saharon shelah- further cardinal arithmetic
TRANSCRIPT
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
1/54
430
re
vision:1996-10-12
modified:1996-10-12
ISRAEL JOURNAL OF MATHEMATICS 95 (1996), 61114
FURTHER CARDINAL ARITHMETIC
BY
Saharon Shelah
Institute of Mathematics
The Hebrew University of Jerusalem, Jerusalem, Israel
and
Department of Mathematics
Rutgers University, New Brunswick, NJ, USA
ABSTRACT
We continue the investigations in the authors book on cardinal arithmetic,
assuming some knowledge of it. We deal with the cofinality of (S0(),)
for real valued measurable (Section 3), densities of box products (Section
5,3), prove the equality cov(,,+, 2) = pp() in more cases even when
cf() = 0 (Section 1), deal with bounds of pp() for limit of inaccessible
(Section 4) and give proofs to various claims I was sure I had already
written but did not find (Section 6).
Annotated Contents
1. Equivalence of two covering properties . . . . . . . . . . . . . . . . . 63
[We try to characterize when, say, has few countable subsets; for a given
(0, ), we try to translate to expressions with pcfs the cardinal
Min
|P|: P S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
2/54
430
re
vision:1996-10-12
modified:1996-10-12
62 S. SHELAH Isr. J. Math.
[We show that if > , = cov(, +, +, ) and cov(,,, 2) (or
), then cov(, +, +, 2) = cov(,,, 2). This is used in [Sh-f, Appendix,1]
to clarify the conditions for the holding of versions of the weak diamond.]
3. Cofinality ofS0() for real valued measurable and trees . . . . . . . 72
[Dealing with partition theorems on trees, RubinShelah [RuSh117] arrive at thestatement: > > 0 are regular, a S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
3/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 63
1. Equivalence of Two Covering Properties
1.1 Claim: If pp = +, > cf() = > 0 then cov(,,+, 2) = +.
Proof: Let = 3()+; choose B : <
+ increasing continuous, such
that B (H(), ,
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
4/54
430
re
vision:1996-10-12
modified:1996-10-12
64 S. SHELAH Isr. J. Math.
a, be
Ch,() =
sup( N1) if N
1,
0 otherwise,
we have: Ch < f a, mod Jbda,
.
[Why? Clearly ChB(N0 ) B+1, so a, B+1, hence there are in B+1
elements b[a,]: pcf(a,) and fa,, : < : pcf(a,) as in [Sh
371, 2.6, 1]. So for some , (, +) we have Ch b+ [a,] < f , so it is
enough to prove a, b+ [a,] is bounded below but otherwise pp() = +
will be contradicted. Let = sup{,: N0}.]
(e) E =: { < +: < & |C| < and > } is a club of .
Now as S is stationary, there is () S E. Remember otp C() = +.
Let C() = {(), : < +} (in increasing order).
Let (for any < +) M0 be the Skolem Hull of
f(), : <
{i : i }, andlet M1 be the Skolem Hull ofa
f(), : <
{i : i }. Note: for < +
non-limit
f(), : <
=
f: C(),
. Clearly M0 : < +, M1 : 0, we have (0) (1) (2) = (3)
and ifcov(, 1, 1, 2) < they are all equal, where:
(0) =: is the minimal such that: ifa Reg +\, |a| then we
can find a: < such that a =
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
6/54
430
re
vision:1996-10-12
modified:1996-10-12
66 S. SHELAH Isr. J. Math.
1.2A Remark: (1) We can get similar results with more parameters: replacing
0 and/or 1 by higher cardinals.
(2) Of course, by assumptions as in [Sh410, 6] (e.g. | pcfa| |a|) we get
(0) = (3). This (i.e. Claim 1.2) will be continued in [Sh513].
Proof:
(1) (2): Trivial.
(2) (3): Let = 3((3))+ and for + we choose B (H(), ,
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
7/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 67
For n = 0 we can define Na0 , Nb0 , An, trivially. Suppose N
am, N
bm, Am,,
Pm, are defined for m n, < and fm (m < n) are defined. Now an is well
defined and Reg +\ B and |an| . So an = an, and an, an,+1
where an, =: an An, and, of course, an, Reg +\ has cardinality .
Note that an, is not necessarily in B but
()1 every countable subset ofan, is included in some subset ofB which belongs
to Pn, and is Reg +\.
By the definition of (3) (see equivalently there), for each n, we can
find an increase sequence an,,k: k < of subsets of an, with union an, and
dn,,k [, (3)] pcf(an,,k), |dn,,k| < such that:
()2 ifb an,,k is countable then b is included in a finite union of some members
of {b[an,,k]: dn,,k} (hence max pcf(b) (3)).
