saharon shelah- colouring and non-productivity of ℵ2-c.c

8/3/2019 Saharon Shelah- Colouring and Non-Productivity of 2-C.C.

1/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

COLOURING ANDNON-PRODUCTIVITY OF 2-C.C.

SH572

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityDepartment of Mathematics

New Brunswick, NJ USA

Abstract. We prove that colouring of pairs from 2 with strong properties exists.The easiest to state (and quite a well known problem) it solves: there are two topo-logical spaces with cellularity 1 whose product has cellularity 2 ; equivalently wecan speak on cellularity of Boolean algebras or on Boolean algebras satisfying the2 -c.c. whose product fails the 2 -c.c. We also deal more with guessing of clubs.

1,2 Done 9/94; 3 written 10/94; 4 written 1/95.I would like to thank Alice Leonhardt for the beautiful typing and Zoran Spasojevic for helpingto proof read it.Latest Revision 96/Jan/19

Typeset by AM S -TEX1


2/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

2 SAHARON SHELAH

Annotated content

1 Retry at 2-c.c. not productive

[We prove P r 1(1 ,2 ,2 ,0) which is a much stronger result].

2 The implicit properties

[We dene a property implicit in 1, note what the proof in 1 gives, andlook at related implication for successor of singular non-strong limit andshow that P r 1 implies P r 6].

3 Guessing clubs revisited

[We improve some results mainly from [Sh 413], giving complete proofs. Weshow that for regular uncountable and < we can ndC : < + , cf() = and functions h , from C onto , such that for

every club E of + for stationarily many < + we have: cf() = and forevery < for arbitrarily large nacc(C ) we have E, h () = .Also if C = { , : < }, ( , increasing continuous in ) we can demand{ < : , +1 E (and , E )} is a stationary subset of . In fact foreach < the set { < : , +1 E, , E and f ( , +1 ) = } is astationary subset of . We also deal with a parallel to the last one (withoutf ) to successor of singulars and to inaccessibles.]

4 More on P r 1

[We prove that Pr 1(+2 , +2 , +2 , ) holds for regular ].

On history, references and consequences see [Sh:g, AP1] and [Sh:g, III, 0].


3/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

COLOURING AND NON-PRODUCTIVITY OF 2 -C.C. SH572 3

1 Retry at 2 -c.c. not productive

1.1 Theorem. P r 1(2 ,2 ,2 ,0).

1.2 Remark. 1) Is this hard? Apostriory it does not look so, but we have workedhard on it several times without success (worse: produce several false proofs). Wethank Juhasz and Soukup for pointing out a gap.2) Remember thatDenition P r 1(,,, ) means that there is a symmetric two-place function d from to such that:if u : < satises

u ,

|u | < ,

< u u = ,

and < then for some < we have

u & u d(, ) = .

3) If we are content with proving that there is a colouring with 1 colours, then wecan simplify somewhat: in stage C we let c(, ) = dsq (h 1 (, )) and this shortensstage D.

Proof.

Stage A: First we dene a preliminary colouring.There is a function dsq : > (1) 1 such that:

if A [1]1 and ( , ) : A is such that > 1 , > 1 , Rang( ) Rang( ) and < 1 then for some < fromA we have: if , are subsequences of , respectively and Rang( ), Rang( ) then

dsq ( ) = .

Proof of . Choose pairwise distinct 2 for < 1 . Let d0 : [1]2 1

be such that:() if n < and , < 1 for < 1 , < n are pairwise distinct and

< 1 then for some < < 1 we have < n = d0({ , , , })(exists by [Sh 261, see (2.4),p.176] the n there is 2).


4/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

4 SAHARON SHELAH

Dene dsq ( ) for > (1) as follows. If g( ) 1 or is constant thendsq ( ) is 0. Otherwise letn( ) =: max {g( () (k ) ) : < k < g ( ) and () = (k)} < .

