Andrzej Roslanowski and Saharon Shelah- The Last Forcing Standing

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    2 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    However, the exact relationships between these and other properness conditions arenot clear.

    While there are some similarities between conditions studied so far, we are farfrom the state that was achieved for CS iterations and the concept of properness.The considered properties are (unfortunatelly) tailored to fit particular forcing no-tions and they do not provide any satisfactory general framework covering all ex-amples. The search for the right notion of propernes is still far from beingcompleted.

    Basic definitions concerning strategically complete forcing notions, their itera-tions and trees of conditions are reminded in the further part of the Introduction.In the second section of the paper we prove our Iteration Theorem 2.6 and in thefollowing section we present the forcing notions to which this theorem applies. Somespecial properties of and relationships between the forcings from the third sectionare investigated in the fourth section.

    1.1. Notation. Our notation is rather standard and compatible with that of clas-sical textbooks (like Jech [2]). However, in forcing we keep the older conventionthat a stronger condition is the larger one.

    (1) Ordinal numbers will be denoted be the lower case initial letters of the Greekalphabet ( , , , . . .) and also by i, j (with possible sub- and superscripts).

    Cardinal numbers will be called , ; will be always assumed to bea regular uncountable cardinal such that

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    THE LAST FORCING STANDING 3

    (6) A tree is a downward closed set of sequences. A complete tree is atree T

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    4 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    pt Prk(t) for t T, and if s, t T, s t, then ps = ptrk(s).

    If, additionally, (T, rk) is a standard tree, then p is called a standard treeof conditions.

    (4) Let p0, p1 be trees of conditions in Q, pi = pit : t T. We write p0 p1

    whenever for each t T we have p0t p1t .

    Note that our standard trees and trees of conditions are a special case of that[6, Def. A.1.7] when = 1.

    2. Semi-purity and iterations

    In this section we introduce a new property of (

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    THE LAST FORCING STANDING 5

    (4) Let Y be an findexing sequence and p Q. A game mainY

    (p,Q, , pr, D)between two players, Generic and Antigeneric, is defined as follows. A play

    of the game lasts steps during which the players construct a sequencep, q : < . At stage < of the play, first Generic chooses a systemp = p, : Y of pairwise incompatible conditions from Q. ThenAntigeneric answers by picking a system q = q, : Y of conditionsfrom Q satisfying

    p, pr q, for all Y.

    At the end, Generic wins the play p, q : < if and only if, lettingq = q, : < & Y,() there is an aux-generic condition p p over q, D.

    (5) A forcing notionQ is fsemi-purely proper over an indexing sequence Y anda filterD if for some sequence pr of binary relations on Q, (Q, , pr) is aforcing with the fcomplete semi-purity and for every p Q Generic has a

    winning strategy in mainY (p,Q, , pr, D). We then say that the sequencepr witnesses the semi-pure properness ofQ.

    (6) IfD is the club filter on , then we omit it and we write mainY

    (p,Q, , pr)etc. If pr=pr for all < , then we write pr instead of pr, like in

    mainY

    (p,Q, , pr). If f() = for all , then we write instead of f (inphrases like complete semipurity etc).

    The proof of the following proposition may be considered as an introduction tothe (more complicated and general) proof of Theorem 2.6 dealing with the itera-tions.

    Proposition 2.4. Assume that f : + 1, + < f() for < and D isa normal filter on . Let Y = Y : < be an findexing sequence. If a forcing

    notionQ is fsemi-purely proper over Y , D, then it is proper in the standardsense.

    Proof. Let pr be a sequence witnessing the semi-pure properness ofQ. AssumeN (H(), ,

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    6 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    play (r, A, , r) : < of

    auxY

    (p, q,Q, , pr, D) in which COM follows herwinning strategy and INC plays:

    r0 = p+, and for > 0 he lets r = r.Let

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    THE LAST FORCING STANDING 7

    So suppose we have defined qj , rj for j < i. Stipulating r1 = p, t = t0, and

    =

    0 we ask if there is a condition q P such that rjti

    prw q for all j < i which

    forces a value to i. If there are such conditions, let qiti be one of them. Otherwise

    let qiti be any prw bound to {r

    jti

    : j < i} (there is such a bound by () + ()). Thenfor t T \ {ti} define qit so that letting s = t ti:

    if < rk(s), then qit() = qiti

    (),

    if rk(s) < rk(t), / w, then qit() is the 1),

    a condition in Q4 is a tree T Tclub such that

    (t T)(root(T) t |succT(t)| > 1).