By the properties of pcf:()3 for each , k < and c Reg +\ such that c Pn, we can find
e = e,kc (3)+ pcfc, |e| |dn,,k| < such that for every dn,,k we
have: cb[an,,k] is included in a finite union of members of{b[c]: ec}.
By [Sh371, 1.4] we can find fn an
such that:
()4 () sup(Nbn ) < fn();
() if c Pn,, , k < , c Reg +\ and e,kc pcf(c) [, (3)]
(where e,kc is from ()3) then for some m < , p + pcf(c)
and p
< p
, (for p
m) the function fn
(b
[c]) is included in
Maxpm fc,p bp [c] (the Max taken pointwise).
Note
()5 if b an,,k is countable (where , k < ) then there is c Pn,, |c| < ,
c Reg +\ such that b c.
By ()4 :
()6 if ,k < , c Pn,, c Reg +\, and dn,,k (3)
+ pcfc\ then
fn b[c] B.
You can check that (by ()2 ()6) :
()7 if b an,,k is countable then there is fn,,kb B, | Dom fn,,kb
| < such
that fn b fn,,kb
.
Let i(i < ) list the Skolem function of (H(), ,
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
8/54
430
re
vision:1996-10-12
modified:1996-10-12
68 S. SHELAH Isr. J. Math.
and Pn+1, = S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
9/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 69
1.3 Claim: Assume 0 < cf < < , pp() and
cov(, +, +, 2) < .
Then cov(,,+, 2) < .
Proof: Easy.
1.3A Definition: Assume = cf > = cf > 0.
(1) (C, P) T[, ] if (C, P) T[, ] (see [Sh420, Def 2.1(1)]), and
S(C) = sup(acc C) (note: acc C C), and we do not allow (viii)
(in [Sh420, Definition 2.1(1)]), or replace it by:
(viii) for some list ai: i < ofS(C) P, we have: S(C), acc C
implies {a : a P, a } {ai: i < }.
(2) For (C, P) T[, ] we define a filter D(C,P)
() on [S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
10/54
430
re
vision:1996-10-12
modified:1996-10-12
70 S. SHELAH Isr. J. Math.
1.3C Claim: Let > = cf > 0, = +, (C, P) T[, ] then the
following cardinals are equal:
(0) = cf (S = cf > 0, = + and (C, P) T[, ]. Let
B1 be a rich enough model with universe 1 and countable vocabulary which is
rich enough (e.g. all functions (from 1 to 1) definable in (H((1)+), , 2
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
11/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 71
of of cardinality is included in the union of < of them (exists by the
definition of = cov(, +, +, )). Let P0 = {Ai: i < }. Let P1 be a family of
cov(,,, 2) subsets of , each of cardinality < such that any subset of of
cardinality < is included in one of them.
Let P =:iaAi: a P1
; clearly P is a family of subsets of each
of cardinality , |P| |P1| = cov(,,, 2), and every A , |A| is
included in some union of < members ofP0 (by the choice ofP0), sayibAi,
b , |b| < ; by the choice of P1, for some a P1 we have b a, hence
A ibAi
iaAi P. So P exemplify cov(,
+, +, 2) cov(,,, 2).
Second we prove the inequality . If 0 then cov(, +, +, 2) =
and cov(,,, 2) = so trivially holds; so assume > 0. Obviously
cov(, +, +, 2) . Note, if is singular then, as cf+ > for some
1 < , we have = cov(, +
, +
, ) = cov(, +
, +
,
) whenever
[1, ]is a successor (by [Sh355, 5.2(8)]); also cov(,,, 2) sup{cov( ,,, 2):
[1, ] is a successor cardinal} and cov( ,,, 2) cov(, , , 2) when < ,
so without loss of generality is regular uncountable. Hence for any 1 < we
have
()1 we can find a family P = {Ai: i < 1}, Ai , |Ai| , such that any
subfamily of cardinality + has a transversal. [Why? By [Sh355, 5.4],
(=+) and [Sh355,1.5A] even for .]
Hence if 1 , cf1 < + (or even cf1 ) then ()1 . Now we shallprove below
(1) ()1 cov(1,,, 2) cov(, +, +, 2)
and obviously
(2) if cf then cov(,,, 2) =
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
12/54
430
re
vision:1996-10-12
modified:1996-10-12
72 S. SHELAH Isr. J. Math.
the definitions ofb[A] and Aa; note a S
and)
cov(, +, +, 2) cov(, +, +, ) + cov(,,, 2)
= + cov(,,, 2) = cov(,,, 2)
and
cov(,,, 2) cov(, +, +, 2) + cov(,,, 2),
hence, cov(,,, 2) = cov(, +, +, 2) + cov(,,, 2).
