MARTIN S AXIOM AND SEPARATED MAD FAMILIES

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1 MARTIN S AXIOM AND SEPARATED MAD FAMILIES ALAN DOW AND SAHARON SHELAH Abstract. Two famles A, B of subsets of ω are sad to be separated f there s a subset of ω whch mod fnte contans every member of A and s almost dsont from every member of B. If A and B are countable dsont subsets of an almost dsont famly, then they are separated. Luzn gaps are well-known examples of ω 1 -szed subfamles of an almost dsont famly whch can not be separated. An almost dsont famly wll be sad to be ω 1 - separated f any dsont par of ω 1 -szed subsets are separated. It s known that the proper forcng axom (PFA) mples that no maxmal almost dsont famly s ω 1 -separated. We prove that ths does not follow from Martn s Axom. 1. Introducton In ths paper we construct a model of Martn s Axom n whch there s a maxmal almost dsont famly of subsets of ω whch has a strong separaton property we call ω 1 -separated. Combnatoral propertes of almost dsont famles are fundamental and well-studed. Two sets are sad to be almost dsont, also orthogonal to each other, f ther ntersecton s fnte, and two famles of sets are orthogonal f each member of one s orthogonal to each n the other. Two famles of subsets of ω are separated f there s a set C whch s orthogonal to the frst whle ω \ C s orthogonal to the other (.e. C mod fnte contans each member of the frst famly). One of the most famous and nfluental papers on orthogonal almost dsont famles s the 1947 paper of Luzn. Proposton 1.1. [3] There exst orthogonal ℵ 1 -szed almost dsont famles of subsets of ω whch can not be separated. Date: September 28, Mathematcs Subect Classfcaton. 03A35. Key words and phrases. maxmal almost dsont famly, Martn s Axom. Research of the frst author was supported by NSF grant No. NSF-DMS The research of the second author was supported by the Unted States- Israel Bnatonal Scence Foundaton (BSF Grant no ), and by the NSF grant No. NSF-DMS Ths s paper number F1090 n the second author s personal lstng. 1

2 2 A. DOW AND S. SHELAH Of course Luzn s method was based on the deas ntroduced by Hausdorff n constructng (ω 1, ω 1)-gaps. Todorcevc [6] ntroduces the termnology of a Luzn gap. A par ({a α } α ω1, {b α } α ω1 ) of famles of countable sets s a Luzn gap f for all α β, a α b α s empty, whle (a α b β ) (a β b α ) s not empty. It s well-known that a Luzn gap can not be separated. Abraham and Shelah [1] study the notons of Luzn sequences and Luzn* sequences. A sequence {a α : α ω 1 } s a Luzn sequence (respectvely Luzn* sequence) f for all ω 1 and n ω, the set {α : a α a n} s fnte (respectvely {α : a α a < n} s fnte). An uncountable almost dsont famly s defned to be nseparable f no uncountable par can be separated. Each Luzn* famly s Luzn, and each Luzn famly s nseparable. One of the results of [1] was to establsh that MA(ω 1 ) mples that each nseparable sequence could be wrtten as a countable unon of Luzn* sequences. We may say that an almost dsont famly s nowhere nseparable f t contans no nseparable subfamly. Regrettably the word separable n the context of almost dsont famles s already defned to mean somethng qute unrelated; specfcally the term completely separable almost dsont famly s defned to mean that every set havng nfnte ntersecton wth nfntely many of the members wll contan a member. There s a partcularly natural example of a Luzn gap ( [5, Lemma 1] and [6, p57]) whch can best be defned as subsets of the tree 2 <ω. Proposton 1.2. If {f α : α ω 1 } 2 ω, and for each α ω 1, let a α = {f α n : f α (n) = 0} and b α = {f α n : f α (n) = 1}. Then, for all α < β, a α b α =, and (a α b β ) (a β b α ) = 1. On the other hand, t s a very well-known result of Slver that MA(ω 1 ) mples that any famly of ω 1 many branches themselves s nowhere nseparable ([4, p162]). Ths s the geness of the followng noton. Defnton 1.3. A famly A [ω] ω s specal f (1) A s an almost dsont famly and (2) there s a functon c : [A] <ω ω and a lnear orderng < (or < c ) of A satsfyng that, for each n ω, f B 1 < < B n and C 1 < < C n are two sequences from A, and f c({b 1,..., B n }) = c({c 1,..., C n }) = k, then for all {1,..., n}, B C s contaned n k. Defnton 1.4. An almost dsont famly A s <λ-specal f each A [A] <λ s specal. We say that A s λ-specal f t s <λ + -specal. It s a well-known generalzaton of Slver s result that

