The character spectrum of

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  • Topology and its Applications 158 (2011) 25352555

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    01doTopology and its Applications

    he character spectrum of (N)

    aharon Shelah a,b,,1

    instein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The Hebrew University of Jerusalem, Jerusalem, 91904, Israelepartment of Mathematics, Hill Center Busch Campus, Rutgers, The State University of New Jersey, 110 Frelinghuysen Road, Piscataway, NJ 08854-8019, USA

    r t i c l e i n f o a b s t r a c t

    dicated to Kenneth Kunen

    SC:imary 03E35condary 03E17, 54A25

    ywords:t theorytraltersrdinal invariantsaracterrcing

    We show the consistency of: the set of regular cardinals which are the character of someultralter on N can be quite chaotic, in particular can have many gaps.

    2011 Elsevier B.V. All rights reserved.


    The set of characters of non-principal ultralters on N, that we call the character spectrum and denote by Sp , is nat-ally of interest to topologists and set theorists alike, see Denition 0.1 below. A natural question is what can this set ofrdinals be? The rst result on Sp is Pospils proof that c Sp .It is consistent that Sp = {20 }, since Martins Axiom implies Sp = {20 }. Nevertheless, Sp = {20 } is not a theoremZFC. Juhsz (see [1]) proved the consistency of the existence of a non-principal ultralter D so that (D) < 20 . Kunen

    n [2]) mentions that 1 Sp in the side-by-side Sacks model.Those initial results show that (D) is not a trivial cardinal invariant. But we may wonder whether Sp is an interesting

    t. For instance, can Sp include more than two members? Does it have to be a convex set? It is proved in [3, 6] thatp | large is consistent, e.g. 20 is large and all regular uncountable 20 (or just of uncountable conality) belong to. It was asked there: among regular cardinals is it convex? Now (proved in [6]) Sp does not have to be convex. In theodel of [6], there is a triple of cardinals (,,) such that < < ,, Sp but / Sp . In the present paper weow that Sp may exhibit much more chaotic behavior.To be specic, starting from two disjoint sets 1 and 2 of regular uncountable cardinals we produce a forcing notionwhich forces the following properties:

    ) no cardinal (of V) is collapsed in VP;) 20 is an upper bound for the union of 1 and 2;c) 1 Sp whereas 2 Sp = .

    Correspondence to: Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The Hebrew University of Jerusalem, Jerusalem, 91904, Israel.E-mail address: author thanks Alice Leonhardt for the beautiful typing. Partially supported by the Binational Science Foundation. Publication 915.

    66-8641/$ see front matter 2011 Elsevier B.V. All rights reserved.i:10.1016/j.topol.2011.08.014

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    (a(b(ce proof requires that each element of 2 be measurable and that 1 satisfy

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    or2) We say D is an ultra I-lter system when in addition:

    ) if s I , A N and A = mod Ds then for some t we have sI t and A l(Dt).

    3) If D is an I-lter system for = 1,2 then we let (D = D,t : t I and):

    ) D1 D2 means I1 I2 (as quasi-orders, so possibly I1 = I2) and s I1 D1,s D2,s) D1 D2 means I1 I2 and s I1 l(D1,s) l(D2,s)c) D1 D2 means I1 I2 and s I1 l(D1,s) = l(D2,s)) D1 = D2 means I1 = I2 and s I1 l(D1,s) = l(D2,s).

    bservation 1.4. Let I be a partial order.0) , and quasi-order the set of I-lter systems and l(Dt): t I is an I-lter system for any I-lter system Dd D1 D2 D1 D2 and D1 = D2 D1 D2 D1 D2 and D1 D2 D1 D1 = D2 and D1 D2 D1 1 = D2.1) If As [N]0 for each s I and At As for sI t then there is an I-lter system D such that s I Ds = {As}.2) If D is an I-lter system then for some ultra I-lter system D we have D D .3) If D is an I-lter system, s I and A and (t)[s I t A = mod l(Dt)], then for some I-lter system D weve D D and A D s .4) If D: < is an -increasing sequence of I-lter systems then some I-lter system D is an upper bound of the

    quence; in fact, one can use the limit, i.e. D,s ={D,s: < }; similarly for -increasing.5) If D is an I-lter system and D = l(Dt): t I then D D .6) If D is an I-lter system and each Dt is an ultralter on then D is an ultra I-lter system and necessarily sI t

    s = Dt .7) If D1 is an ultra I-lter system and D2 is an I-lter system such that D1 D2 then D1 D2.8) Assume P1 P2 and P1 D is an I-lter system for = 1,2. If P1 D 1 D 2 then P2 D 1 D 2; also if P1 1 D 2 then P2 D 1

    D 2.9) If P1 P2 and P D is an I -lter system for = 1,2 and P1 D 1 is ultra and P2 D 1 D 2 then P2 D 1,t

    2,t and (l(D 1,t)+)V[P1] l(D 2,t)


    oof. 0) Easy.1) Check.2) Use parts (3), (4), easy, but we elaborate. We try to choose D by induction on < (20 + |I|)+ such that D is an

    lter system, < D D and for each = +1 for some t, D,t = D,t . For = 0 let D = D , for limit use part) and for = + 1 if D is not ultra, use part (3). By cardinality consideration for some , D is dened but we cannotne D+1 so necessarily D is ultra as required.3)9) Easy, too. 1.4

    aim 1.5. 1) Assume the quasi-order I as a forcing notion adds no new reals. An I-lter system D is ultra iff I {l(Dt): t G I } isultralter on .

