s. shelah and l. soukup- on the number of non-isomorphic subgraphs

8/3/2019 S. Shelah and L. Soukup- On the number of non-isomorphic subgraphs

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On the number of non-isomorphic subgraphs

S. Shelah

Institute of Mathematics

Hebrew University, Jerusalem

L. Soukup

Mathematical Institute of the

Hungarian Academy of Sciences

October 6, 2003

Abstract

Let K be the family of graphs on 1 without cliques or independentsubsets of size 1. We prove that

(a) it is consistent with CH that every G K

has 2

1

many pairwisenon-isomorphic subgraphs,

(b) the following proposition holds in L: () there is aG K suchthat for each partition (A,B) of 1 either G = G[A] or G =G[B],

(c) the failure of () is consistent with ZFC.

1 Introduction

We assume only basic knowledge of set theory simple combinatorics for

section 2, believing in L |= + defined below for section 3, and finite supportiterated forcing for section 4.

The first author was supported by the United States Israel Binational Science Foun-

dation, Publication 370The second author was supported by the Hungarian National Foundation for Scientific

Research grant no. l805

1


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Answering a question of R. Jamison, H. A. Kierstead and P. J. Nyikos[5] proved that if an n-uniform hypergraph G = V, E is isomorphic to eachof its induced subgraphs of cardinality |V|, then G must be either empty orcomplete. They raised several new problems. Some of them will be investi-gated in this paper. To present them we need to introduce some notions.

An infinite graph G = V, E is called non-trivialiffG contains no cliqueor independent subset of size |V|. Denote the class of all non-trivial graphs on1 by K. Let I(G) be the set of all isomorphism classes of induced subgraphsof G = V, E with size |V|.

H. A. Kierstead and P. J. Nyikos proved that |I(G)| for each G Kand asked whether |I(G)| 2 or |I(G)| 21 hold or not. In [3] it wasshown that (i) |I(G)| 2 for each G K, (ii) under + there exists aG K with |I(G)| = 1. In section 2 we show that if ZFC is consistent, thenso is ZFC + CH + |I(G)| = 21 for each G K. Given any G K wewill investigate its partition tree. Applying the weak principle of Devlinand Shelah [2] we show that if this partition tree is a special Aronszajn tree,then |I(G)| > 1. This result completes the investigation of problem 2 of [5]for 1.

Consider a graph G = V, E . We say that G is almost smooth if itis isomorphic to G[W] whenever W V with |V \ W| < |V|. The graphG is called quasi smooth iff it is isomorphic either to G[W] or to G[V \ W]

whenever W V. H. A. Kierstead and P. J. Nyikos asked (problem 3)whether an almost smooth, non-trivial graph can exist. In [3] various modelsof ZFC was constructed which contain such graphs on 1. It was also shownthat the existence of a non-trivial, quasi smooth graph on 1 is consistentwith ZFC. But in that model CH failed. In section 3 we prove that +, andso V=L, too, implies the existence of such a graph.

In section 4 we construct a model of ZFC in which there is no quasi-smooth G K. Our main idea is that given a G K we try to construct apartition (A0, A1) of1 which is so bad that not only G = G[Ai] in the groundmodel but certain simple generic extensions can not add such isomorphisms

to the ground model. We divide the class K into three subclasses and developdifferent methods to carry out our plan.

The question whether the existence of an almost-smooth G K can beproved in ZFC is still open.

We use the standard set-theoretical notation throughout, cf [4]. Given agraph G = V, E we write V(G) = V and E(G) = E. If H V(G) we


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define G[H] to be H, E(G) [H]2

. Given x V take G(x) = {y V :{x, y} E}. If G and H are graphs we write G = H to mean that G and Hare isomorphic. If f : V(G) V(H) is a function we denote by f : G = Hthe fact that f is an isomorphism between G and H.

Given a set X let Bijp(X) be the set of all bijections between subsets ofX. If G = V, E is a graph take

Isop(G) = {f Bijp(V) : f : G[dom(f)] = G[ran(f)]} .

