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
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