    2) Assume the quasi-order I as a forcing notion adds no new 1-sequences of ordinals and P is a c.c.c. forcing notion (or just I is1-complete). If P D t : t I is an I-lter system then PI

    {l(D t): t G I } is an ultralter on N iff P D t : t I is antra I-lter system.

    oof. Easy. 1.5iscussion 1.6. An I-lter system D may be degenerated, i.e. Dt = D is an ultralter, the same for every t I . But in thisse adding a generic set to I will not add naturally a new ultralter, which is our aim here.

    enition 1.7. 1) For D = D: >, each D a lter on N let QD be{T : T > is closed under initial segments, and for some

    tr(T ) >, the trunk of T , we have:(i) g

    (tr(T )

    ) T = {tr(T ) }

    (ii) tr(T ) > {n: ^n T } D }

    dered by inverse inclusion.

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    (a(b2) For p QD let wfst(p, D) be the set of pairs (S, ) such that:

    ) () S { p: tr(p) p}() tr(p) S( ) tr(p) S S

    ) () is a function from S into 1() if are from S then () > ()( ) if S and () > 0 then {k: ^k S} = mod D .

    3) If p QD and p then we let p[] = { p: or }.4) If D = D: >, D = D for > then let QD = QD and wfst(p, D) = wfst(p, D); we may write insteadp when this holds for some p QD with tr(p) = ;wfst stands for well founded sub-tree.

    aim 1.8. Assume >, D is a lter on N for > andY is a subset of = = {: >}. Then exactly onethe following clauses holds:

    ) there is q QD such that() = tr(q)() Y q = , equivalently q+ = q\{tr(q) : < g(tr(q))} is disjoint to Y

    ) there is a function such that (Dom( ), ) wfst(, D) and max(Dom( )) Y ; that is:() Dom( ) is a set satisfying

    (i) {: >}(ii) (iii) if and then

    () (i) Rang( ) 1(ii) () < ()

    ( ) for every at least one of the following holds:(i) Y(ii) the set {n: ^n } belongs to D+ .

    oof. Similar to [4, 4.7] or better [5, 5.4].In full, recall = {: >}. We dene when dp() for by induction on the ordinal :

    = 0: always a limit ordinal: dp() iff rk() for every < = + 1: dp() iff both of the following occurs:

    (i) /Y(ii) the following set belongs to D+ : {n: dp(^n) }.

    e dene dp() Ord{} such that = dp() iff ( Ord)[dp() iff ].Easily

    for every ,dp() 1 {}.

    Case 1: dp() = .For each such that dp() = clearly there is A D such that n A dp(^n) = . Let q be

    { p: or and if then (g()) A


    early q is as required in clause (a) of 1.8.

    Case 2: dp() < .We dene

    = {: and if k [g(), g()) then k /Y and dp(k)> dp((k + 1))}.e dene : 1 by () = dp().Now check. 1.8

    aim 1.9. P1 QD 1 P2 QD 2 when:

    ) P1 P2 and D = D ,: > for = 1,2

    ) D1, is a P1-name of a lter on N

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    emhe(coc) D 2, is a P2-name of a lter on N) P2 D 1, D 2, and moreover (l(D 1,)+)V[P1] l(D 2,)

    + , i.e. for every A P(N)V[P1] we have A l(D 1,) A l(D 2,).

    oof. Like [4, 4] more [5, 5] but we elaborate.Without loss of generality P1 and P2 p for every p P2. Clearly P1 QD 1

    P2 QD 2by clause (d) of the

    sumption and moreover P1 P2 P2 QD 2recalling Denition 1.1(1), (2). Now we can force by P1 so without loss of

    nerality it is trivial, hence we have to prove that QD1 P2QD 2identifying q QD1 with (,q) P2Q D 2

    . By clause (d) ofe assumption, this identication is well dened and QD1 ic P2 QD 2

    because for p1, p2 QD1 , p1, p2 are compatible iffr(p1) p2) (tr(p2) p1). It suces to verify 1.1(3), requirement 2. So let (p2,q 2

    ) P2 QD 2 ; without loss of generalityr some from V we have p2 = tr(q 2), so > and of course:

    )1 P2 q 2 QD2 .

    1.1(3), it suces to nd q QD1 such that

    )2 q q QD1 (p2,q 2),q are compatible; that is, (p2,q 2

    ), (,q) are compatible in P2 QD 2.

    ow we shall apply Claim 1.8 in V with , D1 here standing for , D there. Still Y is missing, so let

    Y = {: >and there is r QD1 such that = tr(r) and(, r), (p2,q 2

    ) are incompatible in P2 QD 2equivalently p2 P2 q 2

    , r are incompatible in Q D 2


    Claim 1.8 below we get clause (a) or clause (b) there.

    Case 1: Clause (a) holds, say as witnessed by q QD1 .We shall prove that in this case q is as required, i.e. q QD1 and [q QD1 r QD1 (p2,q 2

    ) P2 QD 2and r are

    mpatible (in P2 QD 2)].

    Why? Let = tr(r). Clearly ( q) hence by the choice of q, i.e. 1.8(a)() we have / Y so r cannot witnessY hence r, (p2,q 2

    ) are compatible in P2 QD 2as required.

