Saharon Shelah and Juris Steprans- Martin's Axiom is Consistent with the Existence of Nowhere...

8/3/2019 Saharon Shelah and Juris Steprans- Martin's Axiom is Consistent with the Existence of Nowhere Trivial Automorphisms

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MARTINS AXIOM IS CONSISTENT WITH THE EXISTENCE OF NOWHERE

TRIVIAL AUTOMORPHISMS

SAHARON SHELAH AND JURIS STEPRANS

Abstract. Martins Axiom does not imply that all automorphisms of P(N)/[N]


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2 S. SHELAH AND J. STEPRANS

(4) max(ap) Ap

and the relation on Q(S) is defined by p q if and only if

(2.1) a

p

aq

, (max(a

p

) + 1) aq

= a

p

, f

p

fq

,B

p

B

q

,

p

q

(2.2) (Aq \ max(aq)) (aq \ ap) Ap

and, for each B belonging to Bp Bp , the following two conditions hold:

(2.3) ({n, m} [aq \ max(ap)]2)fq(B [m, n)) = Fp(B [m, n))

(2.4) ({n, m} [Aq \ max(aq)]2)Fq(B [m, n)) = Fp(B [m, n)).

If G is generic for Q(S) then define AS[G] =

pG ap and FS[G] to be

pG f

p.

Lemma 2.1. For any tower of permutations S the structure (Q(S), ) is a partial order.

Proof. That (Q(S), ) is reflexive and antisymmetric is obvious. To prove transitivity suppose thatp q and q r. The condition 2.1 for p r is easily seen to be satisfied. To see that condition 2.2 forp r is satisfied note that

Ar \ max(ar) Aq \ max(a

r) Aq \ max(aq) Ap

and that

ar \ ap (ar \ aq) (aq \ ap) Aq \ max(aq) Ap Ap

which shows that (Ar \ max(ar)) (ar \ ap) Ap , as required.To show that conditions 2.3 and 2.4 hold, let B Bp Bp. Given any pair {n, m} [ar \max(ap)]2, it

may, without loss of generality be assumed that n and m are successive elements ofar \ max(ap). Hence,

either {n, m} [(ar

\ max(aq

)]2

or {n, m} [(aq

\ max(ap

)]2

. In the second case, it follows immediatelyfrom the fact that p q that fq(B [n, m)) = Fp(B [n, m)). Since q r it follows that fq fr andso fr(B [n, m)) = Fp(B [n, m)). On the other hand, in the first case f

r(B [n, m)) = Fq(B [n, m))since q r and Bp Bq . Moreover, using condition 2.4 and p q it is possible to conclude thatFq(B [n, m)) = Fp(B [n, m)). Hence, in either case fr(B [n, m)) = Fp(B [n, m)) whichestablishes that condition 2.3 holds for p r.

To see that condition 2.4 holds for p r let {n, m} [Ar \max(ar)]2. It follows from p q and q r

that Fq(B [n, m)) = Fr(B [n, m)) and, since {n, m} Ar \max(ar) Aq \max(a

q), that Fq(B [n, m)) = Fp(B [n, m)). Hence, Fr(B [n, m)) = Fp(B [n, m)) establishing condition 2.4.

Lemma 2.2. Given a tower of permutations S = {(A, F,B)}W, an integer n, B WB and W the following sets are dense inQ(S):(2.5) {p Q(S) : max(ap) > n}

(2.6) {p Q(S) : p }

(2.7) {p Q(S) : B Bp}

Proof. To prove that the set 2.5 is dense let p Q(S) and n be given. Let k Ap

be such that k > n.Using Condition 4 of Definition 2.1 it follows that max(ap) Ap and, hence, Fp [max(a

p), k) is apermutation of [max(ap), k). Letting q = (ap {k}, fp Fp [max(a

p), k), p,Bp) it follows that q p.Observe for later reference, that is has actually been shown that

(2.8) (p Q(S))(n N)(q p)max(aq) > n and q = p and Bp = Bq.


