Chapter 2 Gaps and limits

Chapter 2 Gaps and limits

$1. INTRODUCTION. For an infinite set X the symbol fin(X) will denote the family of all finite subsets of X. We write shortly fin, if it is clear what X we have in mind. Obviously, fin(X) is an ideal in the power-set algebra P ( X ) and hence we can form the factor algebra P(X)/fin(X). Clearly, if card X = card Y then P(X)/fin(X) is isomorphic to P(Y)/fin(Y). In a large part of this book we shall be concerned with the algebra P(X)/fin, for a countable X and usually we take X = w as a fixed representative. By definition, any algebra P(X)/fin consists of abstraction classes of subsets A X according to the congruence

c

A=fi,B

=

AABEfin.

The zero of P(X)/fin is fin:

[A]= CD

=

A E fin.

Let us introduce the following notation, which will be extensively used in the sequel. For sets A, B X: A =* B ( A almost equals B ) , if A A B E fin ( A is almost contained in B ) , if A \ B E fin B A if A B and non A =* B , i.e. if A \ B E fin but A (-&B ,

c

c*

c*

B \ A 4 fin, ( A is almost disjoint from B ) , if A n B =* $3, i.e. if AnBEfin. Thus, we have the following equivalences A =* B = [A]= [B] (in P(X)/fin) A IB

1.1.

A C, B

E

[A]5 [B]

(Boolean ordering in P(X)/fin)

A S;. B

E

[A]< [B]

(the strict ordering)

A IB

=

[A]. [ B ]= 0 31

(Boolean disjointness)

CH.2 GAPSA N D LIMITS

32

Let us assume now that X = w. For A w and n E w the set A \ n is a cofinal segment of A : A \ n = { z E A : i 2 n } (since n = ( 0 , . . . ,n - l}, in our notation). If [A]= a, i.e. if the set A is a representative of an a E P(w)/fin, then each cofinal segment A \ n, for n E w is also a representative of a. Let us note also and

AC,B

=

3n[A\ncB]

A I B

G

3n[(A\n)nB=p].

Now, we prove some simple properties of the algebra P(w)/fin. 1.2. The algebra P(w)/fin has antichains of cardinality c (continuum).' Indeed, from 1.1 we have

[A]>(D

=

A e f i n and

[ A ] . [ B=(D ]

=

A I B

and hence any antichain E of P(w)/fin has a form E = { [ A ] : A E R } , where R is an almost disjoint family (i.e. R consist of infinite subsets of w, any two of them are almost disjoint). It remains to recall a classical result of Sierpiliski - there are almost disjoint families R of cardinality c. To see this let z = (zn: n < w) be a sequence enumerating all the rational numbers. For any irrational number (Y there is an infinite subset A , w such that the sequence slA, = {s, : i E A,} converges to a. Obviously, A , I AD,for (Y # /3 and hence R = { A , : Q E W \ Q} is as required. From 1.2 it follows immediately 1.3. The algebra P(w)/fin has cardinality c. It is almost obvious that 1.4. The algebra P(w)/fin is atomless. Indeed, if [A]> 0,then A 4 fin and hence A splits into two infinite disjoint sets A = B U C. Hence [A]> [ B ]> (D. Any descending chain ( a , : (Y < y) (in an arbitrary Boolean algebra) is called a limit (or a tower) of length (or type) y, if it cannot be further extended, i.e. there is no a > 0 such that a < a,, for each (Y < y. For example, the chain (w \ n : n E w) is a limit of length w in' the power-set algebra P(w).

33

$2. DOMINANCE

1.5. THEOREM (Hausdorff [Hz], [Hs]). In the algebra P(w)/fin there are no limits of length w (and hence of any length y < w1).

PROOF. Let ( a , : n < w ) be a decreasing chain. Choose representatives A , E a,, for each n < w. We have An+1 $ A ,l i.e. *

c*A,,

A,+1

and A ,

\ A,+1

$n!fi

for each n


Replacing, if necessary, An+l by its far enough cofinal segment we may assume A , , for each n < w . Define a sequence (k, : n < w) by induction that An+l as follows. Let LO be an arbitrary element of Ao: k,+1 is the smallest number in

c

A,+1

\ A,

exceeding all the ko, . . . , k,. Thus we have

{k, : i 2 n } c A , Hence if A = {ki : i

for each n

< w.

< w}, then [ A ]> 0 and [A] 5 [A,] = a,, for each n < w . QED

$2. DOMINANCE. Let us consider two relations defined for functions f,g

:w

4

w

as follows

f =* g

f 5. g

c

{i E {i E

w : f(i)

# g(i)}

E fin

(f almost equals g )

w : f(i)

> g(i)}

E fin

( g dominates

f).

Obviously, the relation =* is an equivalence and 5. is a partial pre-ordering (it means f imply f =* g , the reflexive and transitive) on the set w w . Since f g and g relation 5 , is in fact a partial ordering on the set w w / =* of abstraction classes. The g and non f g) can be described corresponding strict ordering <+(that is f thus

=.

f <, g if and only if f(i) 5 g(i) holds for large enough i and f ( i ) < g ( i ) for infinitely many i E w. Let

XA

denote the characteristic function of a set A . Clearly,we have then

for arbitrary A , B

w,

and

cH.2 GAPSAND

34

LIMITS

s

We say that a set F ww is bounded if it is bounded from above with respect to the relation I,,i.e. if there is a g E ww such that f 5. g holds for each f E F (we write shortly F g in this case).

<*

2.1. Every countable set F = { fj : j < w } is bounded.

+

Indeed, if we define g ( n ) = m a { fj(n) : j 5 n} 1, for n E w then F I*g. The smallest possible cardinality of an unbounded (i.e. not bounded) set will be denoted by lb. Thus

lb = mintcard F : F is unbounded}. We call a set F ww dominating, if it is cofinal in ww with respect to the ordering I* i.e. if, for each function g : w + w there is an f E F such that g f. Of course, any dominating set is unbounded. Hence, if we introduce another cardinal coefficient d = min{card F : F is dominating}, then we have inequalities w1

Ilbldlc.

2.2. THEOREM. The cardinal lb is regular and cf(d) 2 lb.

PROOF.Let F be an unbounded set of cardinality Ib. Suppose that Ib is singular. Then we have F= UF;, i€I

where card I < lb and card F, < Ib,for each i E I . Thus, the sets F, are bounded for each i E w , F; g, for some g; E ww. Since card I < lb, there is a bound g of the set {g, : i E w } . It follows that F <* g, contradicting our supposition. Thus, lb must be regular. Similarly, let D be a dominating set of cardinality d and suppose that cf(dl) < lb. Thus, we have a decomposition D= Dj,

u

j€J

where card J < lb and card Dj < d for each j E J . Thus, no of the sets Dj is dominating and hence there are functions gj such that for all j E J : gj non

I*f

for each f E Dj.

$2. DOMINANCE

35

Now, if g is a bound of the set {g, : j E J } , then we have g f for no f E D, i.e. D is not dominating, contrary to the assumption. Hence cf(d) 2 lb, which finishes the proof.

