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Corollary 5.3 [MATH] and [MATH] both hold for any ladder system [MATH] Let us mention that, unlike [MATH] or [MATH] , no [MATH] -tree can have a single colouring that uniformizes all ladder system colourings at once. |
Proposition 5.4 Suppose that [MATH] is an [MATH] -tree and [MATH] . Then for any ladder system [MATH] , there is a monochromatic 2-colouring of [MATH] so that [MATH] is not a [MATH] -uniformization of [MATH] whenever [MATH] is pruned. |
Proof. Take any increasing sequence of limit ordinals [MATH] with supremum [MATH] . List [MATH] as [MATH] and define [MATH] to be constant 0 if and only if there are infinitely many [MATH] so that [MATH] . One can immediately check that if [MATH] uniformizes [MATH] for some downward closed subset [MATH] then [MATH] for... |
Let us present the proofs of the theorems now, starting by some preliminaries. Throughout the rest of this section we only work with subtrees of [MATH] and [MATH] ; hence, we assume the tree to be downward closed and pruned. Now, we say that such a tree [MATH] (of possibly countable height) is strongly pruned if for an... |
Being strongly pruned allows the extension of a countable tree with an extra level (while preserving being strongly pruned). Observation 5.5 |
Suppose that [MATH] is a subtree of countable limit height [MATH] . Then for any [MATH] and [MATH] , there is a [MATH] of height [MATH] so that [MATH] [MATH] and [MATH] |
From this, one readily constructs Aronszajn subtrees of [MATH] . Now, our goal is to define uniformizations i.e., to inductively extend a tree and a colouring at the same time. This motivates the next definition. |
Definition 5.6 We say that a colouring [MATH] is rich if for any [MATH] [MATH] and [MATH] there are [MATH] extending [MATH] with [MATH] and [MATH] arbitrary small. |
In turn, rich maps must be defined on strongly pruned trees. The following observation should make it clear how this definition will be useful. |
Observation 5.7 Suppose [MATH] is a subtree of [MATH] of countable limit height [MATH] [MATH] and [MATH] is rich. If [MATH] for some cofinal, type [MATH] subset [MATH] of [MATH] then there is a [MATH] above [MATH] so that |
(1) [MATH] (2) [MATH] is arbitrary small and (3) [MATH] for almost all [MATH] Proof. Indeed, fix some [MATH] and simply define [MATH] in [MATH] inductively so that for all [MATH] |
(1) [MATH] where [MATH] is the [MATH] th element of [MATH] above [MATH] (2) [MATH] , and (3) [MATH] We can always pick the next [MATH] since [MATH] is rich. In the end, [MATH] is as desired. |
In fact, when proving the theorems, we need a stronger notion. We say that [MATH] is an [MATH] -bramble if [MATH] is a cofinal, rooted, binary subtree and for any branch [MATH] [MATH] is a bounded subset of [MATH] (and so [MATH] ). |
Definition 5.8 A map [MATH] is flush if it is rich and for any limit [MATH] and [MATH] -bramble [MATH] , there is some branch [MATH] so that [MATH] for [MATH] |
We define a flush map on [MATH] similarly however, as [MATH] branches at limit steps, we need a bit more care. Definition 5.9 A map [MATH] is flush if it is rich and for any limit [MATH] and [MATH] -bramble [MATH] , there is some branch [MATH] so that [MATH] [MATH] and [MATH] is arbitrary small. |
We will show that there are flush maps on both [MATH] and [MATH] , and that any flush map witnesses Theorem 5.1 and Theorem 5.2 Lemma 5.10 |
There are flush maps on both [MATH] and [MATH] Proof. We construct flush maps [MATH] for [MATH] by induction, so that [MATH] extends [MATH] for [MATH] . The final union [MATH] is the desired map on [MATH] then. |
We suppose that [MATH] has been constructed already and extend it to [MATH] which, in turn, defines [MATH] (in limit steps, we simply take unions). If [MATH] is successor then being flush compared to being rich gives no extra requirements. Note that any element [MATH] has infinitely many successors [MATH] so that [MATH... |
Now, suppose that [MATH] is limit and list all pairs [MATH] as [MATH] where [MATH] is an [MATH] -bramble and [MATH] . We inductively pick distinct [MATH] which is possible since each [MATH] has continuum many branches. Now, simply set [MATH] for [MATH] where [MATH] ; for all other [MATH] of height [MATH] , we set the v... |
Note that we preserved the map being rich. Indeed, for any [MATH] of height less than [MATH] [MATH] and [MATH] , we can build an [MATH] -bramble [MATH] with root [MATH] so that [MATH] implies that [MATH] . In turn, the branch [MATH] we picked for the pair [MATH] will give the element [MATH] that witnesses richness. Thi... |
Working on [MATH] , almost the same argument gives a flush map again. The only slight difference is that after we selected the appropriate branch [MATH] , we pick for each [MATH] [MATH] extending [MATH] so that [MATH] . Now, we declare [MATH] |
Finally, we are ready to prove the main theorems. Proof of Theorem 5.1 Our goal is to show that any flush map [MATH] will work (the existence of which is ensured by Lemma 5.10 ). |
Fix an [MATH] -colouring [MATH] of some ladder system [MATH] . The corresponding tree [MATH] will be constructed by defining its initial segments [MATH] for [MATH] by induction on [MATH] so that |
(a) [MATH] is a countable, downward closed and strongly pruned subtree of [MATH] of height [MATH] (b) [MATH] is an end extension of [MATH] for [MATH] |
(c) [MATH] is rich, and (d) [MATH] uniformizes [MATH] Clearly, if we succeed then [MATH] is the tree we were looking for. In limit steps of the induction, we simply take unions. So suppose that [MATH] is defined and we will find [MATH] now. If [MATH] is a successor ordinal then the extension is straightforward again. |
Now, suppose that [MATH] is limit. For each [MATH] and [MATH] , we construct an [MATH] -bramble [MATH] with the following properties: |
(1) [MATH] has root [MATH] and [MATH] implies that [MATH] (2) for any [MATH] above [MATH] and [MATH] of height [MATH] [MATH] Why is this possible? We build [MATH] as the downward closure of a set [MATH] so that |
(i) [MATH] (ii) [MATH] if and only if [MATH] (iii) [MATH] for any [MATH] , and (iv) if [MATH] then [MATH] and [MATH] where [MATH] is the [MATH] th element of [MATH] above [MATH] |
Given [MATH] , we can easily pick [MATH] and [MATH] in the right position using that [MATH] is rich. Now, since [MATH] is flush, for each [MATH] and each pair [MATH] , there is some branch [MATH] so that [MATH] for an appropriate [MATH] that extends [MATH] . We simply let [MATH] collect the countably many elements of t... |
First, it is clear that we preserved [MATH] being rich on [MATH] . Second, for any [MATH] and for almost any [MATH] [MATH] by property ( iv ) of the [MATH] -brambles [MATH] . In turn, [MATH] is still a rich uniformization, as desired. |
This finishes the inductive construction and hence the proof of the theorem. Finally, we show how to modify the above argument for [MATH] and to produce uniformizations on non-special trees [MATH] (however, these trees will not be Aronszajn any more). |
Proof of Theorem 5.2 Suppose that [MATH] is flush and take any ladder system colouring [MATH] . The corresponding non-special tree [MATH] will be constructed by defining its initial segments [MATH] for [MATH] by induction on [MATH] so that |
(a) [MATH] is a downward closed and strongly pruned subtree of [MATH] of height [MATH] (b) [MATH] is an end extension of [MATH] for [MATH] |
(c) [MATH] is rich, and (d) [MATH] uniformizes [MATH] In limit steps, we will simply take unions, and in successor steps [MATH] where [MATH] is also a successor, we only aim to preserve the above properties. This so far is the same as the proof of Theorem 5.1 except we allowed [MATH] to have uncountable levels. |
