saharon shelah- forcing axiom failure for any λ > ℵ1

8/3/2019 Saharon Shelah- Forcing Axiom Failure For Any > 1

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FORCING AXIOM

FAILURE FOR ANY > 1

SH784

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityMathematics DepartmentNew Brunswick, NJ USA

Annotated Content1 A forcing axiom for > 1 fails

[The forcing axiom is: ifP is a forcing notion preserving stationary subsetsof any regular uncountable and Ii is dense open subset ofP for i <

then some directed G P meets every Ii.We prove (in ZFC) that it fails for every regular > 1. In our coun-terexample the forcing notion P adds no new sequence of ordinals of length< ).

2 There are {1}-semi-proper forcing notions

I would like to thank Alice Leonhardt for the beautiful typing.Publ. 784 - Revised with proofreading for the JournalLatest version - 04/Mar/2

Typeset by AMS-TEX

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1 A forcing axiom for > 0 fail

David Aspero asks on the possibility of, see Definition below, the forcing axiom

FA(K, 2) for the case K = the class of forcing notions preserving stationarily ofsubsets of 1 and of 2. We answer negatively for any regular > 1 (evendemanding adding no new sequence of ordinals of length < ), see 1.16 below.

1.1 Definition. 1) Let FA(K, ), the -forcing axiom for Kmean that K is a familyof forcing notions and for any P K and dense open sets Ji P for i < there isa directed G P meeting every Ji.2) IfK= {P} we may write P instead ofK.

1.2 Definition. Let be regular uncountable. We define a forcing notion P = P2

as follows:

(A) if p P iff p = (, S, W) = (p, Sp, Cp) satisfying

(a) <

(b) Sp = S : = Sp :

(c) Cp = C : = Cp :

such that

(d) S is a stationary subset of consisting of limit ordinals

(e) C is a closed subset of

(f) if is a limit ordinal then C is a closed unbounded subset of (g) if C then C = C

(h) C S =

(i) for every and C we have S = S

(B) order: naturalp q iff p q, Sp = Sq (p + 1) and Cp = Cq (p + 1).

1.3 Observation: 1) P2

is a (non empty) forcing notion of cardinality 2.2) Ji = {p P2 :

p i} is dense open for any i < .

Proof. 1) Obvious.2) Given p P2 if

p i we are done. So assume p < i and for (p, i]let Sq be S

for any stationary subset S of { < : > i a limit ordinal}which does not belong to {Sp :

p} and let Cq = {j : p < j < } and


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4 SAHARON SHELAH

(iii) pi > i (and i+1 > pi follows from p P Ni+1)

(iv) Spii

= S and Cpii

= {j : j < i}

(v) p

i is the iSq

i

= S and

Cqi

= {j : j < i}

(vii) pi is the 0. If FA(P2), (the forcing axiom for the forcingnotionP2, dense sets) holds, then there is a witness (S, C) to where

1.7 Definition. 1) For regular uncountable, we say that (S, C) is a witness to or (S, C) is a -witness if:

(a) S = S : <

(b) C = C : <

(c) for every < , (, S ( + 1), C ( + 1)) P2.


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FORCING AXIOM FAILURE FOR ANY > 1 SH784 5

2) For (S, C) a witness for , let F = F(S,C) be the function F : defined by

F() = Min{ : S = S}.

3) For < let W(S,C)

= { < : F(S,C)() = }.

Proof of 1.6. LetJi = {p P2 : p i}, by 1.3(2) this is a dense open subset ofP2,

hence by the assumption there is a directed G P2 such that i < Ji G = .DefineS = S

p, C = C

p for every p G such that

p .Now check. 1.6

1.8 Observation: Let (S, C) be a witness for and F = F(S,C).1) If < then F() .2) If < is limit then F() < .

3) If < then WF()

(S,C).

4) If < and i = F() and C then / S = Si.

Proof. Easy (for part (4) remember that each S is a set of limit ordinals < andthat for limit p, p P2 we have = sup(C) and S C S = ).

1.9 Claim. Assume (S, C) is a -witness and S

satisfies S

cf() > 0 and F(S,C) S is constant and S is stationary. Then there is a club E

of such that: (S, C, S, E) is a strong (, )-witness, where

1.10 Definition. 1) We say that p = (S, C, S, E) is a strong -witness if

(a) (S, C) is a -witness

(b) S is a set of limit ordinals and is a stationary subset of

(c) E is a club of

(d) for every club E of , for stationarily many S we have

= sup{ C : < Suc1C

(, E) E}

where

()(i) Suc0C() = Min(C\( + 1)),

(ii) Suc1C(, E) = sup(E Suc0C()).


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2) We say (S, C, S, E) is a strong (, )-witness if in addition

(e) S cf() .

3) For (S, C, S, E) a strong -witness we let C = C : S acc(E), C =

C {Suc1C(, E) : C}; if p = (S, C, S

, E) we write C = Cp and Sp =

S, Cp = C, Sp = S

, Ep = E. We call (S, C, S, E, C) an expanded strong

-witness (or (, )-witness).

