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Our paper is concluded with some open problems and remarks in Section . Let us also mention that the results of this article are summarized in two lecture videos that are accessible on the author’s webpage and, in fact, Theorem 6.1 is presented in detail. |
Notations and terminology We use standard notations from set theory following classical textbooks . In forcing arguments, stronger conditions are smaller. |
For us, a tree [MATH] is a partially ordered set so that [MATH] is well ordered for any [MATH] ; we will usually omit the subscript from [MATH] if it leads to no confusion. Let [MATH] denote the order type of [MATH] , and let [MATH] denote the [MATH] th level of [MATH] i.e., all elements [MATH] with [MATH] (and let [MA... |
An [MATH] -tree is a tree of height [MATH] with countable levels. An Aronszajn tree is an [MATH] -tree with no uncountable chains, and a Suslin tree is an [MATH] -tree without uncountable chains or uncountable antichains. |
The assumption [MATH] denotes the existence of a sequence [MATH] so that for any [MATH] , the set [MATH] is stationary. Similarly, [MATH] asserts the existence of a sequence [MATH] so that [MATH] and for any [MATH] , there is a closed unbounded [MATH] so that for any [MATH] [MATH] |
We will use the following standard fact multiple times. Fact 1.2 Suppose that [MATH] is an [MATH] -tree and [MATH] is a countable elementary submodel of some [MATH] with [MATH] . Let [MATH] be an element of [MATH] and [MATH] . Then |
(1) if [MATH] has no uncountable antichains then [MATH] , and (2) if [MATH] has no uncountable chains then [MATH] Proof. (1) Indeed, take a maximal antichain [MATH] so that [MATH] , and note that [MATH] as [MATH] is countable. Then [MATH] and [MATH] is compatible with some element of [MATH] . So there must be some [MAT... |
(2) If we choose [MATH] to be a maximal chain in [MATH] then again, [MATH] is countable and we must have some element of [MATH] off the branch [MATH] |
Acknowledgments We thank S.-D. Friedman and M. Levine for valuable discussions and helpful comments on some of our results. The author would like to thank the generous support of the FWF Grant I1921 and NKFIH OTKA-113047. |
2. Non-uniformization results for Suslin trees In this section, we will show that if [MATH] then for any [MATH] -tree [MATH] with no uncountable antichains and any ladder system [MATH] |
fails Let us mention that in , where [MATH] -uniformizations were first introduced, the only non-uniformization result that can be immediately extracted is 20 , Lemma 3.3] : under [MATH] , any Aronszajn tree [MATH] that is club-embeddable into all its subtrees and any ladder system [MATH] , there is a monochromatic 2-c... |
, or by using [MATH] to construct a minimal lexicographically ordered Aronszajn tree and then invoke 20 , Lemma 2.9] We start by recalling some definitions and in particular the basis of the result of Devlin and Shelah from point ( II ). |
Theorem 2.1 [MATH] if and only if for any [MATH] there is a [MATH] so that for any [MATH] the set [MATH] is stationary. The latter statement is often referred to as the weak diamond , denoted by [MATH] , which was discovered in search for the minimal assumptions that imply [MATH] . In general, applications of this prin... |
Theorem 2.2 20 , Theorem 3.2] [MATH] implies that for any [MATH] there is a [MATH] so that for any [MATH] there is a countable elementary submodel [MATH] containing [MATH] so that [MATH] |
The function [MATH] from the theorem will be referred to as an oracle for [MATH] By Moore’s result from point ( III ), [MATH] by itself is not strong enough to show the existence of colourings without a [MATH] -uniformization for an arbitrary Aronszajn tree [MATH] . However, [MATH] does have consequences for Suslin tre... |
Theorem 2.3 [MATH] implies [MATH] for any [MATH] -tree [MATH] without uncountable antichains and any ladder system This result generalizes Devlin and Shelah’s theorem from point ( II ), but applies to Suslin trees as well. So, we see that Moore’s model cannot contain Suslin trees, and the current author’s result from |
