t in T q .(3) For all s, t in T p , s ≤ T p t if and only if s ≤ T q t .(4) dom( b p ) ⊂ dom( b q ).(5) For all ξ ∈ dom( b p ), b p ( ξ ) ≤ T b q ( ξ ).(6) d p ⊂ d q . Lemma 3.2.
Assume h q n : n ∈ ω i is a decreasing sequence of condi-tions in Q , m ≤ ω and for each i ∈ m let c i ⊂ S n ∈ ω T q n be a cofinalbranch. Then there is a lower bound q for the sequence h q n : n ∈ ω i in which every c i has a maximum with respect to the tree order in T q .Moreover, for every t ∈ ( T q ) α q either there is i ∈ m such that t is aboveall elements of c i or there is ξ ∈ D = S n ∈ ω dom( b q n ) such that t is aboveall elements of { b q n ( ξ ) : n ∈ ω ∧ ξ ∈ dom( b q n ) } . In particular, Q is σ -closed. Note that the lexicoraphic order here is independent of any structure on Λ if itexists. In other words, this order which we refer to as < lex is determined by thecondition q . ALVIN’S QUESTION 7
Proof.
For each n ∈ ω , let α n = α q n and T n = T q n . If the set of all α n ’shas a maximum, it means that after some n , the sequence q n is constant.So without loss of generality assume α = sup { α n : n ∈ ω } is a limitordinal above all α n ’s. Let T = S n ∈ ω T n . For each ξ ∈ D = S n ∈ ω dom( b q n ),let b ξ be the set of all t ∈ T such that for some n ∈ ω , t ≤ b q n ( ξ ).Observe that b ξ is a cofinal branch in T . Since we are going to put anelement on top of every b ξ , from now on, assume that c i ’s are differentfrom b ξ .Now we are ready to define the lower bound q . We let α q = α andobviously ( T q ) <α = T . We put distinct element t ξ on top of b ξ for each ξ ∈ D . We also put distinct element s i on top of c i for each i ∈ m .Therefore, ( T q ) α = { t ξ : ξ ∈ D } ∪ { s i : i ∈ m } . Let E ⊂ ω \ D suchthat | E | = m . Let b q : D ∪ E −→ ( T q ) α be any bijective function suchthat b q ( ξ ) = t ξ for each ξ ∈ D . For each ξ ∈ D , let d q ( ξ ) = d q n ( ξ )where n ∈ ω such that ξ ∈ dom( d q n ). For each η ∈ E let d q ( η ) = α + 1.We need to show that q is a lower bound in Q . We only showCondition 3 of Definition 3.1 for q ∈ Q . The rest of the conditionsand the fact that q is an extension of all q n ’s are obvious. Let A = { t ξ : ξ ∈ D } , S = { s i : i ∈ m } , and u < lex v be two distinct elementsin ( T q ) α . If u ∈ S , then ∆( u, v ) < α < d q ( η ), where η ∈ E suchthat b q ( η ) = u . If u, v are both in A , and ξ, ξ ′ are in D such that b q ( ξ ) = u, b q ( ξ ′ ) = v , let n ∈ ω such that ξ, ξ ′ are in dom( b q n ) . Then∆( u, v ) = ∆( b q n ( ξ ) , b q n ( ξ ′ )) < d q n ( ξ ) = d q ( ξ ). If u ∈ A, v ∈ S and ξ ∈ D such that b q ( ξ ) = u , fix n ∈ ω such that ξ ∈ dom( b q n ) and α q n > ∆( u, v ). Let u ′ , v ′ be the elements in T α n which are below u, v respectively. It is obvious that b q n ( ξ ) = u ′ . Then ∆( u, v ) = ∆( u ′ , v ′ ) For every ξ ∈ ω , d ( ξ ) = sup { ∆( b ξ , b η ) : b ξ < lex b η } ,and if b = b ξ we sometimes use d ( b ) instead of d ( ξ ).It is worth pointing out that, by Fact 2.1, the role of d is to control < lex so that ( B, < lex ) has no copy of ω ∗ . The behavior of d plays anessential role from the technical point of view, mostly in the densitylemmas for the forcings which we introduce and use in the next section. Lemma 3.4. The function d is a countable to one function, i.e. forall α ∈ ω there are countably many ξ ∈ ω with d ( ξ ) = α . H. LAMEI RAMANDI Proof. Assume that the set A = { ξ : d ( ξ ) = α } is uncountable. Thenfor each pair of distinct ordinals ξ, η in A , b ξ ( α +1) = b η ( α +1). But thismeans that T has an uncountable level