    (4) For = 1, 2, 3, 4 we define a binary relation pr on Q by

    T1 pr T2 if and only if (T1, T2 Q and) T1 T2 and root(T1) = root(T2).

    (5) Let Q1, consists of all conditions T Q1 such that for each branch

    lim(T) the set { : |succT()| > 1} is a club of .(6) Let Q3, consists of all conditions T Q

    3 such that for some club C

    we have if t T and lh(t) C, then |succT(t)| > 1, and if t T and lh(t) / C, then |succT(t)| = 1.

    Observation 3.2. Let T Tclub. Then T Q1, if and only if there exists asequence F : < of fronts of T we have

    if < < , t F, then there is s t such that s F, if < is limit, t F (for < ) are such that t t whenever

    < < , then

    1 if and only if t

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    THE LAST FORCING STANDING 13

    (i) F is a front of T, T+1 T, and F T

    +1,(ii) if is limit, then T =

    i 1, then |succT+1(t)| > 1.

    Then Sdef=

    1} contains a club (foreach i < ). Also the set { < : F} is a club (remember (ii)). So wemay pick a limit ordinal < such that lh(s) < , F and |succTi()| > 1for all i < . Then (by (ii)) also |succT ()| > 1 and hence (by (iv)+(iii)+(i))|succS()| > 1 (and s ). So we may conclude that S Tclub. We will arguethat S Q considering the four possible values of separately.Case = 1Suppose lim(S). Then for each < the set { < : |succT ()| > 1}contains a club and thus the set

    Adef=

    < : is limit and ( < )(|succT ()| > 1) and F

    contains a club. But if A, then also |succT()| > 1 and hence |succS()| >1 (remember (ii)+(iv)).

    Case = 2Suppose that a sequence si : i < j S is increasing and |succS(si)| > 1 for alli < j. Let s =

    i 1 (for all i < j) and

    hence |succT (s)| > 1. By (iv)+(iii)+(i) we easily conclude |succS(s)| > 1 (notethat s t for some t F).Case = 3Let C be a club such that

    C & t T |succT(t)| > 1.

    Set C =

    1 for all < , and hence also |succT(t)| > 1 (by (ii)). Invoking (iv) we see that|succS(t)| > 1 whenever t S, lh(t) C is limit.Case = 4If root(S) s S, then |succT (s)| > 1 for all < and hence |succS(s)| > 1(remember (i)).

    Proposition 3.6. Let be a strongly inaccessible cardinal, Y = for <

    and Y = Y : < . Then the forcing notions (Q, , pr) for {2, 3, 4} are

    -semipurely proper over Y.

    Proof. Let 1 < 4, T Q. Consider the following strategy st of Generic in thegame main

    Y(T,Q, , pr).

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    14 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    In the course of the play, in addition to her innings T, : Y, Genericchooses also sets A Y and conditions T Q so that T

    is decided before the

    stage of the game. Suppose that the two players arrived to a stage < . If = 0 then Generic lets T0 = T and if is limit, then she puts T =

    i

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    THE LAST FORCING STANDING 15

    Case = 3.The winning strategy st of COM in the game aux

    Y(S, S, Q3, , pr) is almost

    exactly the same as in the previous case. The only difference is that now COMshrinks the answers S of INC to members of Q

    3, pretending they were played

    in the game. The argument that this is a winning strategy is exactly the same asbefore (as Q3, Q

    2).

    Case = 4. Similar.

    The forcing notions considered above can be slightly generalized by allowingthe use of filters other than the club filter on . The forcing notions QEE of [7,

    Definition 1.11] and PEE of [7, Definition 4.2] follow this pattern. However, to applythe iteration theorems of [7] we need to assume that the filter Econtrolling splittingsalong branches is concentrated on a stationary co-stationary set. Therefore the case

    ofE being the club filter seems to be of a different character. Putting general filterson the splitting nodes only and controlling the splitting levels by the club filter leadsto Definition 3.7.

    The forcing notion Q2,E was studied by Brown and Groszek [1] who describedwhen this forcing adds a generic of minimal degree.

