FORCING COPIES OF ω 1 WITH PFA(S)

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1 FORCING COPIES OF ω 1 WITH PFA(S) ALAN DOW Abstract. We work n the PFA(S)[S] model. We show that a non-compact sequentally compact space of small character wll contan a copy of ω 1. Droppng the character assumpton, we obtan that there are uncountable free sequences, and no countably tght compactfcaton. 1. No Introducton As usual S s a coherent Sousln tree. Not sure f there s a conventon, but we vew t as a (full branchng) subtree of ω <ω1. It s mpossble to turn S upsde down n our head so we wll use s < s to mean that s s, and so, n forcng statements, s s a stronger condton. We have an S-name of a sequentally compact space X whch we may assume has base set θ. We have an assgnment of S-names of neghborhoods { U(x, n) : x θ, n ω}. We may assume that 1 forces that these are regular descendng. If possble they also form a neghborhood base. Otherwse, we assume smply that 1 forces there s no fnte (countable) cover. 2. the sequental structure Snce S s ccc, t follows that f {ẋ n : n ω} s a sequence of S-names and 1 ẋ n X for each n, then there s an nfnte L ω such that 1 {ẋ n : n L} s a convergng sequence n X. Defnton 2.1. Say that a sequence {ẋ n : n L} s an S-convergng sequence provdng 1 {ẋ n : n L} s a convergng sequence (whch ncludes, for example, constant sequences). We may suppose for what we want to do, that ω 1 has the property that 1 forces that ω 1 does not have compact closure. We defne an S-convergence structure (s what I m callng t). We may thnk of these as unoffcal S-names of ponts n the closure of ω 1. Let Y denote the set of all objects y defned recursvely as follows. For each wellfounded tree T ω <ω, we let max(t ) denote the set of maxmal elements of T. An object y Y wll be a functon nto ω 1 wth doman max(t y ) for some well-founded tree T = T y wth certan propertes whch can be defned by nducton on the rank of T. The tree T should have a root (whch need not be ) and each non-maxmal node above the root should have an nfnte set of mmedate successors. As usual, for t T, T t wll denote the subtree wth root t and consstng of all t T whch Date: Frday at 10:20:42 (est) Mathematcs Subject Classfcaton. 54A20, 54G20, 54A35, 03E35, 54G12. Key words and phrases. proper forcng, Martn s Axom, countable tghtness, compact. and whoever jons hm. 1

2 2 ALAN DOW seql cohere are comparable wth t. The meanng of the rank of T wll really be the rank of T t where t s the root of T. We see no need for beng overly detaled on the noton of rank. A tree T of rank 0 wll have a unque maxmal node and any y wth doman max(t ) (and mage n ω 1 ) s n Y. We thnk of such a y as namng the pont y(max(t )) X. A tree T of rank 1 wll have a root t and max(t ) wll be an nfnte subset of {t l : l ω}. Any functon y from max(t ) nto ω 1 whch s an S-convergng sequence s a member of Y. Agan, we thnk of y as an S-name of ths unque lmt (a member of θ). For any well-founded T, t T, and functon y wth doman max(t ), we let y t denote the functon y max(t t ). Thus, by nducton on rank greater than 0, a functon y wth doman max(t ) s n y f (1) let t be the root of T, (2) let L be the set of n such that t n T, (3) for each n L, y t n Y, (4) the sequence {y t n : n L} s an S-convergng sequence, (5) we treat y as an S-name of the lmt. (6) We adopt the termnology that each y t n (n L) s a predecessor of y. Now that we have dentfed Y as the ponts of our space and ntroduced the noton of S-convergng sequence, we extend the noton to defne a closure operator on any gven fnte power of Y. Defnton 2.2. For each nteger n > 0, and subset B of Y n we smlarly defne the herarchy {B (α) : α ω 1 } by recurson. In addton, we agan (recursvely) vew each b B (α) as namng a pont n X n. The set B wll equal B (0). Naturally the pont