Now, the difference (compared to the previous proof) comes at stages when [MATH] is already defined for some limit [MATH] and we aim to construct the [MATH] th level [MATH] . Recall that we need to make sure [MATH] is non-special i.e., for any partition [MATH] there is some [MATH] such that [MATH] |
First, we need a definition: an [MATH] -control tuple is a 4-tuple [MATH] that satisfies the following: (i) [MATH] is a countable, downward closed and pruned subtree of height [MATH] , and [MATH] |
(ii) [MATH] is rich, [MATH] and [MATH] , and (iii) if [MATH] then there are subtrees [MATH] for [MATH] of height [MATH] so that [MATH] and [MATH] |
[MATH] is rich for all [MATH] for any [MATH] [MATH] is either dense in each [MATH] above [MATH] (we call [MATH] a large colour of [MATH] this case), or empty above [MATH] |
Enumerate all [MATH] -control tuples [MATH] extended by an extra pair of natural numbers [MATH] and [MATH] as [MATH] so that each 6-tuple appears [MATH] times. If [MATH] then we fix [MATH] that witnesses condition ( iii ). |
We will define [MATH] to be [MATH] so that for any [MATH] (e) [MATH] and [MATH] (f) [MATH] and [MATH] (g) for all [MATH] above the height of [MATH] [MATH] |
(h) if [MATH] then for any [MATH] , if [MATH] is a large colour of [MATH] then [MATH] for some [MATH] , and (i) if [MATH] then for any [MATH] , there is [MATH] such that [MATH] is either dense or empty above [MATH] |
If we succeed, then conditions ( )-( ) will ensure that [MATH] is a rich uniformization of [MATH] . The last two conditions will help us prove that the final tree is non-special (although, this might not be clear at this point). |
Consider the tuple [MATH] and let us list [MATH] as [MATH] . We build an [MATH] -bramble [MATH] as the downward closure of a set [MATH] so that |
(1) [MATH] and [MATH] if and only if [MATH] (2) [MATH] for any [MATH] (3) [MATH] for all [MATH] and [MATH] (4) if [MATH] then [MATH] for any [MATH] |
(5) if [MATH] and [MATH] is a large colour of [MATH] then for all [MATH] [MATH] , and (6) if [MATH] and [MATH] then for any [MATH] [MATH] is either dense or empty above [MATH] |
If we succeed, it should be clear that we indeed get an [MATH] -bramble [MATH] . Moreover, for any branch [MATH] [MATH] and [MATH] for all [MATH] above [MATH] . So, we can apply that [MATH] is flush and for each [MATH] , we can pick a branch [MATH] so that [MATH] for [MATH] . We do this for all [MATH] which defines [MA... |
We still need to explain the construction of the binary system [MATH] corresponding to a fixed 6-tuple [MATH] . First, choose [MATH] with [MATH] above [MATH] so that [MATH] . This is possible since [MATH] is rich. |
Suppose that [MATH] has been defined and fix some [MATH] . Our goal is to find the right [MATH] and [MATH] Claim 5.11 There is a [MATH] above [MATH] of height at least [MATH] , so that |
(1) if [MATH] then [MATH] , and (2) if [MATH] and [MATH] then [MATH] Proof. We look at the elements of [MATH] between [MATH] and [MATH] , let these be [MATH] . If, additionally, [MATH] then we look at the minimal [MATH] so that [MATH] and [MATH] . We extend our previous list [MATH] by the elements of [MATH] to get [MAT... |
Now, in a finite induction, we define [MATH] in [MATH] (or just in [MATH] if [MATH] ) so that [MATH] and [MATH] . This is possible since [MATH] is rich. |
Now, suppose that [MATH] is given as above. If [MATH] [MATH] and [MATH] is large for [MATH] then we can find [MATH] above [MATH] in [MATH] so that [MATH] for [MATH] . If [MATH] then we can find [MATH] above [MATH] in [MATH] so that [MATH] is either empty or dense above [MATH] for both [MATH] . This defines [MATH] and h... |