3 MARTIN S AXIOM AND SEPARATED MAD FAMILIES 3 Proposton 1.5. MA ω1 mples that dsont ℵ 1 -szed subsets of an ω 1 -specal famly are separated. Proof. Let A be an ω 1 -specal famly and let A 0 and A 1 be dsont subsets of A, each wth cardnalty at most ω 1. Of course f ether s countable, then the fact that MA ω1 mples that b > ω 1, shows that they can be separated. Otherwse, let c be the functon from [A 0 A 1 ] <ω nto ω whch wtnesses that A 0 A 1 s specal. A poset Q s defned to be the famly of functons q nto ω such that dom(q) s a fnte subset of A 0 A 1 and q satsfes that (a\q(a)) s dsont from (b\q(b)) whenever a dom(q) A 0 and b dom(q) A 1. The poset Q s smply ordered by extenson. It suffces to show that Q s ccc. Let {q : ω 1 } Q. Towards provng that ths s not an antchan, we may assume that there are ntegers n, k and subsets I, I 0, I 1 of n such that, for each η ω 1, (1) dom(q ) = {a l : l < n} as ordered by < c, (2) q(a l ) = q(aη l ) < k for all l < n, (3) a l A 0 f l I 0, (4) a l A 1 f l I 1, (5) n = I 0 I 1, (6) a l = aη l f and only f l I (7) c(a 0,..., a n 1) = c(a η 0,..., a η n 1) < k, (8) a l k = a η l k for each l < n, We check that q q η Q for such, η. Indeed, suppose that a dom(q ) A 0 and a η dom(q η ) A 1. It follows that and so by the hypothess on c, we have that a aη k. In addton, k (a \q (a )) (aη \q η(a η )) s the same as k (a \q (a )) (a \q (a η )) and so s empty. For a collecton A [ω] ω wth the property that no fnte unon from A s cofnte, the well-known poset P A as defned n [4, p153] s ccc and forces an nfnte set a ω whch satsfes that for all Y ω (n the ground model), a s almost dsont from Y f and only f Y s (mod fnte) covered by a fnte unon from A. We make a mnor change so that f A s the empty famly, then P A smply adds a standard Cohen real. Defnton 1.6. For a famly A [ω] ω, we defne P A so that p P A f p = (t p, A p ) [ω] <ω [A] <ω, and p < q f t q t p, max(t q ) < mn(t p \t q ), A q A p, and t p A t q for all A A q. The followng result was the man motvaton for ths paper.

4 4 A. DOW AND S. SHELAH Proposton 1.7. [2, 2.10] PFA mples that every maxmal almost dsont famly contans a Luzn sequence. The prevous result follows easly from Lemma 1.8 whch s new and llustrates some of the key deas. Lemma 1.8. Let U be an ultraflter on ω and suppose that A s an uncountable almost dsont famly satsfyng that, for each U U, all but countably many members of A meet U n an nfnte set. Then there s a ccc poset whch forces that A contans a Luzn sequence. Proof. Clearly we may assume that A has cardnalty ω 1 and fx an enumeraton {a α : α ω 1 }. The poset Q conssts smply of condtons q = (n q, I q ) ω [ω 1 ] <ω where a condton p extends q f I p I q, n p n q, and for β I p \ I q and all α I q β, a α a β n q. Obvously, f G Q s a generc flter, the desred Luzn sequence wll be gven by {a α : α I = p G I p}. There wll be a condton that forces I s uncountable so long as Q s ccc. Assume now that {q : ω 1 } Q. By passng to a subcollecton, we may assume there s an n so that n = n q for all ω 1. In addton, we may assume that the sequence {I q : ω 1 } forms a -system wth root I 0 = {α 1, α 2,..., α m }. For each ω 1, let I q \ I 0 = {β1,..., β l } be lsted n ncreasng order, and assume that α m < β1. One fnal reducton (not really needed) s to choose a sequence of ntegers {k 1,..., k l } so that for all ω 1 and 1 l, we have that k a β. We defne a collecton T ([ω] <ω ) l accordng to t = t 1,..., t l T f J t = { : t a β \ n for all 1 l} s uncountable. Of course we have that the sequence,..., s a member of T. We show that, for each t T and 1 l, the set U( t, ) = {k ω : t 1,..., t 1, t {k}, t +1,... t l T } s a member of U. For each k / U( t, ), J t,,k = { J t : k a β } s countable, thus we may choose J t \ k / U( t,) J t,,k and observe that for each k a β, J t,,k s uncountable. Ths means that ω \ U( t, ) s dsont from uncountably many members of A and so can not be a member of U. Now choose δ ω 1 so large that for all t T, 1 l and k ω, f J t,,k s countable, then t s contaned n δ and, f U( t, ) U, t meets n an nfnte set. If t = t 1,..., t l T and f 1, l, then a β δ there s an k a β δ U( t, ), whch mples that t = t 1,..., t 1, t