3. Cofinality of S0() for Real Valued Measurable and Trees
In RubinShelah [RuSh117] two covering properties were discussed concerning
partition theorems on trees, the stronger one was sufficient, the weaker one nec-
essary so it was asked whether they are equivalent. [Sh371, 6.1, 6.2] gave a partial
positive answer (for successor of regular, but then it gives a stronger theorem);
here we prove the equivalence.
In GitikShelah [GiSh412] cardinal arithmetic, e.g. near a real valued mea-
surable cardinal , was investigated, e.g. {2: < } is finite (and more); this
section continues it. In particular we answer a problem of Fremlin: for real
valued measurable, do we have cf(S = > 0. Then
the following conditions are equivalent:
(A) for every < we have cov(,,, 2) < ,
(B) if < and a S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
13/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 73
3.1A Remark: (1) Note that (B) is equivalent to: if a S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
14/54
430
re
vision:1996-10-12
modified:1996-10-12
74 S. SHELAH Isr. J. Math.
in I[] (see [Sh420, 1.5]) and let S S+, C = C: S+ be such that: C
closed, otp C , [ nacc C C = C ], [otp C = S] and
for S+ limit, C is unbounded in (see [Sh420, 1.2]).
Without loss of generality C is definable in (B, , , ). Let 0 [, )
be minimal such that cov(0, ,, 2) , so 0 > , > cf0. We choose by
induction on < , A, a such that:
() A (H(), ,
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
15/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 75
There is no problem to carry the definition: for n = 0 define N0 by (i)
[trivially (b) holds and also (c), as for (d), note that C A0 A() and
{i: i < } A() as C is definable in B hence {,, : S+, < , and
is the -th member of C} is a relation ofB hence each C+1( < ) is in A()
hence each {i: i < } is and we can compute the Skolem Hull in Aj for j <
large enough].
Next, choose Mn by (k), it satisfies (e) + (a). If Nn : < , Mn are
defined, we can find fn satisfying (f) + (g) + (h) by [Sh371,1.4] (remember ()).
For n + 1 define Nn by (j) and then Mn+1 by (k).
Next by [Sh400, 3.3A or 5.1A(1)] we have
()
n
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
16/54
430
re
vision:1996-10-12
modified:1996-10-12
76 S. SHELAH Isr. J. Math.
3.2 Conclusion: (1) If is real valued measurable then = cf [S = cf > 0, I is a -complete ideal on extending
Jbd and is -saturated (i.e. we cannot partition to sets not in I). Then for
< , cf(S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
17/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 77
() for every successor 2 there is a tree from [Sh355, 3.5]: cf
levels, every level of cardinality < and (cf)-branches,
() for every (, ), there is a tree T of cardinality with
branches of the same height,
() cf cf and even cf > 0 pp(cf)() =+ 2.
(C) Like (B) but we omit () and retain ().
Proof:
First Case: = 0. Trivially (A) holds.
Second Case: is regular uncountable. So and 2 = 2 and [ <
2 < 2] hence 2 ). Try to apply [Sh410, 4.3], its
assumptions (i) + (ii) hold (with here standing for there) and if possibility
(A) here fails then the assumption (iii) there holds, too; so there is as there; so(), () of (B) of 3.3 holds and let us prove (), so assume (, ), without
loss of generality, is regular, and we shall prove the statement in () of 3.3(B).
Without loss of generality is regular and (, )&cf pp() < ;
i.e. is (, +, 2)-inaccessible. [ Why? If is not as required, we shall show how
to replace by an appropriate regular [, ).]
Let (, ) be minimal such that pp() , (so cf ) now
pp() < (by the choice of) and =: pp()+, by [Sh355, 2.3] is as required] .
Let be minimal such that 2
. So trivially < and(2
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
18/54
430
re
vision:1996-10-12
modified:1996-10-12
78 S. SHELAH Isr. J. Math.
(, 2
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
19/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 79
()n n < < n&cf
n ppn() < n hence pp
+n
(n) = pp+(n)
(n)
=
2n
.
Note that 2 m, so Max{cfn, cfm} = Max{n, m} m. As the class of cardinals
is well ordered we get S1 =: {n < : n S0, n n+1} is co-infinite and
S =: {n: n } is finite (so () of 3.4(2)(b) holds).
So for some n() < , S n() hence for every n [n(), ) for some
m (n, ), n < m. Note: n = m n = m (as their cofinalities are distinct)
and [n / S0 n / {m: m < }]. Assume n n(), if n > n+1, letm = mn = Min{m: m+1 > n and m n} (it is well defined as
k n < k
and k < k < = m+1 and ifm+1 / S0
trivially and if m + 1 S0 by one of the demands on m+1 (in its choice) and
[Sh355, 2.3] we have m+1 m; but m < n, so m+1 < n contradicting the
choice of m. So by the last sentence, n n() mn < mn+1. By [Sh355,
5.11] we get the desired conclusion (i.e. also part () of 3.4(2)). 3.4
Remark: It seemed that we cannot get more as we can get an appropriate prod-
uct of a forcing notion as in Gitik and Shelah [GiSh344].