    Case 2: Clause (b) holds as witnessed by the function .By the denition of Y , in V, we can choose q such that:

    (a) q = q : Y (b) q QD1 and tr(q) = (c) q witness Y , i.e. p2 q,q 2

    are incompatible in QD 2.

    e dene a P2-name q as follows:


    = {: or q 2

    and if g() k < g()

    and k Y then qk, hencek g() qk


    Clearly P2 q QD 2

    and tr(q

    ) = and QD 2| q

    2 q


    )3 if Y then and p2 P2 ( q ).

    hy? Otherwise there is p3 P2 such that p2 p3 and p3 P2 q , as tr(q

    ) is forced to be and tr(q) = ,

    cessarily p3 P2 q,q are compatible. But p2 P2 q 2


    , we get a contradiction to the choice of q .]Now we know that Dom( ) and q

    hence S := {: Dom( ) hence and p2 / q } is not

    pty. So as S Dom( ) the set U = {(): S} is not empty, and by the choice of the function we have U 1,nce there is a minimal U and let Dom( ) be such that () = . By the denition, if = 0 then by clauses) and () of 1.8(b), i.e. the choice of () we have Y and, of course, S . By ()3, p2 P2 ( q

    ) we get easyntradiction to S , hence we can assume > 0. By the denition of S there is p P2 such that P2 | p2 p and

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    pmq an

















    (P2 q hence q

    2 and, of course, S . By the choice of the function , in V we have A := {n: ^n Dom( )} =

    od D1, , hence by clause (d) of the assumption of the claim P2 A = mod D 2, and, of course, p P2 {n: ^n } D 2,. Together p P2 there is n such that ^n q Dom( ), so let n and p P2 be such that P2 | p p

    d p P2 ^n q Dom( ).

    So (^n) is well dened, i.e. ^n belongs to Dom( ) hence (^n) < () = and easily ^n S and^n) U , so we get a contradiction to the choice of . 1.9nition 1.10. 1) We say d= (D, F ) is a frame when:

    ) D = D: > and D [N]0 , / l(D) for >) F [N]0 and / l(F ).

    1A) Above let Dd = Dd,: >, Dd, = l(D), Fd = l(F ),Qd = QDd and if D = D for > we may write, Dd instead of D, Dd , respectively.2) We say A is a d-candidate when (d is a frame and):

    ) A is a Qd-name of a subset of N.

    3) We say A is d-null when it is a d-candidate and is not d-positive, see below.4) We say A is d-positive when for some p Qd , for a dense set of p p some quadruple (p, A, S, ) is a localitness2 for (A ,d) or for (, A ,d) when = tr(p) or for (p, A ,d) or for A being d-positive, which means:

    ) p Qd) A F+d) S = Sn: n A and = n: n A) (Sn, n) wfst(p, D) for n A recalling Denition 1.7(2)) if Sn and n() = 0 then p[] n A .

    nition 1.11. 1) For a frame d= (D, F ) let id d = id (d) = {A N: A is a Qd-name which is d-null}.2) If P d is a frame then id d [P] is the P Qd -name of id d .

    aim 1.12. For a frame d,QD id d is an ideal on N containing the nite sets and N / id d; moreover, for every A P(N) from V,e have A = mod Fd iff Qd A id d.

    oof. It suces to prove the following 1 4.

    1 If QD if A1 A2 and A2 id d then A1 id d.

    hy? If (p, A, S, ) is a local witness for (A 1,d) then obviously it is a local witness for (A 2,d).]

    2 if Qd if A1, A2 id d then A1 A2 id d.

    hy? It suces to prove: if Qd A 1 A 2 = A N and A is d-positive then A is d-positive for some {1,2}. Let, A, S, ) be a local witness for (A ,d) and we shall prove that there are {1,2} and a local witness for (tr(p), A ,d);the dense in Denition 1.10(4) this suces.For any n A and Sn such that n() = 0 we choose (n, , n, , Sn, ) such that:

    )2.1 (a) ,n {1,2}(b) (Sn, , n, ) wfst(p[], Dd)(c) if n, () = 0 so Sn, then there is q Qd such that p q, tr(q) = and q n A n, ; let qn, be such q.

    hy (n, , n, , Sn, ) exists? We shall use 1.8; that is for {1,2} let Yn,, = {: p and there is r QD suchat tr(r) = and p r and r n A }.We apply for = 1,2 Claim 1.8 with Dd, ,Yn,, here standing for D, ,Y there. If for some {1,2} clause (b) therelds as witness by the function , easily the desired ()2.1 holds. If for both = 1,2 clause (a) there holds then for = 1,2ere is q Qd such that tr(q) = and q Yn,, = .

    An equivalent version is when we weaken clause (e) to: if Sn and n() = 0 then there is q Qd such that tr(q) = , p q and q n A , see)2.2 in the proof. Moreover, we can omit p q; hence actually only tr(p) is important so we may write tr(p) instead of p.

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    (a(b(Necessarily q := q1 q2 p belongs to Qd and has trunk and is disjoint to Yn,,1 Yn,,2. But Qd | p[] q and] Qd n A = A 1 A 2, hence there are {1,2} and r Qd such that q r and r Qd n A , but then tr(r) Yn,,d tr(r) q q , contradicting the choice of q . So ()2.1 holds indeed.]

    )2.2 without loss of generality n, () = 0 g() > n.

    hy? Obvious.]

    )2.3 for n A there are n, S n, n such that(a) (S n, n) wfst(p, Dd)(b) S n Sn and max(S n) = S n max(Sn)(c) n {1,2} and max(S n) n, = n .

    hy? Easy.]

    )2.4 for n A letting S n ={Sn, : max(S n)} S n , for some n and qn we have:

    (S n, n ) wfst(p, D) {: n () = 0} = {: for some we have S n, n() = 0, Sn, and n, () = 0} q = qn, : n () = 0 n () = 0 p qn, tr(qn,) = qn, n A n .

    hy? Think.]

