saharon shelah and juris steprans- somewhere trivial autohomeomorphisms

8/3/2019 Saharon Shelah and Juris Steprans- Somewhere Trivial Autohomeomorphisms

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Somewhere Trivial Autohomeomorphisms

Saharon Shelah and Juris Steprans

October 6, 2003

Abstract

It is show to be consistent that there is a non-trivial autohomeomor-

phism of N \ N while all such autohomeomorphisms are trivial on some

open set. The model used is one due to Velickovic in which, coincidentally,

Martins Axiom also holds.

1 Introduction

An automorphism of ofP()/[]


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129 to 152 of [4], that it is equiconsistent with ZFC that all automorphisms

of P()/[]


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to be somewhere trivial if there is some A []0 on which it is trivial. A

homomorphism is trivial if it is trivial on .

It has already been mentioned that it was shown in [4] that it is consistent

that all automorphisms of P()/[]


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The notation iX

will be used to denote the constant function whose domain

is X and which has value i at each point in X. Whenever reference is made to

a topology on P() this will be to the Cantor set topology under the canonical

identification of 2 with P() in other words, a natural base for this topology

consists of all sets of the form

{A : 1A 0\A g}

where g is a finite partial function from to 2.

The argument to be presented in the next section will be a modifiction and

combination of arguments from pages 129 to 152 of [4], [5] and [8]. For the

readers benefit, some definitions and lemmas from [4] will be recalled.

Definition 1.3 An 1-oracle is a sequence M = {M : 1} such that

M is a countable, transitive model of ZFC without the power set axiom

M andM |= is countable

{ 1 : A M} contains a closed unbounded set for each A 1

Notice that the existence of an oracle requires that 1 is true.

Definition 1.4 IfM is an oracle then a partial order on 1 (or some set

coded by 1) will be said to satisfy the M-chain condition if there is a closed

unbounded set C such that for every C andA , A M, if A is predense

in the order (, ( )) then it is predense in (1, ).

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Further discussion of these definitions as well as proofs of the following lemmas

can all be found in [4].

Lemma 1.2 Assume that1 holds and (x) is a12 formula possibly with

a real parameter for each 1. Suppose also that there is no r R such

that (r) holds for all 1 and that there is still no such r even after adding

a Cohen real. Then there is an oracle M such that any partial order Q which

satisfies the M-chain condition will not add r R such that (r) holds for all

1.

Lemma 1.3 If {M : 1} are oracles then there is a single oracle M such

that if any partial order satisfies the M-chain condition then it satisfies the M

chain condition for each 1.

The oracle M of Lemma 1.3 is easily decribed. It is the diagonal union of the

oracles {M : 1}. This fact, rather than the statement of Lemma 1.3, will

be used in the proof of Lemma 2.6.

Lemma 1.4 If V is a model of1 then there is, in V, an oracle M such that

ifQ satisfies the M-chain condition then 1 Q R V is second category

Lemma 1.5 IfM is any oracle andQ satisfies the M-chain condition thenQ

satisfies the countable chain condition.

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2 The proof

The following partial order P, was introduced by Velickovic in [8] to add a non-

trivial automorphism of P()/[]


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Proof: It follows from MA

that there are A and B such that

A dom(f) for each

B ran(f) for each

limsupn | (2n+1 \ 2n) \ A | =

limsupn | (2n+1 \ 2n) \ B | =

Let f = f (A (f1B)).

Lemma 2.2 P is countably closed.

Proof: Given a sequence {fn : n } P such that fn fn+1 for each n

choose inductively kn such that f = {fn ( \ kn) : n } is a function. Now

apply Lemma 2.1.

From Lemma 2.1 it follows that, given a sequence {f : 1}, it will be

useful to find an element f P such that f f for each 1. The following

partial order is designed to do precisely this.

Definition 2.2 Given {f : } = F define P(F) to be the partial order

consisting of all g P such that there is some such that g f. The

ordering onP(F) is as opposed to inP.

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Definition 2.3 For anyG which is a centred subset ofP define G

: P()/[]


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then in V[F], not only is there is a non-trivial autohomeomorphism of N \ N,

but MA also holds. It will be shown that a closer analysis of this model yields

that in V[F] all autohomeomorphisms of N \N are somewhere trivial.

Loosely speaking, the following theorem will show that if : P()/[]


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domain pn

is dense in Epn

and so it follows that the domain pn

is dense in U and,

moreover, not being a point to which a function can be continuously extended

is an absolute property. This contradicts the fact that Gn is comeagre.