which is a contradiction. (cid:3) Lemma 3.5. For every t ∈ T and β > ht( t ) , there is an α > β suchthat ( T α ∩ T t , < lex ) contains a copy of the rationals.Proof. We will show that for all q ∈ Q and t ∈ T q , the set { p ≤ q :( Q , < ) ֒ → ( { s ∈ ( T p ) α p : t ≤ T p s } , < lex ) } is dense blow q . Fix r ≤ q with α r > β and t ∈ ( T r ) α r ∩ T t . Let ξ ∈ ω such that b r ( ξ ) = t . Withoutloss of generality we can assume that d r ( ξ ) < α r . Fix X ⊂ Λ \ T r aninfinite countable set and u ∈ X . Let p < r be the condition such thatthe following hold. • α p = α r + 1, and dom( b p ) = dom( r ) ∪ E where E consists ofthe first ω ordinals after sup(dom( r )). • T r ⊂ T p , ( T r ) α r = ( T p ) α r and for all s ∈ ( T r ) α r \ { t } there is aunique s ′ ∈ ( T p ) α p with s ′ > s . • ( T p ) α p consists of the set of all s ′ as above union with X . More-over, for every x ∈ X , t is below x , in the tree order. • Define < lex on X so that X becomes a countable dense linearorder without smallest element and with max( X ) = u . • For every s ∈ ( T r ) α r \ { t } and s ′ > s in ( T p ) α p let b p ( s ′ ) = b r ( s ).Also b p ( u ) = b r ( t ). Extend b p on E such that b p ↾ E is abijection from E to X \ { u } . • The function d p agrees with d r on dom( b r ) and d p ↾ E is con-stantly α p + 1.It is easy to see that p ∈ Q is an extension of r and the set X \ { u } isa copy of the rationals whose elements are above t . (cid:3) There is a well known σ -closed poset which is closely related to ourposet Q and which generates a Kurepa tree. Todorcevic showed thatthe generic Kurepa tree of that poset does not have Aronszajn subtrees.The proof of the following lemma uses the same idea but we includethe proof for more clarity. Lemma 3.6. Every uncountable downward closed subset of T contains b ξ for some ξ ∈ ω . In particular, { b ξ : ξ ∈ ω } is the set of all branchesof T .Proof. Let ˙ A be a Q -name for an uncountable downward closed subsetof T and p ∈ Q forces that ˙ A contains non of the b ξ ’s. Let M ≺ H θ becountable where θ is a regular large enough regular cardinal such that˙ A, p are in M . By Lemma 3.2 for m = 0, there is an ( M, Q )-genericcondition q ≤ p such that α q = δ where δ = M ∩ ω and for each ALVIN’S QUESTION 9 t ∈ ( T q ) δ there is ξ ∈ M such that b q ( ξ ) = t . But then q forces that ˙ A has no element of height δ which is a contradiction. (cid:3) The proof of the following lemma is very similar to the one above. Lemma 3.7. Ω( T ) is stationary. We note that if CH holds then the forcing Q satisfies the ℵ chaincondition. On the other hand, if κ > ω and we consider κ manybranches for T in the definition of Q , then Q collapses κ to ω . This isbecause Q adds a countable to one function from κ to ω .4. Making B Minimal in C In this section we introduce the forcings which make B a minimalelement of C . The idea is as follows. If L ⊂ B is nowhere dense,we make L σ -well ordered. For somewhere dense suborders of B weintroduce a forcing which adds embedding from B to them and whichkeeps B inside C . Definition 4.1. Assume L ⊂ B is nowhere dense. Define S L to be theposet consisting of all increasing continuous sequences h α i : i ∈ β + 1 i in ω <ω such that for all i ∈ β + 1 and t ∈ T α i ∩ ( S L ) there is ξ < α i with t ∈ b ξ . We let q ≤ p , if p is an initial segment of q .It is easy to see that for every nowhere dense L ⊂ B , S L is Ω( T )-complete. Therefore, as long as Ω( T ) is stationary, S L preserves ω .Moreover, S L shoots a club in Ω( L ). So S L forces that L is σ -wellordered. This uses Lemmas 2.7, 2.6, and the fact that L has no copyof ω ∗ . Definition 4.2. Assume U = T x for some x ∈ T and L ⊂ B ( U ) isdense in B ( U ). Define E L to be the poset consisting of all conditions q = ( f q , φ q ) such that:(1) f q : T ↾ A q −→ U ↾ A q is a < lex -preserving tree embeddingwhere A q is a countable and closed subset of ω with max( A q ) = α q ,(2) φ q is a countable partial injection from ω into { ξ ∈ ω : b ξ ∈ L } such that the map b ξ b φ q ( ξ ) is < lex -preserving,(3) for all t ∈ T α q there are at most finitely many ξ ∈ dom( φ q ) ∪ range( φ q ) with t ∈ b ξ ,(4) f q , φ q are consistent, i.e. for all ξ ∈ dom( φ q ), f q ( b ξ ( α q )) ∈ b φ q ( ξ ) ,(5) for all ξ ∈ dom( φ p ), d ( ξ ) ≤ d ( φ p ( ξ ))We let q ≤ p if A p is an initial segment of A q , f p ⊂ f q , and φ p ⊂ φ q . Lemma 4.3. For all β ∈ ω the set of all conditions q ∈ E L with α q > β is dense in E L .Proof. Fix p ∈ E L and let D p = dom( φ p ) and R p = range( φ p ). Wesometimes abuse the notation and use D p , R p in order to refer to thecorresponding set of branches, { b ξ : ξ ∈ D p } and { b ξ : ξ ∈ R p } . Weconsider the following partition of U = T α p ∩ range( f p ). Let U be theset of all u ∈ U such that if u ∈ b ∈ R p then there is a c ∈ B with u ∈ c and b < lex c . Note that if u ∈ U and there is no b ∈ R p with u ∈ b then u ∈ U . We let U = U \ U .First we will show that if u ∈ U then there is α u ∈ ω and X u ⊂ T α u ∩ T u such that:a. α u > max( { ∆( b, c ) : b, c are in A } ∪ { β } ), where A is the set ofall b ∈ R p such that u ∈ b ,b. ( X u , < lex ) is isomorphic to the rationals, andc. { b ( α u ) : b ∈ A } ⊂ X u .In order to see this, let b m be the maximum of A with respect to < lex .This is possible because A is finite. Let c ∈ B such that u ∈ c and b m < lex c . This is possible because we assumed that u ∈ U . Let t m bethe element in c \ b which has the lowest height. By Lemma 3.5, thereis a copy of the rationals X t m in some level α t m > ht( t m ) such that forall x ∈ X t m , t m < T x . In other words, u < T t m < T x for all x ∈ X t m and X t m is isomorphic to the rationals when it is considered with < lex .Moreover, b m ( α t m ) < lex x for all x ∈ X t m . Assume b ′ < lex b are two consecutive elements of ( A, < lex ). Let t b ∈ b \ b ′ which has the minimum height. There is α b ′ b > β such that( T α b ′ b ∩ T t b , < lex ) has a copy of the rationals X b ′ b . Moreover, X b ′ b canbe chosen is such a way that for all x ∈ X b ′ b , b ′ ( α b ′ b ) < lex x < lex b ( α b ′ b ).This is because there is no restriction for branching to the left in thetree T . More precisely, for all γ ∈ ω there is a c < lex b in B suchthat ∆( b, c ) > γ. Similarly, if a is the minimum of A with respect to < lex , there is α a > β and X a ⊂ T α a ∩ T u which is isomorphic to therationals. Moreover α a , X a can be chosen in such a way that if x ∈ X a then u < T x and for all b ∈ A , x < lex b ( α a ).Now let α be above α t m , α a and all of α b ′ b ’s as above. Then T α a ∩ T u contains X u which is a copy of the rationals and { b ( α u ) : b ∈ A } ⊂ X u .Note that U is the set of all u ∈ U such that for some b u ∈ R p , u ∈ b u and if u ∈ c ∈ B then c ≤ lex b u . By the same argument as abovewe can show for all u ∈ U there is α u ∈ ω and X u ⊂ T α u ∩ T u suchthat:d. α u > max( { ∆( b, c ) : b, c are in A } ∪ { β } ), where A is the set ofall b ∈ R p such that u ∈ b , ALVIN’S QUESTION 11 e. ( X u \ { b m ( α ) } , < lex ) is isomorphic to the rationals, where b m isthe maximum