5 MARTIN S AXIOM AND SEPARATED MAD FAMILIES 5 {k}, t +1,... t l T. Therefore a routne fnte recurson shows the exstence of a sequence t = t 1,..., t l T satsfyng that for each 1, l, t a β δ s not empty. It s routne to verfy that for any J t, q and q δ are compatble snce t a β δ n s a wtness to a β δ n for each 1, l. a β 2. An ω 1 -specal mad famly and MA ω1 We produce a model of MA ω1 n whch there s an ω 1 -specal maxmal almost dsont famly. By Proposton 1.5, ths wll complete the proof of the man theorem. Theorem 2.1. It s consstent wth Martn s Axom and c = ω 2 that there s a maxmal almost dsont famly whch s ω 1 -separated, and so contans no Luzn famly. The method to construct the model s to begn wth a forcng to produce a model n whch c = ω 2 and there s an ω 1 -specal maxmal almost dsont famly whch remans maxmal n extensons by ccc forcngs of cardnalty ω 1. Then the standard fnte support teraton of ccc posets of cardnalty ω 1 can be used to extend to a model of Martn s Axom. Let us say that a maxmal almost dsont famly s κ- ndestructble f ts maxmalty s preserved by any forcng of cardnalty less than κ. Lemma 2.2. Suppose that κ s a regular cardnal and A [ω] ω s an almost dsont famly of cardnalty κ wth the property that each Y ω s ether (mod fnte) covered by fntely many members of A or meets all but fewer than κ many members of A. Then A s κ-ndestructble. Proof. Let Q be a poset of cardnalty less than κ. Assume that Ẏ s a Q-name of a subset of ω. For each A A, choose q A Q (f one exsts) and n A ω whch forces that Ẏ A n A. Fx any q Q and n ω so that A q,n = {A A : q A = q, and n A = n} has cardnalty κ. It follows from the assumptons on A then that Ẏ q + = {m ω : ( p < q) p m Ẏ } s mod fnte covered by fntely many members of A. Ths shows that q forces that Ẏ s also covered by fntely many members of A. Lemma 2.3. If A s an ω 2 -ndestructble maxmal almost dsont famly, then t remans maxmal n any extenson by a fnte support teraton of ccc posets of sze at most ω 1. Proof. Let P α, Q α : α λ be a fnte support teraton so that, for each α, Pα Q α s ccc and Q α ω 1.