is optimal. Let us also mention that [MATH] easily implies that for any ladder system and Aronszajn tree [MATH] there is a monochromatic 2-colouring of without a full |
[MATH] -uniformization i.e., a [MATH] -uniformization [MATH] with [MATH] Proof. Fix an [MATH] -tree [MATH] without uncountable antichains and ladder system ; we can assume that [MATH] and so [MATH] for all [MATH] |
We need to find a monochromatic 2-colouring of [MATH] with no [MATH] -uniformization, and to do so, we first define a function [MATH] . In fact, we will only specify the values of [MATH] on maps [MATH] in [MATH] where [MATH] is a subtree of [MATH] for some limit [MATH] . On all other elements of [MATH] , we let [MATH] ... |
Two cases are distinguished. Case 1. Suppose that there is an [MATH] and [MATH] , so that for any [MATH] above [MATH] such that [MATH] , the set |
[EQUATION] is cofinal in [MATH] . Pick such an [MATH] canonically (using some well order of [MATH] ), and define [MATH] Case 2. If there is no such pair [MATH] then we let [MATH] |
Now, let [MATH] be the oracle for [MATH] given by [MATH] . We claim that there is no [MATH] -uniformization for the monochromatic coloring that is defined by taking [MATH] to be constant [MATH] |
Otherwise, let [MATH] be a subtree and suppose [MATH] is a [MATH] -uniformization of . Now, we reach a contradiction: find a countable elementary submodel [MATH] containing [MATH] and [MATH] so that [MATH] . Let [MATH] |
Claim 2.4 We used the second case definition for [MATH] Proof. Otherwise, we used the first case definition of [MATH] with some node [MATH] and [MATH] . But then for any [MATH] above [MATH] so that [MATH] [MATH] for infinitely many [MATH] . However, [MATH] and so for every [MATH] such that [MATH] [MATH] holds for almos... |
In particular, we deduce that [MATH] and so [MATH] . We know that Case 1 failed with any [MATH] and [MATH] , so we can find [MATH] so that [MATH] and [MATH] for almost all [MATH] . Consider the set |
[EQUATION] Note that [MATH] and [MATH] so by Fact 1.2 , we can find some [MATH] . However, any [MATH] is also in [MATH] , so [MATH] for almost all [MATH] (where [MATH] ). This contradicts that [MATH] (i.e., that [MATH] ), which in turn finishes the proof of the theorem. |
In Section , we show that even [MATH] for all is consistent with the existence of Suslin trees (but necessarily, CH will fail in such a model). However, we are not sure if it suffices to assume in the above result that [MATH] has no stationary antichains. |
3. Non-uniformization results for Aronszajn trees Our next goal is to find an assumption that gives ladder system colourings which have no [MATH] -uniformization for any [MATH] -tree [MATH] that is not necessarily Suslin. To prove such a result, we employ the technique of parametrized weak diamonds |
. Let [MATH] stand for the ideal of meager sets in [MATH] Definition 3.1 Let [MATH] denote the following statement: if [MATH] is an [MATH] set of ordinals and [MATH] so that [MATH] for all [MATH] , then there is a [MATH] so that for all [MATH] the set |
[EQUATION] is stationary. Recall that [MATH] is the class of sets constructible from [MATH] (in the sense of Gödel) and [MATH] is the minimal model extending [MATH] which contains [MATH] . See 12 , Chapter 13] for more details on constructibility. So, compared to [MATH] , we changed the range of [MATH] from 2 to [MATH]... |
[MATH] implies the existence of Suslin trees 21 , Theorem 3.1] and follows from the classical diamond principle [MATH] However, [MATH] has the great advantage that it can hold in various models with [MATH] (see |
for more details). Theorem 3.2 [MATH] implies [MATH] for any [MATH] -tree [MATH] and ladder system In turn, [MATH] is inconsistent with quite weak forms of uniformization on trees, but we will see in later sections that, quite surprisingly, we cannot always produce a monochromatic 2-colouring without a [MATH] -uniformi... |
