saharon shelah- more on cardinal arithmetic

8/3/2019 Saharon Shelah- More on Cardinal Arithmetic

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MORE ON CARDINAL ARITHMETIC

SH410

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Mathematics DepartmentNew Brunswick, NJ USA

Done 1990Partially supported by the Basic Research Fund, Israel Academy of Science

I thank Alice Leonhardt for the beautiful typingPubl. 410Latest Revision - 07/July/3

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

An Annotated Table of Contents

1 On Normal Filters

In [Sh 355, 4] and [Sh 400, 3,5] we have computed cov(,,,) when > =

cf > 0, using tcfi


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MORE ON CARDINAL ARITHMETIC SH410 3

6 The Existence of Strongly Almost Disjoint Families

We characterize such existence questions by pps. An example is the question ofthe existence of a family of+ subsets of > 0 , each of cardinality (> 0) such

that the intersection of any two is finite.


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4 SAHARON SHELAH

1 On Normal Filters

The following Lemma 1.2 is similar to [Sh 355, 5.4], [Sh 400, 3.5], but deal with

normal ideals (see [Sh 371, 4], in particular [Sh 371, Definition 4.1,Claim 4.6]).Remember prc is defined in [Sh 371, 4] as:

1.1 Definition. 1) For a regular uncountable cardinal , normal ideal J on , a-sequence of cardinals > , and f Ord, we define:

prcJ(f, ) = Min

|P | :P is a family of-sequences of sets of ordinals,

B = Bi : i < , |Bi| < i or at least

{i < : |Bi| i} J, such that: for every g

Ord,g J f there is a sequence A

: < of members

P satisfying {i < : g(i) / .

() i strictly increasing and for limit i:

if i =j


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1.3 Remark. 1) On getting =+ see [Sh 420, 6.1](C) and [Sh 430, 4]. The problemis when pcf(a) has an accumulation point which is inaccessible.2) In the case () holds, if = i : i < is (strictly) increasing continuous,

supi


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6 SAHARON SHELAH

Proof of the Inequality from the Fact. Let us define a family P :

P = {Bi, : < i and |Bi,| < i} : i < : Bi, : i < , < i P

.Now each member ofP has the right form as each i is regular and Bi, : i we have B,i : i < P; let us enumerate

> as { : < } such that [ { : < }], and let us define

B

= B,i : i < P

, hence by Definition 1.1 it is enough to show thatE =: {i < : g(i) /

the

set X =: {i A : i /j]}

X =: {i < : if i / Y then for some >i we have i X}

belong to J. It suffices to show() for every \X, for some > we have B,i.

Why () holds? Choose by induction on n < , n n such that: An . For

n = 0 remember A = . For n + 1, as An and / X clearly / Xn so

necessarily j


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for the given B and any g, i : i < as there. Let Yi0 = { < i : |Bi,| cf|B

,i|} and Y

i2 = { < i : cf|B

i,| i} (for

each i < , < i).

Clearly Yi0 , Yi1 , Yi2 is a partition of { : < i}, now for Yi1 Yi2 , leti =: cf |B

,i|, and Bi,, : <

i be an increasing continuous sequence of

subsets of Bi, of cardinality < |B,i| and

sup( pcf(a) holds as is a

successor cardinal.

Case 2: Assume assumption. Here we partition a to sets diagonally; i.e. without

loss of generality each i is a cardinal hence by clause () we have i j


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8 SAHARON SHELAH

Proof. For (c) see [Sh 371, 1.6], so we can assume 1 > 0 as otherwise we havethere gotten a conclusion stronger than (a) + (b) + (c).