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NOWHERE TRIVIAL AUTOMORPHISMS 3

To prove that the set 2.6 is dense let p Q(S) and W be given. Since it may as well be assumedthat > p, it is possible to find n so large that A \ n Ap and for all {i, j} [A \ n]

2 and B Bp

F(B [i, j)) = Fp(B [i, j)).

Using the set 2.5 of Lemma 2.2 choose q such that p q and n < max(aq). From 2.8 it can be assumedthat Bq = Bp and that q = p. Now, let r = (aq, fq, ,Bp). That conditions 2.2 and 2.4 for therelation q r is satisfied follows from the choice of n while condition 2.1 is obvious. Condition 2.3 hasno content in the case of q r since aq = ar. Now use transitivity and the fact that p q. Observe forlater reference, that is has actually been shown that

(2.9) (p Q(S))( > p)(q p)q = .

There is no problem in proving that the set 2.7 is dense.

Lemma 2.3. IfS = {(A, F,B)}W is a tower of permutations and p Q(S) then

(2.10) p Q(S) AS[G] \ max(ap

) Ap

(2.11) p Q(S) (B Bp Bp)({n, m} [Ap \ max(a

p)]2)FS[G](B [n, m)) = Fp(B [n, m))

Proof. This is standard using condition 2.2 for 2.10 and condition 2.3 for 2.11.

Let be a regular uncountable cardinal and let C be any set containing 0 and closed underlimits of increasing 1-sequences such that \C is also unbounded. Now define P, as well as a P-namefor a tower of permutations S = {(A, F,B)}C, by induction on . Let A0 [N]

0 and F0 bearbitrary subject to the fact that F0 is a permutation ofN such that F0 [n, m) is a permutation foreach {n, m} A0. Let B0 = P(N) in the sense of the ground model. Then let S1 = {(A0, F0,B0)} andlet P1 = Q0 be Cohen forcing. If is a limit then P is simply the finite support limit of {P} and

S =

CS. If / C then P+1 = P Q where Q is a ccc partial order chosen according to somebookkeeping scheme which will guarantee that Martins Axiom holds at stage . In this case S+1 = S.If C\ {0} then P+1 = P Q(S) and A is defined to be AS [G], F is defined to be FS [G] whereG is the canonical name for the generic set on Q(S). In this case S+1 = S {(A, F, P(N) VP)}.As usual, if p P then p P p() Q.

Definition 2.2. Let Pw P consist of all those p P such that there are k N, [C ]


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Definition 2.3. If p Pw is witnessed by (k, {(a, f)}) then define p+ P by

p+() = p() if / (a, f, max( ),B

p()

) if .Lemma 2.5. If p Pw then p

+ p.

Proof. Proceed by induction on to show that p+ p . Note that the cases = 0 or alimit pose no problem. Given that p+ p let be enumerated, in order, by {1, 2, . . . n}.If follows directly from Lemma 2.4 and the definition of p+ that

p+ P A1 \ k (A2 \ k) . . . (An \ k)

p+ P (B Bj Bi)({m, m

} Aj \ k)Fj (B [m, m)) = Fi(B [m, m

)) ifi j.

In particular, noting that there is some i such that = i,

p+ P A \ k An

and

p+ P (B B Bn)({m, m

} A \ k)F(B [m, m)) = Fn(B [m, m

)).

Hence, p+ + 1 p + 1.

Lemma 2.6. For each the subsetPw is dense inP.

Proof. Proceed by induction on noting that the cases 1 and a limit are trivial. Therefore,suppose that Lemma 2.6 has been established for and that p P+1. Without loss of generality it maybe assumed that C. Choose p p and (a,f, ) such that and p P p() = (a, f , ,B)and, moreover, p P B = {Bi}im and, for each i m the condition p

decides the value of the

least ordinal (i) such that Bi B(i). Furthermore, it may, without loss of generality be assumedthat {} {(i)}im domain(p

) and that p (i) P(i) Bi Bp((i)) for i m. Then use

the induction hypothesis to find p Pw extending p and such that (k, {(a, f)}) witnesses that

p Pw. Without loss of generality, k max(a). Now, the fact that {max(a), k} A implies thatF [max(a), k) is a permutation of [max(a), k). Hence it is possible to define f

= f F [max(a), k)and a = a {k}. Then define q so that

q() =

p() if

(a, f, ,B

Bp()) if =

and note that q Pw+1 and q p.