Now let M = {Dpn \ Epn : n and p C} {E

pn \ dom(f

pn) : n }

and observe that M is meagre. Now recall the following fact: If V is a model

of ZFC and r is a Cohen real and N V[r] is a meagre set then there is a

meagre set N V such that N V N V. Let N be a meagre set such that

Gn P() \ N for each n . Let M = M N. It is true in V that for every

A (P() \ M) V there is some p C such that [pn(A)] = ([A]). Since this

statement is arithmetic in the parameters A and ([A]) and both of these

parameters belong to V this must be true in V also. Now apply Lemma 1.1.

Lemma 2.4 Given 1, a sequence {f : } = F and a countable

elementary submodelA (H(2), ), such thatF A, there is f P which is

A-generic forP(F). Moreover, for any extension {f : } = F ofF such

that 1 and f = f, every D A is predense inP(F) provided that it

is dense inP(F).

Proof: Let {Ek : k } enumerate all dense subsets ofP(F) in A. Construct

sequences {gn : n } and {Kn : n } such that for all n

gn gn+1

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Kn

< Kn+1

gn+1 Kn = gn Kn

there is some i such that 2i+1 = Kn and | (2i+1 \ 2i) \ dom(gn) | n

for each bijection t : Kn Kn there h jn+1Ej such that t gn

( \ Kn) h

It is easy to see that this can be done. Hence, it is possible to define f = {gn :

n }. Notice that {g : g f} is dense in P(F) and definable in A hence

f f for each . To check that f has the desired properties suppose that

g P(F) for some and that f = f. If E is dense in P(F) then there

is some m such that E {Ej : j m + 1} and g ( \ Km) f ( \ Km).

By extending g if necessary, it may, without loss of generality, be assumed that

g Km = t and that t : Km Km is a bijection. It follows that t gm

( \ Km) h for some h E and hence g h E.

Lemma 2.5 Suppose that V is a model of 20 = 1. If is a P-name for a

nowhere trivial automorphism ofP()/[]


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enumerate all possible names for continuous functions from a Borel comeagre

subset of some P(C) to some P(B) so that each name occurs cofinally often. It

suffices to construct F = {f : 1} P by induction so that for every limit

ordinal the following conditions are satisfied

f+n decides, in P, the values of (A+n2) and 1(A+n2) for n 2

there is some C A such that 1 P(F) F(C) = [(C)]

It is possible to construct F inductively because a failure would mean that for

some 1 it must be the case that

f P is trivial on A

contradicting that is a name for a nowhere trivial automorphism ofP()/[]


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there is some C V such that F([C]) = n

([C]) for all n

In the last clause the possibility that C / dom(n) is allowed in the sense that

if C / dom(n) then F([C]) = n([C]).

Proof: The proof will be rely on constructing a particular oracle which will

guarantee that the three clauses are all satisfied. The only wrinkle is that the

oracle and the sequence {f : 1} must be constructed simultaneously.

The sequence {f : 1} will be obtained by diagonalizing across 1 such

sequences.

In particular, let N be any oracle such that forcing with an N-oracle chain

condition partial order preserves the fact that R V is of second category

such an oracle exists by Lemma 1.4. Then construct sequences F = {f :

1} P and M = {M : 1} P for 1 such that

a. f = f if <

b. F is nowhere trivial for 1

c. for 1, ifQ satisfies the M-chain condition and G is Q-generic over

V then, in V[G], for every A V P(), B V P() there do not

exist {n : n } such that n : P(A) P(B) is continuous and for all

C P(A) V there exists n such that F

([C]) = n([C])

d. {f : {, } []2} M for each 1

e. f+1 is P({f : })-generic over M

+1

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f. M

M

if

g. M0 = N

To see that this suffices let M = M+1 and let F = {f = f

+1+1 :

1}. It follows from the remark following Lemma 1.3 that {M : 1} is

an oracle and that any partial order which satisfies the {M : 1}-chain

condition also satisfies each of the M-chain conditions for 1. Since f+1

is P({f : })-generic over M+1 it follows that P({f : 1}) satisfies

the {M : 1} chain condition. In particular, this partial order satisfies the

M0-chain condition and hence the second clause of the theorem will be satisfied.

That the first clause is satisfied follows from Lemma 1.5. So it only remains to

be shown that the last clause is satisfied.

To this end, suppose that A V P(), B V P() and P(F)-names n

are given such that for each n

1 P(F) n : P(A) P(B) is continuous

Since P(F) satisfies the countable chain condition, there is some 1 such

that M models that for each n

1 P({f:}) n : P(A) P(B) is continuous

It now follows that this statement about M = M must be true in M

+1

because M M+1. But it now follows from the fact that f

+1 is generic over

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M

+1that

f+1 P({f:})

(C P(A) V)(\ )F([C]) = \([C])

Since f+1 f+1 = f

for each 1 M

+1 it follows that FM

+1 =

F M+1 and hence

f+1 P({f:}) (C P(A) V)(\ )F([C]) = \([C])

Since the necessary dense sets are definable in M

+1 = M it follows that

f P(F) (C P(A) V)(\ )F([C]) = \([C])

which is what is required.

