Saharon Shelah- Universality Among Graphs Omitting a Complete Bipartite Graph

8/3/2019 Saharon Shelah- Universality Among Graphs Omitting a Complete Bipartite Graph

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UNIVERSALITY AMONG GRAPHS OMITTING A COMPLETE

BIPARTITE GRAPH

SH706

SAHARON SHELAH

Abstract. For cardinals ,, we consider the class of graphs of cardinality which has no subgraph which is (, )-complete bipartite graph. The questionis whether in such a class there is a universal one under (weak) embedding. Wesolve this problem completely under GCH. Under various assumptions mostlyrelated to cardinal arithmetic we prove non-existence of universals for thisproblem. We also look at combinatorial properties useful for those problems

concerning -dense families.

Date: May 5, 2010.Key words and phrases. Set theory, graph theory, universal, bipartite graphs, black boxes.The author thanks Alice Leonhardt for the beautiful typing. This research was supported by

the United States-Israel Binational Science Foundation. Done 1,2 - 5/99; 3 in ?/99First Typed - 12/16/99.

1


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Anotated Content

0 Introduction

1 Some no we-universal

[We define Pr(, ) and using it gives sufficient conditions for the non-existence of we-universal in H,, mainly when = +. Also we givesufficient conditions for no ste-universal when = , 2 .]

2 No we-universal by Pr(, ) and its relatives

[We give finer sufficient conditions and deal/analyze the combinatorial prop-erties we use; Pr(, ) says that there are partial functions f from to for < , which are dense, = otp(Dom(f)) and > otp(Dom(f) Dom(f)) for = .]

3 Complete characterization under G.C.H.

[Two theorems cover G.C.H. - one assume is strong limit and the otherassume = + = 2 + 2


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UNIVERSALITY AMONG GRAPHS OMITTING A COMPLETE BIPARTITE GRAPH SH706 3

We give many sufficient conditions for the non-existence of universals (mainly we-universal) and some for the existence, for this dealing with some set-theoretic prop-

erties. Mostly when we get no G H,, is we/ste-universal we, moreover, getno G H,, is we/ste-universal among the bipartite ones. Hence we get alsoresults on families of bi-partite graphs. We do not look at the case < 0 here.

Rado has proved that: if is regular > 0 and 2


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4 SAHARON SHELAH

(a) f is a one-to-one function from G1 into G2; pedantically from VG1 into

VG2

(b)st for , G1 we haveRG1 f()RG2f().

3) we say f is a weak embedding of G1 into G2 if

(a) above and

(b)we for , G1 we haveRG1 f()RG2f().

4) The -complete graph K is the graph (, R) were R = or any graphisomorphic to it.

{0.2}Definition 0.4. 1) G is a bipartite graph means G = (U,V,R) = (UG, VG, RG)where U, V are disjoint non-empty sets, R U V. For a bipartite graph G, we

would like sometimes to treat as a usual graph (not bipartite), so let G as a graph,G[gr], be (UG VG, {(, ) : , V U and RG RG}). The cardinalityof G is (|UG|, |VG|) or |UG| + |VG|.2) We say f is a strong embedding of the bipartite graph G1 into the bipartitegraph G2 if :

(a) f is a one-to-one function from UG1 VG1 into UG2 VG2 mapping UG1

into UG2 and mapping VG1 into VG2

(b) for (, ) UG1 VG1 we haveRG1 f()RG2f().

3) We say f is a weak embedding of the bipartite graph G1 into the bipartite graphG2 if

(a) f is a one-to-one function from UG1 VG1 into UG2 VG2, f mapping UG1

into UG2 and mapping VG1 into VG2

(b) for (, ) UG1 VG1 we haveRG1 f()RG2f().

4) In parts (2), (3) above, if G1 is a bipartite graph and G2 is a graph then we

mean G[gr]1 , G2.

5) The (, )-complete bipartite graph K, is (U,V,R) with U = {i : i < }, V ={ + i : i < }, R = {(i, +j) : i < , j < }, or any graph isomorphic to it.

{0.3}Definition 0.5. 1) For a family H of graphs (or of bipartite graphs) we say G isste-universal [or we-universal] for H iff every G H can be strongly embedded [orweakly embedded] into G.2) We say H has a ste-universal (or we-universal) if some G H is ste-universal (orwe-universal) for H.

{0.4}Definition 0.6. 1) Let H,, = H

gr,, be the family of graphs G of cardinality

(i.e. with nodes) such that the complete (, )-bipartite graph cannot be weaklyembedded into it; gr stands for graph.

2) Let Hbp,,

be the family of bipartite graphs G of cardinality such that the

complete (, )-bipartite graph cannot be weakly embedded into it. If = (, )we may write (similarly in (3)); bp stands for bipartite.


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3) Let Hsbp,{,}

= Hsbp,,

be the family of bipartite graphs G of cardinality such

that K, (the (, )-complete bipartite graph) and K, (the (, )-complete bi-

partite graph) cannot be weakly embedded into it; sbp stands for symmetricallybipartite.

4) H = Hgr is the family of graphs of cardinality and H

bp

is the family of bipartite

graphs of cardinality .{0.5}

Observation 0.7. 1) The following are equivalent:

(a) in Hsbp,, there is a we-universal

(b) in {G[gr] : G Hsbp,,} there is a we-universal.

2) Similarly for ste-universal.3) If G is ste-universal for H then it is we-universal for H (in all versions).4) Assume that for every G H,, there is a bipartite graph from H,, not x-

embeddable into it, then in Hbp,, and in Hbp,, and in Hsbp,, there is no x-universalmember; for x {we, ste}.

Proof. (1) (a) (b): Trivially.