QED

The next lemma carries more information on unbounded sets.

2.3. LEMMA.Tf F = { f a : a < lb} is an unbounded set of cardinality lb, then there exists an increasing (with respect to 5.) sequence (ga : a < b) dominating the set F (i.e. Va3p(fa 5 g p ) ) . Moreover, we may claim that the terms ga can be taken from a dominating set D, given in advance.

PROOF.The required sequence ga is defined by induction. The first term go is chosen so that go. 2f0 and go E D. If we have already the terms gc E D so that 96. 2{gv : q < 5) and gc+ 2fc, for all [ < a < lb, then we choose ga as any bound of the set {gc :

E < a} U { f a } . Of course, we can find such a bound in D.

QED

Any increasing (in the sense of dominance) unbounded sequence is sometimes called a limit in ww (comp. 1.5). From the lemma it follows immediately

2.4. COROLLARY. The ordinal number lb is the length of a shortest limit with terms in a given dominating set.

A dominating and increasing sequence ( f a : a < y) is called a scale (of length 7). Thus, any scale is also a limit in w w . It is consistent (with set theory) to assert that a scale exists. Indeed, assume CH (the continuum hypothesis). Then we have lb = d = w1. Thus, we may consistently assume that lb = d and then - in the lemma 2.3 - we can take as the F a dominating set of cardinality lb = d. The resulting increasing sequence (ga : a < lb) is dominating (because it dominates a dominating set) and hence - a scale. However, the existence of a scale is independent of set theory (see the theorem 5.11). Let us note that if a sequence ( f a : a < y) is a scale then obviously y 2 (a and there is a subsequence (fca : a < d) of length d, which is a scale too. Indeed, if

D = {ga : a < d} is an arbitrary dominating set, then for each a < d we can find a to < y such that ga 5. fern. The sequence (ft- : a < (a) is dominating, since it dominates a dominating set D and hence it is scale of length 8. 2.5. THEOREM. A scale exists if and only if the equality lb = d holds.

c H . 2 GAPSAND LIMITS

36

PROOF.If there is a scale, then (as remarked above) there is a scale s =

(f, :

(Y

< d)

of length d. Let F be an arbitrary unbounded set of cardinality Ib. Applying the lemma 2.3 to F and D = {fa : (Y < d} we obtain a subsequence t = (fern : (Y < Ib) of length lb, which dominates F and hence t is unbounded, hence cofinal in s, hence dominating, i.e. - a scale. Thus, Ib = d. The converse implication has been proved earlier (after 2.4). QED Now, we shall discus a connection between the cardinal 03 and some gaps in

P( w )/fin. First, let us modify a little the definition of gaps (given in 7.2, Chapter 1) viz. the condition S 5 T will be replaced now by S IT . Thus, a pair of subsets (S,T) is a gap in the new sense, if S I T , i.e. if a . b = 0 for each a E S and b E T . Of course, the difference is completely inessential - replacing the upper class T by -T we pass from one of these definitions to the other. An element c fills (or separates) a gap (S,T ) ,if S 5 c and c I T . A gap K = (S,T) is said to be unfilled, or non-separated, if no c fills K . We often shall be dealing with well ordered gaps, that is gaps K = ( S , T ) where both classes can be represented as increasing chains S = (a, : a! < n) and T = { b p : p < A). Such a gap K is often called a (n, A)-gap. We can always assume that the type (.,A) of K is a pair of regular cardinals, since K contains a cofinal subgap L (in obvious sense) of type (cf(n),cf(A)) (note that K can be filled iff L Can). Finally, if X is an infinite set and S,T

P ( X ) are two families of infinite subsets n B E fin(X) for each A E S and B E T). From 1.1follows that any such K determines (and is determined by) a gap in P(X)/fin

of X , then K = ( S , T ) ,is called a gap on X , if S I T (i.e. A

K* = ( { [ A ]: A E S } ; {[B] : B E 2’)). Clearly, the gap h’* is well ordered in P(X)/fin if and only if A’ is well ordered by the almost-inclusion In this case the types of K* and K are identical. We C and C IT ) if and only note also that a set C X separates the gap K (i.e. S if [C] separates K* in P(X)/fin.

c

c*.

c*

$2. DOMINANCE

37

We shall often make use of the above remarks without specific references. Let us prove now the following characterization of lb.

2.6. THEOREM (Rothberger [RI]). The cardinal Ib is the smallest cardinal K for which there exists a non-separable ( K , w)-gap in P(w)/fin.

PROOF.For an arbitrary function f : w

+w

let us denote

Clearly, we have then

First, we show that there is an unfilled gap of type (lb,w). To see this, let : a < lb} denote an unbounded set of cardinality Ib and let En = { n } x w (a vertical axis). If a set C w x w is almost disjoint from all the axes En, i.e.

F = {f,

c

C n EnE fin(w x w )

for each n E w ,

then we can define a function

g ( n ) = max{j E w

:

(n,j)E C n E n }

for n E w ,

(we assume g ( n ) = 0, if C n En= 9 ). Since F is unbounded, we have f a non <* g , for some a < Ib. From (*) it follows that A(f,) non A(g) and even stronger C, because C C A ( g ) . A(f,)non This means that no set C separates the gap

c*

c*

K = ( { A ( f , ) : a < b} ; {En: n E w } ) of type (lb,w) on the set w x w . Hence, there is an unfilled (K,w)-gap on w and consequently in P(w)/fin as well. Now assume that

L = ( { A , : a < K } ; {B,: n < w } ) is a non-separated (tc,w)-gap on the set w . We have to show that K 2 Ib. We may assume that the sets B, are disjoint (for B, can be always replaced by the sets B,\(BoU...uB,,-l)) and that w = Un
38

cH.2 GAPSA N D LIMITS

to the B’s) because so modified gap remains unfilled. For each n E w we choose a one-to-one function h, mapping B, onto the vertical axis En = {n} x w . Then h = Un
Q


; {h[B,] : n


on the set w x w . From (*) it follows that the functions

form an unbounded family and hence

K

1 lb.

QED

According to Corollary 2.4 the family F in the above proof can be replaced by a limit (i.e. an unbounded increasing sequence) (fa : Q < Ib). Then we have A(f,) &. A(fp), for all Q < /3. Thus, the lower class of the gap K is an increasing sequence. The same can be said about the upper class {En: n < w } for, it is countable and hence each En can be always replaced by Eo U . . . U En. In other words, the gap K can be claimed to be well ordered. Hence, we obtain the following

2.7. COROLLARY. The cardinal Ib is the smallest cardinal K for which there is a well ordered non-separated ( K , w)-gap. From Ib > w (see 2.1) we get immediately

2.8. COROLLARY. In the algebra P(w)/fin every countable (i.e. of type ( w , w ) ) gap is filled. Nevertheless there are always unfilled (q,wl)-gaps. This famous theorem of HausdorfF will be proved in the next section.

$3. HAUSDORFF GAPS. For almost disjoint sets A, B E w denote p(A, B ) = min{ k E w : (A \ k) n B = 9 }. Thus, for disjoint A, B we havep(A, B ) = 0 and if max AnB = k then p(A, B) = k+l.