4. Bounds for pp(1) for Limits of Inaccessibles
4.1 Convention: For any cardinal , > cf = 1 we let Y, Eq be as in
[Sh420, 3.1], is a strictly increasing continuous sequence of singular cardinals
of cofinality 0 of length 1, =i
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
20/54
430
re
vision:1996-10-12
modified:1996-10-12
80 S. SHELAH Isr. J. Math.
4.2 Theorem (Hypothesis [Sh420, 6.1C] ):
(1) Assume
(a) > cf = 1, Y = Y, Eq Eq,
(b) every D FIL(Y) is nice (see [Sh420, 3.5]), E = FIL(Y) (or at least
there is a niceE (see [Sh420, 5.25], E =
E = Min E, E is-divisible
having weak -sums, but we concentrate on the first case),
(c) < < pp+E(), inaccessible.
Then there are e Eq and x: x Y/e, a sequence of inaccessibles < and
a D FIL(e, Y) E nice to , D FIL(e, Y) such that:
()xY/e
x/D has true cofinality ,
() = tlimDx: x Y.
(2) We can weaken (b) to E FIL(Eq, Y) and for D E, in the game
wG(,D,e, Y) the second player wins choosing filters only from E.(3) Moreover, for given e0, D0, 0x: x Y/e0, if
xY/e0
0x/De0 is-directed,
then without loss of generality e0 e, D0 D and x x[e0] .
4.2A Remark: (1) We could have separated the two roles of (in the definition
of Y, etc. and in (, pp+E())) but the result is less useful; except for the
unique possible cardinal appearing later.
(2) Compare with a conclusion of [Sh386] (see in particular 5.8 there):
Theorem: Suppose > 2
1
, (weakly) inaccessible.(1) If 1 < i = cfi < for i < 1, D is a normal filter on 1,
i (x),(B)f,D = tcf
xY/e f(x)/D
.
Let K0 =:
(f, D): D E, f Y/e and conditions (A)f and (B)f,D hold }, so
K0 = . Now if (f, D) K0, for some
(C)f,D, in G(D,f,e, Y) the second player wins (see [Sh420, 3.4(2)])
* I.e.: ifa Reg, |a| < min(a), inaccessible then > sup( pcfa).
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
21/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 81
hence K1 = where K1 =: {(f , D , ) K0 condition (C)f,D, holds}.
Choose (f1, D1, ) K1 with minimal. By the definition of the game
() for every A = mod D1 we have (f1, D1 + A, ) K1.
Let e1 = e(D1).
Case A: {x: f1(x) inaccessible} = mod D1. We can get the desired conclu-
sion (by increasing D1).
Case B: {x: f1(x) successor cardinal} = mod D1. By (), without loss of
generality f1(x) = g(x)+, g(x) a cardinal (so (x)) for every x Y/e. By
[Sh355, 1.3] for every regular (, ) there is f (Y/e) Ord satisfying:
(a) f < f1, each f(x) regular,
(b) tlimD1f = ,
(c)x f(x)/D1 has true cofinality .
By (a) we get
(d) f g.
By (b) we get, by the normality of D1, that for the D1-majority of x Y/e,
f(x) (x); as f(x) is regular (by (a)) and (x) singular (see 4.1) we get
(e) for the D1-majority of x Y/e, we have f(x) > (x).
Let be large enough, let N be an elementary submodel of (H(), ,
sup( N), so for some () < :
h N Y/e1 Ord
x Y/e1: f,()(x) h(x) < f(x)
= mod D1.
[Why? For any such h define h Y/e1Ord by: h(x) is h(x) if h(x) < f(x)
and zero otherwise, so for some h < , h < f,h mod D1. Let () =
sup
h: h N
Y/e1 N
; it is < as N < , and it is as required.]
Let f = f,()
. The continuation imitates [Sh371, 4], [Sh410, 5].
Let
K2 =
(D, B, jx: x Y/e1): D1 D E, player II wins GE (f
1, D),
e1 = e(D), B = < Bx,j : j < j0x (x) > : x Y/e1 N,
|Bx,jx| g(x) and jx < j0x (x),
{x Y/e1: f(x) is in Bx,jx} D
.
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
22/54
430
re
vision:1996-10-12
modified:1996-10-12
82 S. SHELAH Isr. J. Math.
Clearly K2 = . For each (D, B, jx: x Y/e1) K2 :
()1 letting h Y/e1 Ord, h(x) = |Bx,jx|, for some h =
, f1, 0, h
, for
some < and D player II wins in G,E (D, h, e1, Y).
