On the number of non-isomorphic subgraphs

26 Pages·0.24 MB·English

by  Saharon Shelah

4
9 On the number of non-isomorphic subgraphs
9
1

n
S. Shelah ∗
a
J
Institute of Mathematics
5
1 Hebrew University, Jerusalem

]
O L. Soukup †
L Mathematical Institute of the
.
h Hungarian Academy of Sciences
t
a
m
February 1, 2008
[

1
v
0 Abstract
1
LetKbethefamilyofgraphsonω withoutcliquesorindependent
2 1
1 subsets of size ω1. We prove that
0
(a) it is consistent with CHthat every G ∈K has 2ω1 many pairwise
4
9 non-isomorphic subgraphs,
/
h (b) the following proposition holds in L: (∗) there is a G ∈ K such
at that for each partition (A,B) of ω1 either G ∼= G[A] or G ∼=
m G[B],
:
v (c) the failure of (∗) is consistent with ZFC.
i
X
r
a 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 support
iterated forcing for section 4.
∗The first author was supported by the United States Israel Binational Science Foun-
dation, Publication 370
†ThesecondauthorwassupportedbytheHungarianNationalFoundationforScientific
Research grant no. l805
1
February 1, 2008 2
Answering a question of R. Jamison, H. A. Kierstead and P. J. Nyikos
[5] proved that if an n-uniform hypergraph G = hV,Ei is isomorphic to each
of its induced subgraphs of cardinality |V|, then G must be either empty or
complete. 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 = hV,Ei is called non-trivial iff G contains no clique
orindependent subset ofsize|V|. Denotetheclassofallnon-trivialgraphson
ω by K. Let I(G) be the set of all isomorphism classes of induced subgraphs
1
of G = hV,Ei with size |V|.
H. A. Kierstead and P. J. Nyikos proved that |I(G)| ≥ ω for each G ∈ K
and asked whether |I(G)| ≥ 2ω or |I(G)| ≥ 2ω1 hold or not. In [3] it was
shown that (i) |I(G)| ≥ 2ω for each G ∈ K, (ii) under ♦+ there exists a
G ∈ K with |I(G)| = ω . In section 2 we show that if ZFC is consistent, then
1
so is ZFC + CH + “|I(G)| = 2ω1 for each G ∈ K”. Given any G ∈ K we
will investigate its partition tree. Applying the weak ♦ principle of Devlin
and Shelah [2] we show that if this partition tree is a special Aronszajn tree,
then |I(G)| > ω . This result completes the investigation of problem 2 of [5]
1
for ω .
1
Consider a graph G = hV,Ei . We say that G is almost smooth if it
is isomorphic to G[W] whenever W ⊂ V with |V \ W| < |V|. The graph
G 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 models
of ZFC was constructed which contain such graphs on ω . It was also shown
1
that the existence of a non-trivial, quasi smooth graph on ω is consistent
1
with ZFC. But in that model CH failed. In section 3 we prove that ♦+, and
so 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 a
∼
partition(A ,A )ofω whichissobadthatnotonlyG 6= G[A ]intheground
0 1 1 i
model but certain simple generic extensions can not add such isomorphisms
to the groundmodel. Wedivide the class K into three subclasses anddevelop
different methods to carry out our plan.
The question whether the existence of an almost-smooth G ∈ K can be
proved in ZFC is still open.
We use the standard set-theoretical notation throughout, cf [4]. Given a
graph G = hV,Ei we write V(G) = V and E(G) = E. If H ⊂ V(G) we
February 1, 2008 3
define G[H] to be hH,E(G) ∩ [H]2i. 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 H
∼
are isomorphic. If f : V(G) → V(H) is a function we denote by f : G = H
the fact that f is an isomorphism between G and H.
Given a set X let Bij (X) be the set of all bijections between subsets of
p
X. If G = hV,Ei is a graph take
∼
Iso (G) = {f ∈ Bij (V) : f : G[dom(f)] = G[ran(f)]}.
p p
We denote by Fin(X,Y) the set of all functions mapping a finite subset of
X to Y.