Let a Reg have cardinality < , be unbounded in , I a 2-complete

ideal on a, and (

< )[a

I] and = tcf(a/I). As cf() = < < without loss of generality < Min(a), and let i : i < be increasing continuouswith limit , 0 cf(i) < (remember > 0); without loss of generality < 0 < Min)a) and pp(,1)(i) < i+1 and, if i > 0 is a limit ordinal thena i is unbounded in i. Also without loss of generality for every i < :

()1 0 < i & 1 cf() < pp(,1)() < i+1

hence

()2 b (0, i) & |b| < sup pcf1-complete(b) < i+1.

Let b[a] : pcf(a) be a generating sequence (exists by [Sh 371, 2.6]) andwithout loss of generality = max pcf(a). By [Sh 345a, 3.6] (and 3.1(7)) withoutloss of generality b[a] b[a] b[a]. Let i < satisfy cf(i) 1, as|a| < and ()1 we have sup pcf1-complete(i a) < i+1 hence for some ci we

have: ci i+1 pcf(a i), |ci| < 1 and a i ci

b[a] (otherwise we can

find a 1-complete proper ideal J on i a such that

[ < pp+(,1)(i) & pcf(i a) b[a] (i a\i) J],

a contradiction to ()1. Note that ci i+1.Let S0 = { < : cf() 1}, so for some i() < we have:

S1 = { S0 : c i()}.

is a stationary subset of .

By renaming, without loss of generality i() = 0 and so ci (i, i+1), and fori S1 let (i, , ei,) : < i list {(, b[a]) : ci}; so:

()1 a i


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and by 1.6 pcf2-complete

iS1

di

and by ()2 above, for j < implies

/ pcf1-complete iS1

di j

; also = max pcf(d). So we are left with clauses

(a) + (b). Let d = {, : < < 1}, as 1 < , clearly without loss of generalityfor some ()

S3 = { S2 : = ()} is stationary.

For each < () let

P =: {S S2 : max pcf{, : S} < }.

If for some < (), the normal ideal on which P generates is proper, we havefinished. If not, for each < () there are members S,i(i < ) ofP and clubC of such that:

S3 C i


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10 SAHARON SHELAH

2 On measures of the size of D cf() = 0, > 2 and [ < & cf() pp(

) < ] and pp() < cov(,, 1, 2). Then { : < = pp()} hasorder type .

We shall return to this in [Sh 430, 1] so we do not elaborate.

2.2 Claim. Suppose , are regular, 0 < < 1 , and ()[

1 & cf () < pp


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(ii) the function b fb : < max pcf(b) (for b a) is as in [Sh 371, 1];i.e. satisfies:

()1 fb b, we stipulate fb (a\b) = 0a\b and f

b : < is


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12 SAHARON SHELAH

Proof. Let = 3()+, and for + let B be an elementary submodel of(H(), ,


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(2) holds.5) Assume

(a) (+

)V

= (+

)W

, (), ()+ ;(b) there is C : ( + 1)\( + 1), 0 cf

W W satisfying C a clubof for each such that [ acc(C) C = C ] and otp(C) .[nec?]

Then (W, V) satisfies the -strong covering (see [Sh:g, Ch.VII]) which means:

for every ordinal and model M V with universe , with countablevocabulary, there is N M of power< , N an ordinal and the universeof N belongs to W.

6) Moreover, in (5)

+ in the game where a play last moves, in the ith move (for i < ) the first

and second players choose ai, bi []


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14 SAHARON SHELAH

There is no problem to carry the definition. Let N = N =:n


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by [Sh 400, 3.3A] we know M = N , so Y M M W andW |= |M | M < as required. [Saharon: fill details]

Proof of Part (3). As in part (2), we choose () = V1 and get Ni, i, gi, hi(for i < ()). By 0 applied to {hi : i < ()} (again translating to a set ofordinals) there is a set B W such that W |= |B| < and A B infinite,without loss of generality B H (see proof of part (2)). So there is a limit ordinal < () such that = sup{i < : hi B}. Now let M be the Skolem Hull in

(H()W, ,


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16 SAHARON SHELAH

As [i < j hi < hj ] clearly we get:

() i


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Now we shall show that for E, N W. For E we define by inductionon n < , Mn :

M0 is the Skolem Hull in (H()

W

, , and for anya Reg we have: if |a| < , Min(a) > and a , even sup(a) < and I is a-complete ideal on a, then tcf(a/I) (when tcf is well defined).2) If we write instead of we mean for some , < .3) If we omit we mean = 0.4) accessible is just the negation of inaccessible.