Proof. Fix [MATH] and , and let [MATH] denote the increasing enumeration of [MATH] . We will first define [MATH] using [MATH] as a parameter. By a standard coding argument (much like in Theorem 2.2 ), we can associate (in a canonical way) to any [MATH] so that [MATH] is a subtree and [MATH] some code [MATH] . Moreover,... |
Now, suppose [MATH] codes some [MATH] so that [MATH] is a subtree and [MATH] . Then we let [EQUATION] [EQUATION] Otherwise, we let [MATH] ; it is easy to see that [MATH] |
Claim 3.3 [MATH] is a meager subset of [MATH] Proof. It suffice to note that the set [EQUATION] is nowhere dense for any choice of [MATH] and [MATH] |
Now, suppose [MATH] is the oracle for [MATH] witnessing [MATH] , and define the colouring [MATH] by [MATH] for [MATH] . We claim that this (not necessarily monochromatic) 2-colouring [MATH] has no [MATH] -uniformization. |
Otherwise, suppose that [MATH] is a [MATH] -uniformization. Now, we can find some limit [MATH] so that [MATH] . This means that for any [MATH] |
[EQUATION] for infinitely many [MATH] , a contradiction. Next, one would like to find a monochromatic 2-colouring without a [MATH] -uniformization , much like in the case of Suslin trees. We will use [MATH] to do this but if the interested reader looks closely at the preceding proofs for an optimal assumption then he o... |
Theorem 3.4 [MATH] implies that for any [MATH] -tree [MATH] , there is a ladder system [MATH] so that [MATH] Note that, given a tree [MATH] , we are choosing both the ladder system and the monochromatic 2-colouring without [MATH] -uniformization; we will see later that this is best possible from diamond-type assumption... |
We use the following equivalent form of [MATH] (see ): for any [MATH] there is a [MATH] so that for any [MATH] there is a countable elementary [MATH] containing [MATH] so that [MATH] |
Proof. We fix an [MATH] -tree [MATH] and define a map [MATH] first as follows. Take bijections [MATH] for all [MATH] . Given [MATH] for some pruned [MATH] , we distinguish two cases. |
Case 1. Suppose that there is an [MATH] and [MATH] so that for any [MATH] above [MATH] such that [MATH] , the set [EQUATION] is cofinal in [MATH] . Pick such an [MATH] canonically (using some well order of [MATH] ), and define [MATH] |
Now, let [MATH] . We choose the values [MATH] for [MATH] so that (1) [MATH] is a cofinal, type [MATH] subset of [MATH] , and (2) |
for any [MATH] , there are infinitely many [MATH] such that [MATH] In plain words, we use [MATH] to record a colour which contradicts [MATH] on each branch [MATH] restricted to the levels in [MATH] (infinitely often). The set [MATH] is recorded by the restriction [MATH] |
Case 2. If Case 1 fails, we let [MATH] be the constant 1 function. On any other element of [MATH] , we define [MATH] to be constant 1 again. |
Now, let [MATH] be the oracle for [MATH] that witnesses [MATH] . We define the ladder system [MATH] : if [MATH] has type [MATH] and is cofinal in [MATH] then we let this set be [MATH] . Otherwise, [MATH] is any type [MATH] , cofinal subset of [MATH] . To colour [MATH] , we simply let [MATH] be constant [MATH] |
We claim that [MATH] has no [MATH] -uniformization. Otherwise, let [MATH] be a [MATH] -uniformization of . Now, find a countable elementary submodel [MATH] containing [MATH] and [MATH] such that [MATH] . Let [MATH] and [MATH] |
Claim 3.5 We used the second case definition for [MATH] Proof. Otherwise, we used the first case definition of [MATH] with some node [MATH] and [MATH] . But then for any [MATH] above [MATH] so that [MATH] [MATH] for infinitely many [MATH] . However, [MATH] and so for every [MATH] [MATH] holds for almost all [MATH] . In... |
In particular, we deduce that [MATH] and so [MATH] . In turn, by elementarity, there are arbitrary large [MATH] so that [MATH] as well. Hence, for any [MATH] such that [MATH] and any [MATH] , there is [MATH] so that [MATH] . But this implies that Case 1 does hold for [MATH] with [MATH] and any choice of [MATH] . This c... |