So choose (D, B, jx: x Y/e1, 0) such that:
()2 (D, B, jx: x Y/e1) K2, ()1 for 0 holds and (under those restric-
tions) 0 is minimal.
So (as player I can move twice), for every A D+, if we replace D by D + A,
then ()2 still holds.
So without loss of generality (for the first and third members use normality):
()3 one of the following sets belongs to D:
A0, =
x Y/e1: cf|Bx,jx| > (x) and j
0x < }
(for some < 1 such that |Y/e1| < ),A1 =
x Y/e1: cf|Bx,jx| < (x) |Bx,jx|
,
A2, = {x Y/e1: |Bx,jx| and jx < } (for some < 1).
If A2, D then (for x Y/e1)
Bx =:
Bx,j: x Y/e1, j < j0x and |Bx,jx| < and j <
is a set of ordinals and
{x Y/e1: f(x) Bx} D
and Bx: x Y/e1 belongs to N (as (D, B, jx: x Y/e1) K2 and the
definition ofK2), contradiction to the choice of f (see , remember D1 D by
the definition of K2).
If A1 D, we can find B1 N, B1 = B1x,j: j < j
1x (x): x Y/e1,
|B1x,j| g(x) andj (x). Let
a =
cf|Bx,j |: cf|Bx,j| > (x), x Y/e1, j < j0x, j < and (x) >
,
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
23/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 83
so a is a set of regular cardinals, and (remember |Y/e1| < ) we have |a| < Min a,
so let b = b[a]: pcfa be as in [Sh371, 2.6]. So as (by the Definition ofK2),
Bx,j: j < j0x: x Y/e1 N, clearly a N hence without loss of generality
b N. Let = sup[ pcfa], so by Hypothesis [420, 6.1(C)], < , but
N, so + 1 N.
By the minimality of the rank we have for every pcfa,
{x y/e1: cf|Bx,jx| b} = mod D hencex cf|Bx,jx|/D is -directed, hence
we get contradiction to the minimality of the rank of f1.
(2), (3) Proof left to the reader. 4.2
4.2B Remark:
(1) The proof of 4.3 below shows that in [Sh386] the assumption of the existence
of nice filters is very weak, removing it will cost a little for at most one place.
(2) We could have used the framework of [Sh386] but not for 4.3 (or use forcing).
4.3 Claim (Hypothesis 6.1(C) of [Sh420] even in any K[A]): Assume > cf =
1, > > 1, pp(,1)() > , inaccessible. Then for some e Eq,
D FIL(e, Y) and sequence of inaccessiblesx: x Y/e, we havetlimD x =
and = tcf(
x/D) except perhaps for a unique in V (not depending on
) and then pp+(,1)() +.
Proof: By the Hyp. (see [Sh513, 6.12]) for some a Reg , |a| < Min(a),
= max pcf(a), and
( < )(b)[b a & |b| < >]& > sup pcf 1complete
(b) > ],
J = J when
A . Choose A such that L[A] and for every < , there is a one
to one function f from || (i.e. ||V) onto , f L[A], so Card
L[A]
+ 1
=
CardV, and apply 4.2 to the universe K[A] (its assumption holds by [Sh420, 5.6]).
Second assume () in K[A] there is a Ramsey cardinal > when A +
and assume our desired conclusion fails. Let S be stationary [ S cf =
+], a: < , exemplify S I[] (exist by [Sh420, 1]). We can find a, J as
described above. Let f: < exemplify = tcf(a/J), now by [Sh355, 1.3]
without loss of generality = max pcfa. Let A0 be such that a, f: < ,
b[a]: pcfa are in L[A0]. Hence in L[A0] for suitable J, f/J: < is
increasing, and without loss of generality for some c: a: S L[A0],
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
24/54
430
re
vision:1996-10-12
modified:1996-10-12
84 S. SHELAH Isr. J. Math.
we have: for S, cf = |a|+, a a club of and f (a\c): a is
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
25/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 85
4.5 Theorem: If is regular (> 1) A , Z K[A] a bounded subset of
then for some < , Z
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
26/54
430
re
vision:1996-10-12
modified:1996-10-12
86 S. SHELAH Isr. J. Math.
for some = they are equal and we get contradiction by g, g() = 0, g() = 1,
Dom g = {, }].
Also trivial is: for limit, d
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
27/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 87
(D) for everyx
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
28/54
430
re
vision:1996-10-12
modified:1996-10-12
88 S. SHELAH Isr. J. Math.
(E) (G) when = : where { <
: } = mod J: Easy too.
Next assume every is regular, J an ideal on .