At this point, we concluded the construction of the level [MATH] and so the whole tree [MATH] is defined. Why is [MATH] non special? Suppose that [MATH] and we need to find [MATH] so that [MATH] . Take a continuous [MATH] -sequence of countable elementary submodels [MATH] of [MATH] with [MATH] . Let [MATH] [MATH] and n... |
Claim 5.12 There is some [MATH] so that [MATH] is an [MATH] -control tuple. Proof. [MATH] is certainly an [MATH] -control triple so it is enumerated at some step [MATH] . The node [MATH] that we introduced (above [MATH] ) satisfies that for any colour [MATH] [MATH] is either dense or empty above [MATH] by ( ). So this ... |
In turn, at step [MATH] , we enumerated [MATH] as [MATH] for some [MATH] . First, we claim that [MATH] is a large colour of [MATH] . Indeed, [MATH] witnesses that [MATH] cannot be empty above any restriction of [MATH] . However, in this case there is some [MATH] so that [MATH] by condition ( ). In other words, [MATH] a... |
There is another well-investigated class of non-special trees: given a stationary set [MATH] , let [MATH] denote the set of all closed subsets of [MATH] with the end extension relation |
. We believe that a non-essential modification of the above arguments yields the following analogue of Theorem 5.2 : there is a single colouring [MATH] so that any ladder system colouring has a non-special [MATH] -uniformization. |
6. Uniformization results from strong diamonds Recall that [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] |
[EQUATION] Our main results read as follows. Theorem 6.1 [MATH] implies that for any ladder system [MATH] , there is a special Aronszajn tree [MATH] so that [MATH] |
Theorem 6.2 [MATH] implies that for any ladder system [MATH] , there is an Aronszajn tree [MATH] so that any monochromatic [MATH] -colouring of [MATH] has a Suslin [MATH] -uniformization. |
In the latter result, the tree [MATH] is necessarily non-special but we don’t know if it can be made almost Suslin i.e., to contain no stationary antichains. Equivalently, whether [MATH] implies the existence of stationary antichains in [MATH] |
Combined with previous results, we also get the following result. Corollary 6.3 [MATH] implies that for any ladder system , there is an Aronszajn tree [MATH] so that [MATH] holds but [MATH] fails. |
We are not aware of an analogue of this result in the classical context of [MATH] -uniformizations. On a related note, it is proved in |
that consistently, [MATH] holds but [MATH] fails for some [MATH] The idea to prove the above theorems is the following: in the end, no matter how we constructed the tree [MATH] , the [MATH] -sequence [MATH] will allow us to guess initial segments of any ladder system colouring [MATH] club often. So, at some stage [MATH... |
We will use rich maps, strongly pruned trees and some arguments that resemble the work in Section . We state one extra preliminary lemma before proving the theorems. |
Lemma 6.4 Let [MATH] and suppose that [MATH] is a downward closed, strongly pruned subtree of [MATH] of height [MATH] , and [MATH] is a ladder system [MATH] -colouring. If [MATH] is a rich [MATH] -uniformization of [MATH] then there is a rich [MATH] extending [MATH] so that [MATH] is a [MATH] -uniformization of [MATH] |
Proof. Look at the poset [MATH] of all maps [MATH] so that (1) [MATH] where [MATH] is finite, (2) [MATH] is still a function, and |
(3) [MATH] uniformizes : for any [MATH] of limit height [MATH] and for almost all [MATH] [MATH] It is easily checked that for any [MATH] and [MATH] extending some element of [MATH] , there is some [MATH] so that [MATH] and [MATH] . So, one can find a filter [MATH] so that [MATH] is a rich map on a strongly pruned subtr... |
Finally, given a countable set [MATH] , we will say that some set [MATH] is [MATH] -definable if [MATH] whenever [MATH] is a countable elementary submodel of [MATH] with [MATH] , where [MATH] is some fixed well-order on [MATH] |