Corollary 2.1. For each the subsetP is dense inP.

Proof. This is immediate from Lemma 2.6 and Lemma 2.5.

Lemma 2.7. Given that p Pw and that this is witnessed by (k, {(a, f)}) and (C ) \domain(p) then the following condition p extends p and is also inPw :

p()

p() if =

({k}, {(j,j)}jk, 0,

Bp()) if = .

Proof. Notice that since / domain(p) the restrictions on extension do not apply and it is easy to checkthat the condition ({k}, {(j,j)}jk, max( ),Bp()) belongs to Q. Since p Pw it followsthat k A0 and, hence, that p Pw . That p p is immediate from the definition.


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Lemma 2.8. Suppose that p and q are conditions inP such that:

p P is witnessed by (k, {(ap , f

p )}p)

q P is witnessed by (k, {(aq , f

q )}q)

max(domain(p)) = max(domain(q)) = p q (aq , f

q ) = (a

p , f

p )

max(domain(q)) < min(domain(p) \ {0}) p(0) and (0) are compatible.

Under these conditions p and q are compatible.

Proof. Let the maximum member of q \ {}. Define q p by

(q p)() =

q() if domain(q) \ (q )

p() if domain(p) \ (p )

(aq(), fq(), q(), (Bp B0) Bq()) if q

(ap(), fp(), p(),Bp() Bq()) if p(ap(), fp(), q(),Bp() Bq()) if =

and define p q by

(p q)() =

q() if domain(q) \ (q )

p() if domain(p) \ (p )

(aq(), fq(), q(), (Bp B0) Bq()) if q(ap(), fp(), p(),Bp() Bq()) if p(ap(), fp(), p(),Bp() Bq()) if =

(The only difference is to be found in the last lines of the two definitions.) It is easy to check thatp p q and q q p and that both p q and q p belong to Pw . Moreover (p q)

+ = (q p)+.Hence q and p are compatible.

Lemma 2.9. If the cofinality ofC is not1 andS = {(A, F,B)}C is a tower of permutationsthenQ(S) has property K.

Proof. This is standard. If the cofinality of C is less than 1 then choose a countable, cofinal subsetC of C. Using observation 2.9 of Lemma 2.2 it follows that the set of all p Q(S) such thatp C is dense. It follows that Q(S) has a -centred dense subset.

On the other hand, if the cofinality of C is greater than 1 and {p}1 Q(S) it is possible to

choose C such that

p

< for each . Using observation 2.9 of Lemma 2.2 it may be assumed,by extending each condition, that p = for each . Now there are a and f and an uncountable set of such that ap = a and fp = f. Any two of these are easily seen to be compatible.

The fact that the tower of permutations needs to be generic, or at least some other condition mustbe satisfied, in order for Lemma 2.9 to hold has been observed in Theorem 2 of [2].

Lemma 2.10. P has the countable chain condition for each .

Proof. Proceed by induction on . If = 1 the result is immediate and if is a limit the resultfollows from the induction hypothesis and the finite support of the iteration. Therefore consider thecase = + 1 and assume that the countable chain condition has already been established for P.

Next, observe that ifC is not cofinal in then the induction hypothesis is easily applied since, inthis case, C has a maximal element below and so Q has the countable chain condition by Lemma 2.9.


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If C has cofinality different from 1 then, once again Lemma 2.9 implies that Q has the countablechain condition; therefore, in either case, so does P+1. Hence it remains to consider the case that C is cofinal in and has cofinality 1. From the hypothesis on C, and the fact that must be the limit of

C it follows that C. By appealing to Lemma 2.6 and extending the conditions in question, it ispossible to guarantee that each p is in P

and that this is witnessed by (k, {(a

, f

)}) As well, by

thinning out, it may be assumed that there is k such that k = k for each , and there is a pair (a, f)

such that (a , f ) = (a, f) for each and that {domain(p)}1 form a -system with root . Using

the fact that is a limit, choose some C such that there is some uncountable 1 such thatif {, } []2 and then max(domain(p )) < min(domain(p [, ))). It may as well beassumed that / domain(p) for each . Then use Lemma 2.7 to extend each p to some p+.