All that remains to be done is to show that the inductive construction can be

completed. For this, suppose that F = {f : 1} and M = {M : 1}

have been constructed for .

If is a limit then it is easy to use Lemma 2.5 in order to satisfy conditions

(a) and (b). If is a successor then Lemma 2.4 must also be used in order to

satisfy condition (e). To construct M for 1 use Lemmas 1.2 and Lemma

2.3 to satisfy condition (c). It is then easy to enlarge the terms of the oracle to

satisfy conditions (d) and (f).

The proof of the main theorem will require the following definition, which is a

reformulated form of the partial order which appeared in [4] on page 134.

Definition 2.4 Given a sequence {(W, V) : } define Q({(W, V) :

}) to be the partial order which consists of all functions g such that there is

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[]


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Let M be an arbitrary oracle in V. A sequence {(W

, V

) : 1

} will be

constructed in V[G][H] so that

ifQ = Q({(W, V) : }) then Q1 satisfies the M-chain condition

(W, V) V[G] = V (P())2

V W

| W W | < 0 if =

for each p Q and Q-name, Y M, for a subset of

p 1V 0W\V Q+1 F([W]) [Y] = F([V])

the dense subsets ofQ+1 which guarantee that the previous statement is

true are predense in Q1

Before continuing, define (A) arbitrarily to satisfy that [(A)] =

F([A]) for each [A] dom(F). Next, choose an almost disjoint family

{W : 1} in the model V[G]. The set W will be chosen so that, among

other things, W W this will, of course, guarantee that the resulting

family is almost disjoint. If this construction succeeds then it is possible to

proceed as in [5] to prove that forcing with P(F) Q1 adds a set to which the

partial automorphism F can not be extended.

In particular, ifH1H2 is P(F)Q1 generic then, setting X = {f1({1}) :

f H2}, it follows that X W V for each 1 but, in V[G][H1 H2], for

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every Y there is 1

such that (W

) Y (V

) for each .

Just as in [5], it is possible to define a relation R on 1 by R(, ) holds if and

only if either ((W)\(V))

(V) = or ((W)\

(V)(V)) = .

It is easy to see that this is a semiopen relation as defined in [1] and that

moreover, there is no S [1]1 such that [S]2 R = . The reason for the

last statement is that otherwise, letting Y = {(V) : S} would yield

a contradiction to the fact that (W) Y (V) for all but countably

many . Hence, by the results of [1], there is a proper partial order K which

adds a set S [1]1 such that [S]2 R. This makes the fact that F can

not be extended to the set X absolute. The reason for this is that if there is

a set Y such that F([X]) can be defined to be [Y], then it must be the case

that Y (W) (V) for each S. But then there is an uncountable

set S S, as well as J , such that Y (W) \ J = (V) \ J for each

S. It follows that (V) \ J Y and that ((W) \ (V)) \ J \ Y

for each S. Choosing and in S such that (V) J = (V) J and

(W) J = (W) J yields the desired contradiciton.

The iteration D P(F) Q1 K is proper and only 1 dense sets in it need

be met in order to obtain S and the set X such that F can not be extended

to include [X] in its domain. Let f1 be the element ofP obtained by forcing with

P({frakF) and Lemma 2.1 and note that, in V, f1 P does not extend to X

because f1 P F

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Hence it may be assumed that the construction breaks down at some point

1. What can go wrong? First, there are certain predense sets required at

stage which must remain predense in the partial order Q+1. It is shown on

page 134 of [4] that, for each predense set E Q, there is a dense open set

W P(W) such that if W W then, letting W = W, E remains predense

in Q+1 for any V = V W. Note that W is closed under the operation

of taking infinite subsets. Recall that F was chosen so that P(W) V[G] is

of second category in P(W) in the model V[G][H1]. It follows that it may be

assumed that W V[G] and that W O for every open set O P(W)

which is definable from M and {(W, V) : } but note that Q is

definable from {(W, V) : }. Hence the only possible problem is that it is

not possible to find V W satisfying the required properties namely, there

is no V W such that for each p Q and Q-name, Y M, for a subset

of

p 1V 0W\V Q+1 F([W]) [Y] F([V]).

To see that this can not happen it will be necessary to discuss forcing in

Q+1 before Q+1 has been defined namely, before V has been defined.

This will be done by defining as follows: If p Q, X and Y are Q names

for subsets of and f : W 2 is a partial function, define

(p,f) X Y

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if and only if for each p Q

such that p p f is a function and for each

n there is p such that

p p p f is a function

p Q k XY for some k n

Claim 1 If the following conditions are satisfied

(p,f) X X Y

X is aQ name belonging to M

X and Y are subsets of in V[G]

V = f1({1})

then then p fQ+1 X X Y.