    )2.5 there is {1,2} such that A := {n A: n = } = mod Fd .

    hy? Obvious as A F+d .]We now consider the quadruple (p, A, S , ) dened by:

    p = { p: if tr(p) ,n g() and max(S n) then qn,} where S n,qn, are from ()2.4.

    hy p Qd with tr(p) = tr(p)? Recall ()2.2.]So together we have:

    A is from ()2.5, so A F+d S = S n : n A where S n is from ()2.4 = n : n A where n is from ()2.4.

    Now check that (p, A, S , ) is a local witness for (tr(p), A , D) hence 2 holds as said in the beginning of its proof.

    3 Qd id d; moreover if A = mod Fd is from V then A id d.

    hy? Because of clause (b) in Denition 1.10(4).

    4 QD [d] N / id d , moreover if B F+d and B V then B / id d.

    hy? This means that B is d-positive which is obvious: use the local witness (p, A, S, ) where p is any member of Qd ,= B , Sn = {tr(p)}, n(tr(p)) = 0. 1.12bservation 1.13. Assume d1,d2 are frames and Dd1 = D = Dd2 and Fd1 Fd2 then QD id d1 id d2 .

    oof. Should be clear. 1.13aim 1.14.We have P2 id d 1 id d 2 and (id d 1 )

    +[P1] (id d 2 )+[P2] when:

    ) P1 P2) P d is a frame for = 1,2c) P2 Dd1, Dd2, for >

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    H[) if A (D+d 1,

    )V[P1] then A (D+d 2


    ) P2 Fd 1 Fd 2 f) if A (F+d 1)V[P1] then A (F+d 2

    )V[P2] .

    oof. Should be clear by 1.15 below recalling 1.9. 1.14aim 1.15. Let d be a frame and A a QDd -name of a subset of N. We have A is d-null iff for a pre-dense set of p Qd we have(p) p there is no local witness for (p[], A ,d) equivalently, for (, A ,d).

    oof. Straight. 1.15mark 1.16. The point of 1.15 is that the second condition is clearly absolute in the relevant cases by 1.9, i.e. in 1.14.

    nition 1.17. 1) n(I) is the set of nite functions from I to H (0).2) Let K be the set of forcing notions Q such that some pair (I, f ) witness it, i.e. (I, f ,Q) K+ which means:

    ) f is a function from Q to n(I)) if p1, p2 Q and the functions g(p1), g(p2) are compatible then p1, p2 have a common upper bound p with g(p) =g(p1) g(p2).

    2) We dene K=wkK by: (I1, f1,Q1)wkK (I2, f2,Q2) means that:

    ) (I, f) witness Q K for = 1,2) I1 I2) f1 f2) Q1 ic Q2.

    3) We dene stK similarly adding:

    )+ Q1 Q2.

    4) If q K+ let q= (Iq, fq,Qq).

    mark 1.18. We can use much less in Denition 1.17.

    Consistency of many gaps

    We prove the rst result promised in the introduction. Assume = 1 and we like to build a c.c.c. forcing notionof cardinality , such that VP is as required: Sp includes 1 and is disjoint to 2; really we force by P

    T ,

    e T quite complete and translate P-names of ultra systems of lters to ultra-lters. In order to have 1 Sp , weall represent P as an FS iteration P,Q

    : , < , |P | and T is, e.g. >2 and for each 2 we have a = D ,s: s T a P-name of an ultra system of lters for unboundedly many < , increasing with ; in the ende force by P {T : 1}. Toward this for each s T , 1 we many times force by Q+D,s from Section 1.But in order to have 2 Sp = , we intend to represent P as the union of a -increasing sequence P: < and

    r each 2 for stationarily many < , cf() = and P+1 is essentially the ultrapower (P) /E , E a -completetra-lter on , so is a measurable cardinal.To accomplish both we dene a set Q, each x Q consist of an FS iteration of P,Q : +, < + with D s,: s {T : 1} for many < + , increasing with and Q = QDt(), .In the end for suitable x, we shall use P for some < + of conality (e.g. = 1). So why go so high as +? Itlps in the construction toward the other aim; we shall construct x: increasing in Q such that for each 2 for< of conality ,x+1 is essentially (x) /E . In particular, we have to prove Q = , the existence of the ultrapower ande existence of limit which happens to be a major proof here. For this we have to choose the right denition, in particularing id (D,s,D,t ) from Denition 1.11.For this section we assume:

    ypothesis 2.1. 1) We now x two cardinals and as well as two sets, 1 and 2, of regular cardinals in the interval, ] and let = 1 2.

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    DreprOur assumptions are:

    ) is regular and uncountable, = 0 and < ) 1 and 2 are disjoint sets of regular cardinals < from the interval [,) but / 2c) each 1 we have and = /E .

    2) Furthermore (and see 2.4 below so it is not a burden)

    f) T = T : 1,T is a tree of cardinality with levels, such that above any element there are elements of anyhigher level (may add T is 2-complete and even T is -complete, then clause (g) follows)

    ) for every 1, forcing by T := {T : 1\}, the product with Easton support, adds no sequence of ordinals oflength < and, for simplicity, collapses no cardinal and changes no conality; if = 1 add T is 1-complete;let T =Tmin(1)

    ) if 1 then |T | is (1\) except when sup(1) is strongly inaccessible and then the value is sup(1).

    oice 2.2. 1) Without loss of generality T : 1 is a sequence of pairwise disjoint trees.2) Let T be the disjoint sum of {T : }, so it is a forest.3) Let t = ti: i S be a sequence of members of T where S = { < +: cf() = cf()} such that if t T then S: t = t} is a stationary subset of +; let t(i) = ti .4) Furthermore choose