We denote by Fin(X, Y) the set of all functions mapping a finite subset ofX to Y.

Given a poset P and p, q P we write pPq to mean that p and q arecompatible in P.

The axiom + claims that there is a sequence S : < 1 of contablesets such that for each X 1 we have a closed unbounded C 1 satisfyingX S and C S for each C.

We denote by TC(x) the transitive closure of a set x. If is a cardinaltake H = {x : |TC(x)| < } and H = H, .

Let us denote by D1 the club filter on 1.

2 I(G

) can be always largeTheorem 2.1 Asume that GCH holds and every Aronszajn-tree is special.

Then |I(G)| = 21 for eachG K.

Remark: S.Shelah proved, [7, chapter V. 6,7], that the assumption oftheorem 2.1 is consistent with ZFC.

During the proof we will apply the following definitions and lemmas.

Lemma 2.2 Assume that G K, A [1]1 and |{G(x) A : x 1| = 1.Then |I(G)| = 21.

Proof: See [3, theorem 2.1 and lemma 2.13].

Definition 2.3 Consider a graph G = 1, E.


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1. For each 1 let us define the ordinal 1 and the sequence : as follows: put 0 = 0 and if : < is defined, thentake

= min { : < > and ({

, } E iff {

, } E)} .

If = , then we put = .

2. Given , 1 write G iff =

for each .

3. Take TG =

1, G

. TG is called the partition tree of G.

Lemma 2.4 If G = 1, E K with |I(G)| < 21, then TG is an Aronszajntree.

Proof: By the construction of TG, if , 1, < and G() =G() , then G . So the levels ofTG are countable by lemma 2.2. Onthe other hand, TG does not contain 1-branches, because the branches areprehomogeneous subsets and G is non-trivial.

Definition 2.5 1. Let F : (2)


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Consider the Aronszajn-tree TG

= 1, G. Since every Aronszajn-treeis special and I is a countably complete ideal on 1, there is an antichain S

in TG with S / J. Take

A =

1 : S( G )

.

Now property () below holds:

() S (S A) \ + 1

A ({, } E iff {, } / E).

Indeed, if for each A we had {, } E iff {, } E, then G would hold by the construction of TG.Let 1, S, T S and f : G[(A ) T] G be an

embedding. Define F(,,T,f) 2 as follows:

F(,,T,f) = 1 iff x G( A )({x, f()} E iff {, } E).

In case = , under suitable encoding, F can be viewed as a functionfrom (2)


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Proof:Given a set X, AP(X) and F Bijp(X) take

Cl(A, F) =

{B : B A and B0, B1 B f F Y [X]


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Consider a sequence F = f0, . . . , f n1. Given a family F Bijp(X) wesay that F is an F-term provided fi = f or fi = f1 for some f F, foreach i < n. We denote the function f0 fn1 by F as well. We willassume that the empty term denotes the identity function on X. If l ntake (l)F = f0, . . . , f l1 and F(l) = fl, . . . , f n1. Let

Sub(F) =

fi0, . . . , f il1

: l n, i0 < .. . < il1 < n

.

Given f F and x, y X with x / dom(f) and y / ran(f) let Ff,x,y bethe term that we obtain replacing each occurrence of f and off1 in F withf {x, y} and with f1 {y, x}, respectively.

Lemma 3.3 Assume that F Bijp(X), AP(X) is F-closed, F0, . . . , F n1are F-terms, z0, . . . , zn1 X, A0, . . . , An1 A such that for each i < n

() zi /

{FAi : F Sub(Fi)} .

If f F, x X\dom(f), Y [X\ran(f)] with |A Y| < for eachA A, then there are infinitely many y Y such that () remains true whenreplacing f with f {x, y}, that is,

() zi / FAi : F Sub(F

f,x,yi )

for eachi < n.

Proof: It is enough to prove it for n = 1. Write F = f0 . . . , f k1, A = A0,z = z0. Take

YF,A = {y Y : () holds for y} .