(G)(F): (F) is a particular case of (G), because (S
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
29/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 89
Ci i, j Ci Cj = j Ci, otp(Ci) ||+ and S =: {i < : cf(i) = ||+, =
sup(Ci)} stationary: so wlog j Ci
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
30/54
430
re
vision:1996-10-12
modified:1996-10-12
90 S. SHELAH Isr. J. Math.
5.9 Observation: In several of the models of set theory in which we know
strong, singular, limit, 2 > + our sufficient conditions for dcf(, 2) = 2
usually hold by the sufficient condition 5.4(a) (simplest: if GCH holds below ,
cf = 0).
Remark: We could prove this consistency by looking more at the consistency
proofs, adding many Cohen subsets to in preliminary forcing; but the present
way looks more informative.
6. Odds and Ends
6.1 Lemma: Suppose cf() > +, I an ideal on , f Ord for < is
I-increasing. Then there are J, s, f( < ) such that:
(A) s = si: i < , each si a set of ordinals,(B)i
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
31/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 91
For a set a sup
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
32/54
430
re
vision:1996-10-12
modified:1996-10-12
92 S. SHELAH Isr. J. Math.
()4 J is an ideal on extending I, in fact is the ideal generated by I{a,:
(, )}.
As f: < is I-increasing (i.e. ()1):
()5 J decreases with , in fact a,/I increases with , decreases with ,
()6 if D is an ultrafilter on disjoint to J, then f/D is a
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
33/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 93
b a I. [Does this occur? E.g. for I = S fori < ,
J a -complete ideal on and = tcfi
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
34/54
430
re
vision:1996-10-12
modified:1996-10-12
94 S. SHELAH Isr. J. Math.
club E E of , for stationarily many S, g(C , E) E) (remember
g(C , E) = {sup( E): C ; > Min(E)}). Let C
, = { C
: =
Min(C\ sup(E)}, they have all the properties of the Cs and guess clubs in
a weak sense: for every club E of for some S E, if1 < 2 are successive
members of E then |(1, 2] C, | 1; moreover, the function sup(E )
is one to one on C, .
Now we define by induction on < , an ordinal and functions g
i
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
35/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 95
(3) In () of 6.2, < can be replaced by < + (prove by
induction on ).
6.3 Observation: Assume < }. Then there are ,
andT
, satisfying the condition (
) below for = 2 or at least arbitrarily large
regular 2.
() T a tree with levels, (where ) with a set X of -branches, and
for < , .
Case 1:2, T =2 are O.K. (the set of branches
2 has cardinality 2).
Case 2: Not Case 1. So for some < , 2
, but by the choice of , 2
,so 2 = , < and so < 2|| = 2. Note |>2| = as .
Subcase 2A: cf() = cf(). Let >2 =j , for some ordinal j < we have
{ 2: j j} has cardinality .
As cf() = cf() and (by its definition) clearly < , hence |Bj | < .
Let
T = { : < g() and Bj} .
It is as required.
Subcase 2B: Not 2A so cf() = cf(). As ()[ < = 2
cf() = cf(2
) > ], clearly cf() so is regular. If = we get = , so singular. So if < , < i = cf(i) < for i < then (see
[Sh-g, 345a, 1.3(10)]) max pcf{i: i < } i
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
36/54
430
re
vision:1996-10-12
modified:1996-10-12
96 S. SHELAH Isr. J. Math.
is T i . In fact we can get any regular cardinal in (, pp+()) in the same
way. Let = min{: < , cf() = and pp() > }, so (by [Sh355,
2.3]), also has those properties and pp() pp(). So if pp+() = (2)+
or pp() = 2 is singular, we are done. So assume this fails.
If > 0, then (as in 3.4) < 2 cov(, +, +, ) < 2 and we can
finish as in subcase 2A (as in 3.4; actually cov(2
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
37/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 97
Now choose by induction on < |a0|+, c, satisfying > max pcf{: < }.
If we are stuck in , maxpcf{: < } is the desired maximum by ()1. If we
succeed = maxpcf{: < |a0|+} is in pcf{: < } for some < |a0|
+ by
()2; easy contradiction. 6.4A
6.4
6.5 Conclusion: Assume 0 = cf() 0 < , [ (0, ) & cf()
pp() < ] and pp+ () > = cf() > . Then we can find n for n < ,
0 < n < n+1 < , =n
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
38/54
430
re
vision:1996-10-12
modified:1996-10-12
98 S. SHELAH Isr. J. Math.
Let i = maxpcf(b {j : j < i}). For each i let bi = b {j : j < i} and
fb,: < : pcfb be as in [Sh371, 1]. Let
T0i = Max=1,n fb
, bi: pcf(bi), < , n < .