1.11 Observation. In Definition 1.10(3) for S acc(E) we have:

C

is a club of, Min(C

) sup(E Min(C)) and if1 < 2 are successivemembers ofC then C(1, 2) has at most one member (which necessarilyis sup(E 2)) hence acc(C

) = acc(C) and C < Suc

1C

() / C\C and acc(C) = acc(C

).

Proof of 1.9. As in [Sh:g, III], but let us elaborate, so assume toward contradictionthat for no club E of is (S, C, S, E) a strong (, )-witness. We choose byinduction on n sets En, En, An such that:

(a) En, En are clubs of

(b) E0 = (c) En is a club of such that the following set is not stationary (in )

An = { S : acc(En) and

= sup{ C : < Suc1C

(, En) En}}

(d) En+1 is a club of included in acc(En En) and disjoint to An.

For n = 0, En is defined by clause (b).

If En is defined, choose En as in clause (c), possible by our assumption towardcontradiction, also An S is defined and not stationary. So obviously En+1 asrequired in clause (d) exists.

So E =: {En : n < } is a club of and let () be the constant value ofF(S,C) S

, exists by an assumption of the claim. Recall that S() is a stationary

subset of , so clearly E =: { E : = sup( E S())} is a club of . AsS is a stationary subset of , we can choose S E. For each n < we


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1.13 Claim. LetP = PC be as in Definition 1.12.1) P is a (non empty) forcing notion.2) For i < the setJi = {c P : i < sup(c)} is dense open.

Proof. 1) Trivial.2) Ifc P, i < and c / Ji then let c2 = c {i + 1}, clearly (c2\c) S = as S

is a set of limit ordinals hence c2 P and obviously c c2 Ji. 1.13

1.14 Claim. Assume p = (S, C, S, E, C) is an expanded strong -witness.Forcing with P = PC add no new bounded subsets of , no new sequence of

ordinals of length < and preserve stationarity of subsets of .

Proof. Assume p P, and f

is a P-name of a function from to the

ordinals or just to V and S is stationary and we shall prove that there are q, satisfying (the parallel of) of 1.4, i.e.,

(i) p q P

(ii) q = + 1

(iii) S if =

(iv) q forces a value to f ( )

(v) if <

and p :

then q P f() < .

This is clearly enough for all the desired consequences. We prove this by inductionon , so without loss of generality = || and without loss of generality < = cf(), but if < then Sis immaterial so without loss of generality < & S cf() . Also we can shrink S as long as it is a stationary subsetof and recall that F(S,C) is regressive on limit ordinals (see Observation 1.8(2))

so without loss of generality F(S,C) S is constantly say ().

Let be large enough and choose N = Ni : i < such that

(a) Ni (H(), ,


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Let i = Ni and i() = Min{i : i < is a limit ordinal and i S}, it is welldefined as i : i < is (strictly increasing continuous) hence {i : i < } is a clubof , hence < cf(i()) = cf(i())

.

Let1

W =: {i i(): i > 0 and if i < i() and j < i then Ci() i+1 j+1}.

Clearly W i() is a closed subset of i() and as i() = sup(Ci()), also W i()is unbounded in i(). Also as by 1.11 we have ( acc Ci()) Ci()

C = Ci()

clearly

() if i W then Nj : j W (i + 1) Ni+1.

Also note that

() if i < i() is nonlimit, then i > sup(Ci() i) hence i > sup(Ci()

i).

[Why? By 1.8(4) as i() S S() recalling the choice of () clearly

Ci() S() = but by clause (f) we have i S() so i / C . ButCi() is a closed subset of i() hence i > sup(Ci() i), and C

i()

i\ sup(Ci() i) has at most two members (see 1.11) so Ci()

i is a

bounded subset of i so we are done.]

Now by induction on i W we choose pi, pi and prove on them the following:

()(i) pi, pi P Ni+1

(ii) pi is increasing (in P)

(iii) max(pi) > i

(of course i+1 > max(pi) as pi P Ni+1)(iv) pi = p {sup(i (C

i()

i+1)) + 1} if i = Min(W)

(v) if 0 < i = sup(W i) and i = max(Ci() i+1) so i i < i+1 then

pi = {pj : j W i} {i, i + 1}

(vi) if j < i are in W then pj pi pi

(vii) i W,i < i() and j < i satisfies j = Max(W i) and i = max({i} (C i+1)) so i i < i+1 then p

i = pj {i + 1}

(viii) pi is the 0 then W = {i < i() : i Ci()} {i()} is O.K.