Now we prove the lemma by induction on k.Ifk = 0, then YF,A = Y\A. Suppose we know the lemma for k 1. Using

the induction hypothesis we can assume that () below holds:

() Y = YG,F(l)A : l n, G Sub((l)Ff,x,y), G = Ff,x,y .Assume that |YF,A| < and a contradiction will be derived.First let us remark that either fk1 = f or fk1 = f

1 by ().Case 1:fk1 = f

1.

Then YF,AY\A by (), so we are done.


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Case 2:fk1 = f.In this case x A and for all but finitely many y Y we have z =

Ff,x,y(x). Then for each y, y Y take

l(y, y) = max

l n : i < l Ff,x,y(i) (x) = Ff,x,y

(i) (x)

.

By Ramseys theorem, we can assume that l(y, y ) = l whenever y, y Y.

Clearly l < n. Then Ff,x,y(l) (x) = Ff,x,y

(l) (x) but Ff,x,y(l1)(x) = F

f,x,y

(l1) (x), so

fl = f1 and Ff,x,y(l1)(x) = x for each y Y. Thus z = (l1)F

f,x,y(x) for eachy Y, which contradicts () because x A.

The lemma is proved.

We are ready to construct our desired graph.First fix a sequence M : < 1 of countable, elementary submodels of

some H with M : < M for each < 1, where is a large enoughregular cardinal.

Then choose a -sequence S : < 1 M0 for the uncountable sub-sets of 1, that is , { < 1 : X = S} / NS(1) whenever X [1]1.We can also assume that S is cofinal in for each limit .

We will define, by induction on ,

1. graphs G = ,E with G = G[] for < ,

2. countable sets F Isop(G),

satisfying the induction hypotheses (I)(II) below:

(I) {S : } is uncovered by I J where

I = Cl({G() : } ,

F)

andJ = Cl({\G() : } ,

F).

To formulate (II) we need the following definition.

Definition 3.4 Assume that = + 1 and Y. We say that Y is largeif n , fi, xi : i < n, hif


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1. i < n i < fi Fi ,2. i < n i xi < ,

3. i < n ran(fi)Y,

4. i = j < n ran(fi) ran(fj) =

5. h Fin(Y , 2) and dom(h)

{ran(fi) : i < n} = ,

then

y Y [,) such that

6. i < n x dom(fi) ({y, fi(x)} E iff {xi, x} E),

7. z dom(h) {y, z} E iff h(z) = 1.

Take

(II) If = + 1, then is large.

The construction will be carried out in such a way that G : Mand F : < M.

To start with take G0 = , and F = {}. Assume that the construc-tion is done for < .Case 1: is limit.

We must take G = {G : < }. We will define sets F0, F1Isop(G)

and will take F = F0 F1.

Let

F0 = {f Isop(G) M : n : n < sup {n : n < } = ,

fn Fn and fn : Gn = Gn [ran(f)] for each n } .

Take F =


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and f : Gf

=

Gf

[W f

] for some f

< .We want to find functions gW,f f for W, f W such that

(A) gW,f : G = G[W]

(B) taking F1 =

gW,f : W, f W

the induction hypothesis (I) remainstrue.

First we prove a lemma:

Lemma 3.5 If W, f W, g Isop(G, G[W]), g f, |g\f| < , then

(i) for each x W\dom(f) the set

{y W : g {x, y} Isop(G, G[W])}

is cofinal in .

(ii) for each y W\ran(f) the set

{x W : g {x, y} Isop(G, G[W])}

is cofinal in .

Proof: (i): Define the function h : ran(g)\ran(f) 2 with h(g(z)) = 1iff {z, x} E. Choose LW with ran(h) and f . SinceW ( + ) is large, we have a y W [,+ ) such that

1. {y, f(z)} E iff {x, z} E for each z dom(f)

2. {y, g(z)} E iff h(g(z)) = i for each z dom(g)\dom(f).