We now rephrase various old results.

3.2 Claim. 1) For regular, in the definition, and -complete I, tcf(a/I)can be replaced by / pcf-complete(a) and also by = tcf(a/I) for any a Reg (, ), |a| < , I being -complete; also if cf() / [, ) then sup(a) <

is not necessary just / a.2) Assume > > and cf() . Then is ( , ,)-inaccessible iff[ (, ) & cf() < pp(,)(

) < ].3) If = cf() > = cf() > , is ( , ,)-accessible then there is a seta Reg (, ) of ( , ,)-inaccessible cardinals each > , a of cardinality < such that pcf-complete(a).4) If = cf() > > = cf(), and (a)[a Reg (, ) & |a| < &

pcf-complete(a)] then there is a set a of ( , ,)-inaccessible cardinals (, )with |a| < such that = max pcf(a), and pcf-complete(a). If is ( , ,)-

inaccessible then necessarily sup(a) = , Jbda J


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(ii) for limit, = (

(ii) for limit, = (


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20 SAHARON SHELAH

3.6 Claim. Assume |a| < Min(a), then

cf|a|(pcf(a)) max pcf(a).

Proof. More is proved in [Sh 371, 3].

The following answers a question of Gerlits, Hajnal and Szentmiklossy in [GHS].They dealt with -good topological spaces X (i.e. every subset is the union of compact sets) and weakly -good spaces (every Y X of cardinality > contains a compact subset of cardinality > ). [GHS] has the easy implication.

We return to this in [Sh 513, 6].

3.7 Theorem. The following conditions on < < are equivalent: ( is aninfinite cardinal, and are ordinals)

(A),, there are functions f : for < such that:

< i 2, pp+() > 1.

Proof. First note

3.8 Observation. Let < , an infinite cardinal, , are ordinals. If for everyregular 1, 1 < the statement (A),,1 holds and is singular (e.g. > ||),then (A),, holds.

Proof. We prove this by induction on ; if trivial: use the constant functions.

As is singular =


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(B) (A)

First Case: 2 ||.Let {A : < } be a family of || distinct subsets of , let A

= {2i : i

A} {2i + 1 : i / A, i < } and let f : {0, 1} be

f(i) =

0 if i A1 if i / A.

Second Case: ,cf() , > 2, pp+() > , a regular cardinal.So there are regular cardinals i <

for i < (such that i > ) and an ideal

J on , / J such thati


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22 SAHARON SHELAH

hence gi j 2 (otherwise g() = sup(Ni g

())(as Ni (H(), , j(); asH(h) is not well defined, no < satisfies the requirements in (). But of the threedemands on , j trivially satisfy two and a half: < , f (\A) = g

(\A)and f A g

A; so the remaining one should fail, i.e. [h A fj A].So for some A we have h() > fj (); now h Njj()+1 Nj hence h() Njhence h() Nj \fj (), hence (by the definition of gj), gj() h() hence (as

j S) we have g() h() but h G A

g(), so h() < g(), contradiction.

3.7


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24 SAHARON SHELAH

4 Entangled Orders Narrow Order Boolean Algebra Revisited

4.1 Theorem. 1) If +4 cf() < 2 then there is an entangled linear orderof cardinality .2) Moreover, if 0 < we can demand that the linear order has density character 0 (in fact, in every interval of the linear order).

Remark. See more [Sh 462] and [Sh 666].