    ) S0 = { < +: cf() = 0} is stationary) = ,t,n: S0, t T , n N; let (, t,n) = ,t,n) ,t,n: n < is an increasing -sequence of ordinals with limit

    ) ,t,n { S: t = t}) guess clubs, i.e. if E is a club of + then the set { S0: C := {,t,n: t,n} E} is stationary.

    mark 2.3. If |T | < we can nd such , but in general it is easy to force such .

    aim 2.4. Assuming 2.1(1) only, a sequence T as in 2.1, clauses (f), (g), (h) (and also t, s, S , as in 2.2) exists, provided that1 { : = 2,). 2.4enition 2.5. Let Q be the set of objects x consisting of (below , +):

    ) P H (++) and I

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    deaim 2.11, i.e. completing an appropriate T -lter system to an ultra one, e.g. in Case 3 in the proof of 2.11. To help werry a strong induction hypothesis, see clause (i)( )2 in there and then rst nd an R j ,+[P j x]-name, then reect ita i .2) Note that it helps to have not only Q

    = QD , but possibly some related forcing notions. First in proving there is a

    it, see 2.11, in proving the reection discussed above lead us to use some unions. Second, using ultrapower by E , see13, for limit of conality , the ultrapower naturally leads us to use some iterations.3) We may in 2.1 demand / 1, equivalently < min(), but let T be a singleton {t} and T is Tmin(1) T .this case in 2.17 we get PT {} 1 Sp .

    nition 2.7. 1) For x Q, of course we let Qx = Qx = Q[x] = Q,Px = P[x] = P,Px = Px = P = Px+ , etc.2) We dene a two-place relation Q on Q: xQ y iff:

    ) (Ix

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    3aim 2.9. If (A) then (B) where

    ) (a) S0(b) P( ),Q

    ( ), E , etc., are as in Denition 2.5 except that all is up to (c) Q


    ,u ,t are as in clause (h) of Denition 2.5(d) D

    ,t :=

    {D ,t : S E} {u ,t,n: n N and s T satises sT t} so a P Q l()-name

    ) (a) PQ l()D

    ,t : t T is aT -lter system

    (b) PQ l()l(D

    ,t) = l({u ,s,n: sT t and n N})(c) PQ l()

    if t T and A {D,t : S} then u ,t,n A for every large enough n.

    oof. Straight; the point is PQ l() / l(D

    ,t) for t T , which holds as

    )1 if A D (,t,n) then for every large enough k, (,t,n)(k) A

    )2 if A D+ (,t,n) in V

    P then for innitely many k, (,t,n)

    (k) A)3

    is a dominating real. 2.9

    bservation 2.10. 1) Q partially orders Q.2) Px satises the c.c.c. and even is locally 1-centered3 when x Q and + .

    oof. Easy. 2.10aim 2.11 (The upper bound existence claim). If x: < is Q-increasing and is a limit ordinal < + then there is x which iscanonical limit of x: < , see below.

    enition 2.12. We say x= x is a canonical limit of x= x: < when x is Q-increasing, is a limit ordinal < + andor every < +):

    ) x Q) x Q x for < and Ex

    {Ex : < }c) I[x] ={I[x]: < }) if has uncountable conality then

    () Px ={Px : < }

    () Px

    Dx,t =

    {Dx,t : < } for t T if Ex S

    ( ) Qx =

    {Qx : < }

    () gx =

    {gx : < }.

    ) if has conality 0, then() if +\(S Ex )\(S0 Ex ) or S0C Ex then P [x ] Q

    [x] ={Q [x]: < } and similarly g

    [x] ={g

    [x]: < }() if S Ex then P [x ] D ,t[x]

    {D ,t[x]: < }f) in fact |Px | ({|Px |: < })0 .

    oof. Let

    0 (a) I ={I[x]: < } for < +(b) I

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    ree note that:

    2 (a) R, [Q, x] is a partial order(b) above R, [Q, x] is a dense subset of R, [Q, x] where R, [Q, x] is dened like R, [Q, x] when in 1(a) we

    omit subclause ().

    hy? Clause (a) by 3 below and clause (b) is easy.]So below we may ignore the difference between R, [Q, x] and R, [Q, x]

    3 for (, ,Q) as above; if (, p0) is a witness for p = (p1, p2) R, [Q, x] and (, ) then for some q0 P [x ] thepair (,q0) is a witness for (p1, p2) R, [Q, x].

    hy? As we can increase p0 in P [x], without loss of generality (p1) p0, where on recall Denition 2.5, clause (c).(, p0) is a witness for (p1, p2) R, [Q, x] necessarily p0, p2 are compatible in Q hence they have a common upperund q2 Q. As P [x ] Q, there is q0 P [x ] such that q0 q P [x ] q,q2 are compatible in Q. As we cancrease q0 in P [x ] and p0 q2 without loss of generality p0 q0 but (p1) p0 hence (p1) q0. As x x and[x ],Q

    [x ]: < + is FS iteration and p1 P [x]P [x ], clearly q0 q P [x ] q, p1 are compatible. So clearly,q0) is a witness for p R, [Q, x] as required in 3.]