But this means that g {x, y} Isop(G, G[W]).(ii) The same proof works using that + is large for each < .

By induction on n, we will pick points zn and will constructfamilies of partial automorphisms,

gW,fn : W, f W

such that gW,f =

gW,fn : n <

will work.During the inductive construction we will speak about F-terms and

about functions which are represented by them in the nth step.


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If F = h0, . . . hk1 is an F-term and n take F[n] = j0 jk1where

ji =

gW,fn if hi = gW,f,

(gW,fn )1 if hi = (g

W,f)1,hi otherwise.

First fix an enumeration {Wn, fn , un, in : 1 n < } ofW 2 andan enumeration Fn,i : i < ln , jn, An,i : i < ln , Dn n < of the quadru-ples F0, . . . , F k1 , j, A0, . . . , Ak1 , D where k < , F0, . . . , F k1 are F-terms, j 2, D S and either j = 0 and A0, . . . , Ak1 I or j = 1 andA0, . . . , Ak1 J .

During the inductive construction conditions (i)(v) below will be satis-fied:

(i) gW,f0 = f

(ii) gW,fn Isop(G, G[W])

(iii) gW,fn gW,fn1, |g

W,fn \f| <

(iv) zk /

F[n]Ak,i : F Sub(Fk,i)

for each i < lk and k < n

(v) ifin = 1, then un dom(gWn,fnn ),

if in = 0, then either un / Wn or un ran(gWn,fnn ).

If n = 0, then take gW,f0 = f.If n > 0, then let gW,fn = g

W,fn1 whenever W, f = Wn, fn. Assume that

in = 0, W, f = Wn, fn and un / dom(gWn,fnn1 ). Then, by lemma 3.5, the

set Y =

y W : gW,fn1 {un, y} Isop(G, G[W])

is unbounded in .

Since the members of I J are bounded in , we can apply lemma 3.3

to pick a point y Y such that taking gWn,fnn = gWn,fnn1 {un, y} condition

(iv) holds.If in = 1 and W, f = Wn, fn, then the same argument works.

Finally pick a point

zn / Dn\

F[n]An,i : F Sub(Fn,i), i < ln

.

The inductive construction is done.


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Take gW,f

= gW,f

n : n < . By (v),gW,fn : G

= G[W].

By (iv), we have

zk Dk\

Fk,iAk,i : i < lk

and so it follows that {S : } is uncovered by I J.Case 2: = + 1.

To start with we fix an enumeration

fki , xki : i < nk, hk

: k

of

pairs fi, xi : i < n , h satisfying 3.4.15.

If k takeB0k = h

1k {0}

fki () : i < nk, dom(f

ki ) and

, xki

/ E

and

B1k = h1k {1}

fki () : i < nk, dom(f

ki ) and

, xki

E

.

Applying lemma 3.2 -many times we can find partitions (C0k , Ck1 ), k < ,

of such that taking

I+ = Cl((I

C1k : k

,

F)

andJ+ = Cl((I

C0k : k

,

F)

the set {S : } is uncovered by I+ J

+ .

We can assume that BikCik for i < 2 and k < because B

0k J and

B1k I. Take

E = E

{,+ n} : < , n and B1n

and F = .

By the construction of G = ,E, it follows that is large, so (II)holds. On the other hand

I =

X Y : X I+ , Y []


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and J =X Y : X J+ , Y []


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1 to such that f1

{n} is antichain in T for each n . The ordering onQT is as expected: f QT g iff f g. For < 1 denote by T the set ofelements of T with height . Take T


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and K2 = K \ (K0 K1).

4.1 G K0

First we recall a definition of [1].

Definition 4.2 A poset P is stable if

B [P] B [P] p P p p p B b B (pP

b iff pP

b).

We will say that p and p are twins for B and that B shows the stability of

P for B.

Lemma 4.3 P2 is stable.

Proof: First let us remark that it is enough to prove that both C and QT arestable for any Aronszajn-tree for in [1] it was proved that any finite supportiteration of stable, c.c.c. posets is stable.