Given a poset P and p,q ∈ P we write pk q to mean that p and q are
P
compatible in P.
The axiom ♦+ claims that there is a sequence hS : α < ω i of contable
α 1
sets such that for each X ⊂ ω we have a closed unbounded C ⊂ ω satisfying
1 1
X ∩ν ∈ S and C ∩ν ∈ S for each ν ∈ C.
ν ν
We denote by TC(x) the transitive closure of a set x. If κ is a cardinal
take H = {x : |TC(x)| < κ} and H = hH ,∈i.
κ κ κ
Let us denote by D the club filter on ω .
ω1 1
2 I(G) can be always large
Theorem 2.1 Asume that GCH holds and every Aronszajn-tree is special.
Then |I(G)| = 2ω1 for each G ∈ K.
Remark: S.Shelah proved, [7, chapter V. §6,7], that the assumption of
theorem 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 ∈ [ω ]ω1 and |{G(x)∩A : x ∈ ω | = ω .
1 1 1
Then |I(G)| = 2ω1.
Proof: See [3, theorem 2.1 and lemma 2.13].
Definition 2.3 Consider a graph G = hω ,Ei.
1
February 1, 2008 4
1. For each ν ∈ ω let us define the ordinal γ ∈ ω and the sequence
1 ν 1
ξν : γ ≤ γ as follows: put ξν = 0 and if hξν : α < γi is defined, then
D γ νE 0 α
take
ξν = min{ξ : ∀α < γ ξ > ξν and ({ξν,ξ} ∈ E iff {ξν,ν} ∈ E)}.
γ α α α
If ξν = ν, then we put γ = γ.
γ ν
2. Given ν,µ ∈ ω write ν ≺G µ iff ξν = ξµ for each γ ≤ γ .
1 γ γ ν
3. Take TG = ω ,≺G . T is called the partition tree of G.
D 1 E G
Lemma 2.4 If G = hω ,Ei ∈ K with |I(G)| < 2ω1, then T G is an Aronszajn
1
tree.
Proof: By the construction of T G, if ν,µ ∈ ω , ν < µ and G(ν) ∩ ν =
1
G(µ)∩ν, then ν ≺G µ. So the levels of TG are countable by lemma 2.2. On
the other hand, TG does not contain ω -branches, because the branches are
1
prehomogeneous subsets and G is non-trivial.
Definition 2.5 1. Let F : (2ω)<ω1 → 2 and A ⊂ ω . We say that a
1
function g : ω → 2 is an A-diamond for F iff, for any h ∈ (2ω)ω1,
1
{α ∈ A : F(h⌈α) = g(α)} is a stationary subset of ω .
1
2. A ⊂ ω is called a small subset of ω iff for some F : (2ω)<ω1 → 2 no
1 1
function is an A-diamond for F.
3. J = {A ⊂ ω : A is a small subset of ω }.
1 1
In [2] the following was proved:
Theorem 2.6 If 2ω < 2ω1, then J is a countably complete, proper, normal
ideal on ω .
1
After this preparation we are ready to prove theorem 2.1.
Proof: Assume that G = hω ,Ei ∈ K.
1
|I(G)| < 2ω1 and a contradiction will be derived.
February 1, 2008 5
Since 2ω1 = ω , we can fix a sequence {G : ν < ω } of graphs on ω
2 ν 1 1
such that for each Y ∈ [ω ]ω1 there is a ν < ω with G[Y] ∼= G . Write
1 1 ν
G = hω ,E i.
ν 1 ν
Consider the Aronszajn-tree T G = ω ,≺G . Since every Aronszajn-tree
D 1 E
is special and I is a countably complete ideal on ω , there is an antichain S
1
in TG with S ∈/ J. Take
A = α ∈ ω : ∃σ ∈ S(α ≺G σ) .
n 1 o
Now property (∗) below holds:
(∗) ∀σ ∈ S ∀ρ ∈ (S ∪A)\σ +1
∃α ∈ A∩σ ({σ,α} ∈ E iff {ρ,α} ∈/ E).