The maximum is on a non-empty set as g( ) 2 and is not constant, remember 2 were pairwise distinct so () = (k) () (k ) > 2 (is the largestcommon initial segment of () , (k ) ). Let a( ) = {(, k) : < k < g ( ) andg( () (k ) ) = n( )} so a( ) is non-empty and choose the (lexicographically)minimal pair ( , k ) in it. Lastly let

dsq ( ) = d0({ ( ), (k )}).

So dsq is a function with the right domain and range. Now suppose we are givenA [1]1 , < 1 and , > (1) for A such that Rang( ) Rang( ). We should nd < from A such that dsq ( ) = for any subsequences , subsequences of , respectively such that Rang( ) and Rang( ).

For each A we can nd m < such that:

()0 if < k < g ( ) and ( )() = ( )(k) then( )( ) m = ( )( k ) m .

Next we can nd B [A]1 such that for all B (the point is that the values donot depend on ) we have:

(a) g( ) = m0 , g( ) = m1 ,(b) a = {(, k) : < k < m 0 + m1 and ( )() = ( )(k)},(c) b = { < m 0 + m1 : = ( )()},(d) m = m2 ,(e) ( )( ) m : < m 0 + m1 = ,(f) Rang( ) : B is a -system with heart w,(g) u = { : ( )() w} (so u = { : < m 0 + m1 } as Rang( )),(h) = ( )() for u

,(i) if < B then sup Rang( ) < .

For B let =: (

)() : < m 0 + m1 , /u and apply ( ), i.e. the

choice of d0 . So for some < from B , we have

< m 0 + m1 & /u = d0 {( )(), ( )()} .

We shall prove that < are as required (in ). So let , be subsequences of , (so let = v1 and = v2) such that Rang( ), Rang( )and we have to prove = dsq ( ). Let = , so = ( ) (v1(m0 + v2 ))(in a slight abuse of notation, we look at as a function with domain v1(m0 + v2 )and also as a member of > (1) where m + v =: {m + : v}, of course). By the

denition of dsq it is enough to prove the following two things:()1 n( ) m2 (see clause (d) and ()0 above),()2 for every 1 , 2 v1 (m0 + v2) we have

g( (1 ) (2 ) ) [m2 , ) = d0({ (1), (2)}).


5/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


Proof of ()1 . Let 1 v1 and 2 v2 be such that (1) = and (2) = .So clearly 1 , m 0 + 2 b (see clause (c)) and

(2 ) m2 = (2 ) m2 = (1 ) m2 (rst equality as , B

and m = m = m2

(see clause (d) and (e)), second equality as (2 ) = (1 ) since 1 , m 0 + 2 b (see clause (c)). But (2) = = = (1),hence (2 ) = (1 ) , so together with the previous sentence we have

m2 g( (1 ) (2 ) ) = g( (1 ) (m 0 + 2 ) ) < .

Hence n( ) m2 as required in ()1 .

Proof of ()2 . If 1 < 2 are from v1 , by the choice of m2 = m it is easy. Namely,if (1 , 2) a( ) then ( 1 , 2) a( ) and g( (1 ) (2 ) ) = g( (1 ) (2 ) ) and let E = { : a limit ordinal < 2 and ()( < u )}. Clearly E isa club of 2 . For each E S 21 , there is < such that

[ , ) & u (, ) (, ).

Also for S 21 let

=: Min < 1 : A but if u

Rang( (, ))

(so > ) then /A .

Note that < 1 is well dened by the choice of A s. So, by Fodors lemma, forsome < 1 and < 2 we have that

W =: { S 2 () :

= and = }

is stationary. Let h be a strictly increasing function from 2 into W such that < h (). By the demand on (and W )

0 < < W & u (, ) (, ).

Hence

< < S 21 & uh ( ) Min{ : A (, )( ) } =1

Min{ : (, )() = h()},

hence

< < S 21 & uh ( ) 2

h2 (, ) Min{ : A (, )( ) } = ().

Let


7/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


E 0 =: < 2 : a limit ordinal, E and

< h( ) < (hence sup( uh ( ) ) < ) .

For each S 21 there is < such that > and

[ , ) & uh ( ) (, ) (, ).