    4 if , ,Q are as above and (1) + then R, [Q, x] R, (1)[Q, x].

    hy? We check the conditions from Denition 1.1(3), the second alternative. First, if p = (p1, p2) R, [Q, x] we shallove p R, (1)[Q, x]; as p R, [Q, x], some (, p0) witness it, easily it witnesses p R, (1)[Q, x] as P [x] (1)[x].Second, assume R, [Q, x] | p q and we should prove R, (1)[Q, x] | p q, this is obvious by the denition of

    e orders for those forcing notions. Together R, [Q, x] R, (1)[Q, x].Third, we should prove R, [Q, x] ic R, (1)[Q, x] so assume p,q R, [Q, x] has a common upper bound r = (r1, r2)R, (1)[Q, x]. Now easily (r1 , r2) is a common upper bound of p,q in R, [Q, x] as required.Fourth, for p R, (1)[Q, x] we should nd q R, [Q, x] such that if R, [Q, x] | q q then q, p are compatibleR, (1)[Q, x].Now let p = (p1, p2) R, (1)[Q, x] and let (, p0) witness it; without loss of generality P [x] | (p1) p0.Let q1 = p1 P [x], now q := (q1, p2) satises

    q R, [Q, x].

    hy? The pair (, p0) witness it because if p0 q P [x] then rst p1,q has a common upper bound r P (1)[x]nce r P [x] is a common upper bound of q,q1; second q, p2 has a common upper bound in Q as (, p0) witness1, p2). So indeed (, p0) witness q = (q1, p2) R, [Q, x].

    If q q R, [Q, x] then q, p are compatible in P (1)[x].

    hy? Let q = (q1,q2) and let r1 = (p1[ , (1))) q1, easily (r1,q2) R, (1)[Q, x] is a common upper bound of q, p.This nishes checking the last demand for R, [Q, x] R, (1)[Q, x] so 4 holds.]

    5 if Q satises the c.c.c. then R, [Q, x] satises the c.c.c.

    hy? Let pi = (p1,i, p2,i) R, [Q, x] for i < 1. Let (i, p0,i) be a witness for (p1,i, p2,i). As before let qi Q be such that,0, p1,i, p2,i are below it.We can nd an uncountable S such that f [xi ](p1,i): i S are pairwise compatible functions and i: i S is non-creasing. As Q satises the c.c.c., for some i < j from S there is a common upper bound q Q of qi,q j ; let {: < n}t in increasing order {} dom(p1,i) dom(p1, j)\ and let n = .By induction on n we choose r P [x j ] such that:

    if = 0 so = then r0 r P [x j ] r,q are compatible in Q if =m+ 1 then rm r

    P [x j ] | (p1,i) r and (p1, j) r.

    r = 0 use q Q and P [x j ] Q. For = m + 1, we shall choose r Pm+1[xi ] as follows: if / dom(p1,i) then= rm {(, p1, j())}; if / dom(p1, j) similarly; otherwise, i.e. if dom(p1,i) dom(p1, j) use the demands on g calling () of clause (c) and end of clause (d) of Denition 2.5.

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    4Having carried the induction, (rm,q) is well dened. Now let r P [x j ] be above r0 such that r r P [x j ] rm, re compatible. Also r r P [x j ] r0 r P [x j ] r,q are compatible in Q. So ( j, r) witness (rm,q) R, [Q, x]d easily (rm,q) is above pi = (p1,i, p2,i) and above p j = (p1, j, p2, j), so 5 holds indeed.]

    6 for , ,Q as above, Q R, [Q, x] when we identify p2 Q with (, p2).

    hy? Again, rst p Q p R, [Q, x] by the identication, and for p,q Q we have Q | p q R, [Q, x] | q by the denition of the order of R, [Q, x]. So Q R, [Q, x] holds, moreover Q ic R, [Q, x] by the denitionthe order.Lastly, let q R, [Q, x], so by 2 without loss of generality q = (q1,q2) R, [Q, x] and we shall nd p Q such

    at p p Q p, (q1,q2) are compatible.Let p = q2, i.e. (,q2), and the rest should be clear.]

    7 for , ,Q as above we have P [x] R, [Q, x] when we identify p1 P [x] with (p1,).

    hy? Similarly.]

    Now by induction on i + we choose i and P, f (when i and j < i j < ), Q

    , g

    (when < i and

    < i j ) and4 also D i (when i S) such that

    the relevant parts of clauses (a)(e) of Denition 2.12 and of the denition of x Q holds, in particular (all whendened):(a) P H (+) is a c.c.c. forcing notion(b) () Px P and Px P = Px for <

    () (I

  • 2548 S. Shelah / Topology and its Applications 158 (2011) 25352555




















    LePCase 2: i is a limit ordinal.Let = i be { j: j < i}, clearly j: j i is increasing continuous and i E . Below vary on .Let P ={P: < } and f ={ f: < } and from 0 recall I

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    (Let q0 = q1 , it belongs to P [x]; clearly q0 q P [x] p2,q are compatible in P [x]. Also clearly p0 q0 [x] so there is r1 such that q0 r1 P[x] and r1 forces a truth value to n A

    so as r1 is compatible with q1,cessarily r1 n A

    . So p0 q0 r1 P [x].By the denition of A

    and the choice of p0, there is q2 P [x] such that:

    )3.7 (a) P | p2 q2 and q0 r1 q2(b) q2 P n A .

    t (1) < be such that q1 P(1)[x]; as (q1) = q0 r1 q2 and as P ,Q : < is an FS iteration,

    early q1,q2 are compatible in P(1) and let q4 P(1) be a common upper bound of (q1),q2. Let q0 P(1)[x] bech that q0 q P(1)[x] q,q4 are compatible in P(1) , so as (q1) q4, without loss of generality (q1) q0.

    )3.8 q0 P [x] and (,q0) witness (q1,q4) R, [P, x].

    hy? As x Q and q1 = q0 q0 clearly q0 P [x ] q1 P [x]/G P [x ].For proving q0 P [x ] q4 P/G P [x ] recall the choice of q0.]