The way to see this is to note that the following statements are all equivalent

p Q k X XY

k X \ Y and p Q k X or k Y \ X and p Q k / X

k X\ Y and p Q+1 k X or k Y \ X and p Q+1 k / X

so

long as the name X is still a Q+1 name; in other words, the antichains

in Q deciding membership in X remain maximal in Q+1.

This last equivalence is guaranteed by the choice of W, because the relevant

dense sets are definable from X, which belongs to M, and Q. The claim now

follows from the definition of .

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It will be now be shown that it is possible to choose V

W

such that for

each p Q and Q-name, Y M, for a subset of

p 1V 0W\V Q+1 (W) Y

(V)

If this is not possible then it follows from Claim 1 that there is no V such that

V W and such that for each p Q and each Q-name for a subset of ,

Y M

(p, 1V 0W\V) F([W]) [Y] F([V]).

It will be shown that this implies that there is are continuous functions n, for

n , such that for each C W there is n such that n(C) (C),

thus contradicting the fact that M is being assumed to satisfy the conclusion of

Lemma 2.6.

Here is how to conclude this. For each p Q

and A W

define qp

(A) =

(p, 1A 0W\A) and qp(A) = p 1A 0W\A. Given p and r in Q, n and

Y M, a Q-name for a subset of , define p,r,n,Y(A) to be the set

{k (W) \ n : (p)(p p qr(A) is not a function or p

Q k Y}

for A dom(p,r,n,Y) = {A W : qr(A) p is a function }.

It will first be shown that for each A W there are p, r, n and Y such

that (A) p,r,n,Y(A). To see see this recall that there is some rA Q

and some YA M, a Q-name for a subset of , such that

qrA(A) (W) YA

(A)

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fails to be true and qrA

(A) is a function. Hence, there is some pA

Q

such

that pA qrA(A) is a function and there is some nA such that for each p

and k nA either p pA qrA(A) is not a function or p

Q k ((W)

YA)((A)).

Ifk pA,rA,nA,YA(A)\(A) and k > nA then the definition ofpA,rA,nA,YA(A)

implies that p Q k YA whenever p is such that ppA qrA(A) is a func-

tion. But this means that p Q k ((W) YA)((A)) contradicting

the choices of pA, rA, nA and YA.

On the other hand, suppose that k belongs to ((A) (W)) \ nA. If

k pA,rA,nA,YA(A) then there exists p such that p pA qrA(A) is a function

andp Q k YA. It follows that there is p p such that p Q k YA.

If p pA qrA(A) is a function then p Q k (

(W) YA)((A))

once again contradicting the choice of pA, rA, nA and YA. But why should

p pA qrA(A) be a function?

The fact that it is possible to choose p such that p pA qrA(A) is a

function follows from the choice of W. Recall that W was chosen so that

W O for every open set O P(W) which is definable from M and Q.

Moreover, the set O consisting of all W W such that there exists e [W]


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pA

rAQ

k YA

is easily seen to be dense and open in W and to be definable from Q and

YA, pA and rA all of which belong to M. Hence W belongs to this dense

open set O. Now let e [W]


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with those of [6] it is possible to show that that it is consistent (even with

MA1) that every autohomeomorphism of N \ N is somewhere trivial while

any two P-points have the same topological type in the sense that there is an

autohomeomorphism of N \N mapping one to the other. In this model it will

of course follow that there are 2c autohomeomorphisms of N \ N which raises

the following question.

Question 3.1 Is it consistent that there are only20

automorphisms ofP()/[]


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References

1. U. Abraham, M. Rubin, and S. Shelah, On the consistency of some partition

theorems for continuous colorings and the structure of 1-dense real order

types, Annals of Pure and Applied Logic 29 (1985), 123206.

2. K. P. Hart and J. van Mill, Open problem on, Open Problems in Topology

(G. M. Reed and J. van Mill, eds.), North-Holland, pp. 99125.

3. W. Just, A modification of Shelahs oracle-c.c. with applications, Transac-

tions of the American Mathematical Society ??? (1989), ????

4. S. Shelah, Proper forcing, Lecture Notes in Mathematics, vol. 940, Springer-

Verlag, Berlin, 1982.

5. S. Shelah and J. Steprans, PFA implies all automorphism are trivial, Proc.

Amer. Math. Soc. 104 (1988), 12201225.

6. J. Steprans, Martins Axiom and the transitivity of P-points, Submitted to

Israel Journal of Mathematics, 19.

7. B. Velickovic, Definable automorphisms of P()/{\, Proc. Amer. Math. Soc

96 (1986), 130135.

8. B. Velickovic, OCA and automorphisms of P()/{\, preprint, 1990.

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saharon shelah and juris steprans- somewhere trivial autohomeomorphisms

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