For each < 1 , is a regressive function on S 1 , hence for some( ) < the set S =: { S 1 E 0 : = ( )} is stationary.

Let = sup {( ) + 1 : < 1} and note that < 2 . Let

E 1 =: { < 2 : for every < 1 , = sup( S ) and > },

and note that E 1 is a club of 2 (and as S E 0 clearly E 1 E 0) and choose E 1 S 2 () . Then by induction on i < 1 choose an ordinal i such that i : i < 1 is strictly increasing with limit and i S i \ (+ 1). We know

that < i u i and < min( u ), hence for every i < i large enough( u i )((, i )( i ) (, )).

Choose such i (j .

Now for each i < 1 for some (i) < ,

3 ((i), ) & uh ( i ) (,

) (, ).

As =i


8/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

8 SAHARON SHELAH

some i < j < 1 we have dsq ( ) = whenever is a subsequence of i , asubsequence of j such that i Rang( ), j Rang( ). Now for uh ( i ) , u j we have

(, ) = (, ) (, ) (see ()1) and

(, ) is O.K. as .[Why? Because it is a subsequence of i (see the choice of i ) and ibelongs to Rang( (, )) by ()2] and

(, ) is O.K. as

[Why? Because (, ) is a subsequence of j by the choice of j and j belongs to Rang( (, )) by ()3].

Now by ()4 the colour c(, ) is () as required and get the desired conclusion.1.1

Remark. Can we get P r 1(+2 , +2 , +2 , ) for regulars by the above proof? If = (+ ) and , > (+ )).See more in 2.


9/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


2 Larger Cardinals: The implicit properties

More generally (than in the remark at the end of 1):

2.1 Denition. 1) P r 6(,,, ) means that there is d : > such that:if (u , v ) : < satises

u > , v > ,

|u v | < ,

u v Rang( ),

and < and E a club of then for some < from E we have

u & v d( ) = .

2) P r 6S (,,, ) is dened similarly but < are required to be in E S .P r 6 (,,, ) means for some stationary S { < : cf() } we haveP r 6S (,,, ). If is omitted, we mean = . Lastly P r 6nacc (,,, ) is denedsimilarly but demanding , nacc( E ) and P r 6 (,,, ) is dened similarlybut E = .

2.2 Lemma. 0) If P r 6(,,, ) and 1 and 1 then P r 6(,, 1 , 1) (and similar monotonicity properties for Denition 2.1(2)). Without loss of generality u = v in Denition 2.1.1) If P r 6(+ , + , + , ), then P r 1(+2 , +2 , +2 , ).2) If P r 6(+ , + , , ), so + then P r 1(+2 , +2 , +2 , ) provided that

() there is a sequence A = A : < ++ of subsets of such that for every u ++ with u of cardinality < , we have

A \ { A : u, = } = .

3) If is regular and =


10/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

10 SAHARON SHELAH

|u | = < for < + and continue as there, but after the denition of E 1 wedo not choose i , i instead we rst continue choosing i , i for i < + as there aswithout loss of generality =

i sup[{( j ) : j i} { j : j < i }]and then choose i < i large enough (so no need of the club A of + ).3) As in the proof of 1.1, Stage A.4) Combine the proofs here and there (and not used). 2.2

This leaves some problems on P r 1 open; e.g.

2.3 Question. 1) If > 0 is inaccessible, do we have P r 1(+ , + , + , ) (ratherthan with < )?2) If > 0 is regular (singular) and = + , do we have P r 1(+ , + , + , )?[clearly, yes, for the weaker version: c a symmetric two place function from + to +such that for every < + and pairwise disjoint u : < + with u [+ ] > =

, = cf () and P r 6( , , , ). Then P r 1(+ , + , , ).

Proof. Let e = e : < be a club system, S { < + : cf () = } stationarysuch that / ida (e S ) and e cf() = and

= sup( S ) & < = sup { e : cf ( ) > + + , so nacc(e )}

and e S e e (exists by [Sh 365, 2.10]). Letf = f : < , f : + such that every partial function g from + to

(really suffice) of cardinality is included in some f (exist by [EK] or see[Sh:g, AP1.7]).