Finally, we end this section by showing that consistently all trees and all ladder systems fail the uniformization property in the strongest sense. |
Theorem 3.6 After adding [MATH] Cohen subsets to [MATH] , for any [MATH] -tree [MATH] and ladder system [MATH] That is, we will iterate posets of the form [MATH] with countable support in [MATH] stages. Note that this forcing is countably closed and so no new countable sets are introduced. We also use the following obs... |
Observation 3.7 Suppose that [MATH] is a countably closed poset and is a ladder system colouring with no [MATH] -uniformization (for some [MATH] -tree [MATH] ). Then has no [MATH] -uniformization in any forcing extension by [MATH] |
Proof. Suppose that [MATH] is a [MATH] -uniformization of . Construct a decreasing sequence of conditions [MATH] with maps [MATH] so that [MATH] . Note that whether a map [MATH] is a [MATH] -uniformization of is decided on its countable initial segments. In turn, [MATH] must be a [MATH] -uniformization of in the ground... |
Proof of Theorem 3.6 Any [MATH] -tree [MATH] and ladder system of the final extension appears at some intermediate stage of the iteration, so we will assume [MATH] are in the ground model [MATH] and show that if [MATH] is Cohen generic over [MATH] then [MATH] holds in [MATH] . In fact, we prove that if [MATH] is consta... |
Now, suppose that does have a uniformization in [MATH] and we reach a contradiction. Find a Cohen condition [MATH] (i.e., a countable partial function from [MATH] to [MATH] ) and name [MATH] , so that [MATH] is a [MATH] -uniformization of . Take a continuous, increasing sequence of countable elementary submodels [MATH]... |
Let us describe the general step in the construction i.e., finding [MATH] from [MATH] (where [MATH] ). Given [MATH] , we first find an extension [MATH] so that [MATH] and [MATH] is [MATH] -generic; this can be done since the forcing is [MATH] -closed. Now, [MATH] decides [MATH] and decides whether [MATH] or not. In the... |
4. Forcing uniformizations and preserving Suslin trees In , Moore introduced a forcing that uniformizes a given colouring on an Aronszajn tree [MATH] , which furthermore can be iterated without adding reals. In turn, consistently, CH and [MATH] holds for any ladder systems [MATH] and Aronszajn tree [MATH] |
However, we proved in Theorem 2.3 that whenever CH holds, for any Suslin tree [MATH] there are monochromatic 2-colourings of ladder systems which do not have [MATH] -uniformization. In turn, there could be no Suslin trees in Moore’s model. We also proved in |
that one can preserve CH and a single Suslin tree [MATH] , while forcing the existence of [MATH] -uniformizations for all [MATH] -colourings and those trees [MATH] which do not embed [MATH] in a strong sense. |
Our goal in this section is to show that if we do not require CH then even [MATH] for all [MATH] is consistent with the existence of Suslin trees. We will do this by proving that for any ladder system colouring [MATH] , there is a natural ccc forcing [MATH] which introduces an [MATH] -uniformization for [MATH] and such... |
Theorem 4.1 For any ladder system [MATH] and [MATH] -colouring [MATH] of [MATH] , there is a ccc forcing [MATH] of size [MATH] , so that |
(1) [MATH] there is an [MATH] -uniformization of [MATH] , and (2) if [MATH] is a Suslin tree in the ground model then [MATH] [MATH] is Suslin. |
Corollary 4.2 Suppose CH holds. Then there is a proper, cardinality and cofinality preserving forcing [MATH] so that in [MATH] [MATH] holds for any ladder system [MATH] , and any Suslin tree in the ground model remains Suslin in [MATH] |
Proof. An appropriate countable support iteration (in length [MATH] ) of posets of the form [MATH] will force that [MATH] holds for all ladder systems [MATH] . The fact that any ground model Suslin tree remains Suslin follows from Tadatoshi Miyamoto’s preservation theorem |