6 6 A. DOW AND S. SHELAH For convenence, we may assume that for each α there s a P α -name < α of a subset of ω 1 ω 1 so that Q α s forced to be the poset (ω 1, < α ). Let Ẏ be a P λ -name of a subset of ω. Let {M : ω 1 } be an ncreasng ɛ-chan of countable elementary submodels of H(θ) for a suffcently large θ such that P λ and Ẏ are n M 0. Let M = ω 1 M and let P = P λ M. Snce P s a ccc poset of cardnalty ω 1, there s a condton p P and an A A such that p P Ẏ A s nfnte. Assume that p > p P λ s such that there s an n ω wth p Pλ Ẏ A n. We may assume that p satsfes that for each α dom(p), p(α) s an element of ω 1 (and not ust forced to be). There s a δ ω 1 such that p(α) δ for all α dom(p) and so that dom(p) M M δ. It follows by elementarty that, for each α dom( p) dom(p) M δ, p (dom(p) M δ α) p(α) < α p(α). Now choose p P and m A \ n such that p < p M δ and p n Ẏ. It s easly checked that p s compatble wth p (manly snce dom(p ) dom(p) dom(p )). It follows that p Pλ Ẏ A s nfnte; and that A remans maxmal. Now we defne the natural poset for ntroducng a functon to make an almost dsont famly specal. Defnton 2.4. For an almost dsont famly A and a lnear order of A, the poset Q A, wll smply be the set of functons c such that dom(c) = P(A c ) for some fnte A c {a β : β < α} whch satsfy condton (2) of Defnton 1.3. The orderng on Q A, s smple extenson. We use Q A f the choce of s clear from the context. We make the followng observatons about Q A,. Lemma 2.5. Let A [ω] ω be an almost dsont famly whch s specal. (1) If s any lnear orderng of A, then the poset Q A, s ccc. (2) If a [ω] ω s almost dsont from each member of A, then A {a} s also specal. Proof. Let c and < be the wtnesses (as n (2) of Defnton 1.3) that A s specal. To prove tem 2, let us notce that 2c (doublng each value) s also a wtness to A s specal; so we may assume that c takes on only even values. Extend < to all of A {a} by declarng b <a for all b A. Also extend c by defnng c(b 1,..., B n 1, a) as follows. Choose k 0 mnmal so that B a k 0 for all 1 < n and let k 1 = c(b 1,..., B n 1 ). Now defne c(b 1,..., B n 1, a) to be (the odd nteger) 3 k0 5 k 1. Assume that c(b 1,..., B n 1, a) = c(c 1,..., C n 1, a) = k. It follows that c(b 1,..., B n 1 ) = c(c 1,..., C n 1 ) < k and so B C k for dstnct 1, < n. Of course each of B a and C a are

7 MARTIN S AXIOM AND SEPARATED MAD FAMILIES 7 contaned n k because of the choce of k 0. Ths proves that A {a} s specal. Now we prove that Q A, s ccc. Let {q : ω 1 } be a subset. We may assume that the famly {A q : ω 1 } forms a -system, each wth cardnalty n and wth root R. For each, let A q be enumerated n -ncreasng order: {a 1,..., a n}. We may assume that for each, and each 1 n, a R whenever a R. We may also assume that there s an uncountable I 0 ω 1 and a fxed permutaton π of n such that a <a π() π(+1) for each 1 < n for all I 0. Smlarly, there s an uncountable I 1 I 0, an nteger k and a functon c defned on the power set of n satsfyng that, for each ncreasng sequence 1 ρ(1) < < ρ(l) n, and c q ({a ρ(1),..., a ρ(l) }) = c({ρ(1),..., ρ(l)}) < k c({a 1,..., a n}) < k. Now choose < I 1 so that for each 1 n, a k = a k. By vrtue of c, t follows that c q c q s a partal functon on P(A ) (as n there are no dsagreements), where A = A q A q. Extend ths to a functon c wth doman P(A ) so that f c ({B 1,..., B m }) = c ({C 1,..., C m }) then each of {B 1,..., B m } and {C 1,..., C m } are contaned n one of A q or A q. To check that q = (c, A ) s a common extenson of q and q we smply have to show that c satsfes the requrement of beng a specalzng functon. Now the only case to check s f c ({B 1,..., B m }) = c ({C 1,..., C m }) = k < k wth {B 1,..., B m } A q and {C 1,..., C m } A q. Let ρ 0 and ρ 1 be the ncreasng mappngs from m nto n so that {B 1,..., B m } = {a ρ 0 (1),..., a ρ 0 (m) } and {C 1,..., C m } = {a ρ 1 (1),..., a ρ 1 (m) }. It follows that c(ρ 0) = c(ρ 1 ). Therefore c q ({a ρ 0 (1),..., a ρ 0 (m) } = c q ({a ρ 1 (1),..., a ρ 1 (m) }) = k. Therefore, for 1 m, a ρ 0 () a ρ 1 () k. If, for example a ρ 1 () s n the root R, then we have that a ρ 0 () a ρ 1 () = B C k. If nether s n the root, then ρ 0 () ρ 1 () and so we know that a ρ 1 () a ρ 1 () k and so B C = a ρ 0 () a ρ 1 () k = a ρ 1 () (a ρ 1 () k) k. Theorem 2.6. If 2 ω 1 = ω 2, then there s a ccc poset of cardnalty ω 2 whch forces that there s an ω 2 -ndestructble maxmal almost dsont famly whch s ω 1 -specal.