So for some f = f () we have

() for every club E of + for some S we have:(a) e E (b) if < and < then

= sup( { nacc( e ) : f ( ) = and cf( ) > }).

This actually proves id p(e S ) is not weakly +

-saturated.The rest is by combining the trick of [Sh:g, III, 4] (using rst () S then somesuitable nacc( e () )) and the proof for 2 . 2.4


11/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


2.5 Fact. P r 1(+ , + , , cf()) implies P r 6(+ , + , , cf()).

Remark. This is not totally immediate as in P r 1 the sets are required to be pairwise

disjoint.

Proof. Let = cf( ) and f for < + be such that < f < J bd f .Let d : [+ ]2 exemplies P r 1(+ , + , , cf()). Let c : be such that forevery < for undoubtedly many < we have c( ) = . For > (+ ) wedene dsq ( ) as follows.

If g( ) 1 or is constant, then dsq ( ) = 0. So assume g( ) 2 and is notconstant.

For < < + let s(, ) = s(, ) = sup {i + 1 : i < and f (i) f (i)},

s(, ) = 0 ,

s( ) = max {s( (), (k)) : ,k < g( ) (so s is symmetric) },

a( ) = {(, k) : s( (), (k)) = s( ) and < k < g ( )}.

As g( ) 2 and is not constant, clearly a( ) = and a( ) is nite, so let ( , k )be the rst pair from a( ) in lexicographical ordering.

Lastly dsq ( ) = c d({ ( ), (k )}) .

Now we are given < , stationary S { < + : cf() cf()}, u : < +(remember 2.2(0)), |u | < cf(), u > such that {Rang( ) : u }.Let u = {Rang( ) : u } and without loss of generality for some stationaryS S and 0 , we have S 0 = min { + 1 : if 1 < 2 are in u then

f 1 [, cf()) < f 2 [, cf())} < and sup( {u : S

}) <

< +

.Now for some 1 ( 0 , cf()) and stationary S 0 , S 1 S and < we have

u & S 0 f ( 1) < ,

u & S 1 f ( 1) > .

Let { : < } enumerate some unbounded S S in increasing order such that < sup( u 0 u 1 ) < min( u 0 u 1 ).

Lastly apply the choice of d. 2.5


12/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

12 SAHARON SHELAH

3 Guessing Clubs Revisited

3.1 Claim. Assume = + , and S { < + : cf () = and is divisible by 2} is stationary.1) There is a strict club system C = C : S such that + / id p(C ) and [ nacc (C ) cf ( ) = ]; moreover, there are h : C such that for every club E of + , for stationarily many S ,


13/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


S :Y = Y +1

and nacc( E +1 )

and E +1 nacc(C ) /J is not stationary .

If we succeed to dene E , for each < , then E =: sup( E nacc(C )) or E () nacc( C ) J ).Clearly if E E and E

, E are dened then

E E

.Now for any club E E () of + , for S acc(E () ) we denehE

: C by letting hE

( ) = otp( B sup( A A

)) for C \{ } and

hE

( ) = 0.


14/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

14 SAHARON SHELAH

Now for any club E of + for stationarily many S acc(E E ), we have

< : { : E E E () nacc( C ) and A } = mod J = Y ()

(this holds by the choice of ()). Let the set of such S acc(E E ) be calledZ E

E . Now for each Z E

E , the set

B [E, E ] =: < :{ : E E E () nacc(C )

and = E

and = sup( A A

)} = mod J

is necessarily unbounded in . So in the same way we have gotten E () we can ndclub E E () such that for any club E of + , for stationarily many Z E

E wehave B [E, E ()] = B [E , E () ] and E = E

(note the minimality in the choiceof E so it can change +1 times; more elaborately if E : < is a decreasing

sequence of clubs and Z E

E , where E = with limit such that = + = tcf( i


15/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


If = sup( E () nacc( C )) and E () nacc( C ) = mod J then (as J is-complete) choose Y J such that for each i < , < i we have

() ( )[ E ()

nacc( C ) & /Y

& f

(i) = ]

{ : E () nacc(C ) & f (i) = } = mod J .