Proof. Without loss of generality 0 > cf() +4. By [Sh 355, 2.1] there is an

increasing continuous sequence i : i < cf() of singular cardinals with limit

such that tcf(

i cf().

Case I: For i < cf() we have max pcf{+j : j < i} < .

So for some unbounded A cf() we have i A +i > max pcf{+j : j

A i}.So a = {+i : i A} is as required in [Sh 355, 4.12] (with

+, cf() here standing

for , there, noting that 2cf() 2 ).

So we can assume:

Assumption not Case ISo there is , 0 < < , cf() < cf(), pp . Choose a minimal

such , so by 3.2(2):

() a Reg \0 & sup(a) < & |a| < cf() max pcf(a) < .

Clearly (by [Sh 355, 2.3]) in ()s conclusion we can replace < by < i.e.

() a Reg \0 & sup(a) < & |a| < cf() max pcf(a) < .

Let =: cf(), so pp() = pp 0. As said above pp and by [Sh 371, 1.7]

there is a strictly increasing sequence i : i < of regular cardinals satisfying

=i


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So +4 cf(). Let P =: {A : A cf(), otp(A) = +, A is a closed subsetof sup(A) and max pcf{+i : i c} =

+sup(A)}. For any C P , try to choose by

induction on i < +(bi) = bi[C] and i = i[C] such that:

(i) bi Reg \j


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26 SAHARON SHELAH

i() > max pcf{i() : < },

i() is strictly increasing, and Ens(

i(),

i()) for

< (). So applying [\Sh:355] , 4.12] we finish.

Why holds: Let i() < cf() be given. Choose C (i(), cf()) from P suchthat (b[C], [C]) is defined iff < [C] and [C] <

+. By the definition ofPwe have max pcf{+ : C} < . Let d =: {

+ : C}, let b[d] : pcf(d)

be as in [Sh 371, 2.6]. Let be minimal such that otp(b[d]) is . We canfind B C (for < ) such that {+ : B} b[d], otp|B| = and theBs are pairwise disjoint. Clearly max pcf{

+ : B} = as {

+ : B}

is b[d], but is not a subset of any finite union of b [c], < . Now letting

a =: {b[C] : < [C]}, there is (by [Sh 371, 2.6]) a subset a of a

such that = max pcf(a) but / pcf(a\a). Now as pcf{+ : B},

+ pcf(b)

we have (by [Sh 345a, 1.12]) pcf( B

b) hence by the previous sentence

pcf(a B

b). Let c =: a B

b , = , we can apply [Sh 355, 4.12] and

get that there is an entangled linear order of cardinality (which is more thanrequired, see [Sh 345b]); and, of course, i() <

pcf{j : j < cf()}. Theassumptions of [Sh 355, 4.12] holds as the c are pairwise disjoint (by (i) above),

pcf{+ : B{ pcf(B

b) = pcf(c) and [1 a max pcf(a 1) i() > 0.

So holds and we finish Subcase IIIb hence Case III.

Case IV: cf() > +4 and (and not Case I).For each < cf() of cofinality +4 we can apply the previous cases (or the

induction hypothesis on ) and get an entangled linear order of power + . So holds and we finish as in Subcase IIIb. 4.1

4.2 Claim. Assume +4 = cf(), i : i < is a strictly increasing sequence

of regular cardinals, < i 2 and = tcf(

i


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Then (2


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28 SAHARON SHELAH

5 prd: Measuring f(i) by a family of idealsand family sequences Bi : i < , |Bi| < i

In [Sh 371, 4], and here in 1 we have dealt with generalizations of the measuringi


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(iv) A A(v) f : A Ord

(vi) if < > T and y A, f(y) = 0 then f(y) > f(y).

5.5 Remark. 1) Clearly for 1 2 if is 2-suitable then is 1-suitable.2) If 2|Y| its value is immaterial, so we can omit it.