of A with respect to < lex ,f. { b ( α u ) : b ∈ A } ⊂ X u , andg. max( X u , < lex ) = b m ( α ) . Now we are ready to introduce the extension q ≤ p . Let α ∈ ω and α > α u for all u ∈ U . Let A q = A p ∪{ α } , φ q = φ p . If u ∈ U then T α ∩ T u contains X u such that conditions b,c hold. If u ∈ U and f p ( t ) = u , let f q ↾ ( T α ∩ T t ) be any < lex preserving function which is consistent with φ q . If u ∈ U then T α ∩ T u contains X u such that conditions e,f,g hold.In addition, if u ∈ U , f q ( t ) = u, b m = max( A ) and b is mapped to b m by φ q , then d ( b ) < d ( b m ). So b ( α ) = max(( T α ∩ T t ) , < lex ). This meansthat if u ∈ U and f p ( t ) = u , we can find f q ↾ ( T α ∩ T t ) which is < lex preserving and which is consistent with φ q . (cid:3) The proof of the following lemma uses Lemma 3.4 and the sameargument as above. Lemma 4.4. For all ξ ∈ ω , the set of all q ∈ E L with ξ ∈ dom( φ q ) isdense in E L . The following lemma shows that ω is preserved by countable supportiteration of the forcings of the form E L , where L is a somewhere densesubset of B . Lemma 4.5. The forcing E L is Ω( T ) -complete.Proof. Assume x ∈ T and L is dense in B ( T x ) . Let θ > ω + be aregular cardinal, M ≺ H θ be countable such that L, T, x are all in M and M ∈ Ω( T ). Let δ = M ∩ ω , and h p n : n ∈ ω i be a decreasing( M, E L )-generic sequence. We use A n , α n , f n , φ n in order to refer to A p n , α p n , f p n , φ p n .We define a lower bound p = ( f p , φ p ) for h p n : n ∈ ω i as follows.Let A = S n ∈ ω A n , A p = A ∪ δ . By Lemma 4.3, sup { α n : n ∈ ω } = δ ,and A p is closed. We define φ p = S n ∈ ω φ n . Note that by elementarityand Lemma 4.4, S { dom( φ n ) : n ∈ ω } = δ . So for each t ∈ T δ thereis a unique ξ ∈ dom( φ p ) such that t ∈ b ξ . Define f : T δ −→ T δ ∩ T x by f ( t ) = b φ p ( ξ ) ( δ ) where ξ is the unique ξ ∈ δ with t ∈ b ξ . Since φ p preserves the lexicoraphic order, f does too. Let f p = f ∪ S n ∈ ω f n .Since f n is consistent with φ n for each n , f p is a tree embedding.Obviously φ p preserves the lexicographic order. Moreover, by elemen-tarity, for each t ∈ T δ there is a unique ξ ∈ δ with t ∈ b ξ . Since φ p is one to one, for each t ∈ T δ there is at most one ξ ∈ range( φ p ) with t ∈ b ξ . The rest of the conditions for p ∈ E L are obvious. (cid:3) Now we are ready to introduce the forcing extension in which C hasa minimal element. Let’s fix some notation. P = P ω is a countablesupport iteration h P i , ˙ Q j : i ≤ ω , j < ω i over a model of CH suchthat Q = Q , and for all 0 < j < ω , ˙ Q j is a P j -name for either E L or S L depending on whether or not L is somewhere dense. As usual,the bookkeeping is such that if L ⊂ B is in V P , either E L or S L hasappeared in some step of the iteration. This is possible because all ofthe iterands have size ℵ so P satisfies the ℵ chain condition. Thiscan be seen by the work in [5] too. T, B are the generic objects thatare introduced by Q , as in the previous section.Since E L , S L are Ω( T )-complete forcings, by Lemma 2.10, the count-able support iteration consisting of the posets E L , S L do not add newbranches to T . It is worth pointing out that although P is a σ -closedforcing, the posets E L , S L that are involved in the iteration are noteven proper. Lemma 4.6. Assume G ⊂ P is V -generic. Then Γ( B ) is stationaryin V [ G ] .Proof. Assume M is a suitable model for P in V with M ∩ ω = δ and h p n : n ∈ ω i is a descending ( M, P )-generic sequence. Let q n = p n ↾ R = S n ∈ ω T q n . Note that if G ⊂ P is a generic filter over V which contains h p n : n ∈ ω i , then in V [ G ] we have T <δ = R . We willfind an ( M, P )-generic condition p below h p n : n ∈ ω i which forces M [ G ] ∈ Γ( T ), in V [ G ].Before we work on the details we explain the idea how to find sucha condition p . Let q = p ↾ 1. Since h q n : n ∈ ω i is ( M, P )-generic,we have to have that dom( b q ) ⊃ δ . If we allow dom( b q ) = δ , theadvantage is that it is easy to find lower bounds for the rest of thesequences h p n ( β ) : n ∈ ω i , by induction on β . But then, the resultinglower bound is going to force that M [ G ] / ∈ Γ( T ). This means that weneed to find a lower bound in such a way that dom( q ) ) δ . We use adiagonalization argument in T and a Skolem closure argument to findsuch a lower bound for h q n : n ∈ ω i . Then we will use induction on β to find a lower bound for each h p n ( β ) : n ∈ ω i with β ∈ M ∩ ω .Now we return to the proof. Note that for all β ∈ M ∩ ω , and˙ L a P β -name for a nowhere dense subset of B in M , h p n : n ∈ ω i decides ( S ˙ L ) ∩ ˙ T <δ . More precisely, there exists U ⊂ R in V suchthat if G is a a P -generic filter over V with { p n : n ∈ ω } ⊂ G then[( S ˙ L )] G ∩ [ ˙ T <δ ] G = U . This is because [( S ˙ L )] G ∩ [ ˙ T <δ ] G is a countablesubset of R and R ∈ V . Here we use the fact that countable supportiteration of Ω( T )-complete forcings do not add new reals. Let U be the ALVIN’S QUESTION 13 set of all countable U ⊂ R such that for some β ∈ M ∩ ω and ˙ L ∈ M which is a P β -name for a nowhere dense subset of B h p n : n ∈ ω i decides ( S ˙ L ) ∩ ˙ T <δ to be U .Assume β ∈ M ∩ ω and ˙ L is a P β -name for a somewhere densesubset of B and ˙ Q β is a P β -name for the forcing E ˙ L . Let’s denote thecanonical name for the generic filter of E ˙ L by ( ˙ f , ˙ φ ). Similar to thecase of nowhere dense subsets of B , h p n : n ∈ ω i decides ( S ˙ L ) ∩ ˙ T <δ ,˙ f ↾ R and ˙ φ ↾ δ . Moreover, ˙ f G ↾ R is in V , for any V -generic filter G .Now, let F be the set of all finite compositions g ◦ g ◦ ... ◦ g n , suchthat for all i ≤ n , g i or g i − is a partial function on R which is ofthe form ˙ f G ↾ R where for some ˙ φ, ( ˙ f , ˙ φ ) is the canonical name forthe generic filter added by E ˙ L , ˙ L ∈ M is a name for a somewheredense subset of B , and G is a P -generic filter over V which contains { p n : n ∈ ω } . Also let W be the collection of all g [ U ] such that U ∈ U and g ∈ F . Note that every W ∈ W is nowhere dense in R .Fix h W n : n ∈ ω i an enumeration of W and h ξ n : n ∈ ω i an enumer-ation of δ . Let h t m : m ∈ ω i be a chain in R such that • if m = 2 k then W k has no element above t m , and • if m = 2 k +1 then t m is not in the downward closure of { b q n ( ξ k ) : n ∈ ω } .By Lemma 3.2, h q n : n ∈ ω i has a lower bound q such that whenever c is a cofinal branch of R , then there is an element on top of c if andonly if one of the following holds. • For some ξ ∈ δ , { b q n ( ξ ) : n ∈ ω } is cofinal in c . • For some g ∈ F , { g ( t m ) : m ∈ ω } is cofinal in c .We can choose q in such a way that if t is on top of c , c satisfies thesecond condition, and η ∈ ω with b q ( η ) = t then d q ( η ) = δ + 1. Forthe rest of the proof, assuming that p ↾ β is given, we find p ( β ). If β / ∈ M ∩ ω we define p ( β ) to be the trivial condition of the correspondingforcing Q β . For each β ∈ M , since p ↾ β is ( M, P β )-generic, it decides p n ( β ) for all n ∈ ω .Assume ˙ Q β is a P β -name for some S ˙ L ∈ M . We define p ( β ) = { ( δ, δ ) } ∪ S n ∈ ω p n ( β ). Observe that if ˙ L is a P β -name for a nowheredense subset of B , t ∈ ( T q ) δ , c t is the set of all elements of T q that areless than t , and ( b q ) − ( t ) ≥ δ then there is s ∈ c t such that p ↾ β forcesthat S ˙ L has no element above s . This makes p ( β ) a lower bound for h p n ( β ) : n ∈ ω i .Assume ˙ Q β = E ˙ L is a P β -name where ˙ L ∈ M . By Lemmas 4.3sup( S n ∈ ω A p n ( β ) ) = δ . Moreover, Lemma 4.4 implies that for all ξ ∈ δ there is n ∈ ω such that ξ ∈ dom( φ p n ( β ) ) . We define p ( β ) = ( f, φ ) asfollows. Let φ = S n ∈ ω φ p n ( β ) , A p ( β ) = { δ } ∪ S n ∈ ω A p n ( β ) , dom( f ) = T q ↾ A p ( β ) . If ht( s ) ∈ A p n ( β ) for some n ∈ ω , let f ( s ) = f p n ( β ) ( s ). If ht( s ) = δ and c is a cofinal branch in R whose elements are below s , let f ( s ) be theelement on top of the chain { f ( v ) : v ∈ c ∩ dom( f ) } . This makes sense,because there is an element on top of { f ( v ) : v ∈ c ∩ dom( f ) } in T q .In order to see this, first assume that for some ξ ∈ δ , { b q n ( ξ ) : n ∈ ω } is cofinal in c . Let n ∈ ω such that ξ ∈ dom( φ p n ( β ) ), and η = φ p n ( β ) ( ξ ) . Then b q ( η ) is the top element of { f ( v ) : v ∈ c ∩ dom( f ) } . If for some g ∈ F , { g ( t m ) : m ∈ ω } is cofinal in c , then { f ◦ g ( t m ) : m ∈ ω } iscofinal in the downward closure of { f ( v ) : v ∈ c ∩ dom( f ) } . So { f ( v ) : v ∈ c ∩ dom( f ) } has a top element as desired. This finishes defining f, φ . It is obvious that p ( β ) is a lower bound for h p n ( β ) : n ∈ ω i .Now we show that if G ⊂ P is V -generic with p ∈ G , then M [ G ] ∈ Γ( T ). Let c ⊂ R be the downward closure of { t m : m ∈ ω } . Let t ∈ T δ be the element on top of c . Let b ∈ B ( T ) with t ∈ b . It is obvious that c is different from the downward closure of b q ( ξ ) for each ξ ∈ δ . So forall ξ ∈ δ , ∆( b, b ξ ) < δ . It is obvious that M [ G ] ∩ ω = δ . By Lemma3.6 and the fact that Ω( T )-complete forcings do not add branches to T , { b ξ : ξ ∈ ω } is the set of all branches of T in V [ G ]. Therefore forall b ′ ∈ B ( T ) ∩ M [ G ], ∆( b ′ , b ) < δ . This means that M [ G ] does notcapture b in V [ G ]. (cid:3) Now we are ready to show that B is a minimal element of C . Lemma4.6 implies that B is not σ -scattered and hence it is not σ -well ordered.It is obvious that B does not contain any real type. Since Ω( T ) isstationary in V [ G ], B does not contain any Aronszajn type either.Recall that ω ∗ does not embed into B in V [ G ∩ Q ] and Ω( T )-completeforcings do not add new branches to ω -trees. This means that B doesnot have any copy of ω ∗ in V [ G ]. Therefore, every uncountable subsetof B in V [ G ] contains a copy of ω . If L ⊂ B and L ∈ C , then L has tobe somewhere dense. But then the forcing E L has added an embeddingfrom B to L . Hence B is minimal in C . References [1] U. Abraham and S. Shelah. Isomorphism types of Aronszajn trees. Israel J.Math. , 50(1-2):75–113, 1985.[2] J. E. Baumgartner. All ℵ -dense sets of reals can be isomorphic. Fund. Math. ,79(2):101–106, 1973.[3] J. E. Baumgartner. A new class of order types. Ann. Math. Logic , 9(3):187–222,1976. ALVIN’S QUESTION 15 [4] T. Ishiu and J. T. Moore. Minimality of non σ -scattered orders. Fund. Math. ,205(1):29–44, 2009.[5] H. Lamei Ramandi. A minimal Kurepa tree with respect to club embeddings. Fund. Math. , 245(3):293–304, 2019.[6] H. Lamei Ramandi. A new minimal non- σ -scattered linear order. J. Symb. Log. ,84(4):1576–1589, 2019. Department of Mathematics, University of Toronto, Toronto, Canada Email address ::