Saharon Shelah- On full Souslin trees

8/3/2019 Saharon Shelah- On full Souslin trees

1/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

On full Souslin trees

Saharon Shelah

Institute of Mathematics

The Hebrew University of Jerusalem

91904 Jerusalem, Israel

and

Department of Mathematics

Rutgers UniversityNew Brunswick, NJ 08854, USA

and

Mathematics Department

University of Wisconsin Madison

Madison, WI 53706, USA

August 17, 2011

Abstract

In the present note we answer a question of Kunen (15.13 in [Mi91])showing (in 1.7) that

it is consistent that there are full Souslin trees.

We thank the NSF for partially supporting this research under grant #144-EF67.

Publication No 624.

0


2/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 1

0 Introduction

In the present paper we answer a combinatorial question of Kunen listed inArnie Millers Problem List. We force, e.g. for the first strongly inaccessibleMahlo cardinal , a full (see 1.1(2)) Souslin tree and we remark thatthe existence of such trees follows from V = L (if is Mahlo stronglyinaccessible). This answers [Mi91, Problem 15.13].

Our notation is rather standard and compatible with those of classicaltextbooks on Set Theory. However, in forcing considerations, we keep theolder tradition that

a stronger condition is the larger one.

We will keep the following conventions concerning use of symbols.

Notation 0.1 1. , will denote cardinal numbers and ,,,,, willbe used to denote ordinals.

2. Sequences (not necessarily finite) of ordinals are denoted by , , (with possible indexes).

3. The length of a sequence is g().

4. For a sequence and an ordinal g(), is the restriction ofthe sequence to (so g() = ). If a sequence is a proper

initial segment of a sequence then we write (and hasthe obvious meaning).

5. A tilde indicates that we are dealing with a name for an object inforcing extension (like x

).

1 Full -Souslin trees

A subset T of >2 is an tree whenever ( is a limit ordinal and) thefollowing three conditions are satisfied:

T, if

T then T, T implies 0, 1 T, and

for every T and < such that g() there is T suchthat and g() = .

A Souslin tree is a tree T >2 in which every antichain is of size lessthan .


3/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 2

Definition 1.1 1. For a tree T

>

2 and an ordinal we let

T[]def= T 2 and T[ 2.

If is limit then we define

limT[2 of a truehight = (T) < (see 1.1(3); so is a limit ordinal) such that lim(T[2 (so it is the endextension order).


4/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 3

2. For a condition T Q

and a limit ordinal < (T), let (T) be theunique member of lim(T[


5/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 4

2) For our purpose it is enough to show that for each ordinal < and acondition T Q the second player has a winning strategy in the followinggame G(T,Q). (Also we can let Player I choose T for odd.)

The game lasts moves and during a play the players, calledI and II, choose successively open dense subsets D of Q andconditions T Q. At stage < of the game:Player I chooses an open dense subset D ofQ andPlayer II answers playing a condition T Q such that

T Q T, ( < )(T Q T), and T D.

The second player wins if he has always legal moves during theplay.

Let us describe the winning strategy for Player II. At each stage < of the game he plays a condition T and writes down on a side a functionf such that (T , f) Q

. Moreover, he keeps an extra obligation that(T, f) Q (T, f) for each < < .So arriving to a non-limit stage of the game he takes the condition ( T, f)he constructed before (or just (T, f), where f is a witness for T, if this isthe first move; by 1.4(2) we can always find a witness). Then he choosesT Q T such that (T

) = (T) + and (T

)[(T)] = lim(T )(T).

Thus (T )(T

) is not defined. Now Player II takes T+1 Q T

fromthe open dense set D+1 played by his opponent at this stage. Clearly(T )(T+1) is not defined, so Player II may use 1.4(1) to choose f+1 suchthat (T, f) Q (T+1, f+1) Q

.At a limit stage of the game, the second player may take the least upperbound (T, f

) Q of the sequence (T, f) : < (exists by 1)) and then

apply the procedure described above.3) Follows from 2) above.

Definition 1.6 LetT

be the canonicalQname for a generic tree added by forcing withQ:

Q T =

{T : T GQ}.

It should be clear that T

is (forced to be) a full tree. The main point isto show that it is Souslin and this is done in the following theorem.

Theorem 1.7 Q T

is a Souslin tree.


6/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 5

ProofSuppose that A is a

Qname such that

Q A

T

is an antichain ,

and let T0 be a condition in Q. We will show that there are < and acondition T Q stronger than T0 such that T Q A

T

[B B, and B = .

Let S and B (H(), , 2 and( T)( T)( & T Q A

)}.

Claim 1.7.2 I

is a dense subset ofQ.


7/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 6

Proof of the claim: Should be clear (remember 1.5(2)).Now we choose by induction on < a continuous increasing sequence

(T, f) : < Q B.

Step: i = 0T0 is already chosen and it belongs to Q B. We take any f0 such that(T0, f0) Q

B (exists by 1.4(2)).Step: limit Since >B B, the sequence (T, f) : < is in B. By 1.5(1) it has theleast upper bound (T, f) (which belongs to B).Step: = + 1First we take (the unique) tree T of true height (T

) = (T) + such

that T (T)>2 = T and:

if(T) S0 and (T) / rang(f) then (T

)[(T)] = lim(T )(T)\{(T )},otherwise (T )[(T)] = lim(T)(T).Let T Q I

be strictly above T (exists by 1.7.2). Clearly we may choose

such T in B. Now we have to define f. We do it by 1.4, but additionallywe require that

if T then ( T)( f() & T Q A

).

Plainly the additional requirement causes no problems (remember the defi-nition of I

and the choice of T) and the choice can be done in B.

There are no difficulties in carrying out the induction. Finally we let

Tdef=


8/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 7

if (T+1) is defined then = (T) = (T+1).But = (as S

). So look at the inductive definition: necessarily forsome T we have = f(

), i.e. = f(

). Now,

T =2 is a Souslin tree which is full and even S0full .

[So, in VQ, in particular we have:

for every < < , for all T 2 there is T 2 such that , and for a limit ordinal < , lim(T[


9/9

(624)

revision:1996-08-

13

modified:1996-0

8-13

[Sh:624] August 17, 2011 8

ProofBy 1.5 and 1.7.

Of course, we do not need to force.

Definition 1.10 Let S0, S . A sequence (C, ) : < limit iscalled a squared diamond sequence for (S, S0) if for each limit ordinal <

(i) C a club of disjoint to S,

(ii) 2,

(iii) if acc(C) then C = C and ,

(iv) if S then : C S0 is a diamond sequence.

Proposition 1.11 Assume (in addition to 1.2)

(e) there exist a squared diamond sequence for (S, S0).

Then there is a Souslin tree T >2 which is S0full.

Proof Look carefully at the proof of 1.7.

Corollary 1.12 Assume that V = L and is Mahlo strongly inaccessible.Then there is a full Souslin tree.

Proof Let S { < : is strongly inaccessible} be a stationary non-reflecting set. By Beller and Litman [BeLi80], there is a square C :

Saharon Shelah- On full Souslin trees

Documents

Transcript of Saharon Shelah- On full Souslin trees