    Definition 3.7. Suppose that E = E : 1}

    contains a club of ,

    (c)3 for some club C we have

    ( C)(t T)(lh(t) C |succT(t)| > 1),

    (c)4 (t T)(root(T) t |succT(t)| > 1).(3) For = 1, 2, 3, 4 we define a binary relation pr on Q

    by

    T1 pr T2 if and only if (T1, T2 Q and) T1 T2 and root(T1) = root(T2).

    Proposition 3.8. LetE = E :

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    16 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Proof. Same as 3.4, 3.6.

    Close relatives of the forcing notions Q,E

    were considered in [6, Section B.8]and [7, Definition 4.6]. The modification now is that we consider trees branchinginto less than successor nodes (but there are many successors from the point ofview of suitably complete filters).

    Definition 3.9. Assume that

    is strongly inaccessible, f : is a increasing function such that eachf() is a regular uncountalble cardinal and

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    THE LAST FORCING STANDING 17

    4. Are Q2, Q1 very different?

    The forcing notions Q1 and Q2 appear to be very close. In this section we will

    show that, consistently, they are equivalent, but also consistently, they may bedifferent.

    Lemma 4.1. Assume T Tclub and consider (T,) as a forcing notion. Let

    bea Tname for the generic branch added by T. Suppose that

    T the set { < : |succT()| > 1} contains a club .

    Then there is T T such that T Q2.

    Proof. Let C

    be a Tname for a club of such that

    T

    C

    |succT(

    )| > 1

    ,

    and putS =

    t T : lh(t) is a limit ordinal and t T

    < lh(t)

    C

    (, lh(t)) =

    }.

    One easily verifies that

    (i) ift S, then t T lh(t) C

    and hence |succT(t)| > 1,(ii) if a sequence t : <

    S is increasing, < , then

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    18 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Proof. Assume (a). By the ( 1} contains a club of ,

    where

    is a Tname for the generic branch. It follows now from Lemma 4.1that (a) holds.

    Proposition 4.4. Assume Ax+C . ThenQ2 is a dense subset ofQ

    1 (so the forcing

    notions Q1,Q2 are equivalent).

    Proof. Let T Q1

    and let us consider (T,) as a forcing notion. Let S be aTname given by

    T S

    = { < : |succT()| > 1}

    where

    is a Tname for the generic branch. Ask the following question

    Does T S

    contains a club of ?

    If the answer is yes, then by Lemma 4.1 there is T T such that T Q2.So assume that the answer to our question is not. Then there is t T such

    that

    t T \ S

    is stationary .

    The forcing notion T is isomorphic with C, so (by Ax+C

    ) there is a directed setdownward closed set H T such that

    H {t T : lh(t) > } = for all < , and ( \ S

    )[H] is well defined and it is a stationary subset of .

    Then =

    [H] =

    H lim(T) and the set { < : |succT()| = 1} is stationary,

    contradicting T Q1.

    Proposition 4.5. It is consistent thatQ2 is not dense inQ1.

    Proof. We will build a (

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    THE LAST FORCING STANDING 19

    Now we inductively define a ( 1.

    Claim 4.5.1. (1) If p, p P , then p P p if and only if dom(p) dom(p),

    p(0) Q0 p(0) and (i dom(p) \ {0})(p(i) = p(i) (p + 1)).

    (2) |P | = ||

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    20 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    q(0) = c {q}.

    One easily verifies that q P and it is stronger than p.

    Case is limit and cf() < Let p P. Fix an increasing sequence : < cf() cofinal in and thenuse the inductive assumption (and properties of an iteration) to choose inductivelya sequence p : < cf() such that for each < < cf() we have

    p P

    , < p < p and p P p P p.

    Then define a condition q P as follows. Declare that dom(q) =

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    THE LAST FORCING STANDING 21

    q(i) =

    n

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    22 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Proof of the Claim. By induction on < choose a sequence T : < so thatfor all < < we have

    ()5 T T, root(T) root(T) and()6 T+1 decides the value of

    3lh(root(T)), and

    ()7 if is limit then T =

    1 for all < . Then also, by ()7, = root(T) and clearly (by ()6) T forces a value to

    3.

    Claim 4.6.3. Let {1, 2}, T Q. Then there is T+ T such that for every

    t T, for some

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    THE LAST FORCING STANDING 23

    For = 1, 2 and t T+ let (t)