It is clear that C is stable. Assume that T is an Aronszajn tree andB [QT]

. Fix a countable ordinal with {dom(p) : p B} T


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(i) G,A,A0, A1 N0,(ii) N : < N for < 1.

For x 1 \ A take

rank(x) = min{ : x N}.

Fix a partition (S0, S1) of 1 with |S0| = |S1| = 1. Take

Vi = Ai rank1Si

for i 2.We show that the partition (V0, V1) works.

Assume on the contrary that P is a stable c.c.c. poset, f is a P-name ofa function, p0 P and

p0f : G = G[V0].

Without loss of generality we can assume that p0 = 1P. Now for eachc A0 choose a maximal antichain Jc P and a function hc : Jc V suchthat qf1(c) = hc(r) for each q Jc.

Take B =

{Jc : c A0} and pick a countable B P showing the

stability of P for B.For b P define the partial function dtb : 1 2A0 as follows. Let

x 1. If there is a function t 2A0 so that

(a) t(c) = 1 for each q Ic if q and b are compatible conditions, then{x, hc(q)} E,

(b) t(c) = 0 for each q Ic if q and b are compatible conditions, then{x, hc(q)} / E,

then take dtb(x) = t. Otherwise x / domdtb.

Sublemma 4.4.1 If pf(x) = y, then there are p p and b B suchthat b and p are twins for B and dtb(x) = tpG(y, A0).

Proof: By the choice ofB, we can find a p p and a b B so that p and bare twins for B. Let c A0. For each q Jc, ifq and p are compatible in P,then {y, c} E iff{x, hc(q)} E , because, taking r as a common extension

of q and p, we have rf(x) = y and f( hq(c)) = c. So {y, c} E iff foreach q Ic if q and p are compatible, then {x, hc(q)} E. But p and b aretwins for

{Jc : c A0}, so dtb(x) = dtp(x) = tpG(y, A0).


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Sublemma 4.4.2There is a b B

such that() |{t randtb : |rl

G(t)| }| = 1.

Proof: Let G be a P-generic filter over V. Put

F = {tpG(y, A0) : y V0 \ A0, |[y]G,A0| }.

Then |F | = 1, so we can write F = {t : < 1}. Fix sequencesp : < 1 G, x : < 1 1 and y : < 1 1 such thatpf(x) = y and tpG(y, A0) = t. By sublemma 4.4.1,

bB randtb F.

But B is countable, so we can find a b B satisfying () above.

Fix b B with property (). Consider the structure

N = P(B B), B , B, Jc, hc : c A0 .

By CH, there is a < 1 with N N. Pick S1 \ . Since G, N, b N,it follows that dtb N N. By () and (ii), there is a

t randtb (N \


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Lemma 4.5Assume CH. If G K

0, thenVC |= there is a partition (V0, V1) of 1 so thatfor each stable c.c.c. poset P we have:

VCP |= G is not isomorphic to G[Vi] for i 2.

Proof: . Fix a set A [1] witnessing G K0 and a bijection f : A

in V. Let r : 2 be the characteristic function of a Cohen real from VC.Take Ai = (f r)1{i} for i < 2. Then (A0, A1) is a partition of A. Using atrivial density argument we can see that x G,A y implies x G,Ai y for i < 2and for x, y 1 \ A. Thus VC |= A0 and A1 witness G K0. Applying

lemma 4.4 in VC we get the desired partition of 1.

Lemma 4.6 In VP2 , if G K0 is quasi-smooth, then G K0.

Proof: Choose a set A [1] witnessing G K0 and a bijection f : A .

Pick < 2, is even, with A, f, G VP. From now on we work in VP.Let {[x]G,A : < 1} be an enumeration of the equivalence classes of G,A.Fix a partition (I0, I1) of 1 into uncountable pieces. Let r : 2 be thecharacteristic function of a Cohen real from VPC. Take Ai = (f r)1{i}for i < 2. Then (A0, A1) is a partition ofA. Using a trivial density argument

we can see that x G,A y implies x G,Ai y for i < 2 and for x, y 1 \ A.For i 2 put

Bi = Ai {x : Ii} {[x]G,A \ {x} : I1i}.