$3. HAUSDORFFGAPS

39

For a set A E w and a family R s P(w) we say that A is n e a to R (abbr. A nt R) if the following condition is satisfied 3.1.

the set

{ B E R : p ( A , B ) < k} is finite for each k E w .

Obviously, if R is a finite family, then we have A nt R, for each set A C w . Note that A nt R holds if 3.1. is true for large enough A E w . Hence 3.2.

H A nt R and A C, C, then C nt R as well.

Now, we shall prove the following 3.3. THEOREM (HausdorfT [H3]). The algebra P(w)/fin contains well ordered non-separable gaps of type (w1,wl).

PROOF.We define by induction two increasing, in the sense of

sothatA,nBa=(9,foreacha
c*,sequences

and

is a gap on the set w having the following property

(*)

A, nt {B,

: 9

< a} for each a < w1.

From (*) it follows that the gap L cannot be filled. Indeed, assume that a set C separates L, i.e. A, C, C and C n B, =, Q, , for each a < w1. Then by (*) and 3.2 we have 3.4.

C nt {a,

: 9

< a } for each /3 < w1.

On the other hand, the function f(v) = p ( C , B , ) must be constant on some uncountable set 2 w1. Let us take a p < w1 so large that the set 2 n /? is infinite. Obviously, we have C non nt {B, : 9 E 2 n p } , and hence C non nt {B, : 9

< p } , which contradicts 3.4.

We pass to the construction of the gap L. Let Ao, Bo be arbitrary infinite disjoint sets such that the difference w \ (A0 U Bo) is still infinite. The latter condition will be assumed at each stage of the induction. Assume that we have defined already the sets A t , B, for 6,9 < a,as required. If a is a successor a = B 1, then, since

+

c H . 2 GAPSA N D

40

LIMITS

s*

s1

\ ( A p u Bp) 4 fin, there are disjoint sets A,, B, such that Ap A,, Bo B, and A , U B, has an infinite complement. Obviously, the condition (*) holds too.

w

Now, assume that a is a limit ordinal. Then the sequence @\(At U B t ) : [ < a ) is almost decreasing and, in view of the theorem 1.5, there is an infinite set H C w such that for each [ < a. H ( w I \* ( A t U Bt)

If we denote E = w \ H , then we have At,

B, c*E

for all ( , q < a ,

and w \ E is infinite. Since there are no unfilled countable gaps (see 2.8), there exists a set C separating the family { A t : ( < a } from {B, : 7 < a } , i.e. 3.5.

AcC.C

and C I B ,

E replacing - if necessary

C

Of course, we may assume that cn E. According to 3.2 we have

C nt {B,

: 7

forall[,q
the set C by

for each ,f? < a.

< ,f?}

However, this does not mean that C nt {B, : r] larger. In order to do this let us consider the sets

< a } . We have to make C even

and define inductively an increasing sequence

so that

Cn+lnt {B,

: r] E

Wn}

for each n

< w.

Assume that Co,. . . C, are defined. If W , is finite, we can put Cn+l = C,. If W , is infinite, then it must be cofinal in a and its order type is w . Indeed, each initial segment W , n p, for a ,f? < a is finite, since we have C nt {B, : q < p } . Hence, we can write

W , = {q, : i < w }

where

90

< 71 < . . . and supq, = a. i
$4.THEPAROVICENKO THEOREM.

41

Clearly, each of the sets

Zi = Bqi \ (BqoU . . . U Bgi-,) intersects the set E on an infinite part. Thus, we can form a diagonal set

D

= { jn : n


where j , E Zn n E

and j,

> n for each n < w .

Now, let us define Cn+l= C, U D. From the construction it follows that D I B,,,, for each n < w . Hence D I B,, for each 9 < a and consequently we have that Cn+lI B,, for each 9 < a. Since obviously D nt {B,,, : n E w } holds, hence also

Cn+lnt { B , : 9 E W,}, which finishes the inductive step. Similarly as in 3.4we can separate the sets {C, : n < w } from { B E: ( < a} within E: there exists a set A such that C, A , for each n < w and A I BE,for each ( < a. Moreover A E . Additionally, we may demand that CO A (enlarging if necessary A and E by a finite number of points). Now, it is clear that we have

c*

c

c

A nt { B , : 9 < a}. Indeed, suppose that for some k the set

W = (9 < a

:

p(A,B,) < k }

wk

w

wk

is infinite. Then also the set (9 E : p ( A , B , ) < k} is infinite, since C (because CO A). This means that A non nt { B , : 9 E W k } holds, which is impossible, since we have

c

Thus, if we define A , = A and B , = E \ A , then (*) holds, A , n B , = (3 and the complement w \ ( A , u B,) is infinite. This finishes the proof of the theorem. QED

3.6. R E M A R K . Let us note that any two elements a I b of P(w)/fin satisfying

a

+ b < 1 are in different classes of some unfilled (w1,w1 )-gap.

$4. THEPAROVICENKO THEOREM. Let us write down some basic properties of

the algebra P(w)/fin. It has the cardinality c (1.2) and

c H . 2 GAPSA N D LIMITS

42

1. it is atomless (1.3), 2. it has no countable limits (1.4), 3. it has no countable unfilled gaps (2.8). Any Boolean algebra A having the above properties 1 - 3 is called a Parovitenko algebra. Thus, the algebra P(w)/fin is a ParoviEenko algebra of cardinality c. We prove below that - assuming the continuum hypothesis - it is a unique Parovitenko algebra of cardinality c. First, we show a simple 4.1. LEMMA.Everyfinitefamily {ao,. . .an},ofnon-zeroelementsofan atomless Boolean dgebra A, has a refining antichain (i.e. antichain {bo, ... ,b,} such that bo 5 ao, .... b, 5 a,). In addition, if A has no countable limits, then the same holds for countable families {an : n E w } . PROOF. By induction on n. If we have a family { aC , ~ . . . ~ U , } then , by the inductive hypothesis there is an antichain {bo, .... b n - l } refining {ao, .... u , - l } . Let us define bn = a, - (bo ... b , - i )

+ +

provided b,

> 0.If b, = 0,then we have an

5

+ ... + bn-1

and hence a , . b,

>0

for some i

< n.

Since the algebra A is atomless, we have a splitting

an*bi=c+d

wherec,d>0

and

c.d=0.

We replace the element b, by c and we take d as the n-th element of our antichain. Now assume that A has no limits of type w and let { a , : n E w } be a countable family of positive elements of A. Using the already proved part the lemma we define induction a matrix a0

a1

1 b: 2 by 1 .. 1 ... b: 2 .. . b', 2 ..

... an 1 ... ...

. . . . .. . . . . . . . . . . . b; 1 .. . . . . .. . . . . .

$4.THEPAROVICENKO THEOREM.

43

in which b: = ao, the column { b: , bi } is an antichain refining { b t , a1 } and, generally, for each n E w the column . . . b:} is an antichain refining {b:-,, . . . b,-l, n-1 u,}.