named s the pont of X n named coordnatewse by b. For lmt α, B (α) (whch could also be denoted as B (<α) ) wll equal β<α B(β). The members of B (α+1) for any α, wll consst of a sequence b k : k ω where, for each k ω, b k wll be a member of B (α), and we requre that ths sequence s S-convergng n each coordnate. There s a subtle, but nessental, dfference between ponts of Y and members of B (α) for B Y. For n > 1 we may vew (Y n ) (ω1) as an S-sequental structure. Notce that ths S-sequental structure on Y n s defned n the ground model. The next lemma should be obvous. Lemma 2.3. For each A Y, 1 forces that A (ω1) s a sequentally compact subset of X. Defnton 2.4. For each S-name A and s A Y n, we defne the S-name ( A) (ω1) accordng to the property that for each s < t and t y ( A) (<ω1), there s a countable B Y n such that t B A and y B (<ω1). For an S-name A and s A Y n, we wll also vew ( A) (ω1) as naturally namng, n the forcng extenson, a subset of Y n. Ths may need some further clarfcaton. Lemma 2.5. Suppose that y s a member of B (α) for some B Y n and some α < ω 1. Also suppose that {s : < l} S and that Ẇ s an S-name for a neghborhood of (the pont named by) y. Then there s a b B such that for each

3 FORCING COPIES OF ω 1 WITH PFA(S) 3 < l, there s an s s forcng that b Ẇ. Proof. We may suppose that B s a countable set and we may proceed by nducton on α. By the defnton of B (α), there s a sequence y, whch s smply a sequence y k : k ω wth the property that y k B (<α) for each k. For each < l, choose s s so that s forces a value, W, on Ẇ (B { y k : k ω}). Snce ths sequence s assumed to be S-convergng to a pont n W, there s a k such that the pont named by y k s n each W. Of course the result now follows by the nducton hypothess. Let us now be gven a maxmal free flter F 0 of S-sequentally closed subsets of Y. 3. A new dea n PFA(S) As you well know, coherent Sousln trees have specal forcng propertes. Assume that g s (the) generc flter on S vewed as a cofnal branch. For each s S, let o(s) denote the level (order-type of doman) of s n S. For any t S, defne t s to be the functon s t [o(s), o(t)). We wll only use ths notaton when o(s) o(t). One of the propertes of S ensures that t s S for all s, t S. We smlarly defne g s to be the branch {t s : t g}. Defnton 3.1. Let bs denote the set of ω 1 -branches of S. Lemma 3.2. In the extenson V [g], bs = {g s : s S}. Furthermore, for each s S, V [g s] = V [g]. The flter F 0 may not generate a maxmal flter n the extenson V [g] and so we wll have to extend t. Lookng ahead to the PFA(S) step, we would lke (but probably can t have) ths (name of) extenson to gve the same flter n V [g s] as t does n V [g]. We adopt a new approach. We wll defne a flter (of S-sequentally closed) subsets of the product structure Y bs. We try to make ths flter somehow symmetrc. We ntroduce some notatonal conventons. Let S <ω denote the set of fnte tuples s : < n for whch there s a δ such that each s S δ. Our conventon wll be that they are dstnct elements. We let Π s:<n denote the projecton from Y bs to Y n (whch we dentfy wth the product Y {g s:<n} ). Defnton 3.3. Suppose that A s an S-name of a subset of Y n for some n, n partcular, that some s forces ths. Let s be any other member of S wth o(s ) = o(s). We defne a new name A s s (the (s, s )-transfer perhaps) whch s defned by the property that for all y <n Y n and s < t S such that t y <n A, we have that t s y <n A s s. Lemma 3.4. For any generc g S, val g s ( A) = val g s ( A s s ). Theorem 3.5. There s a famly F = {(s α, {s α : < n α}, F α ) : α λ} where, (1) for each α λ, {s α : < n α} S <ω, s α S, o(s α 0 ) o(s α ), (2) F α s an S-name such that s α F α = ( F α ) (ω1) Y nα (3) for each s S and F F 0, (s, {s}, ˇF ) F, (4) for each s S o(s α ), (s, {s α : < n α}, ( F α ) sα s ) F, (5) for each generc g S, the famly {Π 1 s α:<nα (val g( F α )) : s α g} s prevoustem fntely drected; we let F 1 be the S-name for the flter base t generates