(b) (a): Assume G is we-universal in {G[gr] : G Hsbp,,} and let Ai : i < i

be its connectivity components. Let Ai be the disjoint union of Ai,0, Ai,1 with

no G-edge inside Ai,0 and no G-edge inside Ai,1 (exists as G = G[gr] for some

G Hsbp,,, note that {Ai,0, Ai,1} is unique as G Ai is connected). Let A

m,i,

for i < i, < 2, m < 2, < be pairwise disjoint sets with |Am,i, | = |Ai,k| when

m + = k mod 2. Let G be the following member ofHbp,,: let UG be the disjoint

union of A,i, for i < i, < 2, < and VG

be the disjoint union of A1,i, for

i < i, < 2, < and RG

= {Ri, : i < i, < 2} where Ri, are chosen suchthat (A0,i,0 , A

1,i,1 , R

i,0)

= (Ai,0, Ai,1, RG Ai,0 Ai,1) = (A

0,i,1 , A

1,i,0 , R

i,1). Easily

G Hbp,, is we-universal.2) The same proof.3), 4) Easy. 0.7


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6 SAHARON SHELAH

1. Some no we-universal

We show that if = 2 then in H,, there is no ste-universalgraph (in 1.5); for we-universal there is a similar theorem if = +, Pr(, ), see1.4 + Definition 1.2, (this holds when = = cf(), S ).

{1.1}Claim 1.1. Assume is regular and S+

(see Definition 1.2 below). Then there

is no we-universal inH+,+,.{1.2}

Definition 1.2. 1) For regular < let S = { < : cf() = cf()}.2) For regular and stationary subset S of let S means that for some A = A : S, limit we have

(a) A is an unbounded subset of

(b) if A is an unbounded subset of then for some (equivalently stationarily

many) S we have A A.

2A) For < let ,

mean that for some family A [] of cardinality we have

(B [])(A A)(A B).3) Pr(, ) for cardinals > means that some F exemplifies it, which means

(a) F is a family of functions

(b) every f F is a partial function from to

(c) if f F then = otp(Dom(f)) and f strictly increasing

(d) f = g F > |Dom(f) Dom(g)|

(e) if g is a partial (strictly) increasing function from to such that Dom(g)has cardinality , then g extends some f F.

4) Pr(, ) is defined similarly for a limit ordinal but clauses (c) + (d) are replacedby

(c) if f F then = otp(Dom(f)) and f is one to one

(d) if f = g F then Dom(f) Dom(g) is a bounded subset of Dom(f) andof Dom(g).

{1.2A}Observation 1.3. 1) We have

(i) = cf() and S Pr(, )

(ii) Pr(, ) Pr(, ) ,

(iii) for any cardinal we have Pr(, ) Pr(, ).

2) If we weaken clause (c) of 1.2(3) to

(c) f F |Rang(f)| = = |Dom(f)|

we get equivalent statement (can combine with 1.3(3)).3) The one to one in Definition 1.2(4), clause (c) is not a serious demand, thatis, omitting it we get an equivalent definition.


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Proof. Easy.1) E.g., clause (iii) holds because for any one to one f : Ord, for some A []

the function f A is strictly increasing (note it first to regular ).2) Left to the reader.3) Why? Let pr: be 1-to-1 onto and pr1,pr2 : be such that = pr(pr1(), pr2()).Let F be as in the Definition 1.2(4) old version. Then {pr1 f : f F} willexemplify the new version, i.e. without the 1-to-1.

For the other direction, just take {f F : f is one to one}. 1.3

Proof. Proof of 1.1 It follows from 1.4 proved below as S

+

easily implies Pr(+, )

for regular (see 1.3(1)). 1.1{1.3}

Claim 1.4. If Pr(, ), so > then inH,+, there is no we-universal.Moreover, for every G H,+, there is a bipartite G H,+, of cardinality

not we-embeddable into it.Proof. Let G be a given graph from H,+,; without loss of generality V

G = .For any A let

()0 (a) Y0A =: { < : is G-connected with every A}

(b) Y2A =: { < : is G-connected with members of A}

(c) Y1A =: { < : is G-connected with every A except possibly

< of them}.Clearly

(d) A B Y0A Y0B and Y

2A Y

2B and Y

1A Y

1B

(e) |A| Y0A Y1A Y

2A.

We now note

()1 if A [] then |Y0A| .

[Why? Otherwise we can find a weak embedding of the ( , +)-complete bipartitegraph into G]

()2 if A [] then |Y1A| .

[Why? If not choose pairwise disjoint subsets Ai of A for i < each of cardinality, now easily Y1A |{i < : / Y

0Ai

}| < so Y1A i

then for some i < , Y0Ai has cardinality > , contradiction by ()1.]Let F = {f : < } exemplify Pr(, ). Now we start to choose the bipartite

graph G:

0 UG = , VG = , RG =


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8 SAHARON SHELAH

[Why? As for any (, ) VG the set { < : RG(, )} is equal to Dom(f)which has cardinality which is < +; note that we are speaking of weak embedding

as bipartite graph, side preserving]

3 the (, +)-complete bipartite graph cannot be weakly embedded into Gprovided that for each < ,3 K,+ cannot be weakly embedded into (U

G, VG, RG ).

[Why? Let U1 UG, V1 V

G have cardinality , + respectively and let V1 ={ : (, ) V1 for some }. If |V

1 | 2 choose 91, 1), (2, 2) V1 such that

1 = 2, so { < : (, (, )) RG for = 1, 2} include U1 hence by the

definition of RG we have |Dom(f1) Dom(f2)| |U1| = , but 1 = 2 |Dom(f1) Dom(f2)| < by clause (d) of Definition 1.2(3) - think.]

4 G cannot be weakly embedded into G provided that for each < :

4 there is no weak embedding f of (UG, VG, RG ) into G

extending f.

[Why? Assume toward contradiction that f is a one-to-one mapping from UG VG

into VG

= mapping edges of G to edges of G. So f UG is a one-to-onemapping from to hence by the choice of F = {f : < } to witness Pr(, )see clause (e) of Definition 1.2(3) there is such that f f UG. So clearlyRG(, ) and Dom(f) implies {f(), f(, )} = {f(), f((, ))} EG

,

hence (, (, )) RG {f(), f((, ))} EG . This clearly contradicts 4

which we are assuming.]So we are left with, for each < , choosing R {(, (, ) : Dom(f), for every B [B].

Case 2: For some < 2, |Y2B | > and for some Z:

(i) Z Y2B

\Y1B

(ii) |Z|

(iii) for every 0 (Y2B

\Y1B

)\Z there is 1 Z such that > |{ B :

is G-connected to 0 but is not G-connected to 1}|.

So we choose such = () < 2, Z = Z and then we choose a sequence B, : Z such that:


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6 (a) B, is a subset of B

(b) |B,| =

(c) 1 = 2 Z B,1 B,2 =

(d) is not G-connected to any B,

(this is possible as Z / Y1B).

Now we can find a sequence C, : < + satisfying

7 () C, B

() |C,| = moreover Z |C, B,| =

() for < we have |C, C,| <

(e.g. if = cf() > 0 by renaming B = , each B, is stationary, choose

nonstationary C, inductively on ; if > cf() reduce it to construction onregulars, if = 0 like = cf() > 0).

Lastly we choose R = {(, (, )) : A, < + and f() C,}.Now 3 is proved as in the first case, as for

4, if f is a counter-example then

clearly for < +, f((, )) Y2B , so as |Y2B

| > by 5 and |Z| and

|Y1B

| (by ()2) necessarily for some < +, 0 =: f((, )) Y2B\Y1

B\Z.

Let 1 Z be as guaranteed in clause (iii) in the present case. Now 0 is G-

connected to every member of C, as 0 = f((, )). Hence 0 is G-connected

to members of B,1 (see clause () above and the choice of R); but 1 is notG-connected to any member of B,1 (see clause (d) above). Reading clause (iii),we get contradiction.

Case 3: Neither Case 1 nor Case 2.Recall that is fixed. For {0, 1} we choose Z, by induction on <

+,such that

8 (a) Z, a subset of Y

2B

of cardinality

(b) Z, is increasing continuous in

(c) Y1B

Z,0

(d) if = + 1 then there is , Z,\Z

,\Y

1B

such that

for every Z,\Y1B

we have

= |{ B : is G-connected to , but not to

}|

(e) if = + 1 and Z, hence = |{ B : is connected to }|

(e.g. = ,), then Y1{B: is G-connected to } is included in

Z,

(f) Z0, (Y2B0

Y2B1) = Z1, (Y

2B0

Y2B1).

Why possible? For clause (c) we have |Y1B

| by ()2, for clause (d) note that

not Case 2 trying Z,\Y1B

as Z, and for clause (e) note again |Y1{B: is G-connected to ,}|

by ()2.Having chosen Z, : <

+, < 2, we let


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10 SAHARON SHELAH

R = {(, (, )) : for some < 2 we have:

A and f() is G-connected to = ,2+ so < +}.

Now why 3 holds? Otherwise, we can find A A, |A| = and B +, |B| = +

such that A and B R(, (), ) where = () mod 2, so for some

< 2 we have |A A| = , and let

A = {f() : A A}.

Easily |B| = + and |A| = and A and B RG

,, contradiction to

K,+ is not weakly embeddable into G.

Lastly, why 4 holds? Otherwise, letting f be a counterexample, let < +

and < 2. Clearly f((, )) is G-connected to every B which is G-

connected to ,2+ hence f((, )) cannot belong to Z,2+\Y

1B

(by the demand

in clause (d) of8), but it has to belong to Z,2++1 (by clause (e) of8), so

f((, )) (Z,2++1\Z,2+) Y

1B

. Putting together = 0, 1 we get f((, ))

((Z0,2+1\Z0,2) Y

1B0

) ((Z1,2+2\Z1,2+1) Y

1B1

) hence f((, )) Y1B0 Y1B1

,

but |Y1B

| < +; as this holds for every < + this is a contradiction to f is one

to one. 1.4{1.4}

Claim 1.5. 1) Assume 2 and = (e.g. = 2).Then in H,, there is no ste-universal (moreover, the counterexamples are

bipartite).2) Assume Pr(, ), , 2 . Then the conclusion of (1) holds.

Proof. 1) By the simple black box ([Sh:300, Ch.III,4]) or [Sh:e, Ch.VI,1], i.e.

[Sh:309]) there is f = f :

, f a function from { i : i < } into suchthat for every f : > for some we have f f.

Let G H,, and we shall show that it is not ste-universal in H,,, withoutloss of generality VG

= . For this we define the following bipartite graph G:

1 (i) UG = > and VG =

(ii) RG = {RG : and f is a one-to-one function} where

RG {( i, ) : i u} where u is defined as follows

2 for we choose u such that if possible

(),u for no < do we have (i < )[f( i)RG i u].

If for every for which f is one to one for some u we have (),u holds,then clearly by we are done.

Otherwise, for this , f is one to one and: there is u < satisfying(i < )(f( i)RG

u i u) for every u . But then A =: {f( 2i) : i cf() in 2.10.

{2.1}Convention 2.1. 0.

{2.2}Claim 2.2. If is a limit cardinal and Pr(, ), then there is no we-universalgraph inH,, even for the class of bipartite members.

Proof. Like the proof of 1.4, except that we replace cases 1-3 by:for every < we let R = {(, (, )) : Dom(f) and < |Y0Dom(f)|

+}.

Now 3 holds as |Y

0Dom(f)| < (by ()1 there) hence |Y

0Dom(f)|

+

< as is alimit cardinal. Lastly 4 holds as for some we have f f hence the functionf maps {(, ) : < |Y0Dom(f)|

+} into Y0Dom(f) but f is a one to one mapping,contradiction. 2.2

Recall{pr.22}

Definition 2.3. For a cardinal and a limit ordinal , Pr(, ) holds when forsome F:

(a) F a family of functions

(b) every f F is a partial function from to

(c) f F otp(Dom f) = , and f is one to one

(d) f, g F, f = g (Dom f) (Dom g) is a bounded subset of Dom(f) andof Dom(g)

(e) if g : is a partial function, one to one, and |Dom g| = , then gextends some f F.

{2.4}Claim 2.4. Assume

(a) Pr(, ), = , ordinal product 1

(b) = +.

Then there is no we-universal inHgr,, even for the class of bipartite members.

Proof. Let G H,, and we shall prove it is not we-universal; let without loss ofgenerality VG

= .Let F be a family exemplifying Pr(, ), let F = {f : < } let A =

Dom(f) and let it be {,,i : i < , < } such that [,(1),i(1) < ,(2),i(2) (1) < (2) ((1) = (2) and i(1) < i(2))] and for i < let A,i = {,,i : < },so clearly

()1 A,i [] and (1, i1) = (2, i2) |A1,i1 A2,i2 | < .

1this is preserved by decreasing


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12 SAHARON SHELAH

[Why the second assertion? As {,,i : < } is an unbounded subset of A (oforder type ).]

For (, i) let f,i = f A,i let B,i = Rang(f,i) so |A,i| = |B,i| = and let Y0,i = { < : is G

-connected to every member of B,i}, so as

G H,, clearly |Y0,i| < . As = + > , clearly for each < for some

< we have2: X := {i < : |Y

0,i| } has cardinality . As < also

:= + is < + = , so + = |X|. We choose by induction on <

+ an

ordinal i, X such that:

()2 i, / {i

, : < }.

Recall that is a successor, hence a regular cardinal. So if S+ = { <

+ :

cf() = } recalling = + > |Y0,i,

| there is < such that (Y0,i, {Y0, :

< } {Y0, : < }). Let g() be the first ordinal having this property, so g

is a well defined function with domain S+

. Clearly, g is a regressive funciton.By Fodors lemma for some stationary S S

+ , and for some B

of cardinality

we have S B Y0,i,

{Y0,j : j < i,}, in fact: B

= {Y

0,i,

:

< } where g S is constantly is O.K., we can decrease B but immaterial

hereNow

()3 for = from S there is no < such that: is G-connected to every B,i

,

is G-connected to every B,i,

is not in B.

Let (, j) : j < + list S in an increasing order. Let G be the bipartite graph

UG =

VG =

RG =

(, (, )) : < and < + and A,i,(,2) A,i

,(,2+1)

.

1 G is a bipartite graph of cardinality

2 the (, )-complete bipartite graph ( Hbp(,)

) cannot be weakly embedded

into G.

[Why 2 holds? So let ((1), (1)) = ((2), (2)) belongs to VG, the set set { UG : connected to ((1), (2)) and to ((2), (2))} is included in

(1),(2){0,1}

(A(1),i((1),2(1)+(1)

A(2),i((2),2(2)+(2)

) which is the union of four sets each of cardinal < (by ()1)

hence has cardinality < .]

3 the (, )-complete bipartite graph cannot be weakly embedded into G.

2in fact by 2.2 without loss of generality is a successor cardinal, so without loss of generality+ =


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UNIVERSALITY AMONG GRAPHS OMITTING A COMPLETE BIPARTITE GRAPH SH70613

[Why? Toward contradiction assume U1 UG, V1 V

G have cardinality , respectively and U1 (, ) V1 R(, ).

Let (, ) V1, clearly U1 RG

(, ) A,i,(,2) A,i,(,2+1) .So U1 Ai

,(,2) Ai

,(,2+1).

Now if (1, 1), (2, 2) V1 and 1 = 2 then i,j < |A1,i A2,j| < by the choice of F, so necessarily for some < we have V1 {} . But if(, 1) = (, 2) V1 then (, 1), (, 2) has no common neighbour, contradiction.

4 there is no weak embedding f of G into G.

[Why? Toward contradiction assume that f is such a weak embedding. By thechoice of F and f : < we can choose < such that f fUG. Asf is a weak embedding < < RG(, ) f()RG

f(, ), hence

i

A,i = Dom(f) < RG(, ) f()R

Gf(, ). Hence if <

then f(, ) is G-connected to every B,(,2) B,(,2) {,} hencef(, ) Y0,(,2) Y

0,(,2+1) which implies that f(, ) B

. So the function

f maps the set {(, ) : < } into B. But f is a one-to-one function and B

has cardinality < , contradiction. 2.4{2.4A}

Definition 2.5. 1) For < and < we define Ps(, ) and Ps(, ) similarlyto the definition of Pr(, ), Pr(, ) in Definition 1.2(3),(4) except that we replaceclause (e) by

(e) if g is a one to one function from to , then g extends some f F

(so the difference is that Dom(g) is required to be ).2) Let Pr3(,,,) means that for some F:

(a) F a family of partial functions from to

(b) |F|

(c) f F otp(Dom(f)) =

(d) if g (f F)(f g).{2.5}

Claim 2.6. 1) Assume is strong limit, > and cf() > cf().Then Pr(, ) holds if < has cofinality cf().2) If = + = 2, cf() = cf(), < then Pr(, ) holds.3) If < is a limit ordinal and = || then Ps(, ).4) If = cf(), < < , = and Pr3(,,,) for every < , thenPs(, ).5) Pr(, ) Ps(, ), Pr(, ) Ps(, ) and similarly with instead of .

6) If 2 then = U() Pr(, ).7) If Ps(, ) then Ps(, ).

Remark 2.7. Recall U() = UJbd (), and for an ideal J on , UJ() = Min{|P| :P [] is such that for every f for some A P we have {i < : f(i) A} = mod J}.

Proof. 1) Let i : i < cf() be increasing continuous with limit such that < 0 and 2

i < i+1, hence for limit , is strong limit cardinal of cofinality

cf(). For Scf()cf() = { < cf() : cf() = cf()}, let f, : < 2

list the


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14 SAHARON SHELAH

partial one-to-one functions from to with domain of cardinality . We chooseby induction on < 2 a subset A, of Dom(f,) of order type

unbound in

such that < sup(A, A,) < ; possible as we have a tree with cf()levels and 2 cf()-branches, each giving a possible A, Dom(f,) and eachA,( < ) disqualifies + || of them.

Now F = {f,A, : Scf()cf()

and < 2} is as required because if f is

a partial function from to such that |Dom(f)| = and f is one to one then{ < cf() : ( i < )(i Dom(f) f(i) < )} contains a club of cf().2) This holds as S for every stationary S { < : cf() = cf()}, see [Sh:108],and without any extra assumption by [Sh:922].3) By the simple black box (see proof of 1.5, well it was phrased for but the sameproof, and we have to rename , )to :

()4 if i < , < and Dom(gi,(i)) then f0 (( i, )) = gi,(i)().

Let h be a one to one function from (>) onto such that (if cf() thenalso in ()5(b) we can replace = by < < h((, )) < h(, )

(b) > < < Dom(gg(),(g())) h((, )) =h((, )).

Let f be the following partial function from to satisfying f() = f0 (h1())

so it suffices to prove that F = {f : and f is one-to-one} exemplifiesPs(, ).

First, clearly each f is a partial function from to . Also for each i < and

< the function gi, has domain of order type i+1

i , hence by ()5(a) also

i < Dom(f{h(i, ) : < } has order type i+1 i . By ()5(b)

also Dom(f) has order type =i

(i+1\i ).

Now if f F then f is one-to-one by the choice of F. Second, let f : be a one-to-one function and we shall prove that for some , f F f f.We choose i

i by induction on i such that j < i j = ij andi

f{h(j, ) : j < i, Dom(gj,i(j))} f.

For i = 0 and i limit this is obvious and for i = j + 1 use ()+3 . So and

f f hence is one-to-one hence f F so we are done.


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5) Easy (recalling 1.3(3)).6) Easy.

7) Easy because iff : is one-to-one then for some u of order type , fuis increasing (trivial if is regular, easy if is singular). 2.6

{2.5A}Claim 2.8. 1) Each of the following is a sufficient condition to Pr3(,,,),recalling Definition 2.5(2):

(a) || = = >

(b) = > || and (1 < )(||1 < )

(c) = (||).

2) If1 2, 1 2, 1 2, 1 2 then Pr3(1, 1, 1, 1) implies Pr3(2, 2, 2, 2).

Proof. 1) If clause (a) holds, this is trivial, just use F = {f : f a partial functionfrom to with = otp(Dom(f))}. If clause (b) holds, note that for everyf , for some i1, i2 < we have otp({j < i1 : j < and f(j) < i2}) andlet F = {f : f a partial function from to with bounded range and boundeddomain if = such that = otp(Dom(f))}. If clause (c) holds, use [Sh:460].2) Trivial. 2.8

{2.5B}Claim 2.9. 1) In 1.4, 1.5(2) and in 2.2 we can weaken the assumption Pr(, ) toPs(, ).2) In 2.4 we can weaken the assumption Pr(, ) to Ps(, ).

Proof. The same proofs.We can get another answer on the existence of universals. 2.9

{2.5C}Claim 2.10. If is strong limit, cf() > cf() and > ( ), then inH,,

there is no we-universal member even for the class of bipartite members.Proof. Let := + (recalling is a limit cardinal) by 2.6(1) we have Pr (, )hence by 2.4 we are done. 2.10

Note that Ps may fail.

Claim 2.11. Assume < , cf() cf() and < ||cf() < .Then Ps(, ) fail (hence also Pr(, ), Pr(, ), Ps(, ).

Proof. Toward contradiction let F witness Pr(, ) so F is of cardinality . Letf : < list F; choose an increasing sequence i : i < cf() such thati = (i)cf() and = {i : i < cf()}. We choose Ui by induction on i such that:

()1i (a) Ui (b) i Ui

(c) |Ui| = i

(d) if f F and Dom(f) Ui is unbounded in Dom(f), then Dom(f) Rang(f) .

For clause (d) note that if Dom(f) Ui is unbounded in Dom(f) then there isu Dom(f) Ui unbounded in Dom(f) of order-type cf() and such u determinesf in F uniquely


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16 SAHARON SHELAH

Now choose f : such that f maps [

j


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3. Complete characterization under GCH

We first resolve the case is strong limit and get a complete answer in 3.5 bydividing to cases (in 3.1, 3.2, 3.4 and 2.10), in 3.4 we deal also with other cardinals.This includes cases in which there are universals (3.1, 3.2) and the existence ofwe-universal and of ste-universal are equivalent. In fact in 3.4 we deal also (in part(2)) with another case: = cf(), =


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18 SAHARON SHELAH

G H|G|,,? Toward contradiction assume A0, A1 B, A0 A1 RG and

{|A0|, |A1|} = {, }. If < 2 and cf(|A|) = cf(i) then for some j < i, |ABj | =

|A| so without loss of generality A Bj , but then by clause (b) no x B\Bjis G-connected to members ofBj and |A| Min{, } = , hence A1 Bj , so we get contradiction to the induction hypothesis. So the remaining caseis cf(|A0|) = cf(i) = cf(|A1|) hence cf() = cf() = cf(i); so as we are assumingcf() cf() cf(i), we necessarily get cf(i) = cf() = cf() = cf(). By thelast assumption of the claim (i.e. < cf() > cf()) we get that < andwithout loss of generality |A0| = , |A1| = , so for some j < cf() we have j < iand |A1 Bj | , so as above A0 Bj , hence again as above A1 Bj andwe are done.]

We let G = {G : T}. Now we shall check that G is as required. First

assume toward contradiction that A, B VG

and A B RG

and {|A|, |B|} ={, }. A set C VG

will be called A-flat if it is included in some B, T T.

Easily above if B is not A-flat then A is A-flat. So without loss of generality forsome T we have A B but then x G\B (for some i(x) < and Ti(x) we have x B\B) |{y B : x is G

-connected to y}| < , so as

we get B B , hence A, B VG and we get contradiction to clause (d) of.So K, does not weakly embed into G

; also |VG

| = so G H,,. Lastly, theste-universality follows from the choice ofVGi : i < for any G H,,. That is,we can choose by induction on i, i Ti and an isomorphism fi from G VGi ontoGi if i > 0, with fi increasing continuous (and i increasing continuous) using forsuccessor i clause (c) of. 3.1

In the previous claim we dealt with the case of , < . In the following claimwe cover the case of = :

{2.7

} Claim 3.2. Assume is strong limit, cf() cf() < = ; hence issingular.Then inH,, there is a ste-universal member.

Proof. Similar to the previous proof and [Sh:26, Th.3.2, p.268]. Let = cf()and = i : i < , Ti : i , T

s, T be as in the proof of 3.1. For any graphG H,, let hG : (VG) be defined by: if|{x : < }| = then hG(x) is thefirst i < such that i |{y VG : y is G-connected to every xi, i < = g(x)}|,otherwise hG(x) is not defined. Now we choose VGi : i < as an increasingcontinuous sequence of subsets of VG with union VG such that if i < then|VGi | i and x

(VGi+1) |Rang(x)| = hG(x) i + 1 (y VG)(y is

G-connected to xi for every i < y VGi+1).

Then when (as in the proof of 3.1) we construct G : Ts

we also constructh : Ts, h : (B) with the natural demands. In the end we have tocheck that K, is not strongly embeddable into G; if cf() = we need to lookat slightly more (as in the end of the proof of 3.1). 3.2

Remark 3.3. More generally see [Sh:829].{2.8}

Claim 3.4. 1) Assume is strong limit, = , cf() = cf(), then inH,,there is no we-universal graph, even for the members from Hsbp,,.

2) Assume


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(a) =

(b) ( < )( < )[|||| ], (recall = )

(c) cf() cf().

Then inH,, there is no we-universal graph, even for the Hsbp,,.

Proof. 1) By part (2).2) Let = cf(). Let i : i < be (strictly) increasing with limit = .

Without loss of generality

|i+1 i| is (finite or) a cardinal (< ) with cofinality = cf().

[Why? If is a limit cardinal, trivial as if a cardinal < fails its successor is O.K.;if is a successor cardinal, i = i is O.K. and also i = i or i = 1i is O.K.]

Given a graph G in H,, without loss of generality VG

= .

For i < let Ti = {f : f is a partial one-to-one mapping from i into = VG

with bounded range such that j < i {0, 1} |j+1 j|/2 = 1, j j+1and mod 2}| and 2, 2+1 Dom(f) f(2)RG

f(2+1)}. Let T =i


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20 SAHARON SHELAH

If (U1 (T+ )) = = (V1 (T

+ )) then we can choose (1, 1) U,(2, 2) V1. [Why? As T+ sup Rang() < , see ()2 recalling i = g() < ,

Rang() i < .]

(c) |UG| |T1| = and |VG| |T2| =

hence

(d) |UG| = = |VG| so

(e) G Hsbp,,.

[Why? Being bipartite is obvious; so toward contradiction assume (U1 UG

V1 VG) (U1 V

G V1 UG) have cardinality (recall that = ) and

U1 V1 RG. If (, ) V1 and i < cf() = be such that < i, then U1

{ : } {(, ) UG VG : = and < i} which clearly has cardinality

< , contradiction. Hence V1 (T+ ) = and by symmetry U1 (T+ ) = ,hence V1 T, U1 T hence U1 V1 (set of unordered pairs) is disjoint to R,contradiction.]

(f) G is not weakly embeddable into G.

[Why? Iff is such an embedding, we try to choose by induction on i < , a memberi of Ti, increasing continuous with i such that ( Dom(i))(j < i)[j cf() by 2.10 the family H,, has no we-universal.If = and cf() > cf() then Pr(, ) holds by 2.6(1) (and recall that Pr(, )

is equivalent here to Pr(, ), since is a cardinal, see 1.3(1)(iii)) hence by 2.2 thefamily H,, has no we-universal member.

If cf() < cf() and < by 3.1 the family H,, has a ste-universal member;the second statement in Theorem 3.5 holds as: if < easy, if then = hence cf() < cf() = cf().

If cf() < cf() (hence = so < ) and = (so < ) by 3.2 the familyH,, has a ste-universal member.


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So the remaining case is cf() = cf(). If < < by 3.1 in H,, there is anste-universal; if = by 3.4(1) in H,, there is no we-universal member. If

< = by 3.2 in H,, there is a ste-universal member. We have checked allpossibilities hence we are done. 3.5

We turn to successor cardinal. In the following case, possibly the existenceof we-universal and ste-universal are not equivalent, see 3.12(2) + 3.12(4) andTheorem 3.15.

{2.10}

Claim 3.7. Assume ( 0 and)

(a) = +

(b) < and =

(c) = .

Then inH,, there is a we-universal member.Proof. If G H,, (and without loss of generality V

G = ) and < , then

() { < : is G-connected to elements < } is bounded in say by < .

Hence there is a club C = CG of such that:

(i) cf() = cf(), C, [, ) > otp{ < : is G-connected to}

(ii) cf() = cf(), C, [, ) otp{ < : is G-connected to}

(iii) if C then / and > sup(C ) cf() = cf().We shall define G with VG

= below. For each < divisible by let ai : i < list P = {a : a , and |a| < or otp(a) = and = sup(a)}, each appearing times, possible as || = = , and let

RG

=

{, + i} : < is divisible by , < ,i < , ai

{ + i, +j} : i = j < and < is divisible by

.

Now clearly we have + < > |{ < : is G-connected to }|

hence K, (which is K, by the assumptions) cannot be weakly embedded intoG. On the other hand if G H,, without loss of generality VG = and letCG be as above, and let : < list in increasing order CG {0}, and we canchoose by induction on , a weak embedding f of G into G ( ). So G

is as required. 3.7{2.11}

Claim 3.8. Assume ( 0 and)

(a) = 2 = +, is a singular cardinal

(b) < and 3 <

3in fact, = is O.K., but already covered by 3.4(2)


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22 SAHARON SHELAH

(c) for every P [] of cardinality for some 4 B [], for sets A Pwe have B A

(d) cf() = cf().

Then inH,, there is no we-universal member (even for the family of bipartitegraphs).

Proof. Let G H,,, without loss of generality VG

= and we shall construct

a G Hsbp,,, not weakly embeddable into it. Now we choose UG = , VG = \.

Notice that = (by (a)), so let (f, B) : < list the pairs (f, B)such that f : is one to one, B [] and f B is increasing such that eachpair appears times. Let B = sup{+ 1 : < is G-connected to membersofB} for B [] (and we shall use it for B [\], i.e. for subsets of VG). NowB is < by clause (c) of the assumption. We shall now choose inductively C for [, ) such that

(i) C B is unbounded of order type

(ii) no \Rang(f) is G-connected to every f(), C

(iii) < |C C| < .

In stage choose B B of order type such that ()[ < sup(B

C) < ), that is B C is bounded in equivalently C equivalently in B for

< ; this is possible by diagonalization, just remember cf() = cf() and > and clause (i).

Now there is C satisfying

()C C B is unbounded of order type such that no \Rang(f) is

G-connected to every f() for C.

[Why? Otherwise for every such C there is a counterexample C and we can easilychoose C,i by induction on i < such that:

(i) C,i B

(ii) sup(C,i) = sup(B) =

(iii) otp(C,i) =

(iv) (j < i)[ > |C,i C,j|]

(iv)+ moreover, if j < i then > |C,i { < : f() is G-connected to

C,0C,j}|).

This is easy: for clause (iv)+ note that for C = C,j C,0 by the choice of C wehave C Rang(f) hence by the choice of Rang(f) clearly DC =: {i < : f(i)

is well defined and G

-connected to C} has cardinality < , so we can really carrythe induction on i < , that is any C B unbounded in of order type suchthat j < i |C DC,jC,0| < will do.Let A0 = C,0, A1 = {C,0C,1+i : i < } they form a complete bipartite subgraphof G by the definition of C,0C,i and |C,0| = = |A0| (by (iii) of ) and|A1| = (the last: by (iv)+), contradiction. So there is C such that ()

C.]

Choose C as any such C such that ()C. Lastly define G

4note that if() this clause always holds; and if 2 it is hard to fail it, not clear ifits negation is consistent


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UG =

VG = \

RG = {(, ) : VG, C}.

Clearly G Hsbp,, recalling < and 1 = 2 |C1 C2 | < . Suppose

toward contradiction that f : is a weak embedding of G into G, hence theset Y = { < : and f = f } is unbounded in and without loss ofgenerality Y Rang(f) =

, i.e. is constant. As f is one to one for every Y large enough, f() (, ) and we get easy contradiction to clause (ii)for and we are done. (Note that we can add nodes to UG). 3.8

For the next claim, we need another pair of definitions:{2.12m}

Definition 3.9. 1) H is the class of G = (VG, RG, PGi )i cf() or

(ii) < or

(iii) < cf() and there are C of order type for < such that u [] otp[{C : u}] > .

Then inH,, there is no we-universal even for the bipartite graphs inH,,.2) In part (1) if we replace clause (d) from the assumption by (d)1 or (d)2 where

(d)1

(d)2 = and among the graphs or cardinal there is no ste-universal

(d)3 = and inH there is no ste-universal, then still there is no ste-universal,

even for the bipartite graphs inH,,.

3) If (a) + (b) of part (1) and 2


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24 SAHARON SHELAH

{2.12d}Remark 3.13. Note that part (2) is not empty: if is strong limit singular, 2 =

=

+

, =

otp{ < : is G-connected to}

(iii) 2 divides for every C.

Let S =: { C : cf() = cf()}; as we have S , see [Sh:922], so let f = f : S, f be a one-to-one function such that (f )(stat S)[f is one-to-one f = f ]. For S let = Min(C\( + 1)), and for i [, ) we leta,i = { < : f() is G-connected to i}, so otp(a,i) by the choice of C, andlet B = {i : i < and |a,i| }.

Now for S we choose a unbounded of order type such that (i B)(a a,i).

[Why? First assume (d)(i), i.e. > cf() let [, ) =

< cf()

A,, |A,| |{y U

Gi : y is G-connected to x}|.

We leave 3.12(3),(4) to the reader. 3.17


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28 SAHARON SHELAH

4. More accurate properties{4.1}

Definition 4.1. Let Q(,,,) mean: there are Ai [] for i < such that for

every B [] there are < ordinals i such that B Ai.{4.2}

Definition 4.2. 1) For let set(, ) = {A : A is a subset of ofcardinality such that i < > |{A ( i)}| and let set(, ) = [] for < .2) Assume , . Let Qrw(,,,) mean that some A exemplifies it,which means

(a) A = Ai : i <

(b) Ai set(, ) for i <

(c) A is (, )-free which means (A set(, ))( and there is F {f : f a partial one to one function from to , |Dom(f)| = } of cardinality such that for every f there 5 isf F, f f.

Proof. 1), 2) Let x = w for part (1) and x = st for part (2).Now suppose that G H,, and without loss of generality VG

= and we

shall construct G Hsbp,, not x-embeddable into G (so part (b) will be proved,

and part (a) follows).

Case 1: < so set(, ) = [].

5we can add there are functions f F, f f, with pairwise disjoint domains, andpossibly increasing F we get it


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Similar to the proof of 1.5. Let f = f : be a simple black box for one

to one functions. It means that each f is a one to one function from {i : i < }

into , such that for every f :>

for some

we have f f. We definethe bipartite graph G as follows:

() (a) UG = > and VG = ()

(b) RG = {RG : } where RG {( , (, i)) : < ,i < }.

Now for each , we choose ,i : i < listing without repetitions theset { < : is G-connected to members of Rang(f)} and without loss ofgenerality = || and A,i = { < : ,i, f( ) are G-connected}.

As G H,, clearly A [] |{i < : A A,i}| < hence = || 2 + but (see Definition 4.2(2)) we have assumed and (in 4.3) =

so 2 hence ; next let A = A,i : i < and as A cannot be awitness for Qrx(,,,) but clauses (a), (b), (c), (d) of Definition 4.2(1) hold,hence clause (e) fails so there is A = A,i : i < exemplifies the failure of

clause (e) of Definition 4.2 with A, A here standing for A, A

there.Let

RG = {( , (, i)) : < ,i < and A

,i}.

The proof that G cannot be x-embedded into G is as in the proof of 1.5.

Case 2: < (and = , < ).First note that by the assumptions of the case

1 there is f = f : such that

(a) f is a function from) to then we can find : such that

(i) g()

(ii) 1 2 1 2

(iii) if < < and > then =

(iv) f f for .

We commit ourselves to

2 (a) UG = (>) and VG = {(, i) : ,i < }

(b) RG = {RG : } where

(c) RG {(( , j), (, i)) : j < ,i < , < }.

We say is G-reasonable iff is one to one and for every < and y VG

the set {( , j) : < , j < and f(( , j)) is G-connected to y} has

cardinality < . We decide

3 (a) if is not G-reasonable then RG =


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30 SAHARON SHELAH

(b) if is G-reasonable let ,i : i < list without repetitions theset { < : is G-connected to at least members of

Rang(f)}; and let A,i = {(, j) : < , j < and f(( , j))is G-connected to ,i}; clearly A,i set(, ) andlet A = A,i : i <

(c) as A cannot guarantee Qrx(,,,) necessarily there isA = A

,i : i <

exemplifying this so

and let

RG = {(( , j), (, i)) : i < , and (, j) A

,i}.

The rest should be clear; for every f : > letting : be as in above, for some , is G-reasonable.

Case 3: = .Left to the reader (as after Case 1,2 it should be clear).

3) As in the proof of 2.6(4), it follows that there is f as needed. 4.3

{4.4} Claim 4.4. 1) NQrst(2, , 2, ).

2) If = = 2 then inH,, there is no ste-universal even for members ofHsbp,,.

Proof. 1) Think.2) By part (1) and 4.3. 4.4

{4.5}Claim 4.5. 1) Assume < andQrx(, 1, , ) andH

sbp(,),, = , thenH

sbp(,),, =

Hbp(,),,

has a x-universal member.

Proof. Read the definitions. 4.5


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5. Independence results on existence of large almost disjointfamilies

This deals with a question of Shafir{3.1}

Definition 5.1. 1) Let Pr2(,,,) mean: there is A [] such that

(a) |A| =

(b) if Ai A for i < and i = j Ai = Aj then |i


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32 SAHARON SHELAH

()1 P is -complete and (P, pr) is +-complete, in both cases the union of anincreasing sequence forms an upper bound

()2 for each p,P {q : p apr q} is +-c.c. of cardinality = and -complete

()3 if p r then for some q, q we have p apr q pr r and p pr q

apr r;moreover q is unique we denote it by inter(p,r)

()4 if p

Ord then for some q we have(a) p pr q

(b) if < and q r and r P

() = then inter(q, r) P

() =.

This gives that clauses (a),(b),(c),(d) of the conclusion hold. As for clause (e),2 follows from ()5 + |P| = and 2 , too.

Define:

()5 (a) f

:= {p : p GP}

(b) P f

= GP is a function from to {0, 1}

(c) for < let A = { < : f

( + ) = 1}.

Also easily

()6 if p

then for some u [] and -Borel funtion B : u2 2and q we have

p pr qq

= B(f

u)

()7 A moreover < [, + ) A

= and [, + ) A

and

= A = A

.

[Why? By density argument.]()8 A := {A

: < } exemplifies Pr2(,,+, ).

Why? By ()7 + ()5(c), A [], |A| = so we are left with proving clause (b) of

5.1; its proof will take awhile. So toward contradiction assume that for some p Pand

: < + we have

1 p P < ,

=

for < < + and |


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lemma, recalling =


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34 SAHARON SHELAH

Let u = {/E : Dom(p)} so u [] and Dom(r,i) u for i < .

For some Y [++]++

and < + and stationary S and = i : i S

we have

2 if Y then otp(u) = and S = S and ,i : i S = .

Let g, be the order preserving function from u onto u.Again as 2 = + without loss of generality

3 For , Y(a) p = p g, and p,i = p,i g, and r,i = r,i g, for i <

(b) u u = u for < < ++

(c) g, is the identity on uand

4 if p apr r, p apr r, r = r g, and < then

(a) r B r B

(b) r / B r / B

.

Choose : < ++ an increasing sequence of ordinals from Y.Let p =


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6. Private Appendix

Continued from pg.36:

(b) Q = P {q : p apr q} satisfies the +-c.c.

(c) if p r and r


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36 SAHARON SHELAH

7. Private Appendix 2{2.3}

Claim 7.1. Assume

(a) = cf() > , 2 > (and )

(b) A [], |A| = , and (X [])(A A)(A X)

(c) for every < we have > |{A A : A }|

(d) cf(, .

Then there is no we-universal inH,,.

Proof. Saharon, recall.

References

[EH74] Paul Erdos and Andras Hajnal, Solved and Unsolved Problems in set theory, Proc. of theSymp. in honor of Tarksis seventieth birthday in Berkeley 1971 (Leon Henkin, ed.), Proc. Symp

in Pure Math., vol. XXV, 1974, pp. 269287.[KP84] Peter Komjath and Janos Pach, Universal graphs without large bipartite subgraphs, Math-ematika 31 (1984), 282290.

[Sha01] Ofer Shafir, Two cases when d and d are equal, Fundamenta Mathematicae 167 (2001),1721.

[Sh:e] Saharon Shelah, Nonstructure theory, vol. accepted, Oxford University Press.[Sh:26] , Notes on combinatorial set theory, Israel Journal of Mathematics 14 (1973),

262277.[Sh:88r] , Abstract elementary classes near 1, Chapter I. 0705.4137. 0705.4137.[Sh:108] , On successors of singular cardinals, Logic Colloquium 78 (Mons, 1978), Stud.

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[KoSh:492] Peter Komjath and Saharon Shelah, Universal graphs without large cliques, Journalof Combinatorial Theory. Ser. B 63 (1995), 125135.

[Sh:829] Saharon Shelah, More on the Revised GCH and the Black Box, Annals of Pure andApplied Logic 140 (2006), 133160, math.LO/0406482.

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Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The He-

brew University of Jerusalem, Jerusalem, 91904, Israel, and, Department of Mathe-

matics, Hill Center - Busch Campus, Rutgers, The State University of New Jersey, 110

Frelinghuysen Road, Piscataway, NJ 08854-8019 USA

E-mail address: [email protected]: http://shelah.logic.at

Documents

Saharon Shelah- Universality Among Graphs Omitting a Complete Bipartite Graph