Similar models that combine features of the constructible universe [MATH] and consequences of MA or PFA have received considerable attention |
. However, we could not find a reference for a model combining the existence of Suslin trees and the uniformization property.For example, in |
, a forcing axiom for stable posets is proved consistent: although the model satisfies [MATH] , it has no Suslin trees. More recently, the centre of attention has been on models of [MATH] (or [MATH] ) i.e., the forcing axiom for ccc partial orders (or proper posets, respectively) that preserve a fixed (coherent) Suslin... |
Corollary 4.3 Suppose [MATH] is a Suslin tree. Then [MATH] implies [MATH] for all [MATH] This should be compared with a result of P. Larson and S. Todorcevic: if [MATH] is a Suslin tree and is an [MATH] -name for a ladder system then forcing with [MATH] will introduce a colouring of with no [MATH] -uniformization 15 , ... |
Now, let us prove the theorem. There are various ways to introduce a uniformization for a given ladder system colouring by a ccc forcing. These techniques include posets of |
(1) countable partial functions [MATH] from [MATH] to [MATH] that are defined on finite unions of the ladders, and so that [MATH] agrees with [MATH] almost everywhere on its domain 27 , Chapter II, Theorem 4.3] |
(2) finite partial maps [MATH] from [MATH] to [MATH] so that [EQUATION] are pairwise compatible In both cases the order on the posets is extension as functions. |
It can be shown that both the above posets preserve Suslin trees (see the remarks at the end of the section). However, we take this opportunity to present another natural variant: |
(3) forcing with finite partial maps defined on what we call [MATH] -closed sets . In our case, the uniformization property is coded into the ordering. |
We present the details below. Proof of Theorem 4.1 Given a ladder system [MATH] , we say that [MATH] is -closed if for any [MATH] [MATH] . Let [MATH] denote the smallest -closed superset of [MATH] |
We show that a finite set [MATH] has finite -closure, and even more: Observation 4.4 Suppose that [MATH] is a ladder system. (1) |
Any finite set [MATH] has finite [MATH] -closure and [MATH] (2) any initial segment of a [MATH] -closed set is [MATH] -closed; (3) |
if [MATH] is [MATH] -closed and [MATH] is finite then [EQUATION] Proof. First, let [MATH] be finite and we show, by induction on [MATH] , that [MATH] has finite [MATH] -closure. We can assume that [MATH] otherwise [MATH] and we are done by induction. Let |
[EQUATION] and note [MATH] and that [MATH] . So [MATH] must have size [MATH] which is finite by induction. Now suppose that [MATH] is [MATH] -closed, let [MATH] and pick any [MATH] . Then |
[EQUATION] So the initial segment [MATH] is [MATH] -closed too. Finally, we prove the last statement by induction on [MATH] . Since [MATH] is [MATH] -closed, we can apply induction to see that |
[EQUATION] and by adding [MATH] to both sides we get the desired equality. Now, given a ladder system [MATH] and [MATH] -colouring [MATH] , let [MATH] consist of all finite maps [MATH] so that [MATH] is -closed. The extension in [MATH] is defined by [MATH] if [MATH] and |
[MATH] ) for all [MATH] and all [MATH] [MATH] First, we prove a few simple facts on [MATH] which show that a generic filter gives an [MATH] -uniformization of [MATH] |
Claim 4.5 Suppose that [MATH] is generic and let [MATH] (1) [MATH] is dense in [MATH] for all [MATH] (2) for any [MATH] and [MATH] [MATH] for all [MATH] |
(3) [MATH] and for all [MATH] and almost all [MATH] [MATH] Proof. ) Given [MATH] and [MATH] , we form [MATH] . This is a finite set and we need to show that there is some [MATH] in [MATH] so that [MATH] i.e., that the extension property ( [MATH] ) is satisfied. To this end, note that for any [MATH] , there is at most o... |
) is a simple consequence of ( [MATH] ), and ( ) follows from the genericity of [MATH] and the previous statements. Finally, we prove that [MATH] is ccc and preserves Suslin trees simultaneously. |