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(ix) pi \j


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Proof. 1) Recall that by Claim 1.13(2), Ji is a dense open subset of P. Now ifG PC is directed not disjoint to Ji for i < , let E = {p : p G}. Bythe definition of PC and Ji clearly E is an unbounded subset of and by the

definition ofPC and G being directed, p G E (max(p) + 1) = p and (p isclosed) hence E is a closed unbounded subset of. So E contradicts the definitionof (S, C, S, E, C) being a strong -witness.2) Follows from 1.14 and direct checking. 1.15

1.16 Conclusion: Let be regular > 1.Then there is a forcing notion P such that:

() P of cardinality 2

() forcing with P add no new sequences of ordinals of length <

() forcing with P preserve stationarity of subsets of (and by clause () alsoof any = cf() [1, ))

() FA(P, ) fail.

Proof. We try P2, it satisfies clause (), (), () (see 1.3(1), 1.5, 1.6). If it satisfiesalso clause () we are done otherwise by Claim 1.6 there is a -witness (S, C). LetS { < : cf() > 0} be stationary, so by 1.9 for some club E of , thequadruple p = (C, S, S, E) is a strong -witness (see Definition 1.10), and letC = Cp.

Now the forcing notion P = PC

(see Definition 1.12) satisfies clauses (), (), ()by claims 1.15(2) and also clause () by claim 1.15(1). So we are done. 1.16


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2 There are {1}-semi-proper not proper forcing notion2

By [Sh:f, XII,2], it was shown that if remnant of large cardinal propertiesholds (e.g. 0#) then every quite semi-proper forcing is proper, more fully UReg-semi-properness implies properness. This leaves the problem

() is the statement

for every forcing notion P, P is proper follows from P issemi-proper, i.e., {1}-semi proper

consistent or is the negation provable

in ZFC.

David Asparo raises the question and we answer affirmatively: there are such forcingnotions. So the iteration theorem for semi proper forcing notions in [Sh:f, X] is notcovered by the one on proper forcing notions even if 0# does not exist.

2.1 Claim. There is a forcing notionP of cardinality 22 which is not proper butis {1}-semi proper. This follows from 2.2 using = 2.

2.2 Claim. Assume = cf() > 1, = 2. Then there is P such that

(a) P is a forcing notion of cardinality 2

(b) if > , p P N (H(), ), N countable, then there is q P above psuch that

q N N[GP] ( means initial segment); this gives P is {1}-semiproper and more

(c) there is a stationaryS []0 such thatP S is not stationary

(d) P is not proper.

Proof. We give many details.

Stage A: Preliminaries.Let M = (, Fn,m)n,m


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S = a []0 : for some b []0 we have() a b are closed under Fn,m for n, m < ,

() sup(a ) S1, sup(b ) S2

() (a ) (b ) ( is being an initial segment)

P = PS =

a :a = ai : i is an increasing continuous sequence

of members of []0\S of length < 1Clearly clause (a) of 2.2 holds.

Stage B: S is a stationary subset of []0 .Why? Let N be a model with universe and countable vocabulary, it is enoughto find a S such that N a N. Without loss of generality N has Skolemfunctions and N expands M. Choose for < , N N, N < , < N N, N, N increasing continuous.So C =: { < : a limit ordinal and N = } is a club of . Choose 1 < 2

from C such that 1 S1, 2 S2. Choose a countable c1 1 unbounded in 1,and a countable c2 2 unbounded in 2.

Choose a countable M N2 such that M N1 N1 and c1 c2 . Leta = M N1 , b = M N2 . As N

expands M, clearly a, b are closed under thefunctions of M. Also c1 M 1 = M (N1 ) = a N1 = 1 hence1 = sup(c1) sup(a ) 1 so sup(a ) = 1. Similarly sup(b ) = 2.Lastly, obviously a b so b witnesses a S, as required.

Stage C: P S is not stationary.Why? Define a

= {a : a G

P, g(a) > }. Clearly

()0 P = .[Why? Trivial.]

()1 for < 1,I1 = {a P : g(a) > } is a dense open subset ofP.[Why? If ai : i j P, j < < 1 we let ai =: aj for i (j,] and thenai : i j P ai : i .]Also


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()2 for < ,J2 = {a P : a for some < g(a)} is a dense open

subset ofP.[Why? Given a = ai : i j. Choose S2 such that > sup( (aj

{}) let c be countable unbounded in and let aj+1 = aj {} c; sotrivially sup(aj+1 ) = S2 hence aj+1 / S. Now let a+ = ai : i j + 1. Now check.]

So

()3 P

ai : i < 1 is an increasing continuous sequence of members of

([]0)V\S whose union is hence

()4 P ai : i < 1 witness S is not stationary (subset) of []0.

So we have finished Stage C.

Stage D: Clauses (c),(d) of 2.2 holds.Why? By Stage B and Stage C.

Stage E: Clause (b) of 2.2 holds.So let > , N a countable elementary submodel of (H(), ,


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now be clear but we elaborate]. Let M be the Skolem Hull in (H(), ,


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REFERENCES.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford Logic Guides.Oxford University Press, 1994.

[Sh:f] Saharon Shelah. Proper and improper forcing. Perspectives in MathematicalLogic. Springer, 1998.

saharon shelah- forcing axiom failure for any λ > ℵ1

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