8 8 A. DOW AND S. SHELAH Proof. We defne a fnte support teraton P α, Q α : α ω 2 and a sequence ċ α, ȧ α : α ω 2 so that, for each α < ω 2, (1) for each β < α, ȧ β s a P α -name of a subset of ω, (2) ċ α s a P α+1 -name such that P α+1 forces that ċ α wtnesses that A α = {ȧ β : β < α} s specal (3) Q α s a P α -name of a product, Q 0 α Q 1 α, of cardnalty at most ω 1, (4) Q 0 α s the P α -name of P Aα (as n Defnton 1.6), (5) Q 1 α s the P α -name of the poset Q Aα as n Defnton 2.4 whch adds ċ α. To prove that the poset P ω2 s ccc t s suffcent to prove by nducton on α, that each Q 1 α s forced to be ccc. Before dong so, let us assume that P ω2 satsfes the above condtons and check that ths wll prove the statement of the theorem. In partcular, we check that the collecton {ȧ α : α ω 2 } of P ω2 -names s forced to be a maxmal almost dsont famly whch s ω 1 -specal and ω 2 -ndestructble. Clearly condton 2 mples that t s forced to be ω 1 -specal (whch mples almost dsont). To show that t wll be ω 2 -ndestructble (and therefore maxmal) we check that t wll satsfy the hypothess of Lemma 2.2. If Ẏ s any P ω 2 - name of a subset of ω, there s an α < ω 2 such that Ẏ s a P α-name. By condton 4 and the remarks precedng Defnton 1.6, we have that t s forced that Ẏ meets ȧ β for all β α. Now assume, by nducton on α, that P α s ccc and let G α be a P α - generc flter. If α s a successor then Lemma 2.5 shows that Q Aα s ccc, so we assume that α s a lmt ordnal. Workng n V [G α ], consder an uncountable collecton {q : ω 1 } Q Aα. If α had countable cofnalty, then there would be a µ < α such that J = { : q Q Aµ } s uncountable. Agan, by Lemma 2.5, Q Aµ s (stll) ccc and so two members of {q : J} would be compatble n Q Aµ and also n Q Aα. Therefore we may assume that α has cofnalty ω 1. By passng to a subcollecton, we may assume that there s some n ω such that A q has cardnalty n for all. For each, fx {β : < n} α so that A q = {a β : < n}. Fx a functon c from P(n) nto ω so that for some uncountable set Γ ω 1, and all ncreasng sequences 1 ρ(1) < < ρ(l) n, c({ρ(1),..., ρ(l)}) = c q ({a β,..., a ρ(1) β }). ρ(l) For smplcty of notaton, we agan ust assume that Γ s all of ω 1. Choose m n maxmal (possbly m = 0) so that there s a µ < α such that β m < µ for uncountably many (and, by another reducton, all). Further, choose a suffcently large k and arrange that for each