Choose i() < such that

B 0 =: {f (i()) : E () nacc( C ) and /Y }

is unbounded in i .Let = be the -th member of B 0 , for < . For each such < for some

j = j (i() + 1 + , ) we have B 1, =: {f ( j ) : f (i()) = and E () nacc( C ) and /Y } is unbounded in j .

Let h, be a one to one function from [j


16/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

16 SAHARON SHELAH

in ). We shall prove below that for some sequence of functions h = h : S ,h : we have

() h for every club E of +

for stationarily many S acc(E ),the following subset of is stationary:

A,E =: < : E and some ordinal in { : < h ( )}

belongs to E .

The proof now breaks into two parts.Proving () h suffices .

For each club E of , let Z E =: { S : = sup( E nacc( C 0 ))}, and note that

this set is a stationary subset of (by the choice of C 0). For each such E and Z E let f ,E be the partial function from to dened by

f ,E ( ) = Sup { : < h ( ) and E }.

So if there is no such , then f ,E ( ) is not well dened (i.e. if the supremum is onthe empty set) but if = f ,E ( ) is well dened then E, h ( ) (because is increasing continuous in and E is a club of ). LetY E =: { Z E : Dom(f ,E ) is a stationary subset of }. So by () h , we know that

for every club E of +

the set Y E is a stationary subset of +

.

Also

1 if E 2 E 1 are clubs of + then Z E 2 Z E 1 and Y E 2 Y E 1 and for

Y E 2 , Dom( f ,E 2 ) Dom( f ,E 1 ) and Dom( f ,E 2 ) f ,E 2 ( ) f ,E 1 ( ).

We claim that

2 for some club E 0 of + for every club E E 0 of + for stationarily many

S we have(i) = sup( E nacc C ),

(ii) { < : Dom( f E, ) (hence Dom f E 0 , ) andf E, ( ) = f E 0 , ( )} is a stationary subset of .

If this fails, then for any club E 0 of there is a club E (E 0) E 0 of , such that

AE0

= : S, = sup( E (E 0) nacc( C )) and for some club

eE 0 , of we have

eE 0 , Dom( f E (E 0 ) , ) f E (E 0 ) , ( ) = f E 0 , ( )


17/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


is not a stationary subset of = + . By obvious monotonicity we can replace E (E 0)by any club of + which is a subset of it, so without loss of generality AE 0 = .

By induction on n < choose clubs E n of + such that E 0 = + and

E n +1 = E (E n ).Then E =:

n


18/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

18 SAHARON SHELAH

Note that sup( en ) is well dened (as Min( en ) Min(nn() sup( en ) = sup( e

n () ) and call this ordinal so that en ()+1 and

hn () () = hn ()+1 (), so we get a contradiction for n() + 1.

So () h holds for some h, which suffices, as indicated above. 3.3

3.5 Discussion. 1) We can squeeze a little more, but it is not so clear if with muchgain. So assume

()0 is regular uncountable, = + , S { < : cf() = } stationary, I anideal on S , C = C : S a strict S -club system, J = J : S with

J an ideal on C extending J bdC + (acc( C )), such that for any club E of we have { S : E C = mod J } = mod I .

2) If we immitate the proof of 3.3 we get

()1 if for S, J is not -regular (see the denition below) and thenwe can nd e = e : S and g = g : S such that

() 1 e is a club of , e acc(C ), g : nacc(C )\ (min( e ) + 1) e is denedby g ( ) = sup( e ) and for every club E of

S : E nacc( C ) = mod J and

Rang( g (E nacc( C ))) is a stationary subset of = mod I.

3) Denition: An ideal J on a set C is -regular if there is a set A C ,A = mod J and a function f : A [ ]< 0 such that < { x A : /f (x)} = mod J .If = |C |, we may omit it.[How do we prove () 1? Try times E , e

: S (for < )].