4 4 ALAN DOW (6) for each generc g S and each s : < n S <ω, the famly {val g ( F α ) : s α g and {g s : < n} = {g s α : < n α }} s a maxmal flter on the famly of S-sequentally closed subsets of Y n Proof. Straghtforward recurson or Zorn s Lemma argument over the famly of symmetrc flters (those satsfyng (1)-(5)). Defnton 3.6. For any s : < l S <ω, let F s:<l denote the flter on Y l nduced by Π s:<l ( F 1 ). Defnton 3.7. Let A denote the famly of all (s, s : < l, A) satsfyng that o(s) o(s 0 ), s : < l S <ω, and s A F + s :<l. As usual, for a famly G of subsets of θ, G + denotes the famly of sets that meet each member of G. Lemma 3.8. For each (s, s : < n, A) A, the object (s, s : < n, A (ω1) ) s n the lst F. In ths next Lemma t s crucal that there are no dots on the sequence y M (s) : s S δ. Melement Lemma 3.9. Suppose that M H(κ) (for sutably bg κ) s a countable elementary submodel contanng Y, A. Let M ω 1 = δ. There s a sequence y M (s) : s S δ such that for every ( s, {s : < n}, A) A M, and every s S δ wth s < s, there s a B Y n M such that s B A and s y M (s s ) : < n B (δ+1). Proof. Let {(s m, {s m : < n m }, A m ) : m ω} enumerate the famly A M. Also, fx an enumeraton, {s δ m : m ω}, of S δ. Let {α m : m ω} be an ncreasng cofnal sequence n δ. At stage m, we let β m be large enough so that s δ 0 [β m, δ) = s δ [β m, δ) for all < m. Replace the lst {(s j, s j : < l j, A j ) : j < m} wth (abuse of notaton of course) {(s j, s j : < l j, A j ) : j < L m } so that for all, j < m wth s j < s δ (from orgnal lst) the new lst ncludes (sδ β m, s δ sj β m : < l j, A j ); and so that for all j < L m n the new lst s j and s j are all n S β m. Nothng else s added to the new lst (n partcular, the new lst s contaned n A M). Now we have s δ 0 s j k = sδ sj k for all < m, j < L m and sutable k < l j. Also, whenever s j < s δ (sδ for < m s unque) we have that sδ A j F + s j :<lj ; and so we also have that s δ 0 ( A j ) sδ F + (because they are essentally the same s δ 0 s j :<lj sets). Let Σ = {σ k : k K} lex-enumerate {(s δ 0 s j ) β m : j < L m, < l j } M. Let Π Σ be the projecton map from Y bs onto Y Σ. Let Π Σ be defned s j :<lj by the equaton Π Σ Π s j :<lj Σ = Π s j :<l. We consder the flter (name) F Σ. For each j < L m and < m such that s ( ) j 1 Π Σ s (( A j :<lj j ) sδ s δ 0 < s δ, t s forced by sδ 0 that the set ) (<ω1) ) s a member of F Σ and all are n M. Select any y m M Y Σ wth the property that Π Σ s j :<lj ( y m) A (<ω1) j for all j < L m. Choose a sequence {B j : j < L m } of countable subsets of Y <ω (n fact B j Y lj ) whch are n M and satsfy that, for each j < L m, s δ 0 B j ( A j ) s s 0 (where s j < s δ ), and so that Π Σ s j :<lj ( y m) B (<δ) j. Note that f s j < s δ, then sδ B j A j.

5 FORCING COPIES OF ω 1 WITH PFA(S) 5 If we now return to the orgnal lst, we have that for all, j < m and s j < s δ, s δ y m s δ 0 s j k β m : k < l j B (<δ) j. Now suppose we have so chosen y m for each m ω. We assert the exstence of an nfnte set L ω wth the property that for all j, ω, s δ forces that the sequence { y m (s δ j sδ α m) : m L} s defned and S-convergng on a cofnte set. For each, y M (s δ ) s the S-name n Y whch s equal to the lmt of ths S-convergng sequence. 