Let us prove the theorems now. Proof of Theorem 6.1 Let [MATH] denote the [MATH] sequence, and let [MATH] for [MATH] The tree [MATH] will be constructed by defining its initial segments [MATH] for [MATH] by induction on [MATH] , along with countable sets [MATH] and so called sealing functions |
[EQUATION] In the end, these functions will tell us how to continue a partial [MATH] -uniformization. We assume that these objects satisfy the following properties, which we preserve throughout the induction: |
(a) [MATH] is a countable, downward closed strongly pruned and uniquely [MATH] -definable tree, (b) [MATH] is an end extension of [MATH] for [MATH] |
and for any limit [MATH] (c) [MATH] is the set of all [MATH] -definable rich maps [MATH] so that [MATH] is strongly pruned, (d) [MATH] is uniquely [MATH] -definable, and |
(e) for all [MATH] (i) [MATH] is a downward closed, strongly pruned subtree of [MATH] , and (ii) [MATH] for all [MATH] and almost all [MATH] |
These assumptions ensure that for any [MATH] , there is some rich extension [MATH] of [MATH] that is defined on [MATH] . Moreover, if [MATH] was a uniformization for some colouring [MATH] of [MATH] , then [MATH] is a uniformization of [MATH] whenever [MATH] is constant [MATH] |
First, the less interesting cases: in limit steps, we simply take unions (which preserves being strongly pruned and the definability requirements). If [MATH] is successor and [MATH] is defined then we canonically extend [MATH] by a level [MATH] so that [MATH] remains strongly pruned. |
Now, we describe the construction of [MATH] for a limit [MATH] when [MATH] is defined already. To this end, fix some [MATH] [MATH] and [MATH] . Consider the poset [MATH] of all finite, non empty chains [MATH] above [MATH] so that for all [MATH] |
(1) if [MATH] and [MATH] then [MATH] , and (2) [MATH] Using that [MATH] is rich, it is easy to check that for any [MATH] [MATH] is dense in [MATH] . Now, take a finite support product [MATH] so that for all [MATH] as above, there are infinitely many [MATH] so that [MATH] . It is easy to find a sufficiently generic filt... |
[EQUATION] The sealing function [MATH] is defined as follows: given [MATH] and [MATH] , let [EQUATION] Since the poset [MATH] is uniquely definable from [MATH] , we can make a canonical choice of [MATH] (using the well order [MATH] ), and so [MATH] and [MATH] are uniquely definable from [MATH] as well. |
This finishes the construction of our (special) strongly pruned Aronszajn tree [MATH] Suppose now that is a monochromatic [MATH] -colouring of [MATH] , and we would like to find a [MATH] -uniformization. There is a club [MATH] so that [MATH] implies that [MATH] . Let [MATH] be the increasing enumeration of [MATH] , and... |
We will define a [MATH] -increasing sequence of functions [MATH] so that (1) [MATH] i.e., [MATH] is a rich [MATH] -definable map on a strongly pruned subtree of [MATH] , and |
(2) [MATH] uniformizes [MATH] If we succeed, then [MATH] is the desired [MATH] -uniformization of [MATH] Start by taking the [MATH] -minimal rich map [MATH] which uniformizes [MATH] ; note that this is possible by Lemma 6.4 and that [MATH] as [MATH] |
Now, suppose we constructed some [MATH] . In turn, [MATH] is defined (where [MATH] is the constant value of [MATH] ). So, we can take a [MATH] -minimal map |
[EQUATION] that is a rich extension of [MATH] . By the properties of the sealing function, [MATH] remains a uniformization of [MATH] . Now, take the [MATH] -minimal rich extension [MATH] of [MATH] that uniformizes [MATH] ; this exists by Lemma 6.4 . The minimal choices and that [MATH] implies that [MATH] |
In limit steps [MATH] , we simply take unions: [MATH] . This map is uniquely definable from the parameters [MATH] (due to the canonical choices we made along the way), and these parameters are all in [MATH] . Hence [MATH] is [MATH] -definable as desired i.e., [MATH] |