Let Ti = {f T0i : for every j < i, f bj T0j moreover for some f
j
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
39/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 99
6.6B Observation: Let < be regular uncountable, 2 < i < (for i < ),
i increasing in i. The following are equivalent:
(A) there is F such that:
(i) |F| = ,
(ii) |{f i: f F}| i,
(iii)
f = g F f =Jbd g
;
(B) there be a sequence i: i < such that:
(i) 2 < i = cf(i) i,
(ii) max pcf{i: i < } = ,
(iii) for j < , j maxpcf{i: i < j};
(C) there is an increasing sequence ai: i < such that pcfi
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
40/54
430
re
vision:1996-10-12
modified:1996-10-12
100 S. SHELAH Isr. J. Math.
()3 if {j: j < i} / Ji, then for every f F, f i || for < ,
and for each < , = : < is a sequence of ordinals. Then for every
X , X = mod D there is : < (a sequence of ordinals) and w
such that:
(a) \w cf( ) ,
(b) if and [ w
=
], then { X: for every < we have
and [ w
=
] } = mod D.
Proof: Essentially by the same proof as 6.6C (replacing i by Min{ X: for
every Y Ni D we have Y}). See more [Sh513, 6]. 6.6D
6.6E Remark: We can rephrase the conclusion as:
(a) B =: { X: if w then = , and: if w then
is <
but > sup{ : < , <
}} is = mod D.
(b) If < for w then { B: if w then >
} =
mod D.
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
41/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 101
(c) w cf() is but .
6.6F Remark: (1) If |a| < min(a), F a, |F| = = cf pcf(a) and even
> = sup(+ pcf(a)) then for some g a, the set {f F: f < g}
is unbounded in (or use a -complete D as in 6.6E). (This is as a/J
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
42/54
430
re
vision:1996-10-12
modified:1996-10-12
102 S. SHELAH Isr. J. Math.
[exists by [Sh345a, Def. 3.3(2)b + Fact 3.4(1)]].
Let = (sup a)+, |a| < = cf < Min a (without loss of generality
there is such ) and N = Ni: i < be an increasing continuous sequence of
elementary submodels of (H(), ,
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
43/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 103
()4 bi,j,(i,j) J[a].
For this end we define by induction on < |a|+ functions fa,, with domain
bi,j, for every < a, such that < f
a,, f
a,, , so the domain
increases with .
We let fa,,0 = fa, bi,j , fa,, =
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
44/54
430
re
vision:1996-10-12
modified:1996-10-12
104 S. SHELAH Isr. J. Math.
Now if < sup(N ) then for some (1), < (1) N , so letting
b =: bi,j, b[a] bi,j, , it belongs to J[a]J
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
45/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 105
(i) b[a] increases with .
This will be proved below.
6.7B Claim: In 6.7A we can also have:
(1) if we let b[a] = b[a] = |a|, c Ni , c a (hence |c| < Min(c) and c a), then
for some finite d (pcfc) a we have c d b[a]. Similarly for
-complete, < cf() (i.e. we have clauses (h), (h)+ for = ).
(4) We can have continuity in when cf() > |a|, i.e. b = +3, let = +2,
let N, N, a, b (as a function), i: =: |a|+ be as in 6.7A but also
i: i < N0. So for j < , cj =: {i: i < j} N0 (and cj a0) hence
(by clause (h) of 6.7A), for some finite dj a1 pcfcj = Ni1 pcfa pcfcj we
have cj
djb1[a]. Assume j(1) < j(2) < . Now if a dj(1)
b1[a]
then for some 0 dj(1) we have
b1
0 [a
]; now 0 dj(1) pcf(
cj(1))
pcf(cj(2)) pcfdj(2)
b1[a]
=dj(2)
pcf(b1[a]) hence (by clause (g) of
6.7A as 0 dj(0) N1) for some 1 dj(2), 0 b11 [a]. So by clause (f)
of 6.7A we have b10 [a] b11 [a] so remembering b
10 [a], we have b
11 [a].
Remembering was any member ofadj(1)
b1[a], we have a
dj(1) b
1[a]
adj(2) b
1[a] (holds without a but not used). So a
dj
b1[a]: j <
is a non-decreasing sequence of subsets of a, but cf() > |a|, so the sequence is
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
46/54
430
re
vision:1996-10-12
modified:1996-10-12
106 S. SHELAH Isr. J. Math.
eventually constant, say for j j(). But
maxpcf
a
djb1[a]
maxpcf
dj
b1[a]
= maxdj
maxpcf(b1[a])
= maxdj
max pcf{i: i < j} < j
= max pcf
a
dj+1
b1[a]