4. S-preservng proper forcng Now we are ready to defne our poset P. Recall that we have a fxed assgnment { U(x, n) : x θ, n ω} of S-names of neghborhoods (regular descendng for each x). Defnton 4.1. A condton p P conssts of (M p, S p, m p ) where M p s a fnte -chan of countable elementary submodels of H(κ) for some sutable κ. We let M p denote the maxmal element of M p and let δ p = M p ω 1. We requre that m p s a postve nteger and S p s a fnte subset of S δp. For s S p and M M p, we use both s M and s M to denote s (M ω 1 ). It s helpful to smultaneously thnk of S p as nducng a fnte subtree, Sp, of S equal to {s M : s S p, and M M p }. For each s S p and each M M p \ M p we defne an S-name Ẇp(s M) of a neghborhood of y M (s M). It s defned as the name of the ntersecton of all sets of the form U(s M, m p ) where s S p, M M p M p, and s M s M and y M (s M) U(s M ). We adopt the conventon that Ẇ p (s M) s all of X f s M / Sp. The defnton of p < q s that M q M p, m q m p, S q Sp and for each s S p and s S q, we have that s forces that y M (s M) Ẇq(s M ) whenever M M p \ M q and M s the mnmal member of M q (M q \ M). It s a notatonal convenence that we make no requrements on sets of the form U(s, m q ) for s S q. We prove a knd of densty lemma. Lemma 4.2. If P M for some countable M H(µ), then for each p P and each α M ω 1, such that M H(κ) M p, there s an M M such that α M, r = (M p {M }, S p, m p ) P, and r < p. Proof. Let M 0 = M H(κ) and let S 0 = {s : < l} enumerate the set {s M : s S p } n the lexcographc order. In ths proof we adopt the conventon that we wll enumerate S q for any condton q n ncreasng lexcographc order. Let M be the maxmum element of M p M and set S = {s M : s S p }. We defne p M to be (M p M, S, m p ). It s routne to verfy that p < p M. By ncreasng α we may assume that s [α, M ω 1 ) = s j [α, M ω 1 ) for all, j < l and that ω 1 (M p M) < α. For each < l, let s = s α and let S = { s : < l}. It s easly checked that r = ({M 0 } (M p M), S 0, m p ) s n P and s an extenson of p M. Notce that ths mples that S r s equal to s 0 S = {s 0 s : < l}. Defne the S-name A as {(s q 0, ymq (s q ) : < l ) : q < p M, M q M q = M p M, m q = m p, S q = s q 0 S} snglem

6 6 ALAN DOW forproper Ths set A s a member of M 0 and, by vrtue of r, (s 0, y M0 (s ) : < l ) s an element of A. We show, by a densty and elementary submodel argument, that we have that s 0 forces that A s n F + s :<l. Frst of all, there s a dense subset of S each member of whch decdes the statement ( F F s:<l ) ( A F = ). Snce ths dense set s n M 0, there s an F M 0 such that s 0 F F s:<l, and ether s 0 A F s:<l or s 0 A F =. However, t s clear that there s a β M 0 such that s 0 β F F s:<l and so, by Lemma 3.9, y M0 (s ) : < l F. It follows then that there s an s M 0 below s 0 whch also forces that A s n F + s. :<l Well, what all ths proves s that (s, { s : < l}, A) s a member of the collecton A and s n M H(κ). Apply Lemma 3.9 and select B M Y l so that s 0 B A and s 0 y M (s 0 s ) : < l B (δ+1). What ths actually means (see Defnton 2.2) s that there s a b k : k ω of elements of B (δ) whch s S-convergng coordnatewse to ths element. Now, by Lemma 2.5, there s a b B satsfyng that each s S p forces that b s n the product neghborhood Ẇp(s 0 ) Ẇp(s l ). Ths b B s of course equal to y Mq (s q ) : < l for some q as n the defnton of A. It follows that M q s the desred value for M. That was a warm-up. What we really should have proven s Lemma 4.3. For all (s, r) S P such that s / M r, and any M 0 M r and (s j, s j : < l j, A j ) A M 0 wth s j < s, there s an a Y lj M 0 such that s a A j and for each s S r and < l j, s a() Ẇr((s s j ) M 0). Proof. Let s 0 = s M 0. By defnton of A, we have that s j A j F + s j :<lj. By Lemma 3.9, there s a countable B M 0 such that s 0 B A j and s 0 y M0 (s 0 s j ) : < l j B (δ0+1), where δ 0 = M 0 ω 1. Apply Lemma 2.5 to conclude there s a b B satsfyng that each s S r forces that b s n the product neghborhood Ẇr(s 0 s j 0 ) Ẇr(s 0 s j l 1 ). All we have to do now s to prove that Theorem 4.4. The poset S P s proper. Proof. As usual, we assume that M H(µ) (for some sutably large µ) s countable, and that M 0 = M H(κ) M p for some condton p. Let D M be a dense subset of S P and assume