Clearly (B0, B1) is a partition of 1 and

Bi [x]G,Ai = Bi [x]G,A = {x}.

So G[Bi] K. But G is quasi-smooth, so G = G[Bi] for some i 2 in VP2 .Thus G K0 is proved.

4.2 G K1

We say that a poset P has property Priff for each sequence p : < 1 Pthere exist disjoint sets U0, U1 [1]

1 such that whenever U0 and U1we have pPp.


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Lemma 4.7C has property Pr.

Indeed, C has property K.

Lemma 4.8 If T is an Aronszajn-tree, then QT has property Pr.

Proof: Let p : < 1 P be given. We can assume that there are astationary set S 1, p QT, < 1, n and {zi : i < n} T suchthat for each S

(a) x dom(p) for each x dom(p) with heightT(x) ,

(b) pT


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Lemma 4.9Assume that the poset P has property Pr and V

P

|= the posetQ has property Pr. Let {p, q : < 1} P Q. Then there aredisjoint sets U0, U1 [1]

1 such that for each U0 and U1 theconditions p, q and p, q are compatible, in other words, p and p havea common extension p, in P with p, q Q q. If P is well-met, thenwe can find conditions {p : U0 U1} in P with p

p such that

p pq Q q for each U0 and U1.

Proof: Let U be a P-name for the set U = { : p GP}, where GP is the P-generic filter. Since P satisfies c.c.c., there is a p P with p|U| = 1.Since VP |=Q has property Pr, there is a condition p p and there are

P-names such that p Vi = {i : < 1} [U]1, for i 2, and q0and q1

are compatible whenever , 1. Choose conditions p p and

ordinals 0, 1, with p

i =

i for i < 2.

Now consider the sequence A = {p : < 1}. Since P has propertyPr, there are disjoint, uncountable sets C0, C1 A such that p

and p

are

compatible whenever C0 and C1. Take Ui = {i : Ci} fori 2. We can assume that U0 U1 = . Let C0 and C1 and letp be a common extension of p and p

. Then p

0, 1 U, that is, p0

and p1

are in GP, so p, must be a common extension of p0 and p1 . So

p0 V0 and 1 V1, thus p

q0 and q1 are compatible in Q, sop0 , q0 PQ p1 , q1.Suppose that P is well-met. Take p0 = p

p0 and p

1

= p p1 .

It works because we can use p p as p

in the argument of the previousparagraph.

Lemma 4.10 If R : , S : < is a finite support iteration suchthat VR |=S has property Pr for < , then R has property Pr, aswell.

Proof: We prove this lemma by induction on . The successor case is covered

by lemma 4.9. Assume that is limit. Let p : < 1 R. Without lossof generality we can assume that supp(p) : < 1 forms a -system withkernel d. Fix < with d . By the induction hypothesis, the posetR has property Pr, so there exist disjoint sets U0, U1 [1]

1 such thatwhenever U0 and U1 we have pRp. But pRp implies pRpbecause supp(p) supp(p) , so R has property Pr, as well.


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The previous lemmas yield the following corollary.Lemma 4.11 P2 has property Pr.

Given G = 1, E K1 and , , 1 with take

DG (, ) = { : {, } E iff {, } / E}.

Lemma 4.12 If G K1, then

() 1 G() 1 , 1 \ G() |DG (, )| < .

Proof: Since G K1, we have an x 1 with |1 \ [x]| < 1. ChooseG() 1 \ with 1 \ [x] G(). It works because , > G() implies, [x].

The bipartite graph 1 2, {{, 0 , , 1} : < < 1} will be de-noted by [1; 1].

Lemma 4.13 If G K1, then neither G nor its complement may have a not necessarily spanned subgraph isomorphic to [1; 1].