(last equality as bj [a] b1[a] mod J maxpcf{j: j < i},
by 6.7C(3A), we will be stuck at some i, and by the previous sentence (and
choice of (a, b, ), i is limit, so pcf({j : j < i}) but it is pcf(b) +,
so = maxpcf{j : j < i}. For each j, by the minimality condition for some
bj b, we have |bj| |a|, j pcf(bj). So pcf{j: j < i} pcf(j
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
47/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 107
(instead of strict inequality) and
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
48/54
430
re
vision:1996-10-12
modified:1996-10-12
108 S. SHELAH Isr. J. Math.
Proof of (h): So let , , c be given; let b[a]: pcfc( Ni+1) be a
generating sequence. We define by induction on n < , An, c, : An such
that:
(a) A0 = {}, c = c, = maxpcfc,
(b) An n, |An| < ,
(c) if An+1 then n An, c cn, < n and = max pcf(c),
(d) An, c, : An belongs to Ni+1+n hence Ni+1+n ,
(e) if An and pcfcomplete(c) and cb+1+n
[a] then
()[ An+1 & = 0] and c0 = c\b+1+n
[a] (so 0 =
maxpcfc0 < = max pcfc),
(f) if An and / pcfcomplete(c) then
c =
bi [
c]: i < in < , i An+1
,
and if = i An+1 then c = b [c],
(g) if An, and pcfcomplete(c) but c b+1nn
[a], then ()[
An+1].
There is no problem to carry the definition (we use 6.7F(1) below, the point is
that c Ni+1+n implies b[c]: pcf[c] Ni+1+n and as there is d as in
6.7F(1), there is one in Ni+1+n+1 so d a+1+n+1). Now let
dn =:
: An and pcfcomplete
(c) and c b+1+n
[a]
and d =:n
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
49/54
430
re
vision:1996-10-12
modified:1996-10-12
Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 109
If = 0 (i.e. clause (h)) we should havenAn finite; the proof is as above
noting the clause (f) is vacuous now. Son |An| = 1 and
nAn = , so
nAn
is finite. Another minor point is d Ni++1 ; this holds as the construction is
unique from Nj : j < i+, ij : j + , (ai(), b: ai()): + ;
no outside information is used so (An, (c, ): An): n < Ni++1 ,
so (using a choice function) really d Ni++1 . 6.7A
6.7E Proof of 6.7B: Let b[a] = b =
-
8/3/2019 Saharon Shelah- Further Cardinal Arithmetic
50/54
430
re
vision:1996-10-12
modified:1996-10-12
110 S. SHELAH Isr. J. Math.
|d| < such that {i: i < } d
b1[a]; hence by clause (g) of 6.7A
and 6.7F(0) we have a1 pcfcomplete({i: i < }) d
b1[a]. So for
< < |a|+, d a1 pcfcomplete{i: i < } a1 pcfcomplete{i: i sup pcfcomplete({i: i < }), whereas d
pcfcomplete{i: i < }, hence > sup d hence
()2 > supd max pcfb1[a] sup pcfcomplete d b
1[a] a .
On the other hand,()3 pcfcomplete{i: i < + 1} pcfcomplete
d+1
b1[a] a
.
For = () we get contradiction by ()1 + ()2 + ()3.
(5) Assume a, c, form a counterexample with minimal. Without loss of
generality |a|+3 < Min(a) and = maxpcfa and = maxpcfc (just let a =:
b[a], c =: c pcf[a
]; if / pcfcomplete(c) then necessarily pcf(c\c)
(by 6.7F(0)) and similarly c\c pcfcomplete(a\a) hence by 6.7F(2),(3)
pcfcomplete(a\a),
contradiction).Also without loss of generality / c. Let , , N, i = i(): ,
a = ai: i be as in 6.7A with a N0, c N0, N0, = |a|+, = |a|+3 > 0 using ranks and normalideals, in Cardinal Arithmetic, Chapter V, Oxford University Press, 1994.
[Sh400] S. Shelah, Cardinal arithmetic, in Cardinal Arithmetic, Chapter IX, Oxford
University Press, 1994.
[Sh410] S. Shelah, More on cardinal arithmetic, Archive for Mathematical Logic
32 (1993), 399428.
[GiSh412] M. Gitik and S. Shelah, More on ideals with simple forcing notions,
Annals of Pure and Applied Logic 59 (1993), 219238.
[Sh420] S. Shelah, Advances in cardinal arithmetic, Proceedings of the Banff Con-
ference in Alberta; 4/91, Finite and Infinite Combinatorics in Sets and
Logic (N. W. Saure et al., eds.), Kluwer Academic Publishers, Dordrecht
1993, pp. 355383.
[Sh460] S. Shelah, The Generalized Continuum Hypothesis revisited, submitted to
Israel Journal of Mathematics.
[Sh513] S. Shelah, PCF and infinite free subsets, submitted to Archive for Mathe-
matical Logic.
[Sh589] S. Shelah, PCF theory: applications, in preparation.