{a:,

In particular, any row a,, b t , b;+,, . . . is a decreasing sequence and - by our assumption - has a lower bound b,. The last column { b , : n < w } , consisting of that bounds, is then a required antichain. QED We can prove now the announced characterization of the algebra P(w)/fin. 4.2. THEOREM (ParoviEenko [PI]). Every Boolean algebra of cardinality 5 w1 be embedded into any ParoviEenko algebra. If CH (the continuum hypothesis) is assumed, then any two ParoviEenko algebras of cardinality c are isomorphic. can

PROOF.Let A be a Boolean algebra with card A 5 algebra. Clearly, A can be represented as a union

w1

and B a ParoviEenko

of an increasing chain of countable subalgebras fulfilling the following conditions =

4.3.

{a,1) for some a E A and each a < w1

A,+1 = A,[4

A, =

u

for each limit ordinal a

Ap

< w1.

B<,

Indeed, the equations 4.3 can be regarded as an inductive definition of the sequence (A, : a < w1). The generator a in the successor case should be defined thus: a = inf(A \ A,), where inf refers to a fixed well ordering of A in type w1. Obviously, we have an embedding fo :

A0 -+

B

viz.

fo((D) = 0,

f o p ) = n.

We shall extend fo successively to A,, for each 01 < w1. Assume that for a limit ordinal a < w1 we have a sequence (fp : p < a),where for each p < a , fp : Ap -i B is an embedding and f p C f-,, for all /3 < y < a. Setting -f, = Up<, f p we get an embedding of A, = Up<, Ap into B.

Thus, it remains to show how to extend an embedding f = f, of A, into B to : A,+1 + B because then the common extension Uo
CH.2 GAPSA N D

44

LIMITS

shall apply the Sikorski theorem. According to 7.4 of Chapter 1 we have to find an element b E B such that for each z E A, the following equivalences hold

Consider the gap determined by the new generator a and the embedding f

L

= ((f(5):

5

5 a } ; {f(y)

: y . a = 0)).

For the implications on the right in (S) it suffices to define b as any element filling the gap L. For the converse implications we have to choose b more carefully. In order to do this let us suppose that an element z cuts a , i.e. z - a > 0 and z . a > 0.Then we have

> CD, f(z) - f(z) > 0, f(z) - f ( y )

and

if y . a = CD if z 5 a .

There exists non-zero elements u ( z ) and w(z) such that 4.4.

5 f(z) - f(y) w(z) 5 f(z) - f(z) u(z)

and

for each y Ia for each L 5 a .

Indeed, let {yn : n < w } be an enumeration of the set {y E A, : y l. u } (recall that A, is countable). The elements f(z) - (f(y0) . . . f(yn)) form a decreasing chain and since there are no countable limits in B, we define u ( z ) > 0 as a lower bound of this chain. The element w(z) is defined similarly. Since the family

+ +

is at most countable, we may - in view of the lemma 4.1 - claim that R is disjoint (because for any refinement of R, 4.4still holds). We enlarge now the gap L putting the elements u ( z ) to the lower and w(z) to the upper class of L , respectively. Since the resulting gap L* is still countable, there is an element b 6 B separating L* (and hence also L ) . For each z cutting a we have u(z)

5 b and

b=0

~ ( 2 ) .

and since u ( z ) + W ( Z ) 5 f(z), hence f(z) cuts b: f(z) - b > 0 and f(z). b > 0.

$4.THEPAROVICENKO THEOREM.

45

This means that the element b satisfies the equivalences (S),which finishes the proof of the first part of the theorem. We prove now the second part. Let A and

B be two ParoviEenko algebras both

< w l } and {bp : ,6 < w1} be fixed enumerations of A and B,respectively. We define an isomorphism f : A -+ B as a union f = Ua
are the units of A and B, respectively). We extend f, in the following way. Suppose that a,+l E A,. If b,+l E B, then nothing happens but if b,+l @ B, then the inverse isomorphism filhas a n extension (as in the proof of the first part) to an embedding g of B,+1 = B,[b,+1] into A. We define A,+1 = g[B,+1] and f,+l = 9-l. Now suppose that a,+l $! A,. Extend fa to an embedding g : A,[a,+l] + B and if b,+l E g[A,[am+lll then define A,+1 = A,[a,+l], B,+I = s[A,+11 and f,+l = 9; otherwise extend g-' to an embedding h of B,+l = g[A,[a,+l]][b,+l] into A and define A,+1 = h[B,+l] and f,+1 = h-'. In either case we have extended fa to f,+l so that a,+l is in the domain A,+] and b,+l is in the range B,+1 of the isomorphism f,+l, which finishes the successor case of the induction. In the limit case a we are given an increasing sequence ( f p : ,6 < a) of isomorphism such that for each ,6 < a we have u p E Ag = dm (fa) and bp E Bp = rg ( f p ) . We take the elements a,, b, and proceed as in the non-limit case with Up<, f p , Up<, Ap and Up<, Bp in place of fa, A, and B,, respectively.

The sequence

a <

w1,

(fa :

we have a, E

< w l ) (just defined) has the following property: for each dm (f,) and b, E rg (fa). Hence, the common extension a

f = Uo
QED

We prove later (see Chapter 8) that the Parovitenko characterization of P(w)/fin is in fact equivalent to CH. That is, if c > w1 then there are non-isomorphic ParoviEenko algebras of cardinality c. The first part of the Parovieenko theorem can be reformulated as follows: 4.5.

if CH holds, then each Boolean algebra of cardinality at most c can be embedded into P(w)/fin.

CH.2 GAPS A N D

46

LIMITS

This implication cannot be reversed. In Chapter 8 we prove that there is a model of set theory, in which c > w1 and the conclusion of 4.5 still holds.

$5. TYPESOF GAPS A N D LIMITS. If A is an atomless Boolean algebra, then for each element a > 0 there are limits (towers) (Q( : E < y) in A, in which a occurs as - say - the first term a = ao. Indeed, the family R of all descending sequences f = (f([) : ( < a) of any type a and such that f(0) = a is partially ordered by inclusion f C g (g is an extension of f). If RO C R is a linearly ordered subfamily, then U Ro is an upper bound of Ro. Thus, by the Kuratowski-Zorn lemma, there are maximal elements in R and any such is a limit. For any limit f of length a there is a limit g of length c ~ ( ( Y )which , is cofinal in f. Hence, without loss of generality we may assume that the types of limits under consideration are regular cardinals. A similar remark applies to (well ordered) gaps.

5.1. Assume that CH holds. Then all limits in P(w)/fin are length c = w1 (we restrict ourselves to regular cardinals, cf the remark above). Since Ib = c in this case, the (essential) types of unfilled gaps are ( c , w ) and (c, c), (comp. 2.7 and 3.3). Thus, there exist limits and unfilled gaps of all possible types.

Now we discuss the situation under MA (the Martin axiom) and c we prove 5.2.

LEMMA.