Proof: Let G = 1, E. Write E() = { 1 : {, } E}. Assume

on the contrary that A, B [1]1

are disjoint sets such that {, } Ewhenever A and B with < . Without loss of generality we canassume that (A \ + 1) () = for each A. Write A = { : < 1}.Then for 1 the set F() = (A ) \ E(+1) is finite because +1 >() and (A ) \ E() = for all but countable many B. By Fodorslemma, we can assume that F() = F for each S, where S is a stationarysubset of 1 containing limit ordinals only. Let T = { S : F } andtake W = {+1 : T}. Then G[W] is an uncountable complete subgraphof G. Contradiction.

Lemma 4.14 If G K1 and VC

|=Q has property Pr, then

VCQ |= G = G[f1{i}] for i < 2,

where f : 1 2 is the C-generic function over V.


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Proof:Assume on the contrary that

p,q h : G = G[f1{0}].

To simplify our notations, we will write E for E(G), D(, ) for DG (, )

and () for G().Let C0 = { < 1 : < implies () < }. Clearly C0 is club. Take

C1 = { < 1 : p,q h = f1{0} }. Since C Q satisfies c.c.c, theset C1 is club. Put C2 = C0 C1.

Now for each < 1 let = min(C2 \ + 1) and choose a conditionp, q p,q and a countable ordinal such that

p, q

h(

) = .

Since > () for each 1, we can fix a stationary set S 1and a finite set D such that D(, ) = D for each S. Since C is well-met, applying lemma 4.9 we can find disjoint uncountable subsets S0, S1 Sand a sequence p : S0 S1 C with p

p such that p

p

q Q q for each S0 and S1.We can assume that the sets {dom(p) : S0} and {dom(p

) : S1}

form -systems with kernels d0 and d1, respectively.Take Y0 = { S0 : {, } E} and Y

1 = { S1 : {, } / E} for

< 1. Write Yi = { < 1 : |Yi | = 1} and Zi = 1 \ Yi for i < 2.

By 4.13, the sets Zi are countable. Pick C2 with Dd0d1Z0Z1 . Let = min(C2 \ + 1) and = min(C2 \ +1) . Since d0 d1 and|Y0 | = |Y

1 | = 1, we can choose i Y

i \

with dom(pi) [, ) = for

i = 0, 1. The set W = D(0, 1) [, )is finite because i i

>() for i < 2. Choose a C-name q such that p0 p

1

q is a commonextension of q0 and q1 in Q and take

r =p0 p

1

{, 1 : W}, q

.

Since W (dom(p0) dom(p1

)) = , r is a condition.

Pick a condition r r from C Q and an ordinal such that rh() =

. Now [,

) because ,

C1. Sincerh(i) = i, h() = and h is an isomorphism,

so {0 , } E and {1 , } / E imply that {0 , } E and {1, } / E.But Di(i , i) = D and D so {0 , } E and {1 , } / E, that is, W. But rran(h) = f1{0} and f1{0} W = , contradiction.


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4.3 G K2

Given a non-trivial graph G = V, E with V [1]1 define

(G) = { 1 : V and |twinG(, V )| }.

The following lemma obviously holds.

Lemma 4.15 If G0 and G1 are graphs on uncountable subsets of 1, G0 =G1, then (G0) = (G1) mod NS1.

Lemma 4.16 Given G K \ K0 and S 1 there is a partition (V0, V1) of

1 such that (G[V0]) S mod NS1 and (G[V1]) 1 \ S modNS1.

Proof: Let be a large enough regular cardinal and fix an increasing,continuous sequence N : < 1 of countable, elementary submodels ofH = H, such that G, S N0 and N : N+1 for < 1.Write = N 1 and C = { : < 1}. Take V0 =

S

(+1 \ ) and

V1 = 1 \ V0 =

1\S(+1 \ ).