5.6 Claim. 1) Y,,, if = cf() > 0 is closed, restriction closed, ofcharacter and is suitable. ForP P(Y),

cY,,(P

) =

Z : for some < , Ai P

for i < we have Z i T).

Let T =: { T : < g() () w}, as is regular uncountable, clearly|T| < . Let

B =:{

B

:

T

} {A

:

T and A

I} {

f1

({

0}

) :

T

}.

Now B is the union of< members ofI and I is -complete, so B I. Choosey \B.

We now choose by induction on n, n such that:


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30 SAHARON SHELAH

0 =

n T,

g(n) = n

n n+1

y An.

For n = 0, 0 = so y Y = A. For n + 1, y Ann, n T and y / B

hence y / Bn , so by () there is n wn such that y An so n+1 = nis as required.

In the end for each n we have fn(y) > 0 as f1n ({0}) B, (remember n T

)hence by condition (vi) from 5.4, fn(y) : n < is a strictly decreasing sequenceof ordinals, contradiction.2) Left to the reader.

3) Again we leave the proof of restriction closed and closed and having character to the reader and prove suitability. The proof is similar but use diagonal union. Soby part (2) and condition (iii) of 5.4 for each T for some B J and functionh from to ordinals such that h(i) T for i < , we have

() Y = B {\A0\A1\ . . . \Ag} {x Y : x

i and each h the identity function.

Let I be the normal ideal on Y generated by J {f1n ({0}) : T}, so we

assume I is a normal proper ideal and we shall get a contradiction.Now define Y

Y = {x Y : (a) (x) a limit ordinal <

(b) if T >(x) then x / B

(c) if T >(x) and A I then x / A}.

ClearlyY

Y

mod I, hence we can find x() Y

. Now we choose by inductionon nn >(x()) as in the proof of part (1) and get similar contradiction.

4) Left to the reader. 5.6


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5.7 Definition. 1) For = i : i < , J an ideal on Y and f : Y ord wedefine

prdJ(f, ) = Min

|P| :P is a family of sequences of the form Bx : x Y,

each Bx a set of ordinals, Bx of cardinality

< (x) such that for every g xord satisfying

g f, we have :

P() = c[J {{x Y : g(x) Bx} : Bx : x Y P}]

.

2) If above J I, I an ideal on

prdJ,I(f, ) = Min

|P| :P is a family of sequences Bx : x Y,

each Bx a set of ordinals of cardinality < (x)

and for every g xord satisfying q f,

we have :

I c(J {{x Y : g(x) Bx} : Bx : x X P})

.

3) If is constantly , we may write instead. We can use also = x : x Y,(but usually do not) with the obvious meaning.

5.8 Claim. 1) If {x Y : f(x) < } J we can in 5.7 demand g |Y|, {x : f(x) > } J then prdJ(f, ) cf().

4) If f1 J f2 or just {x : |f1(x)| > |f2(x)|} J then prdJ(f1, ) prd

J(f2, )

(and the other obvious monotonicity properties).

5) Ifi = , cf() > |Y| we can in Definition 5.7 demand

xYBx = B0 forB P

(i.e. without changing the value).6) If is -complete and restriction closed, () < , A : < () is a partitionofY (J, as in 5.7) then

sup

prdJ+(\A)(f, ) prdJ(f, )

prdJ+(\A)(f, ).

7) If is normal and restriction closed, A {x Y : (x) > } for < , A : < a partition ofY\1({0}) then

sup


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32 SAHARON SHELAH

for eachi, cf(1i ) < 1i &

21 = (

1i )+. Then for (any , J and f YOrd) we

have prdJ(f, 1) = prdJ(f,

2).10) If = i : i < is increasing continuous with limit , and is normal,

J then prdJ(, ) and even prd

J(, ) = cf().

11) Ifi |Y|, = Y,, then for any , prd{}(, ) =

cov(||, , |Y|+, ).