> w l . First,

Martin’s axiom implies the following property called P(c):

if S P ( w ) is uniformly centered (i.e. is infinite, for any finite So C S ) and has cardinality less then c, then there exists an infinite X w such that X A, for each A E S.

s

s*

PROOF.Let P = P(S) consists of elements p of the form

where z p is a zero-one sequence of some length l ( p ) < w and S, is a finite subfamily of S. The ordering on P is defined thus: p 5 q if and only if zp _> zg and S, _> S, and for each i, I(q) 5 i < Z(p), the condition z p ( i )= 1 implies i E S,. It is easy to see that P has the ccc, i.e. each antichain in P is countable. Even stronger, P is a-linked. Indeed, if z p = zP then p is compatible with q , since then the

n

55. TYPES OF

GAPS A N D LIMITS.

47

element (z, S, US,) is below p and q. Since there is only countably many sequences xp, the claim follows. We prove now that the sets and

D , = { p E P : 3i 1 n(zp(i) = 1))

for n E w,

E A = { ~ E PA: E S , }

for A E S

are dense in P. Fix a q = (z,,S,) E P and an n E w. By the assumption, the set S, is infinite and hence there is an i E S, and i 2 max{n, I , ( q ) } . Define now an z E (0,l}i+lby setting

n

n

zIl(q) = 5,

and z ( k ) =

{ il

if k = i if l ( q ) 5 k < i .

Then, clearly, the element (z,S,) is 5 q and is in D,. Now, if an A E S is given, then obviously the element (zg,S, U { A } ) is in EA and below q, which proves our claim. The family { D , : n < w} U { E A : A E S} has cardinality less that c. Thus, the Martin axiom implies the existence of a filter G C P intersecting all the D’s and all the E’s. Let us define

X = { i E w : 3 p E G ( z p ( i )= 1)). Then, X is infinite since G n D , # Q, , for each n < w . Let A E S. There is a q E G n E A . Suppose now that i E X and i > l ( q ) . Then we have zp(i) = 1 for some p E G . Thus, there is an r E G such that r 5 p , q and consequently i E nS,, for zr(i)= 1. Since A E S,, we infer i E A , i.e. X \ l ( q ) A , which proves the lemma. QED The principle P(c) can be also expressed as follows: every centered family R lower bound.

E P(w)/fin of cardinality less that c has a non-zero

Obviously, any descending chain is a centered family. Hence, we obtain the following 5.3. COROLLARY. If MA holds (or a weaker principle P(c)), then m y limit in P(w)/fin has (essential) length c.

c H . 2 GAPSA N D

48

LIMITS

Now let us consider a gap on the set w

L = ({A,

:

< IC}

(Y

; {Bp : /3

< A}),

where IC and X are arbitrary regular cardinals. With any such an L we associate a forcing Q = Q ( L ) ,introduced by K. Kunen, consisting of elements of the form

where u p

IC

and w p

c X are finite sets and xp is a zero-one sequence of some length

I(p) < w . In addition, we assume that

The ordering on Q is defined thus:

5 q if and only if up _> uq and w p _> w q and xp _> x q and for each i, Z(q) 5 i < l ( p ) the following implications hold:

p

if i E

u

A,, then zp(Z)= 1,

a€U,

and

if i E

u

B p , then z p ( i )= 0.

PEW,

It is easy to see that Q ( L ) adds a set X , which separates the gap L . However, a forcing Q ( L ) need not have the ccc. We select below some particular cases when it does.

5.4. I f L is separated, then Q ( L ) has the ccc. Indeed, let a set C separate L. Thus, for any q E Q there is a number k ( q ) E w such that

u

A,

QEU,

Observe that the subset

\qq)cc

and (C \ k(c7)) n

u

PEW,

BB = Q,.

$5. TYPES OF

GAPS A N D LIMITS.

49

is dense in Q,since the sequence x q can be always extended arbitrarily. Now, Q' can be divided into countably many parts

Q' = uu{qE Q' : zq = x z

and

k(q) = k }

k

so that any two p , q in the same part are compatible. In fact, we have ( u pU u q , z, w pU w q ) 5 p , q

(here z = z p = z q ) .

We have shown that Q' is o-linked, which immediately implies the claim. 5.5. If L is a well ordered gap and n

# w1, or (A # w1) then Q(L)has the ccc.

Suppose the opposite, that Q has an antichain E of cardinality w1. If n > wl, then we choose an ordinal y < K so that UqEEu q E y. If we denote

L , = ( { A , : a < 7); {Bg : P < A)), then E is still an antichain in Q(L-,), which is impossible since Q(L7) is separated (by any set A t , with y 5 ( < K ) . Similarly, if K < w1 (i.e. K = w ) , then we can find a y < K for which the inclusion u q y holds for uncountably many q E E (since the function maxuq must be bounded on an uncountable set). Again the gap L, has an uncountable antichain, contradicting 5.4. Now we prove the following

s

5.6. THEOREM (Kunen [K]). If MA and c > w1 hold, then each well ordered gap of type (n,A), where K , X < c and K # w1 (or X # w1)can be separated.

PROOF.From 5.5 we know that the forcing Q = Q(L)has the ccc. Thus, we may apply the Martin axiom: there is a filter G Q,which intersect the following dense subsets for n E w D, = { q E Q : 32 2 n ( x q ( z )= 1)) and for (Y < K and ,B < A. E,,g = { q E Q : (Y E u q A /3 E w q } Clearly, the set X = { z E w : 3q E G ( z q ( i )= 1)) separates then the gap L. QED

Let us define now a so-called dominating forcing. For a family F w w let D = D(F) consists of all elements p = (zp,Fp),where x p is a sequence of nonnegative integers of some length I(p) < w and Fp is a finite subset of F . The ordering on D is defined thus

c H . 2 GAPSAND

50

LIMITS

5 q if and only if x p _> x q and Fp_> Fqand z p ( i )> f ( z ) , whenever f l ( q ) I i < !(PI. p

E

Fq and

It follows immediately that D is a-linked: if z p = z q = z, then (z, Fp U Fq)is below p , q. Clearly, D = D(F) adds a function g , which dominates the set F(i.e. f <* g , for each f E F). Using

ID we can prove easily the following remark.

5.7. MA implies that lb = c.

Indeed, for any F C ww define the sets and

D, = { p E D(F) :

I(p) 2 n}

E j = {PE D(F) : f E F p }

for n E w , for f E F,

which are easily seen to be dense in D(F). Thus, if card F G D(F) intersecting all the D’s and the E’s. If we define

<

c, we have a filter

<*

then g : w + w (since G n D, # $4 for each n E w ) and F g : if f E F then for some q E G, f E Fq (since G n E j # (3 ) and if i > Z(q) then we have g(i) > f(i) (because z p ( i )> f(i) for some p E G, p <_ q). Thus, we have proved that any set F C ww of cardinality < c is bounded, which means that lb = c. Summarizing the above discussion we may say that -- assuming MA plus c > w1 - there are, besides HausdorfFnon-separable (w1, w1)-gaps (which always exist, see 3.3), unfilled Rothberger’s (c,w)-gaps (see 2.7 and 5.7), while all the gaps of type ( K , A) with K , X < c and K # w1 (or X # w1) are separated (theorem 5 . 6 ) . It remains to settle the case of (c,c)- and (c,wl)-gaps, which will be done later in Chapter 8. We prove there that the existence of unfilled gaps of type (c,c) or (c,w1) is both consistent and independent of set theory (with MA plus c > w1). Thus, in the general case (i.e. if no additional assumptions are made) we have two kinds of unfilled gaps only: that of Hausdorff (of type (w1,wl)) and that of Rothberger (of type ( l b , ~ ) ) Note . that, in the latter, the coefficient lb may assume different values, as we have seen above.