4) We can try to get results like 3.1. Now

()2 assume ,,S,I, C, J are as in ()0 and e, g as in () 1 and < and for S, J 0 =: {a e : { Dom( g ) : g() a} J } is weakly normaland satises the condition from [Sh 365, Lemma 2.12]. Then we can ndh : e such that for every club E of ,{ S : for each < the set { nacc( C ) : h (g ()) = } is= mod J } = mod I .

[Why? For each S, acc(e ) choose a clubd, e such that for no club d e of do wehave ( < )( acc(e ))[d d, ]. Now for every club E of letS E = { : E nacc( C ) = mod J , and g (E nacc(C )) is stationary } and for E and < , we choose by induction on < , (, ) as the rst e suchthat: >


19/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


in the interval [ 0 and S { < + : cf () = } is stationary, and C = C : S is an S -club system satisfying + / id p(C, J b[ ]) where J b[ ] = J b[ ]C : S and J b[ ]C =: {A C : for some < , we have > sup{ A : cf ( ) > }}. Then we can nd a strict -club system e = e : < such that

() for every club E of + , for stationarily many S , for every < and < for some we have

()E, nacc (C ) and > and cf ( ) > and { e : E and min (e \ ( + 1)) belongs to E }is a stationary subset of .

3.7 Remark. 1) We know that for the given and S there is C as in the assumptionby [Sh 365, 2]. Moreover, if > 0 then there is such nice strict C .2) Remember J b[ ] = {A C : for some < and < we have

( C )( < cf( ) < nacc( C ))}.

Proof. Let e = e : < be a strict -club system where e = { : < cf( )}is a (strictly) increasing and continuous enumeration of e (with limit ). Now weclaim that for some h = h : < , limit with h a function from e to e and

e

h () > , we have

() h for every club E of + , for stationarily many S acc(E ), AE /J b[ ]C

where AE is the set of all C such that the following subset of e isstationary (in ):

{ e : E and min( e \ ( + 1)) E }.

The rest is like the proof of 3.3 repeating + times instead and using J b[ ]C is( )-based. 3.6

3.8 Claim. Suppose is inaccessible, S is a stationary set of inaccessibles,C an S -club system such that / id p(C ). Then we can nd h = h : S with h : C C , such that < h ( ) and

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

{ C : E and h( ) E } is a stationary subset of .


20/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

20 SAHARON SHELAH

So for some C = { , : < } C , , increasing continuous in we haveh( , ) = , +1 .

Remark. Under quite mild conditions on and S there is C as required - see[Sh 365, 2.12,p.134].

Proof. Like 3.3.

3.9 Claim. Let = cf () > 0 , S stationary, D a normal + -saturated lter on , S is D -positive (i.e. S D + , \ S /D ).1) Assume (C , I ) : S is such that

(a) C = sup( C ), I P (C ),(b) for every club E of ,

{ S : for some A I we have > sup( A\ E )} D + .

Then for some stationary S 0 S, S 0 D + we have(b)+ for every club E of

{ S : for no A I do we have > sup( A\ E )} = mod D.

2) Assume P : S is such that (here really presaturated is enough)

() for every D -positive S 0 S for some D -positive S 1 S 0 and (C , I ) : S we have (C , I ) P , C = sup( C ), I P (C ) and

for every club E of { S 1 : for some A I , > sup( A\ E )} = mod D .

Then

() for some (C , A ) : S we have (C , I ) P , C = sup( C ),I P (C ) and for every club E of

{ S : for no A I , > sup( A\ E )} = mod D.

Remark. This is a straightforward generalization of [Sh:e, III, 6.2B]. IndependentlyGitik found related results on generic extensions which were continued in[DjSh 562] and in [GiSh 577].

Proof. The same.


21/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


3.10 Lemma. Suppose is regular uncountable and S { < + : cf () = } isstationary. Then we can nd (C , h , ) : S and D such that

(A) D is a normal lter on + ,(B ) C is a club of , say C = { , : < }, with , increasing continuous

in ,(C ) h is a function from C to , ,(D ) if A D + (i.e. A + & + \ A /D ) and E is a club of + , then the

following set belongs toD + :

BE,A =: : A S, acc(E ) and for each i <

{ < : , +1 E and h ( , ) = i

(and , E )} is a stationary subset of

(hence, for some < + and < , the set BE,A, =: { BE,A : = , } is in D + ).