9 MARTIN S AXIOM AND SEPARATED MAD FAMILIES 9 and ncreasng sequence 1 ρ(1) < < ρ(l) m c µ ({a β ρ (1),..., a β ρ(l) }) = c µ({a β ρ (1),..., a β ρ(l) }) < k. Further arrange that k s suffcently large (and by a futher reducton) that there s a fxed sequence {t 1,..., t n } P(k) so that for all and for dstnct 1, n, a β a β k and a β k = t. Fnally for <, we may assume that βn < βm+1. Now we return to the extenson V [G α P µ ] and fx elements p P α for ω 1 so that p µ G α and p forces that q has the desred propertes developed above. We may assume several thngs about p : (1) p has determned the sequence {β : 1 n} and the value of c (2) β dom(p ) for each m < n, (3) for all γ dom(p ), (a) p γ forces that p (γ) = ((t γ, A γ), c γ) Q 0 γ Q 1 γ (b) t γ s a member of [ω] <ω and t γ k, (c) there s a fnte set Iγ γ dom(p ) such that A γ = {ȧ δ : δ Iγ}, (d) the doman of c γ s P(A γ) and the nteger values have been determned. Now we may choose an uncountable J ω 1 so that the sequence {dom(p ) : ω 1 } forms a -system (and by possbly ncreasng µ) wth root contaned n µ. Snce P α s ccc, t s routne to fnd < such that (1) p and p are compatble, (2) max(dom(p )) < mn(dom(p ) \ µ), (3) t = t end-extends t β β for each m < n. Now we show that p and p have a common extenson whch forces that q and q are compatble. Ths wll complete the proof that Pα Q 1 α s ccc. Defne p so that dom( p) = dom(p ) dom(p ). The defnton of p µ s any member of G µ whch s below each of p µ and p µ. For each m < n, and γ = β, and for γ = β p(γ) = ((t γ, A γ {ȧ β p(γ) = ((t γ, A γ {ȧ β : 1 m}), c γ) : 1 n}), c γ).

10 10 A. DOW AND S. SHELAH For other γ dom(p ) \ µ, defne p(γ) = p (γ), and smlarly, for other γ dom(p ) \ µ, p(γ) = p (γ). We show that p forces that c q c q, as a partal functon on P(A q A q ), does not have any volatons of the requrements on a specalzng functon. As was shown n Lemma 2.5 ths ensures that there s a sutable extenson, thus showng that q and q are compatble. Let us frst observe that p forces that a β a β s contaned n k for all dstnct 1, n. If m < n then t follows that ths ntersecton s contaned n a β t because of the fact that β β was added approprately n the defnton of p(β ). Then, snce t = t, we β β have that a β a β = a β n k. Ths s also the case f m <, snce β a β, whch was forced by p to be contaned was approprately added n the defnton of p(β ). If 1, m, then the fact that c µ ({a β, a β }) = c µ ({a β, a β }) < k ensures that a β a β k, t also follows that for dstnct 1, n, a β (2.1) a β a β = a β s contaned n k. Snce p also forces that a β k = (a β k) = a β (a β k) = a β a β Suppose that ρ 0 and ρ 1 are ncreasng functons from {1,..., l} nto {1,..., n}. Assume that c({ρ 0 (1),..., ρ 0 (l)}) = c({ρ 1 (1),..., ρ 1 (l)} = k. By the choce of c, not only s ρ 0 and ρ 1 codng an arbtrary nstance where c q wll equal c q but also, f ρ 0 ρ 1, where c q wll agree wth tself. From ths latter fact, we can see that f, then ı = ρ 0 () ρ 1 () =, snce a β s requred to be almost dsont from a ı β. Now suppose that s n the range of ρ 0 and s n the range of ρ 1. We must show that a β a β s contaned n k. We know already, by vrtue of c q, that a β a β a β a β s contaned n k as requred. s contaned n k; and so, by equaton 2.1, References [1] Ur Abraham and Saharon Shelah, Lusn sequences under CH and under Martn s axom, Fund. Math. 169 (2001), no. 2, MR (2002k:03069) [2] Alan Dow, Sequental order under PFA, Canadan Math Bulletn (2011), to appear. [3] N. N. Luzn, On subsets of the seres of natural numbers, Izvestya Akad. Nauk SSSR. Ser. Mat. 11 (1947), MR (9,82c).

11 MARTIN S AXIOM AND SEPARATED MAD FAMILIES 11 [4] D. A. Martn and R. M. Solovay, Internal Cohen extensons, Ann. Math. Logc 2 (1970), no. 2, MR (42 #5787) [5] Jurs Steprāns, Strong Q-sequences and varatons on Martn s axom, Canad. J. Math. 37 (1985), no. 4, MR (87c:03106) [6] Stevo Todorčevć, Analytc gaps, Fund. Math. 150 (1996), no. 1, MR (98:03070) Unversty of North Carolna at Charlotte, Charlotte, NC E-mal address: adow@uncc.edu Department of Mathematcs, Rutgers Unversty, Hll Center, Pscataway, New Jersey, U.S.A Current address: Insttute of Mathematcs, Hebrew Unversty, Gvat Ram, Jerusalem 91904, Israel E-mal address: shelah@math.rutgers.edu