$ 5 . TYPES OF

51

GAPS A N D LIMITS.

We may ask now what types of gaps and limits are consistently possible. It turns out that there are here no limitations except that of the theorems 1.5 and 2.8. Actually, we can consistently assert that there is a limit of any type K 5 c, provided c f ( ~ )> w . Similarly, it is consistent that there is a gap of any type (6,A), where K , A 5 c and c f ( ~ )> w (or cf(A) > w ) . In the proof we shall use the Hechler forcing defined as follows. For any partial ordering ( X ,
< l(p)}

where d ( p ) is a finite subset of X and l ( p ) E w . The ordering on P(X) is defined as follows: p q if and only if p _> q and p(x,i) 5 p(y, i), whenever Z(q) 5 i < l ( p ) and I s x y.

<

It is easy to see that P(X) has the ccc. Indeed, suppose the opposite and let E be an uncountable antichain. We may assume that Z(p) =const. for p E E . Applying the A-system lemma we find an uncountable EO 5 E and a d, so that dm (p) n dm (4) = d

for any two p, q E Eo.

If p , q are equal on the common part of their domains

then they are compatible, since p U q 5 p , q. On a finite set d there is only finitely many zero-one functions and hence EO must be finite, contrary to the assumption. How does the forcing P = P(X) work ? Let V be a model of set theory,

(X,
a generic over V ultrafdter. Then we prove the

following general (Hechler [H4]). In the model V[G] there is an order embedding 5.8. THEOREM of X into P(w)/fin.

PROOF.Define in V[G] the set A,

= {z

Ew

: 3p

E G(p(z,i) = 1))

for x E X .

CH.2 GAPSAND LIMITS

52

Clearly, the family { A , : x E X } is in V [ G ] .We prove that, when partially ordered by C,,it is order isomorphic with X . First, let us note that the sets A , are all infinite. Indeed, the sets

0," = { p E P : 3 i 2 n(p(x, i ) = 1))

for x E X and n E w ,

are in V (since they are definable in V ) and are dense in P : if q E P define a p 5 q by setting pldm ( q ) = q (and p ( x , k ) = 0 for k < l ( q ) , if x # dm ( q ) ) and p(y,k) =

{

0, if l ( q ) 5 k < i 1, if k = i and y 2 x 0, if k = i a n d y n o n 2 x

where i = max{l(q),n}. Obviously, we have then p E 0,". Also the sets

E,, = { p E P : z , y

E dm (p)}

for s , y E X ,

are in V and are dense in P (given a q with x,y # d(q), define a p 5 q setting d ( p ) = d(q) U {x,y} and arbitrary values p(x, k), p(y, k ) , for k < l ( g ) ) . Now, let x < y and choose a p E E,,yflG. We claim that A , \l(p) A,. Indeed, if i 2 Z(p) and i E A , , then we have q(x,i) = 1

for some q E G and q 5 p .

Hence also q(y,i) = 1 and thus i E A,. In order to check that A , to notice that the sets

F,={pEP:

3z>n(p(x,i)=OAp(y,i)=l))

s,

A , it suffices

fornEw,

(which obviously are in V )are dense in P (for a q E P define p 5 q setting d ( p ) = d(q), l ( p ) = i 1, where i = max{l(q), n } and p(z, i) = 1, for z 2 y and = 0, otherwise).

+

Finally, we have to show that if x,y are incomparable (i.e. x non 5 y and 5 x), then A , , A , are also such (i.e. A , non C, A , and A , non A,). This will hold, if we prove that the set

c,

y non

H , = { p E P: 3 2 n(p(~,i)=OAp(y,i) = 1 ) A 3 j 2 n((p(x,j) = 1 Ap(y,j) = 0)}, defined for n E w (obviously in V ) ,are dense in P. To check this, let q E P be arbitrary. w e may assume that x,y E dm ( q ) and define a p 5 q by setting pldm ( q ) = q , d ( p ) = d(q), l ( p ) = j 1, where j = i 1 and i = max{l(q), n } and

+

+

1, if k = i a n d z 2 y 1, if k = j a n d 2 2 x 0, otherwise

$ 5 . TYPES OF GAPS A N D LIMITS.

Note that p 5 q holds, since

I

53

and y are incomparable. Obviously, p is in H, and

QED

the proof is finished.

We shall apply now the above theorem to some specified X. Let e.g. X = K * , i.e. K ordered by the inverse inclusion, where K is a fixed cardinal in V with V I= " c ~ ( K )> w " . We prove that the corresponding decreasing family { A , : a < K } in V[G] is in fact a limit. Let C E V[G] be any infinite subset of w . We have to show that

C non C, A , Let us denote (for y

for some a

< K.


P, = { p E P :

d ( p ) 5 y}

and G, = G n P,.

Clearly, P, is a complete subforcing of the Hechler forcing P = P(X). Hence G, P, is a complete ultrafilter over V . Since P has the ccc and c f ( ~ )> w , the set C has a canonical name C E V(P,) for some large enough y < K and thus C E V[G,]. Now, if an a satisfies y 5 01 < K , then the sets

D, = { p E PIG,

:

3i 2 n(i E C A p ( a , i ) = 0 )

for n E w

are dense in PIG,. Indeed, we have

Let y E PIG, be an arbitrary element. We may assume that a E d(q). Define p 5 q by setting pldm (4) = q , d ( p ) = d(q), Z(p) = i 1 where i 2 max{l(q), n} and i E C and 1, if Z(q) 5 j < l ( p ) and p < y and j E Ap

+

p(P,j) =

{

0 , if Z(q) 5 j < Z ( P ) 0 , if l ( q ) 5 .i< Z ( P )

and and

P < Y and j $ Ap P 2y

Then p E D,, which proves the claim. It follows that C non Now, let X be an ordinal sum

K

c f ( ~> ) w in V . Thus, we have a gap

+ X*,

where

K,

c*A,.

X are fixed cardinals in V and

L = ( { A , : a < ~ } { ;B g : / ? < A } )

c H . 2 GAPSAND

54

LIMITS

in V [ G ]( the lower class of L lies below the upper). In order to show that L is unfilled take any set C E V [ G ]and assume that C Bg,for each p < A. As above, there is a (complete) subforcing Q P such that C E V [ Gn Q] and there is an a E n \ U { d ( q ) : q E Q}. For any q the set Bg \ C is infinite and hence the sets for n E w En= { p E P/G n Q : 3i L n(p(a,i) = 1))

c*

c

nBEd(s)

are dense, which in turn implies that A,non

E t C.