Indeed, if for each α ∈ A ∩ σ we had {σ,α} ∈ E iff {ρ,α} ∈ E, then
σ ≺G ρ would hold by the construction of T G.
Let ν ∈ ω , σ ∈ S, T ⊂ S ∩ σ and f : G[(A ∩ σ) ∪ T] → G be an
1 ν
embedding. Define F(ν,σ,T,f) ∈ 2 as follows:
F(ν,σ,T,f) = 1 iff ∃x ∈ G (∀α ∈ A∩σ)({x,f(α)} ∈ E iff {σ,α} ∈ E).
ν ν
In case ωσ = σ, under suitable encoding, F can be viewed as a function
from (2ω)<ω1 to 2.
Since S ∈/ J, there is a g ∈ 2ω1 such that for every ν ∈ ω = 2ω, T ⊂ S
1
∼
and f : G[A∪T] = G , the set
ν
S = {σ ∈ S : g(σ) = F(ν,σ,T ∩σ,f⌈σ)}
T
is stationary. Take T = {σ ∈ S : g(σ) = 0}. Choose an ordinal ν < ω and a
1
∼
function f with f : G[A∪T] = G . For each σ < ω with σ = ωσ it follows,
ν 1
by (∗), that
σ ∈ T iff ∃x ∈ ω ∀α ∈ S ∩σ ({x,f(α)} ∈ E iff {σ,α} ∈ E).
1 ν
Thus g(σ) = 0 iff F(ν,σ,T ∩ σ,f⌈σ) = 1, for each σ ∈ S, that is, S = ∅,
T
which is a contradiction.
February 1, 2008 6
3 A quasi-smooth graph under ♦+
Theorem 3.1 If ♦+ holds, then there exists a non-trivial, quasi-smooth
graph on ω .
1
Proof: Given a set X, A⊂P(X) and F⊂Bij (X) take
p
Cl(A,F) = {B : B ⊃ A and ∀B ,B ∈ B ∀f ∈ F ∀Y ∈ [X]<ω
\ 0 1
{B ∪B ,f′′B ,B △Y}⊂B}.
0 1 0 0
We say that A is F-closed if A = Cl(A,F). Given A,D⊂P(X), we say that
D is uncovered by A if |D\A| = ω for each A ∈ A and D ∈ D.
Lemma 3.2 Assume that F⊂Bij (X) is a countable set, A0, A1⊂P(X) are
p
countable, F-closed families. If D⊂P(X) is a countable family which is
uncovered by A0∪A1, then there is a partition (B ,B ) of X such that D is
0 1
uncovered by Cl(Ai∪{B },F) for i < 2.
i
Proof: We can assume that F is closed under composition. Fix an enumer-
ation {hD ,k ,F ,i ,A i : n ∈ ω} of D×ω×F<ω×{hi,Ai : i ∈ 2,A ∈ Ai}.
n n n n n
By induction on n, we will pick points x ∈ X and will define finite sets, B0
n n
and B1, such that B0 ∩B1 = ∅ and Bi⊂Bi .
n n n n n+1
Assume that we have done it for n−1. Write F = hf ,...,f i. Take
n 0 k−1
B = B0 ∪B1 and
n−1 n−1 n−1
B− = B ∪ f′′B : j < k .
n n−1 [n j n−1 o
Pick an arbitrary point x ∈ D \(A ∪B−). Put
n n n n
Bin = Bin
n n−1
and
B1−in = B1−in ∪{x }∪ f−1(x ) : j < k .
n n−1 n n j n o
Next choose a partition (B0,B1) of X with Bi ⊃ ∪{Bi : n < ω} for i < 2.
n
We claim that it works. Indeed, a typical element of Cl(Ai ∪ {Bi},F) has
the form
C = A∪ f′′Bi : j < k ,
[n j o
February 1, 2008 7
where A ∈ A, k < ω and f ,...,f ∈ F. So, if D ∈ D, then
0 k−1
D\C ⊃ {x : D = D,A = A,i = i and F = hf ,...,f i}
n n n n n 0 k−1
because x ∈/ A and f−1(x ) ∈ B1−i by the constuction.