(E ) If < + and satises one of the conditions listed below, then S , = { S : = Min (C ) and = } D + where

() = + ,

( ) is inaccessible not strongly inaccessible, < and there is T such that

(a) T is a tree with < nodes and a set of branches, || = ,

(b) if T T, T downward closed and ( )

( a branch of T ) then T has an antichain of cardinality ,

( ) is inaccessible not strongly inaccessible and = Min { : for some we have },

() is strongly inaccessible not ineffable; i.e. is Mahlo and we can nd A = A : < is inaccessible ,A so that for no stationary { < : inaccessible }and A do we have: A = A .

3.11 Remark. We can replace + in 3.10 and any = cf( ) > , as if > + wehave even a stronger theorem.


22/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

22 SAHARON SHELAH

Proof. Let for = cf( ) > 0 ,

= = : if S { < + : cf() = } is stationary

then we can nd (C , h ) : S such that(a) C is a club of of order type ,(b) h : C ,(c) for every club E of + for stationarily many

S acc(E ) we have:i < BE = { C : E, h () = i and

min( C \ ( + 1)) E }

is a stationary subset of .

Now we rst showfor each of the cases from clause (E), the belongs to .

Proof of sufficiency of . We can partition S to + stationary sets so we cannd a partition S , : and < + of S to stationary sets. Without loss

of generality, Min( S , ) and let (C 0 , h

0 ) : S , be as guaranteed by for the stationary set S , . Now dene C , h for S by:

C is C 0 {}\ if S , and < , h ( ) is h ( ) if C C 0 and is zerootherwise. Of course, = if S , .

Lastly, let

D = A + : for some club E of + , for every

S acc(E )\ A for some i < ,

the set { C : E, h ( ) = i and min( C \ ( + 1) E }is not a stationary subset of .

So D and (C , h , ) : S have been dened, and we have to check clauses(A)-(E).Note that = and the proof which appears later does not rely on the intermediateproofs.

Clause ( A): Suppose A D for < , so for each there is a club E of +

() if S , and S acc(E )\ A then{ C : E, Min(C \ ( +1)) E and h () = i } is not stationary in.


23/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


Clearly clubs of + belong to D .Clearly A A A D (by the denition), witnessed by the same E .Also A = A0 A1 D as witnessed by E = E 0 E 1 .

Lastly, A =


24/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

24 SAHARON SHELAH

3.13 Claim. Let = cf () > . A sufficient condition for is the existenceof some < + such that

in the following game of length , rst player has no winning strategy:in the -th move rst player chooses a function f : and second player chooses < . In the end, rst player wins the play if { < : for every < , f () = } is a stationary subset of .

(If we weaken the demand in from stationary to unbounded in , we can weaken it here too).


25/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


4 More on P r 6

4.1 Claim. P r6(+ , + , + , ) for regular.

Proof. We can nd h : + + such that for every < + the setS =: { < + : cf() = and h() = } is stationary, so S : < is a partitionof S =: { < + : cf() = }. We can nd C = C : S such that C is aclub of of order type . For any > (+ ) we dene:

(a) for < g( ), if () S then leta = a, = {otp( C () (k)) : k < g( ) and (k) < ()},

(b) is the < g( ) such that

(i) () S ,(ii) among those with sup( a, ) is maximal, and

(iii) among those with minimal,

(c) if is well dened let d( ) = h( ( )) otherwise let d( ) = 0.

Now suppose (u , v ) : < + , < + and E are as in Denition 2.1 and weshall prove the conclusion there. LetE = { E : is a limit ordinal and < >

sup[{Rang( ) : u v }]}.Clearly E E is a club of + .

For each S let

f 0() =: sup[ {Rang( ) : u v }].