Now, using the induction hypothesis, find {, } []2 such that and p+ + 1 is

compatible with p+ + 1. Choose r P+1 extending both p+ + 1 and p+ + 1.Let G P+1 be generic over V and containing r. Observe that P/P+1 as interpreted in V[G] is a

partial order like P

in V except that the first factor ofP

/P+1

is Q({(A

, F

, P(N) V[G])}). Sincer P+1 k A it follows that k A, in V[G]. It follows that (p

+ [, )) and (p+ [, ))

belong to (P/P+1). Now use Lemma 2.8 in V[G] to conclude that (p

+ [, )) and (p+

[, )) are compatible. Hence, so are p and p.

Lemma 2.11. LetS = {(A, F,B)}W be a tower of permutations and suppose that A [N]0 and

is a one-to-one function from A to N. ThenQ(S) forces that there are infinitely many a A suchthat FS[G](a) = (a).

Proof. Let p Q(S) and suppose that p Q(S) (a A \ k)(a) = FS[G](a) for some integer k.From Lemma 2.2 it can be assumed that k max(ap). Now choose m Ap so large that there existdistinct integers i and j in [max(ap), m) A such that i B if and only if j B for all B Bp. It is

possible to choose a bijection g : {i, j} {Fp(i), Fp(j)} such that g = {i, j}. However, lettingf = Fp ([max(a

p), m) \ {i, j}) g it follows that f(B [max(ap), m)) = Fp(B [max(ap), m)) foreach B Bp. Hence p q = (ap {m}, fp f, p,Bp) and q Q(S) (i) = FS[G](i).

Lemma 2.9 and Lemma 2.11, in the case case when C has the maximal element, together implythat there is a trivial automorphism of P(N)/[N]


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where G is P generic over V. Then let be the automorphism of P(N)/[N]


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where XG is a name for the generic subset of added by S. Let g = Z be the characteristic functionof Z. Using the fact that S P is sufficiently Suslin, find (s, p) S P such that (s, p) (s, p) and

(s, f) S 2|(q S P)q(0) s and q (s, p) and q SP g = fis analytic. Let A be an infinite set of integers disjoint from the domain ofs such that domain(s)A hasinfinite complement in the integers. It follows that ifR is defined to be

(t, f) 2A 2 : (q (t s, p))q S

then R is analytic.There are two possibilities. First, suppose that the domain of R is all of 2A. In this case it is possible

to find a continuous function S defined on a comeagre subset of P(A) such that R(t, S(t1{1})) holdsfor all t in the domain of R. Now let : domain(S) P((A)) be defined by

(W) = {n (A) : S(W)(n) = 1}

and observe that is continuous. Furthermore, (W) (W). To see this, let q S P be anycondition witnessing that R(W A, S(W)) holds. Observe that not only does q force XG A = W but

also q SP Z

(A) = (W). Hence, q SP (W) =

(XG)

(A)

(XG A)

(W).Therefore, since (W) and (W) are sets in the ground model, it follows from the absoluteness of

that (W) (W). But now it follows that is trivial on A by Lemma 1 of [4].In the other case there is some t : A 2 such that there is no f such that R(t, f) holds. This

implies that for every q (s t, p) there are infinitely many integers in (A) whose membership inZ is not decided by q. Genericity implies that (s t, p) SP Z

(A) (t1{1}) which is acontradiction to the assumption that (s, p) SP

(XG) = Z.

Theorem 3.1. Let V be a model of 20 = 1 If P2 is the countable support iteration of partial ordersas in Lemma 3.1 such that P+1 = P S for a stationary set of 2 then V[G] is a model whereevery automorphism of P(N)/[N]

Saharon Shelah and Juris Steprans- Martin's Axiom is Consistent with the Existence of Nowhere...

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