n j n
Consider a sequence F = hf ,...,f i. Given a family F⊂Bij (X) we
0 n−1 p
say that F is an F-term provided f = f or f = f−1 for some f ∈ F, for
i i
each i < n. We denote the function f ◦ ··· ◦ f by F as well. We will
0 n−1
assume that the empty term denotes the identity function on X. If l ≤ n
take F = hf ,...,f i and F = hf ,...,f i. Let
(l) 0 l−1 (l) l n−1
Sub(F) = f ,...,f : l ≤ n,i < ... < i < n .
nD i0 il−1E 0 l−1 o
Given f ∈ F and x,y ∈ X with x ∈/ dom(f) and y ∈/ ran(f) let Ff,x,y be
the term that we obtain replacing each occurrence of f and of f−1 in F with
f ∪{hx,yi} and with f−1 ∪{hy,xi}, respectively.
Lemma 3.3 Assume that F⊂Bij (X), A⊂P(X) is F-closed, F ,...,F
p 0 n−1
are F-terms, z ,...,z ∈ X, A ,...,A ∈ A such that for each i < n
0 n−1 0 n−1
(∗) z ∈/ {F′′A : F ∈ Sub(F )}.
i [ i i
If f ∈ F, x ∈ X\dom(f), Y ∈ [X\ran(f)]ω with |A ∩ Y| < ω for each
A ∈ A, then there are infinitely many y ∈ Y such that (∗) remains true when
replacing f with f ∪{hx,yi}, that is,
(∗∗) z ∈/ F′′A : F ∈ Sub(Ff,x,y)
i [n i i o
for each i < n.
Proof: It is enough to prove it for n = 1. Write F = hf ...,f i, A = A ,
0 k−1 0
z = z . Take
0
Y = {y ∈ Y : (∗∗) holds for y}.
F,A
Now we prove the lemma by induction on k.
If k = 0, then Y = Y\A. Suppose we know the lemma for k−1. Using
F,A
the induction hypothesis we can assume that (†) below holds:
(†) Y = Y : l ≤ n,G ∈ Sub( Ff,x,y),G 6= Ff,x,y .
\n G,F(′l′)A (l) o
February 1, 2008 8
Assume that |Y | < ω and a contradiction will be derived.
F,A
First let us remark that either f = f or f = f−1 by (†).
k−1 k−1
Case 1:f = f−1.
k−1
Then Y ⊃Y\A by (†), so we are done.
F,A
Case 2:f = f.
k−1
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(x) = Ff,x,y′(x) .
n (i) (i) o
By Ramsey’s theorem, we can assume that l(y,y′) = l whenever y,y′ ∈ Y.
Clearly l < n. Then Ff,x,y(x) 6= Ff,x,y′(x) but Ff,x,y(x) = Ff,x,y′(x), so
(l) (l) (l−1) (l−1)
f = f−1 and Ff,x,y(x) = x for each y ∈ Y. Thus z = Ff,x,y(x) for each
l (l−1) (l−1)
y ∈ Y, which contradicts (†) because x ∈ A.
The lemma is proved.
We are ready to construct our desired graph.
First fix a sequence hM : α < ω i of countable, elementary submodels of
α 1
some H with hM : γ < αi ∈ M for each α < ω , where λ is a large enough
λ γ α 1
regular cardinal.
Then choose a ♦-sequence hS : α < ω i ∈ M for the uncountable sub-
α 1 0
sets of ω , that is , {α < ω : X ∩α = S } ∈/ NS(ω ) whenever X ∈ [ω ]ω1.
1 1 α 1 1
We can also assume that S is cofinal in α for each limit α.
α
We will define, by induction on α,
1. graphs G = hωα,E i with G = G [ωβ] for β < α,
α α β α
2. countable sets F ∈ Iso (G ),
α p α
satisfying the induction hypotheses (I)–(II) below:
(I) {S : γ ≤ α} is uncovered by I ∪J where
ωγ α α
I = Cl({G(ν)∩ν : ν ∈ ωα}, F )
α [ β
β≤α
and
J = Cl({ν\G(ν) : ν ∈ ωα}, F ).
α [ β
β≤α
February 1, 2008 9
To formulate (II) we need the following definition.
Definition 3.4 Assume that α = β +1 and Y⊂ωα. We say that Y is large
if ∀n ∈ ω, ∀hhf ,x i : i < ni, ∀h
i i
if
1. ∀i < n ∃α < β f ∈ F ,
i i αi
2. ∀i < n ωα ≤ x < ωβ,
i i
3. ∀i < n ran(f )⊂Y,
i
4. ∀i 6= j < n ran(f )∩ran(f ) = ∅
i j
5. h ∈ Fin(Y ∩ωβ,2) and dom(h)∩ {ran(f ) : i < n} = ∅,
i
S
then
∃y ∈ Y ∩[ωβ,ωα) such that
6. ∀i < n ∀x ∈ dom(f ) ({y,f (x)} ∈ E iff {x ,x} ∈ E ),
i i α i α
7. ∀z ∈ dom(h) {y,z} ∈ E iff h(z) = 1.
α
Take
(II) If α = β +1, then ωα is large.
The construction will be carried out in such a way that hG : β ≤ αi ∈ M
β α
and hF : β < αi ∈ M .
β α
To start with take G = h∅,∅i and F = {∅}. Assume that the construc-
0
tion is done for β < α.
Case 1:α is limit.
We must take G = ∪{G : β < α}. Wewill define sets F0,F1⊂Iso (G )
α β α α p α
and will take F = F0 ∪F1.
α α α
Let
F0 = {f ∈ Iso (G )∩M : ∃hα : n < ωi⊂α sup{α : n < ω} = α,
α p α α n n
∼
f⌈ωα ∈ F and f⌈ωα : G = G [ran(f)] for each n ∈ ω}.
n αn n αn αn
Take F− = Fβ ∪ F0, I− = I and J− = J . Clearly F−⊂M
α α β α β α
βS<α βS<α βS<α
with F− ∈ M , so M |= “|F−| = ω”. Obviously both I− and J− are
α+1 α+1 α α
F−-closed and S = {S : β ≤ α} is uncovered by them.
ωβ
February 1, 2008 10
¿From now on we work in M to construct F1. For W⊂ωα write
α+1 α
L = {ν < α : W ∩(ων +ω) is large}.
W
Take
W = {hW,fi ∈ (P(ωα)∩M )×(∪ F ) : L is cofinal in α
α α β<α β W
∼
and f : G = G [W ∩ωγ ] for some γ < α .
γf γf f f o
We want to find functions gW,f ⊃ f for hW,fi ∈ W such that
α
(A) gW,f : G ∼= G [W]
α α
(B) takingF1 = gW,f : hW,fi ∈ W theinductionhypothesis(I)remains
α n αo
true.
First we prove a lemma:
Lemma 3.5 If hW,fi ∈ W , g ∈ Iso (G ,G [W]), g ⊃ f, |g\f| < ω, then
α p α α
(i) for each x ∈ W\dom(f) the set
{y ∈ W : g ∪{hx,yi} ∈ Iso (G ,G [W])}
p α α
is cofinal in ωα.
(ii) for each y ∈ W\ran(f) the set
{x ∈ W : g ∪{hx,yi} ∈ Iso (G ,G [W])}
p α α
is cofinal in ωα.
Proof: (i): Define the function h : ran(g)\ran(f) → 2 with h(g(z)) = 1
iff {z,x} ∈ E . Choose β ∈ L with ran(h)⊂ωβ and γ ≤ β. Since
α W f
W ∩(ωβ +ω) is large, we have a y ∈ W ∩[ωβ,ωβ +ω) such that
1. {y,f(z)} ∈ E iff {x,z} ∈ E for each z ∈ dom(f)
α α
2. {y,g(z)} ∈ E iff h(g(z)) = i for each z ∈ dom(g)\dom(f).
α