Proof. E.g.6) The first inequality should be clear. Also the second: assume it fails, let =

prdJ+(\A)(f, ), , i > |Yi|, = i


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5.10 Conclusion. 1) Suppose , , , J are as in 5.9, f Y ord and {x : cf[f(x)] =f(x) (x)} J then

+ prdJ(f, ) = + sup{prdJ(g, ) : g


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34 SAHARON SHELAH

5.12 Definition. Let 1 cfJ(ax : x Y) be

suptcf xY

x/I : J I and x ax for x Y.

5.13 Claim. = i : i < is non-decreasing, is a suitable restriction close

family of ideals onY, J , f Y ord andxY

f(x) (x) = cf (x).

1) If prdJ(f, ) is regular, then for some ax : x Y we have:

(i) ax Reg f(x)+\(x)

(ii) | xY

ax| < if () of 5.9(1) and |ax|+ < (x) when 5.9(2)s assumptions

hold

(iii) = 1 cfJ(ax : x Y).

2) If is inaccessible, , -complete and [ < cov(, 0, , ) < ] then

without loss of generality |xY

ax| < 0.

Proof. 1) Like the proof of 1.1.2) Straight.

5.14 Claim. Assume the hypothesis of 5.9.If g YOrd, and each g(s) is an ordinal (x) and f(i) = g(i) and let

= prdJ(g, ) +


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6 The Existence of Strongly Almost Disjoint Families

See [Sh 355, 0] on the history of the subject.

6.1 Theorem. Assume J is an ideal on , not the union of 0 members ofJ, > 0.Then

T3J() = T2J() T

1J() T

4J()

where

T1J() = TJ() = sup{|F| :F is a family of functions from

to such that for f = g fromF we have f =J g}.

T2J() = sup : there are ni < for i < and regulari, >


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36 SAHARON SHELAH

2) The supremum in the definition of T1J() is always obtained.[Why? IfF1,F2 are as there, F1 maximal |F1| < |F2| then for every f F2 thereis gf F, etc.].

Proof. T2J() = T3J().

Trivially T3J() T2J(); for the other direction let appear in the sup defining

T2J(), as exemplified by < i, : < ni >: i < ; as max pcf{i, : i A, < ni}is always regular, without loss of generality is regular.

By [Sh 355, 1]; more elaborately, for some a a =: {i, : < ni, i < }we have [a = = max pcf(a)] and / pcf[a\a] and a = a : i < as in the definition of T

3J(), let

a = {i, : i < , < ni} and f : < be a sequence of members of a whichis and we shall get acontradiction, without loss of generality F M0 . Clearly for every f

we have:{g F : g =J f} has cardinality


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(h) ifb an, max pcf(b) and |b| < thenfn b {Max{fc, b : < n} : n < , < max pcf(b) and

pcf(an) and c {b[an] : pcf(an)}}

where b fb : < max pcf(b) for b an is a definable function (in (H(), ,


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Proof of 6.1A(1). T4J() T1J(). Let An , An An+1, =

n 0, > 0 and I = [] . ThenT4I() is T

2I() hence is T

1J().

Proof. Apply 6.1 (+ here corresponds to there), more exactly by 6.1(A)(2).

6.4 Remark. Asking on almost disjoint sets is an inessential change.

Assignments

1) Improve 1 ( less inaccessible by ), 0 < cf = < , for < large enough pp()() <

+().2) Introduction on 3 and on 4.3) cf() = 1, = < 2

1 ! No inaccessible below ? Or first such fix point?

Similarly below 3(1) for more complicated cases.4) |a| < Min(a), acc[pcfa] has inaccessible member?5) Check that trivial computations on pp appear.6) Localization for a property ??

> scite{4.5} undefined7) sup{|F| : f : , ({i : f(i) = g(i)}) < } for < ; appeared?8) has small stationary set = +, < , find for .9) In V1 we get -like demands for every > 3(1) from there to (old end of4)?.