that (s, r) D. We may assume that there s some elementary submodel M ncludng r, that s / M, and that s M r S r. Ths means that for all s M and x M θ, s s forces a value on U(x, m r ) M r. Let M : l enumerate M r \M n ncreasng order. By Lemma 4.2 we can assume that M r M s not empty, and let M denote the maxmum element. Let α = M ω 1 and δ 0 = M 0 ω 1. We may addtonally assume that s [α, δ 0 ) = s [α, δ 0 ) for all s, s S r. The plan now s to fnd q M P so that (s, q) D and q r. Usual argung wll arrange that q < r M and, loosely speakng, that, for some easly chosen expanson S r,q of S r, (M q M r, S r,q, m r ) wll be a condton n P extendng q. The challenge s to ensure that such a condton s also an extenson of r, whch requres that, for each M M q, we have that s y M (s M) Ẇr(s M) for all s, s S r. Some standard elementary submodels as sde-condtons reasonng,

7 FORCING COPIES OF ω 1 WITH PFA(S) 7 together wth Lemma 4.3 do the trck. We have appled Lemma 4.2 above so we also have that s = s 0 s for each < l. One thng we have ganed s that n checkng f q r, we need only check on membershp n sets of the form Ẇr(s 0 s ). Let us say that q r provdng (1) M r M 0 s an ntal segment of M q, (2) M q \ M r = {M q : < l} has cardnalty l = M r \ M, (3) the tree structure (S q, ) s somorphc to (S r, ), (4) {s q : < l} = {s M q 0 : s S q} s also equal to {s q 0 s : < l}, For q r, and k < l, let y q k denote the tuple y M q k (s M q k ) : s S q (wth {s M q k : s S q} ordered lexcographcally as usual). Thus we have the l-tuple y q k : k < l assocated wth each q r. Recursvely defne a collecton of sets and names. Frst just notaton: For q r and k < l, l q,k = y q k and s q,k : < l q,k = dom( y q k ). Of course, l q,k s equal to l r,k for q r. Also, for each k < l, let k denote the ndex wth the property that s r,k < s. Then we have the S-name: Ẏ l = {(s, y k q : k < l ) : (s, q) D and q r} As usual, we have that Ẏ l M 0 (q r can be descrbed easly wthn M 0 ). Now defne, for k { l 1, l 2,..., 0} (4.1) A(q, k) = {( s, y k q ) : ( s, y j q : j k ) Ẏk+1, and let (for k > 0) s q k < s and y j q : j < k = y j q : j < k } Ẏ k = {(s, y m q : m < k ) : s A(q, k) F + s q,k :<l q,k } Thus A(q, l 1) contans the top element of the sequence y q k : k < l. Of course s A(q, k) F + s equvalent to (s, s q,k sq,k :<l q,k : < l q,k, A(q, k)) beng a member of A. A standard argument, walkng down from s, shows that, for each k < l, there s a β k Mk r such that (s β k, y r m : m < k ) s a member of Ẏ k+1. Now we have that there s some β 0 M 0 such that (s β 0, ) s a member of Ẏ 1 ; and more mportantly that (s β 0,, A(r, 0)) A M 0. By Lemma 4.3, there s a y 0 M 0 such that s y 0 A(r, 0) and, for each s S r and < l r,0, s y 0 () Ẇr((s s) M 0 ). Now, we do not really mean y 0 () but rather y 0 (s 0 ) for a sutable vector s0 : < l r,0. Indeed, there s some q 0 r such that q y 0 = y 0 0. If we let γ 0 = M q0 0 ω 1, then we have that s q0 = s γ 0, and for each 0 < l r,0, s 0 = s q0 = s q0 q γ 0. 0 By elementarty, there s a α 1 M 0 such that (s α 1, s 0 : < l r,0, A(q 0, 1)) A. We apply Lemma 4.3 agan and obtan q 1 M 0 such that s q y 1 1 A(q 0, l 1) and, for each s S r and < l r,1, s y 1 () Ẇr((s s 1 ) M 0). Unlke n the

8 8 ALAN DOW frst step, t may happen that s s 1 s not n S r. If we let γ 1 = M q1 1 ω 1, then we have that s q1 = s γ 1 ; and for each < l r,1, let s 1 = sq1. 1 Well, we just repeat ths argument for l steps untl we fnd q = q l r wth the property that (s, q) D and, for each k < l, and for each s S r and < l r,k, s y q k ((s s q ) M q k ) Ẇr((s s q ) M 0). Agan, for larger values of k, t may happen that (s s k ) M 0 s not n Sr, and so Ẇr((s s k ) M 0) would smply equal X. Department of Mathematcs, UNC-Charlotte, 9201 Unversty Cty Blvd., Charlotte, NC E-mal address: adow@uncc.edu URL: