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Aronszajn trees

Uri AbrahamDepartment of Mathematics

Ben-Gurion University

May 31, 2009

Abstract

This survey describes the contribution of N. Aronszajn to set the-

ory: the construction of an Aronszajn tree. We would like to explain

to the non-specialist in set theory what is an Aronszajn tree, how it

is constructed, and why it occupies such a central place in set theory.

In fact, we describe two constructions: the classical one, and a newer

construction of a canonical Aronszajn tree due to Stevo Todorcevic.

1 Introduction

It is easy to describe the contribution of N. Aronszajn to set theory sinceit consists of a single article: the construction of what is now known as anAronszajn tree. A rather simple construction by todays standards which isoften taught in undergraduate set theory courses. Yet by some reason whichwe will try to explain, the notion of an Aronszajn tree is of prime importancein advanced set theory and it continues to contribute in essential ways toits development. (This contribution of Aronszajn was never published in apaper, and the closest reference is an article of Kurepa [?].)

The best way to introduce the notion of an Aronszajn tree is by contrast-ing it with the D. Konigs Lemma which says that any tree has an infinitebranch if has infinitely many levels and each one is finite. This lemma isessentially about , the set of natural numbers, and an Aronszajn tree canbe seen as a counterexample to the proposition that the lemma can be ex-tended to 1, the first uncountable cardinal. It is this tension between the

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compactness at and the incompactness at 1, discovered by N. Aronszajn,

that makes Aronszajns construction and vision so important in set theory.In order to explain this, we must define first some basic concepts (such astrees). This will be done in the rest of our introduction. Then, in the secondand third sections we will present two proofs of the existence of an Aronszajntree. The second proof, due to S. Todorcevic, exemplify some of the methodof todays combinatorial set theory.

The term tree often refers to graphs without cycles, but in set theoryparlance (and in this paper) it refers to a certain type of partial orderings.So, in order to avoid any possible confusion, we shall begin with a definitionof that notion. A treeis a partially ordered set T = (A,


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Aronszajn tree. In todays terminology, what N. Aronszajn constructed was

a specialtree. For this notion we need the definition of an antichain. For agiven treeT, we say that XTis an antichain if no two members ofXarecomparable in the ordering of the tree. That is, for every distinctu, v Xwe have u< v and v < u. A tree T is said to be specialif and only if it is acountable union of antichains, that is there exist antichains Aifor i suchthat T =

iAi. In this case the function taking x Tto the first i such

that x Ai is one-to-one on chains. This gives an equivalent definition: atree is special iff there is a function ffrom the tree into that is one-to-oneon chains. That isx < y in the tree implies that f(x) = f(y). We note inthe following lemma that such a tree has no uncountable branch. The tree

originally constructed by N. Aronszajn is a special tree.

Lemma 1.2 A special tree has no uncountable branch.

Proof. Suppose T is special and T =

iAi is its representation as acountable union of antichains. If b T is any branch, then b Ai is asingleton sinceAi is an antichain. Hence, b is countable. q.e.d

We end this introduction with some notational remarks concerning theordinals. An ordinal represents a well-order type. In standard developmentof set-theory, any ordinal is in fact the set of all ordinals of smaller order-type. In fact the membership relation well-orders any ordinal, and we

write either or < to denote the fact that is smaller than . Wesay that a set Awith a well-ordering


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2 Constructions of Aronszajn trees

It is easy to obtain a tree of height 1with no uncountable branches: the treeconsisting of all one-to-one countable functions from a countable ordinal into with the function extension ordering is such a tree. It is not an Aronszajntree of course since already its level is uncountable (there are continuummany permutations of).

We begin by proving Aronszajns theorem that a special Aronszajn treeexists.

Lemma 2.1 There exists a special Aronszajn tree. That is, a tree of height1 with all levels countable and with a decomposition into countably many

antichains.

Proof. By (transfinite) induction we define for every countable ordinal a tree T() and a function f() from T() into . The following propertiesare required to hold for :

1. The height of the tree T() is + 1 (the successor of), and each ofits levels is countable. The fact that the height ofT() is a successormeans that it has a last level. The function f() is such that thepre-image of{i}(for every i ) is an antichain in T(). We think ofT() as the initial segment of the Aronszajn tree that we are about to

construct, and f() is the corresponding specialization of that initialpart.

2. For every < , T() is an end-extension ofT(). By this we meanthat T() is the initial segment of T() determined by its first + 1levels. (Namely T() T(), the ordering of T() agrees with thatof T() on the nodes of T(), and if x T() has height , thenx T().)

3. For every < the function f() is the restriction off() to T().

4. For every node y T(), let by = {x T() | x y} be the branchconsisting ofy and its predecessors. Then the image off() on by isco-infinite. That is, there are infinitely many n that are not inthat image.

5. For every node y T() such that ht(y) < and for any finite setE disjoint to the image under f() ofby there is a node zabovey

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and of height in T() such that the image under f() ofbz is disjoint

to E. (Hence there is such a node zof hight when ht(y)< < .)

6. For every node y T() such that ht(y) < and for every n notin the image of by under f() there is an immediate successor z of ssuch that f()(z) =n.

Item ?? is the most intriguing and is the one that contains the main idea ofthe proof. We shall see first that if this inductive construction can be car-ried on, then the resulting tree T(1) =

1

T() with the correspondingordering is indeed a special Aronszajn tree. But this is clear, since items ??and ?? ensure that the resulting tree is special of height 1, and the fact

that T(1) is end-extended by T(2) when 1 < 2 ensures that all levels ofT(1) are countable.We start with T(0), a tree of height 1. For definiteness assume that it

consists of a single node r which is going to be the root of the tree, and letf(0)(r) be arbitrarily chosen.

Let be a countable ordinal. Assume that a sequence of trees T() andfunctions f() is given for < so that properties ?? to ?? hold, and weshall define T() and f(). There are two cases: a successor ordinal and a limit. The successor case is easy. If is the successor of0and T(0) isdefined then we obtainT() by adding to each node y of level 0 inT(0) successors and allowing all possible values for the f() function on the newnodes. The fact that the range off(0) on any by is not all of allows us tofind a free value for the new nodes and to keep the requirement that thespecializing function is one-to-one on branches.

The interesting case is when is a limit (countable) ordinal. Let i |i be an increasing sequence cofinal in . Let T =

iT(i) be the

resulting tree formed as the union of all the trees along the cofinal sequence,and let f =

if(i) be the specializing function. Then T is an end-

extension of each T(i) and hence of each T() for < . Similarly, fextends all the functions f(i). We must add a last level, the -th level, toT in order to obtain T(), and extend f over that new level (in order to

define f()). We must define this -th level and the function f() so thatthe inductive properties 4 and 5 hold (the other properties are evident).

A branch ofTof order-type is said to be an branch. Such a branchcontains nodes of every level < . We shall define a countable family B of branches ofT, and then define the level of the tree T() by assigning anode above each branch in B.

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For every nodey T, and for every finite set E disjoint to the image

of by under f, we shall build a branch b = b(y, E) that contains y and sothat the image ofb underfis co-infinite and disjoint toE. Viewing the pair(y, E) as a mission we can arrange all missions in a countable sequence,and deal with each one in turn. Given a mission (y, E) we shall define thebranch b(y, E) so that it differs from each of the (finitely many) branchesconstructed so far (to simplify the definition). Let i0 be the first sothat ht(y) < i0. We shall define an increasing sequence of nodes above yyi | i i0, where yi is of height i, and a sequence of finite subsets of, Ei | i i0 with E Ei Ei+1 so that the range of byi under f isdisjoint to Ei. There is no problem in constructing such a sequence using

item ?? at each step. The resulting branch bis as required: its image underf is co-infinite since more and more values were excluded from this range.Namely,

ii0

Ei is infinite and disjoint to the range offapplied to b.Let B be the collection of all branchesb(y, E) constructed in this fashion,

where y T and E a finite subset of disjoint to the image of by underf. For every b B add a new point to T that is above all the nodes of b.Together with the tree Tthis yields T(). The specializing function f() isdefined by extending fover these new points. For every new point x addedabove the branchb, we haveb = b(y, E) for some uniquely determined (y, E),and we define f()(x) \ Eas any value outside of the range off over b(for example, pick the first value in this infinite set). Check now that all theinductive properties hold. q.e.d

2.1 A canonical construction of a special Aronszajntree

In this subsection we present another construction of a special Aronszajn treewhich is radically different from the former construction in that no inductionis involved: the tree is defined at once. This new proof is due to S. Todor-cevic and it employs his miraculous theory of walks over ordinals ([?]). Atheory that has allowed Todorcevic and other researchers to resolved many

open problems and which continues to be a source of deep investigation. Inpresenting part of this theory here we restrict our attention to 1, but thetheory applies to infinite successor cardinals in general.

Fix a sequenceC | 0 < 1 such that, for every countable >0,

1. C is an unbounded subset of of order-type if is a limit ordinal,

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and

2. C={0}if is the successor of0.

We say that C | 0 < 1 is a ladder system. Intuitively, a laddersystem enables a short access to the countable ordinals: if = 0+ 1 is asuccessor ordinal then C is the singleton consisting of the maximal ordinalbelow , but if is a limit ordinal then C is a cofinal subset of of theshortest possible length.

Let Q be the set of all finite sequences of members of. So a Q iff forsomen (called the length ofa),a is a function from{0, . . . , n1}into.We write a= a0, . . . , an1. Ifa, b Q then a

bdenotes the concatenation

ofa and b.A lexicographical ordering is defined on Q as follows. For a, b Q

wherea= a0, . . . , an1 and b= b0, . . . , bm1 we define a b iff either

1. n > m and for every i < mwe have ai=bi, or else

2. for some i < min(n, m), ai = bi and for the least such i we have thatai < bi.

Observe that any sequence in Q is smaller than all of its proper initial seg-ments; this is natural in our context. So the empty set is the maximum ofQ. Omitting the empty set, we get a countable dense set without first or lastelement, and hence an ordering isomorphic to the rational.

We now define the walk from down to for every < 1:walk(, ) = 0, . . . , n1 so that 0 = , 0 > 1 . . . and n1 = .This walk will be instrumental in defining the trace of the walk, trace(, )and its shadow, the famous 0(, ) Q.

The definition of the ordinals i that constitute the walk from down to is by the following procedure. We start with 0 = . Supposethat 0 > > i have been defined. Ifi = the procedure stops, but ifi > let i+1be the first ordinal in Ci that is not below (there is such anordinal since Ci is unbounded in i). In case i is a successor ordinal, i+1

is the predecessor ofi. The procedure must stop since there is no infinitedescending sequence of ordinals, and hence we must arrive to some i = .The ordinals i are said to be on the walk.

If walk(, ) =0, . . . , n1then (0, 1) is the first step, (1, 2) thesecond step etc. So there are n1 steps and the trace of the walk is thesequence of length n 1 of finite subsets of defined as follows.

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trace(, ) =Ci | 0 i < n 1. (1)If we collect the cardinalities of the finite sets appearing in the trace we

obtain 0(, ):

0(, ) =|Ci | |0 i < n 1.

Notice that for i < n 1, i > so that Ci is finite and its cardinality|Ci |is in . So that0(, ) Q, namely it is a sequence of length n 1of natural numbers.

A recursive definition of the walk, its trace and rho functions can be of

help in a formal proof because it allows induction.walk(, ) =.

Correspondingly, in no step we get from to , trace(, ) =, and 0(, ) =. For < ,

walk(, ) =walk(, min(C\ )).

Correspondingly

trace(, ) =

C

trace(, min(C\

)). (2)

0(, ) =|C |0(, min C\ )). (3)

We first explain the notation employed in this definition, and then weshall clarify the definition itself. Here, is the sequence of length 1 whosesole member is . Assume that < . Viewing the ordinal as the set ofall smaller ordinals, C\ is the subset ofCconsisting of those membersthat are not below . If Cthen the first member ofC\ is itself,but otherwise min(C\ ) is the first member ofCthat is strictly above .Anyhow, C , which is the set of all members ofCthat are below , is a

finite set. Ifis a successor ordinal then this set is empty, and ifis a limitordinal thenC is finite because Cis of order type and unbounded inand < .

Suppose that walk(, ) = 0 > > n 1. Then for every 0 i j n1 we have walk(j, i) = i > > j. We rephrase this in thefollowing lemma.

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Lemma 2.5 Suppose that1 1 and1

2 are countable ordinals such

that trace(1, 1) = trace(1, 2). Then for every 0 1 we also havetrace(0,

1) = trace(0, 2).

Hence, iftrace(1, 1) = trace(1,

2)thenx0(x, 1) 1= x0(x,

2) 1.

Proof. The lemma is proved by induction on max{1, 2}. If1 = 1 thentrace(1,

1) = and hence trace(1, 2) =, which implies that 1=

2 aswell. So we may assume that 1 <

1, 2.Let 1 > 11 be the first step in the walk from

1 to 1, and let 2 > 2

1

be the first step in the walk from 2 to 1. That is, 11

= min(C1\ 1), and

2

1 = min(C2

\ 1).It follows from the recursive definition of the trace formula (??) that

trace(1, 1) =C1 1

trace(1, 1

1).

trace(1, 2) =C2 1

trace(1, 2

1).

Since these two traces are assumed to be equal, we get C1 1 = C2 1, and trace(1,

21

) = trace(1, 11

). Since max{11 , 21} < max{

1, 2},the inductive assumption implies that trace(0,

21) = trace(0,

11). Let

T0 denote this common trace, and let X denote the common value X =C1 1= C2 1. There are two cases now.

1. X [0, 1) = . That is, X 0. It follows that C1 0 =C20 = X, and the walks from

2 and 1 to 0 lead to 21 and

to 11

(respectively), and then trace(0, 1) = trace(0,

2) =XT0follows as desired.

2. X [0, 1)= and we let 1 be the first ordinal in this intersection.Then the first steps in the walks from both 1 and 2 to 0 lead to 1(and continue from that place in unison). So that the conclusion of thelemma is obvious. q.e.d

The definition of the special Aronszajn tree of Todorcevic can now begiven by the following formula:

T ={x0(x, ) | < 1}.

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The nodes of this tree are order preserving functions from countable ordinals

into Q. The previous lemma implies immediately that every level of T iscountable. Namely

T = {x0(x, ) | < 1}

is countable by Lemma ?? since the number of possible traces is countable.Since the nodes of T are order preserving functions into Q, T is clearly anAronszajn tree. In fact, it is a special Aronszajn tree. To prove this fact, wedefine a specializing map :

{T | < 1 limit} Q that is defined on

the limit levels of the tree (a standard argument then shows that the tree isspecial). Given a node t T, where is a limit ordinal, we find such

that t= x0(x, ) . We define (t) =0(, ). We must prove that iswell defined (its definition does not depend on the choice of) and that isorder-preserving.

So suppose that 1 is another ordinal such that t= x0(x, 1) . Weprove 0(, 1) = 0(, ) by Lemma ??. In details the argument runs asfollows. Since is a limit ordinal, we can pick an ordinal < that boundsboth trace(, ) and trace(, 1). Then Lemma??implies that trace(, ) =trace(, )trace(, ), and trace(, 1) = trace(, 1)

trace(, ). Since0(, ) =0(, 1), we get 0(, ) =0(, 1) as required.

The fact that is order preserving is a direct consequence of Lemma ??.

3 Todorcevic theorem

Theorem 3.1 (Todorcevic) 1 [1]21. That is, there exists a functionf : [1]2 1 such that for every uncountableA 1 f[A] =1.

In fact, we will find a function g : [1]2 1 such that, for every A

[1]1, g[A] contains a club set. This entails the theorem by partitioning

1 =

i1Si into 1 pairwise disjoint stationary sets, and then defining

f(a, b) =i iffg(a, b) Si.

Let T be a special Aronszajn tree, with s : T as its specializingfunction (one-to-one on chains). Let {a | 1} be an uncountableantichain in T such that for < the level of is below the level of .Given < define g(, ) as follows. Let p Tbe the splitting point ofaand aand suppose thats(p) =k . Let x be the first point below a(if thereis any) such that the level of x is above the level of a and s(x) < k. Let

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be the level ofx on the tree, and define g(, ) =(an arbitrary value is

chosen if there is no such x).Let A 1 be an arbitrary uncountable set. We must prove that g[A]

meets every stationary set S. For this, take M H(2) be a countableelementary substructure containing T,s, andA and such that M 1=S. Pick some A with > . ThenaMand the level ofais above .Let x < abe that node at level belowa, and suppose that s(x) =n. Weshall find a M such that g(, ) = . On the branch determined by x,{y T |y < x}, the function s is one-to-one and there is a finite number ofys withs(y)< n. Lety0 < xbe above all of those ys. SinceaA is abovey0 and is not in M, the set{a A | y0 < a} M is uncountable. Lets say

that y T is heavy ify0 y and {a A | y < a} M is uncountable.Ify is heavy and y0 y y then y is heavy too. The collection of heavy

nodes is in Mand is uncountable (because every level ofT is uncountable).Since there are no uncountable chains in Twe can find two heavy nodes thatare incomparable and hence there is a heavy node incomparable with x andwe can have it in M. So let y > y0 be some heavy node in Mand so that yand x split, and let p be their splitting node (which is evidently the splittingnode ofy and a as well). Say s(p) =k . By the choice ofy0 (and as p < x)k > n. Now as y is heavy, we can find a A above y and in M such thatthe level ofa is above all the (finitely many) nodes below x whoses value isbelow k. It is now easy to check that x is the first point below asuch thatthe level ofx is above the level ofa and s(x) < k, and thus g(, ) =asrequired. q.e.d

Concluding remarks. The bibliography reported below is a partial listof papers that either directly investigate Aronszajn trees or employ this no-tion in substantial ways. The reader can get an effective impression that thisnotion continues to be an important driving force in set theory since the daysof its invention by N. Aronszajn. The interest in Aronszajn trees stem fromthe tension between the compactness and incompactness phenomenon. Gen-eralizing the question on Aronszajn trees to arbitrary cardinals, [?] definesthat an infinite cardinal has the tree propertyiff every tree of height and

with all levels of size < necessarily has a branch of length . So, for ex-ample, has the tree property but 1 has not. Investigating the question ofwhich uncountable regular cardinals have the tree property is an importantdirection and a continuing challenge in set theory which is directly connectedwith Aronszajns work. (See [?] for more information on large cardinals andweakly compact cardinals.)

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Investigations of the combinatorics of1is a main issue in set theory and

Aronszajn trees and their linearizations are instrumental in these investiga-tions (see for example recent work of J. T. Moore).

References

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property. Ann. Math. Logic 5 (1972) 2146.[24] J. T. Moore. A five element basis for the uncountable linear orders. To

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[25] J. T. Moore. Proper forcing, cardinal arithmetic, and uncountable linearorders. The Bulletin of Symbolic Logic, 11 (2005) 5160.

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[26] J. T. Moore. Structural analysis of Aronszajn tree. Proceedings of the

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velle Serie, tome 27(41) 259265.[38] S. Todorcevic, On the Lindelof property of Aronszajn trees, Proc. Sixth

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[39] S. Todorcevic, Partitioning pairs of coutable ordinals, Acta Math. 159(1987) 261294.

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[40] S. Todorcevic, Special square sequences, Proc. AMS, 105 (1989) 199

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