> scite{4.5} undefined

10) Bound on pcf(a); or

11) Better bound of when i : i < is increasing continuous +i pcf(a) (like

the one of 4.4 rather than 2.1, 2.2.12) strong limit of cofinality 1, below it 1 inaccessible, (bound better than< (21 )+ inaccessibles.13) If |pcf(a)| |a|+4 then we have large fD for some f : |a|

+4 ordinals, Dimplicit in 2.1, 2.2.14) Localization for -complete ([Sh 345]).15) = first (of cofinality 0 or 1) strong limit. Let pp() =

+, fix point oflarge cofinality. Say 1 2 if (x (1 < 2 <

+ and (R [1, 1]) < ).16) pcf(a) large implies large rank.17) If =

, < < pp() then (R ) is -directed mod {A R :

for every a A, |a| cf we have max pcf A < } at least if in [, ] there is noinaccessible.

May 10, 1992


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18) Proof of 1.6 say more?19) Quotations of [Sh 400, 3.3a,5.1](A)(1) accurate? I make distinction.20) Where is prdp(f, ) prd

21) Note: more on proof of 2.2, 2.4?

Revised Dec 2001:

Compare 3-6, (1-2 earlier)

4 - read, 5.1-5.6


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REFERENCES.

[GHS] J. Gerlits, Andras Hajnal, and Z. Szentmiklossy. On the cardinality ofcertain Hausdorff spaces. Discrete Mathematics, 108:3135, 1992. Topo-logical, algebraical and combinatorial structures. Froliks Memorial Vol-ume.

[Sh:b] Saharon Shelah. Proper forcing, volume 940 of Lecture Notes in Mathe-matics. Springer-Verlag, Berlin-New York, xxix+496 pp, 1982.

[Sh 345] Saharon Shelah. Products of regular cardinals and cardinal invariants ofproducts of Boolean algebras. Israel Journal of Mathematics, 70:129187, 1990.

[Sh 420] Saharon Shelah. Advances in Cardinal Arithmetic. In Finite and Infi-

nite Combinatorics in Sets and Logic, pages 355383. Kluwer AcademicPublishers, 1993. N.W. Sauer et al (eds.). 0708.1979.

[Sh 371] Saharon Shelah. Advanced: cofinalities of small reduced products. InCardinal Arithmetic, volume 29 of Oxford Logic Guides, chapter VIII.Oxford University Press, 1994.

[Sh 355] Saharon Shelah. +1 has a Jonsson Algebra. In Cardinal Arithmetic,volume 29 of Oxford Logic Guides, chapter II. Oxford University Press,1994.

[Sh 345a] Saharon Shelah. Basic: Cofinalities of small reduced products. In Car-

dinal Arithmetic, volume 29 of Oxford Logic Guides, chapter I. OxfordUniversity Press, 1994.

[Sh 400] Saharon Shelah. Cardinal Arithmetic. In Cardinal Arithmetic, volume 29ofOxford Logic Guides, chapter IX. Oxford University Press, 1994. Note:See also [Sh400a] below.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 ofOxford Logic Guides.Oxford University Press, 1994.

[Sh 345b] Saharon Shelah. Entangled Orders and Narrow Boolean Algebras. InCardinal Arithmetic, volume 29 of Oxford Logic Guides. Oxford Univer-

sity Press, 1994. Appendix 2.

[Sh 430] Saharon Shelah. Further cardinal arithmetic. Israel Journal of Mathe-matics, 95:61114, 1996. math.LO/9610226.

[Sh 462] Saharon Shelah. -entangled linear orders and narrowness of productsof Boolean algebras. Fundamenta Mathematicae, 153:199275, 1997.math.LO/9609216.

[Sh 666] Saharon Shelah. On what I do not understand (and have some-thing to say:) Part I. Fundamenta Mathematicae, 166:182, 2000.math.LO/9906113.

saharon shelah- more on cardinal arithmetic

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