    )3.9 (q1,q4)R, [P ,x] n A B \[0,n).

    hy? First, q1 P [x ] n B by the choice of q1 hence (q1,q4)R, [P ,x] n B recalling P [x] R, [P, x] by 7.Second, q4 P n A because q2 P n A and q2 q4,P P and so (q1,q4) R, [P ,x] n A because P , [P, x] by 6.Third, n n recalling the choice of n. So ()3.9 holds.]Together we have proved ()3.Lastly, clearly i E and let D = D

    . If = i / S we are done. So assume S; by the induction hypothe-

    s = j < P j+1 D j+1 is ultra T -lter system, and D increases with , also necessarily cf() = henceP

    {D ,t : < }: t T is ultra hence D is ultra so we are done.

    Case 3: i = j + 1, j / S S0.Let ( j, +] and R = R j , [P j , x], recalling 5 we know R satises the c.c.c., by 6 we know P j R and by

    7 we know < P [x] R. For t T , let D j , ,t

    ={D ,t[x]: < } D j ,t , noting D j ,t ={D ,t : j}, so

    the induction hypothesis, R / l(D j , ,t

    ) so D j , ,t

    = D j , ,t

    : t T is a R j , [P j ]-name of a T -lter system.ence there is D

    i ,

    = D i , ,t : t T , a R-name of an ultra T -lter system above D i , , without loss of generalityi , ,t

    = l(D i , ,t) for t T . In particular this holds for = + hence Ei is a club of + where

    )4 Ei = { < +: is a limit ordinal from E and if < then D i ,+,t P(N)V[P ]: t T is a R j ,1 [P j , x]-namefor some 1 < }.

    we can choose i = (i) Ei E S\( j + 1).Let Pi = R j ,i [P j , x] and similarly P = R j ,[P j , x] for ( j, i) and D i = D

    i ,

    +,t P(N)V[P(i)]: t T .Also the choice of Q

    , g

    for [ j, i) is dictated by clause (g) of hence also of f and it is easy to check that alle clauses in the induction hypothesis are satised.

    Case 4: i = j + 1, j S .So P j D j is an ultra P j -lter system. Let = j .Let Q

    = Q

    D ,P+1 = P QD . By Claim 1.9, P+1[x] = P [x] Q D [x ]

    P QD = P j+1 for < . So+1, [P+1, x] is well dened for [ + 1, +].For t T let D

    +1,s be the dual of id d t(),s [P ], a P+1-name.

    )5 R+1, [P+1,x] / l({D ,s[x]: < } D

    ,s) for (,+].

    ote that for (, ) we know the parallel statements.

    )6 convention: we write (p1, p2, p 3) = (p1, (p2, p 3

    )) for members of R+1, [P+1, x], where we treat P+1 as P Q D,

    so p2 P and P p3 QD and tr(p3) is an object not just a name.

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    (a(b(ce need

    )7 if (A) then (B) where(A) (a) p = (p1, p2, p 3

    ) R+1, [P+1, x](b) t T(c) A is a P+1-name of a member of D

    ,t that is, P+1 N\A is d t(),t-null(d) < and B is a P [x]-name of a member of D ,t[x](e) n N

    (B) p R+1, [P+1,x] A B [0,n).

    rst note

    )7.1 (a) let (, (p0, p3)) where (p0, p3) P [x] QD [x ]

    witness p R+1, [P+1, x](b) let q2 P be above p0, p2(c) let q0 P be such that q0 q P [x] q2,q are compatible(d) let B

    be the following P+1[x+1]-name {n: there is q P [x]/G P+1[x ] forcing n B above p1 when p1 P [x]/G P+1[x ]}.

    Next consider

    )7.2 P+1 A B [0,n).

    hy is ()7.2 true? Note that P+1[x ] B (id d(t ,s))

    + by clause (g) of Denition 2.5, as P [x ] B D ,s and B B .w apply Claim 1.14 for P+1[x] = P [x] QD ,t()[x ] and P+1 = P QD ,t() .Why is ()7.2 enough for proving ()7? As in the proof of Case 2, only much easier.

    Case 5: i = j + 1, j S0.Let = j ; and let P+1 = P Q l()

    so Q

    = Q l()

    , and again P [x] Q l() P Q l()

    by 1.9. Clearly

    +1, [P+1, x] is well dened for [ + 1, +]. We let D +1,t ={D ,t : S E} {u ,s,n: s T t and n N},P+1-name.

    We have to prove the parallel of ()5, i.e.

    )8 R+1, [P+1,x] / l(D,t) for [ + 1, +] and t T .

    2.9 it suces to prove

    )9 R+1, [P+1,x] / l({u ,s: sT t}) for t T .

    w it is like Case 4 only easier. 2.11aim 2.13. If x Q and 2 then we can nd a pair (y, j) such that

    ) xQ y) j is an isomorphism from (Px) /E onto Py extending j1 where j is the canonical embedding of Px into (Px ) /E) j maps (Px) /E onto P

    y for any <

    + satisfying cf() = ) note that j maps j(Px) to a -subforcing of P

    y for + satisfying cf() = .

    Before proving 2.13 recall:

    nition 2.14. 1) For a c.c.c. forcing notion P and P-name A of a subset of N we say that p = (pn,m, tn,m): m,n < presents A when:

    ) pn,m P and tn,m is a truth value) for each n, pn,m: m < is a maximal antichain of P) for n,m < we have pn,m P n A iff tn,m.

    2) For p as in part (1) let Ap be the canonical P-name represented by p.