We shall discuss now a problem of a scale. We know already that a scale exists if the continuum hypothesis is assumed (see the remark after 2.4) and as easily seen, it exists also if we assume MA plus c > w1. Indeed, we have then lb = c (see 5.7) and consequently Ib = d = c. The claim follows now from 2.5. Note that in both mentioned above cases there is a scale of length c. Now, we prove a useful technical lemma.

5.9. LEMMA. Let s = (f, : a < n) E V, where IC > w is a regular cardinal of V , be an increasing unbounded sequence in the model V . If P E V has the ccc and V ”card P < tc” holds, then in the extension V [ G ]the sequence s is still unbounded.

PROOF.By a contradiction. Suppose that we have f.

>s, for some function

f E V[G].Thus, for each a < n there is an integer n, E w such that f(i) 1 fa(;)

for each i 2 n,.

Since n is regular and > w , the value na must be constant on a set X cardinality K in V [ G ] there : is an integer N such that

n, = N

K,

of

for each a E X .

The set X need not be in V but it has always a subset Y C X , Y E V , of cardinality K in V . Indeed, let us denote

Yp = {a < n : p IF ”a E X”}

for p E P,

where X is any name of X : X [ G ]= X . Thus,

Y,, E V

for each p E P

and X =

u PEG

Y,,.

55. TYPES OF

55

GAPS A N D LIMITS.

Obviously, we have card G < n, by the assumption of the lemma, and hence card Yp= n for some p E G, which proves the claim. Denote this Yp by Y . Thus, we have

(*I

fu(i) I f(i)

for each i 2 N

and

Q

E Y.

Since card Y = n, the sequence (fu : Q E Y ) E V is still unbounded in V (since, it is cofinal in s). Let A, = {fa(z) : cr E Y } ,for i 2 N . Thus, the family { A , : i N } is in V and from (*) it follows that

>

V [ G ] " V i 2 N ( A i E fin)". Clearly, the same holds with V replacing V [ G ] Therefore, . the function

is in V and obviously g is a bound of s, a contradition.

QED

Note that 5.9 can be restated as follows: if P E V has the ccc and card P < Ib in V , then each unbounded set in V remains unbounded in V [ G ] . A Cohen forcing @ ( X )(where X is an arbitrary set) consist of finite zero-one functions p : d m ( p ) + (0, l}, where the domain dm ( p ) is a finite set X , ordered by the inverse inclusion. @ ( X )has always ccc. If X = A U ( X \ A ) is any partition of X , then @ ( X )is isomorphic to the product @ ( A )x @ ( X \ A ) . If X E V then also @ ( X )E V and card C ( X ) = card X . Models of the form V [ G ]where , G C @(X) is a complete (over V ) ultrafilter and X E V , are called Cohen models (or Cohen extensions). From the lemma 5.9 it follows easily 5.10. COROLLARY. In any Cohen model V [ G ]the set of old functions ww n V is unbounded but not dominating.

PROOF.If f E V [ G ]is any function: w -+w , then there is a countable subset A C X so that f E V [ Gn @ ( A ) ] .Since @ ( A )is countable, we may apply 5.9. to any unbounded sequence s = (fu :

Q

< Ib")

E

v.

Thus, we have fnon, 2 s and consequently fnon, >ww n V . Hence, no function f E V [ G ]is a bound of ww n V, which proves the first assertion.

56

c H . 2 GAPSAND LIMITS

A forcing Q consisting of all functions g : dm ( 9 ) + w , where the domain dm (9) w is finite, ordered by the inverse inclusion, is countable and hence equivalent to a @(w). Thus, if h = U H where H Q is a generic over V ultrafilter, then h is in V[G].Clearly, the sets

c

03 = { q

EQ:

3i 2 n ( q ( i ) > f(i))}

for n E w and f E wW n V ,

are dense in Q (and belong to V ) . This implies that for each f E ww n V we have

h(i) > f ( i )

for infinitely many i E w ,

i.e. h s , f holds for no old function f, which proves the second assertion.

QED

Finally, let us note the following 5.11. THEOREM. Assume that V GCH holds and that IE > w1 is a regular cardinal in V , such that nw = n. Then, in the Cohen extension V [ G ]via C(n) we have lb = w1 and d = c = K . In particular, lb < d and hence there is no scale.

”lb = w l ” , since CH is true in V . It follows from 5.10 PROOF.We have V that lb = w1 holds also in V [ G ]since , the set ww n V of old functions has cardinality w1 and is unbounded. Clearly, V [ G ] ”c = n”, since nw = K . If a set F ww from V[G]is of power < c, then F E V [ Gn @(7)],for some 7 < n. Now, @(n) is equal to @(y)x @(n \ 7) up to isomorphism and V [ G ]= V [ Gn @(y)][G n @(n \ 7 ) ] .Again from 5.10. it follows that the set H = ww n V [ Gn C ( y ) ]cannot be dominating in ww n V[G].Hence, F is not dominating as a subset of H. We have proved that no set F of cardinality < c can be dominating and consequently d = c, which finishes

+

the proof.

QED

PROBLEMS

A. THEMARTINAXIOM 1. Assume MA. If R, S

c P ( w ) are two families of cardinality < c such that

S U {A}is uniformly centered for each A E R,

c

c.

then there is a set Y w such that Y B for each B E S and Y n A is infinite for each A E R. (This is a strengthening of P(c) and the proof can be as that of 5.2)

PROBLEM s

57

2. Let X be a topological space with countable weight and suppose that the sets N,, for 01 < n, where n is a fixed cardinal < c, are nowhere dense in X . a) Let { U , : n < w } be an enumeration of a fixed base, in which each basic set occurs infinitely many times. If Y g w is as in Ex. 1 above with respect to the families

R = {A,

: n Ew}

and S = { B , :

01

< K},

where

A,

=

{i E w

:

Ui

g U,}

and B, = {i E w : U , n N , = q } ,

then the sets 2, =

x\ u{Ui :

iEY

A

i > n)

for n E w ,

are nowhere dense. b) With the notation as in a) we have an inclusion

Hence, if MA holds then the union of less than c nowhere dense sets is meager (in any space with countable weight).

3. Let X be the Lebesgue measure in the real line JR (or in Rn).For an E > 0 let a forcing P,consist of open sets p with X(p) < E , ordered by the inverse inclusion. a) If E & P,is uncountable, then there is an uncountable subset Eo an integer n so that X(p)

+ 1<

E

GE

and

for each p E Eo.

b) For each p there is a finite union of rational intervals w ( p ) G p such that we have X ( p \ w ( p ) ) <

i.

c) If w ( p ) = w ( q ) for some p , g E Eo, then X(p u q )

X(U G) 5 E . (W is hereditarily Lindelof).

d) If G

P, is a filter, then

e) If an A

C JR has measure zero, then

is dense in P,.

< E . Hence P, has the ccc.

the set

CH.2 GAPS A N D LIMITS

58

Hence, MA implies that the union of less than c A-null sets is A-null.

B. GAPS IN P(w)/fin 1. There exists an almost disjoint family R = { A , :

(Y

< w1} such that for each

(Y < w1 the set A , is near to the segment {Ac; ( < a}. (Define R by induction). If S,T C_ R are uncountable and disjoint, then the gap L = (S,T ) is not filled. (Lusin).