It is enough to prove that (G[V0]) S mod NS1 . Assume that (G[V0]), = , , V0 and |twinG[V0](, V0)| = . Since G, ,

V0 N+1 and |G/ G,V0 | , we have tpGG[V0](, V0) N+1and so twinG[V0](, ) N+1 as well. Thus +1 \ . Hence V0implies n = S which was to be proved.

Lemma 4.17 If G K \ K0 and (G) = mod NS1, then G is not quasi-smooth.

Proof: Assume that S = (G) is stationary and let (S0, S1) be a partition ofS into stationary subsets. By lemma 4.16, there is a partition (V0, V1) of 1with (G([Vi]) S Si. Then G[Vi] and G can not be isomorphic by lemma4.15.

Let us remark that G K2 iff G K \ K0 and there is an A [1] and

x 1 \ A such that |[x]G,A| = |1 \ [x]G,A| = 1.Given G K2 we will write G K2 iff there are two disjoint, countable

subsets of 1, A0 and A1, and there is an x 1, such that |[x]G,A0| =|1 \ [x]G,A0 | = 1 and [x]G,A0 \ A1 = [x]G,A1 \ A0.


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Lemma 4.18If G K2, then G (K

2)VC

.

Proof: Assume that A [1] and x 1 witness G K2 in the ground

model. Fix a bijection f : A in V. Let r : 2 be the characteristicfunction of a Cohen real from VC. Take Ai = (f r)1{i}. By a simpledensity argument, we can see that [x]G,A0 = [x]G,A = [x]G,A1 . Thus A0, A1and x show that g K 2.

Lemma 4.19 Assume that every Aronszajn tree is special. If G K2, thenthere is a partition (V0, V1) of 1 such that (G[Vi]) is stationary for i < 2.

Proof: Choose A0, A1 and x witnessing G K2. Let A = A0 A1. TakeC0 = [x]G,A0 \ A, C1 = (1 \ [x]G,A0) \ A and consider the partition trees Ti ofG[Ci] for i 2 (see definition 2.3). These trees are Aronszajn-trees becauseG is non-trivial. Fix functions hi: Ci specializing Ti. We can findnatural numbers n0 and n1 such that the sets Si = { : h

1i {ni} (Ti) = }

are stationary, that is, h1i {ni} meets stationary many level of Ti. TakeBi = h

1i {ni} and Yi = {c Ci : b Bi c Ti b}.

Pick any Si. Let b Bi (Ti). If c Yi \ (Ti)


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Cohen reals in VP+1

has the property that VP2

|=G is not isomorphic toG[Vi] for i < 2.

Finally assume that G (K2)VP

. By lemma 4.18, we have G (K2)VP+1 .

Since P2 satisfies c.c.c, it follows that G (K2)VP2 . So applying lemma 4.19

we can find a partition (V0, V1) of1 such that both (G[V0]) and (G[V1]) arestationary. Thus, by lemma 4.17, neither G[V0] nor G[V1] are quasi-smooth.So G itself can not be quasi-smooth.

References

[1] U. Abraham, S. Shelah, Forcing with stable posets, Journal of Sym-bolic Logic 47 (1982) no 1, 3742

[2] K. J. Devlin, S. Shelah, A weak version of which follows from 2 < 21,Israel J. Math 28 (1978) no 23, 239247

[3] A. Hajnal, Zs. Nagy, L. Soukup, On the number of non-isomorphic sub-graphs of certain graphs without large cliques and independent subsets, toappear in A tribute to Paul Erdos , Oxford University Press

[4] T. Jech, Set Theory, Academic Press. New York, 1978

[5] H. A. Kierstead, P. J. Nyikos, Hypergraphs with Finitely many Isomor-phism Subtypes, preprint.

[6] D. Macpherson, A. H. Mekler, S. Shelah, The number of Infinite Sub-structures, preprint

[7] S. Shelah, Proper Forcing, Springer-Verlag, Berlin Heilderberg New York,1982

s. shelah and l. soukup- on the number of non-isomorphic subgraphs

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