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    [Wct 2.15. 1) If P is a c.c.c. forcing notion and A is a P-name of a subset of N then some (pn,m, tn,m): n,m < represents A .2) If P is a c.c.c. forcing notion and A, A

    are P-names of subsets of , both represented by (pn,m, tn,m): n,m < then P = A.

    3) For a sequence t= tn,m: n,m < of truth values, for some formula = 0t (x) L1,1 ( ), = {}where x= xn,m: n < e have: for every c.c.c. forcing notion P and pn,m P (n,m < ) we have:

    P | (pn,m: n,m < ) iff (pn,m, tn,m): n,m < represents a P-name of a non-empty subset of .

    4) For k < , sequences t = tn,m: n,m < of truth values for k for some L1,1 ( )-formula = kt0,...,tk (y, x0, . . . , xk)here x = xn,m: n,m < we have:

    for every q, pn,m P (n,m < , k), P a c.c.c. forcing notion we have: P | [q, p0n,m: n,m < , p1n,m: n,m < ,. . . , pkn,m: n,m < ] iff (pn,m, tn,m): n,m < represents a P-name of a subset of which we call A , for k and q P A kand N\A k do not almost include A 0 A 1 A k1.

    oof. Easy. 2.15mark 2.16. In 2.15 we can treat any other relevant properties of such P-names.

    oof of 2.13. Let be large enough, x H () and B = (H (),) /E and let j the canonical embedding of (H (),)to B.We now dene

    (a) P is (Pj(x)j())

    B if +, cf() = and P ={P : < } if < + cf() = (b) I

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    t u geso




    soaofSo by

    T {Dy,t : t

    } is an ultralter on N.

    is suces for by 1.5 so we are done. 2.13We lastly arrive to the desired conclusion.

    nclusion 2.17. There is P such that (for our T see 2.1(g), 2.4):

    ) P is a c.c.c. forcing notion of cardinality and P 20 = ) T has cardinality 1 + , add no new sequence of length

  • S. Shelah / Topology and its Applications 158 (2011) 25352555 2553


















    [WarthSo fullling the second promise from Section 0 (the rst was dealt with in Section 2, i.e. 2.17) the main result of thisction is:

    nclusion 3.1. 1) If u {1,2, . . . ,n, . . .} and n 1 n u n + 1 u and in V there are innitely many measurable cardinals,en for some forcing notion P in VP we have Sp = {n: n u}.2) Assume in V there are innitely many compact cardinals. Then in part (1) we can use any u [1,).

    oof. Straightforward from 3.2, 3.4 below. 3.1aim 3.2. Assume GCH for simplicity, Hypothesis 2.1 and 2 > sup( ) and T is -complete for 1, = cf() formplicity; let f be a function with domain 2 such that > f() > sup( ), f() > 1 is regular (so f()

  • 2554 S. Shelah / Topology and its Applications 158 (2011) 25352555


















    5)4 there are (r,q,q, A): < such that:

    (a) r T

  • S. Shelah / Topology and its Applications 158 (2011) 25352555 2555

    Without loss of generality p2 : U1 are pairwise compatible hence by os theorem for some

    ()13 U1 so < and p2, p2 has a common upper bound p3 Px ()+1 , hence p1, p3 has a common upper boundp4 Px() .

    So recalling q is from ()4,

    ()14 (p4, r,q ) forces(a) A

    D(b) (C 0 , . . . , C n1) D(c) (C 0 , . . . , C n1)

    (A )

    [t], (A )

    [t] .

    Contradiction. 3.2Cl










    [5[6aim 3.4. In 3.2 (and 1.6) instead of E is -complete (so is measurable) we may require that there is 2 2 such that:

    ) ( 2, f) are as in 3.2) dening Q we use 2 if 2 then E is -completec) if 2\ 2 then = max( 2 ) is well dened, [, ] 1 = and E is a uniform -complete ultralter on so is a

    -compact cardinal.

    oof. Similar to 3.2. 3.4mark 3.5. The situation is similar for any set {: u} of successor of regular cardinals.

    aim 3.6. In 3.1 above the sucient conditions for / Sp in VP are sucient also for ()(cf()) = ( / Sp ).

    oof. The same. 3.6So we can resolve Problem (6) from Brendle and Shelah [3, 8].

    nclusion 3.7. If GCH and 1 < = cf() < = , is measurable, then there is a forcing notion P of cardinality col-psing the cardinals in (, ) but no others such that in VP , for every cardinal (,) of conality , we have / Sp =p(Sp ).


    ] Istvan Juhsz, Cardinal Functions in Topology Ten Years Later, second edition, Math. Centre Tracts, vol. 123, Mathematisch Centrum, Amsterdam, 1980.] Kenneth Kunen, Ultralters and independent sets, Trans. Amer. Math. Soc. 172 (1972) 299306.] Joerg Brendle, Saharon Shelah, Ultralters on their ideals and their cardinal characteristics, Trans. Amer. Math. Soc. 351 (1999) 26432674,math.LO/9710217.

    ] Saharon Shelah, Two cardinal invariants of the continuum (d < a) and FS linearly ordered iterated forcing, Acta Math. 192 (2004) 187223, also knownunder the title Are a and d your cup of tea?, math.LO/0012170.

    ] Saharon Shelah, Long iterations for the continuum, Arch. Math. Logic, submitted for publication, math.LO/0112238.] Saharon Shelah, The spectrum of characters of ultralters on , Colloq. Math. 111 (2) (2008) 213220, math.LO/0612240.

    The character spectrum of (N)0 Introduction1 Preliminaries2 Consistency of many gaps3 The n's and collapsingReferences