This finishes the induction and hence the construction of the [MATH] -uniformization [MATH] To summarize, we constructed an Aronszajn tree [MATH] that satisfies [MATH] |
Our next goal is to prove the second theorem about Suslin uniformizations, which will be quite similar. We will construct [MATH] as a subtree of [MATH] instead of [MATH] , and make sure that the sealing functions preserve maximal antichains that are guessed by the diamond sequence. |
We need a small definition before the proof: suppose that [MATH] is a countable tree of height [MATH] and [MATH] is a maximal antichain. We say that [MATH] is [MATH] -reflecting if the set |
[EQUATION] is cofinal in [MATH] Proof of Theorem 6.2 We still use to denote the [MATH] -sequence. [MATH] will be built by constructing its initial segments [MATH] along with [MATH] and [MATH] with the following properties: |
(a) [MATH] is a countable, downward closed, normal tree which is uniquely definable from [MATH] (b) [MATH] is an end extension of [MATH] for [MATH] |
and for any limit [MATH] (c) [MATH] is the set of all [MATH] -definable rich maps [MATH] so that [MATH] is downward closed and normal, |
(d) [MATH] is uniquely definable from [MATH] , and (e) for all [MATH] (i) [MATH] is a downward closed subtree of [MATH] (ii) [MATH] for all [MATH] and almost all [MATH] , and |
(iii) if [MATH] is an [MATH] -reflecting maximal antichain in [MATH] then any [MATH] is above some element of [MATH] As before, the interesting case in the construction is when [MATH] is constructed already for a limit [MATH] and we aim to define [MATH] . For any [MATH] [MATH] and [MATH] , we now consider the poset [MA... |
We take a finite support product [MATH] so that for all [MATH] as above, there are infinitely many [MATH] so that [MATH] . In order to find the right generic branches, we need a new density lemma. |
Claim 6.5 Suppose that [MATH] is an [MATH] -reflecting maximal antichain. Then [EQUATION] is dense in [MATH] Proof. Fix some [MATH] -reflecting maximal antichain [MATH] and suppose that [MATH] is arbitrary. Let [MATH] and find some limit [MATH] above [MATH] so that [MATH] is a maximal antichain in [MATH] . List the ele... |
Now, either [MATH] already (which means [MATH] ) or we can find some [MATH] above [MATH] so that [MATH] . Note that no new elements of [MATH] appear between the heights of [MATH] and [MATH] , so we can define [MATH] which is an extension of [MATH] in [MATH] |
Now, it is easy to find a sufficiently generic filter [MATH] so that we get pairwise different branches [MATH] in [MATH] from the [MATH] th coordinate of [MATH] and [MATH] for any [MATH] and [MATH] -reflecting maximal antichain [MATH] so that [MATH] (where [MATH] ). |
We define [MATH] by adding unique upper bounds [MATH] for each branch [MATH] , just as in the proof of Theorem 6.1 . The sealing function [MATH] is defined as follows: given [MATH] and [MATH] , let |
[EQUATION] Since the poset [MATH] is uniquely definable from [MATH] , we can make a canonical choice of [MATH] , and so [MATH] and [MATH] are [MATH] -definable as well. |
This finishes the construction of an Aronszajn tree [MATH] Suppose that we are given some monochromatic [MATH] -colouring of [MATH] . The diamond sequence and sealing functions provide a unique way to construct a sequence of maps [MATH] along a club [MATH] so that |
(1) [MATH] (2) [MATH] where [MATH] is the constant value of [MATH] , and (3) [MATH] uniformizes [MATH] So [MATH] is a [MATH] -uniformization of [MATH] . We need to show that [MATH] is Suslin; this amounts to proving that if [MATH] is a maximal antichain then [MATH] is countable. The set [MATH] of those [MATH] so that [... |
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