Lemma 4.6 [MATH] is ccc and for any Suslin tree [MATH] [MATH] [MATH] is Suslin. Proof. We can assume that [MATH] . Suppose that [MATH] , and [MATH] are conditions below [MATH] . We will find [MATH] and a common extension [MATH] of [MATH] and [MATH] so that [MATH] is not an antichain in [MATH] |
First, by extending each [MATH] , we can suppose that there are [MATH] of height at least [MATH] , so that [MATH] Now, take a continuous sequence of elementary submodels [MATH] of a large enough [MATH] so that [MATH] contains all the relevant parameters (e.g. [MATH] ). Fix some [MATH] , and find a large enough [MATH] s... |
[EQUATION] and note that [MATH] is finite, so [MATH] and [MATH] as well. Let [MATH] and note that [MATH] by Claim 4.4 ). We can extend [MATH] to a condition [MATH] in [MATH] , and we let [MATH] . Note that [MATH] |
Claim 4.7 There is [MATH] and an extension [MATH] with [MATH] such that (1) [MATH] (2) the unique monotone [MATH] is an isomorphism between [MATH] and [MATH] which fixes [MATH] |
(3) if [MATH] then [MATH] and [MATH] for all [MATH] , and (4) [MATH] Proof. Indeed, [MATH] contains the set [MATH] of all [MATH] such that [MATH] has some extension [MATH] which extends [MATH] and is isomorphic to [MATH] in the above sense i.e., conditions ( ) and ( ) are satisfied. Apply Fact 1.2 to the set [MATH] to ... |
Now, we can amalgamate [MATH] and [MATH] Claim 4.8 [MATH] is [MATH] -closed, and [MATH] is a condition in [MATH] extending both [MATH] and [MATH] |
Proof. First, lets see why [MATH] is [MATH] -closed: let [MATH] , and note that the interesting cases are when [MATH] and [MATH] . We distiguish two cases: first, assume that [MATH] . In this case, [MATH] , so any [MATH] is in [MATH] actually. |
Second, if [MATH] then [MATH] and the latter is a subset of [MATH] Finally, the definition of [MATH] and the choice of [MATH] made sure that no element of [MATH] is at critical levels given by [MATH] for some [MATH] |
Hence, [MATH] is a common extension of both [MATH] and [MATH] , forcing that [MATH] is not an antichain. Finally, the theorem is proved. |
Remark. First, one can also prove that an arbitrary finite support iteration of posets of the form [MATH] preserve Suslin trees, and so we can show the above consistency result together with arbitrary large values for the continuum. |
Second, let us sketch an argument that the forcing from 27 , Chapter II, Theorem 4.3] (see at the beginning of the section) also preserves Suslin trees. This is a non-essential modification of Lemma 4.6 . First, recall that each condition [MATH] is now defined on a set of the form [MATH] where [MATH] is finite. Now, su... |
5. Uniformization results in ZFC The goal of this section is to present positive uniformization results, provable in ZFC, for some trees of height [MATH] . Our previous results show that we must use trees with uncountable levels. We will mostly work with [MATH] , the set of all well-ordered, bounded sequences of ration... |
There is a notable special subtree of [MATH] , denoted by [MATH] , namely the set of those [MATH] which have a maximum. This tree was used by Duro Kurepa to construct the first (special) Aronszajn tree |
We start by stating our results and then proceed with the proofs. Theorem 5.1 There is [MATH] so that for any ladder system [MATH] -colouring [MATH] , there is a (necessarily) special Aronszajn tree [MATH] so that [MATH] uniformizes [MATH] |
Theorem 5.2 There is [MATH] so that for any ladder system [MATH] -colouring [MATH] , there is a non-special tree [MATH] so that [MATH] uniformizes [MATH] |
That is, for both trees [MATH] and [MATH] , there is a single master colouring which witnesses that any ladder system colouring has a [MATH] or [MATH] -uniformization, respectively. In fact, we isolate a relatively simple combinatorial assumption on functions [MATH] that guarantees [MATH] to witness the theorems. |
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