As cf() = > |u v | and the sequences are nite clearly f 0() < . Hence byFodors lemma for some , S 1 =: { S : f 0() = } is a stationary subset of + (note that is xed here). Let =

i


26/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

26 SAHARON SHELAH

b =: otp( C ) : < S and both

are in a2 ,i {} {Rang : u v } .

So b is a subset of of cardinality < hence =: sup( b ) < , hence for some

S 3 =: { S 2 : = }

is a stationary subset of + . Choose such that

() S 3 E and = sup( S 3 E ).

As C has order type , (and is a club of ) for some S 3 E we haveotp( C ) > .We want to show that , are as required. Obviously < , E and E . So assume u , v and we shall prove that d( ) = , whichsuffices. As h( ) = (as S 3 S ) it suffices to prove that ( )( ) = .Now for some 0 , 1 we have (0) = , (1) = (as u , v ) and sinceotp( C ) > , by the denition of it suffices to prove

if ,k < g( ), ( )()

S, ( )(k) < ( )() then(i) otp[C ( )( ) ( )(k)] or

(ii) ( )() = .

Assume , k satisfy the assumption of and we shall show its conclusion.

Case 1: If ( )() and ( )(k) belong to

a2,i { } {Rang( ) : u v }

then clause ( i) holds because

( ) otp( C ( )( ) ( )(k)) b (see the denition of b ) and( ) sup( b ) = (see the denition of ) and( ) = (as S 3 and see the choice of and S 3 ).

Case 2: If ( )() and ( )(k) belong to

a2,i {Rang( ) : u v }

then the proof is similar to the proof of the previous case.

Case 3: No previous case.


27/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


So ( )() and ( )(k) are not in a2,i , hence (as {, } (u v ), and{, } S 2 S 1 )

m {, k} & m < g ( ) ( )(m) = (m)

,

m {, k} & m g( ) ( )(m) = (m g( )) .

As E and > clearly sup(Rang( )) < , but also( )(k) < ( )() (see ).

Together necessarily k < g( ), (k) [, ), [g( ), g( ) + g()) and( g( )) [ , + ). If () = then clause ( ii ) of the conclusion holds.Otherwise necessarily () > hence

otp( C ( )( ) ) ( )(k)) = otp( C ( g ( )) (k)) otp( C ( g ( )) ) sup( a )

so clause (i) of holds. 4.1

4.2 Conclusion. For regular, P r 1(+2 , +2 , +2 , ) holds.

Proof. By 4.1 and 2.2(1). 4.2

4.3 Denition. 1) Let P r 6( , , ) means that for some , an unbounded subset

of { : < , is a cardinal (nite or innite) }, there is a d : > ( ) suchthat if < and are given and (u , v ) : < satises

(i) u > ( )\ 2 ( ),(ii) v > ( )\ 2 ( ),

(iii) |u | = |v | = ,(iv) u (g( ) 1) = , ,(v) u (0) = , ,

(vi) u v ()(() = , )

then for some < we have

u & v ( )[d( )] = , .

2) Let P r 6(, ) means P r 6(,, ).

4.4 Fact. P r 6(,,, ), implies P r 6( , , ).

Proof. Let c be a function from > to exemplifying P r 6(,,, ). Let e be aone to one function from onto .

Now we dene a function d from > ( ) to :

d( ) = Min { : c( e( (m)) : m < g ( ) ) = e( ())}.

4.4


28/29

(572)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5

28 SAHARON SHELAH

4.5 Claim. If P r 6(+ , ), regular and then P r 1(+2 , +2 , +2 , ).

Proof. Like the proof of 1.1.

4.6 Remark. As in 4.1, 4.2 we can prove that if > cf() + thenP r 6(+ , + , + , ), hence P r 1(+2 , +2 , +2 , ), but this does not give new infor-mation.


29/29

2)

r e v

i s

i o n :

1996 -

09 -

05

m o

d i f i

e d

: 1 9 9 6

- 0

9 -

0 5


REFERENCES.

[DjSh 562] Mirna Dzamonja and Saharon Shelah. On squares, outside guessingof clubs and I

saharon shelah- colouring and non-productivity of ℵ2-c.c

Documents