2. The separating forcing Q = Q ( L ) , where L is a Lusin gap from Ex. 1 above, never has the ccc.

3. Consider the Cantor space C, = (0,l}". a) There are unfilled well ordered gaps in C of the form

L = ({fa :

a:


; {gp :

P
<* fp and go* r g p for a: 5 P < w1 and f, <* g p for any a,P < w1. h <. gp} are all F,-sets. Intervals [fn,gp] = { h E C , : fa

where fa b)

c) The Cantor space C, is an increasing union

U

H,, where each Ha is a

,
Ga-set. (Hausdorff).

c. ORDERINGS MODULO FILTER For an arbitrary filter F on w , we define two relations on the set w w as follows and

f IFg

{ i E w : f ( i )I g(k)} E F

f =F

{i E W

g

:

f ( Z ) = g(Z)} E F.

1. The relation < F is a partial pre-ordering, i.e. a partial ordering on the set w " / = F of all abstraction classes. 2. The ordering

IFis linear if and only if F is an ultrafilter.

<*

3. The ordering is identical with the < F , where F is the Frkchet filter on the set w , F = { A E w : w \ A E fin}. Sometimes, it is convenient to use the strong ordering defined thus

f

g

3

{ z E w : f(i) < g ( 2 ) ) E

implies the condition f

F.

S F g (i.e. f < F

g and

5. The cardinal lb remains unchanged, when - in its definition - the ordering is replaced by <*. The same applies for the cardinal d.

<*

PROBLEMS

59

D. GAPSIN ww Consider the strong ordering

f <* g

{ i E w : f ( i ) 2 g(i)} E fin.

E

We say that f and g are far away, if

).(fI

- g(n)l = +m.

lip 1. If fo

<* go are far away, then there exists a far away pair

f1

<* g1 such that we

have fo <* f1 <* g1 <* go.

2. There exist ordered countable gaps

L = ({frl :

12

< w } ; {gm : m < w } ) (the lower class of L below the upper)

between any far away pair fo

<.

go.

3. Let L be a countable gap as in Ex. 2 above. a) There is an increasing sequence ko i 2 km we have mm{fo(i),

<

Icl

< . . . such that for each m and

. . . ,fm(i)} + m < min{go(i), . . . ,gm(i)}.

b) There is a far away pair f <* g filling the gap L (i.e. fn <* f <* g <* g m for each n,m < w ) . Hence, every countable gap in ww is filled.

4. There exists an ordered gap

of type

(WI

,w~1.

5. There exist non-separated (wl,wl)-gaps L as in Ex. 4 above (Kunen). (This is

a fairly nontrivial construction).

6. It is consistent that there are non-separated gaps L, of a given type ( K , A), where K , X 5 c and c ~ ( K ) ,cf(A) > w . (Modify the Hechler forcing so that the functions p may assume arbitrary values from w ) .

CH.2 GAPSA N D

60

LIMITS

E. CHAINS A N D ANTICHAINS

< n) in ww is regular, then we have n 5 d. 2. It is consistent that there are limits in ww of a given type n I c with cf(n) > w 1. If the length n of a limit ( f a :

(Y

(cf D.6).

3. It is consistent that there are maximal antichains {a, : (Y < n} in P(w)/fin, of a given cardinality n 5 c, cf(n) > w. (Modify the Hechler forcing so that it produces almost disjoint sets). 4. The property P(c) implies that each infinite maximal antichain in P(w)/fin has cardinality c. (If R P(w) is an infinite almost disjoint family, then the family {w \ A : A E R} is uniformly centered). 5. Each infinite maximal antichain in P(w)/fin has always cardinality

2 Ib.

F. DOMINATING FAMILIES 1. The cardinal d remains unchanged when the dominance ordering by the pointwise ordering

f5g

=

I*is replaced

vi E w[f(i) I g(i)].

(Each function f from a dominatingfamily replace by functions fn(i) = f ( n + i ) ) .

2. The cardinal d is the minimum number of compact sets covering the space of irrationals ww.

3. It is consistent that there is a scale of a given length n 5 C, cf(n) > w. (Iterate IE times with finite supports the dominating forcing I D).

G.THECARDINAL t Let t denote the length of a shortest limit in P(w)/fin. 1. The cardinal t is regular and satisfies w1 5 t 2. Let

fA

denote the increasing enumeration of a set A

a) If A C, B, then

b)

I c.

fA+

c w.

2f~.

an arbitrary (infinite) A w and any f : w + w there is an infinite subset B C_ A such that f <* f B . For

Hence, t satisfies w1 5 t 5 Ib.

3. We have the equality 2<' = C. (In P(w)/fin define inductively a binary tree of height t ) .

PROBLEMS

4. It is consistent that t < Ib. (Let V c (see F.3.). Apply Ex. 3 above).

"w1

61

< c < 2w1" and force a scale of length

H.THECARDINAL b A family S P(w)/fin is called splitting, if for each a > (D there is an I E S such that both a . r > (D and a - I > 0.Let s denote the minimal power of a splitting family. 1. Inequality t I s holds. (For a n define a decreasing chain (a, : a for each a < n).

< t and any R = {z, : a < n} P(w)/fin < n) such that either a , 5 I , or a , . I, = (D

2. We have s I d. (If f + > f ~( f = ~ increasing enumeration of A ) then f~ assumes a value in any interval {i E w : f"(h) I i < f " + ' ( k ) } ,for n E w where h is fixed and f ( k ) > 0).

3. It is consistent that s = w1 < c. (Consider a Cohen model).

4. It is consistent that t < s and 2' < 2', simultaneously. (Let V " C = wz < 2,1 < 2,z" and iterate with finite supports wz-times a centered forcing P(F,) determined at a - t h stage by a fixed ultrafilter Fa).

+

5 . Each set X in the Cantor space C, = (0,l}, of cardinality < s has the Lebesgue measure zero. (If g assumes the value 1 infinitely many times, then the set

{f E C ( w ) : I.

g

5. f} is a countable union of null sets).

SEQUENTIALLY COMPACT SPACES

A topological HausdorfF space X is called sequentially compact if any sequence I = { I " : n E w } C X contains a subsequence zlA = { I " : n E A } converging to a point of the space X . 1. Let X be a compact (Hausdorff) space. For any { I " : n E w } C X and any A w denote by SA the set of all cluster points of I ( A= {I,, : n E A } . Assume that I = { I , : n E w } has no convergent subsequences. a) Each subsequence IIA has two subsequences slAo and z)A1 for which we ~8. have S A n~S A = b) There is a binary tree T in P(w)/fin of height t so that the sets S,, corresponding to elements on any level of T , are pairwise disjoint. Hence, each compact space of cardinality less than 2' is sequentially compact.

62

CH.2 GAPSA N D LIMITS

2. A Cantor space C, = {0,1}'( is sequentially compact if and only if the weight K. is less than s. (If R is splitting and card R = s then we have (in ( 0 , l } R )a sequence 2 = {I,,: n E w } , where zn(A) = 1 iff n E A ) .