PProfinite tree sets
Jakob KneipSeptember 30, 2019
Tree sets are posets with additional structure that generalize tree-like objectsin graphs, matroids, or other combinatorial structures. They are a specialclass of abstract separation systems.We study infinite tree sets and how they relate to the finite tree sets theyinduce, and obtain a characterization of infinite tree sets in combinatorialterms.
This paper is a sequel to, and assumes familiarity with, two earlier papers [2, 5]. Thefirst of these [2] introduced finite abstract separation systems, whereas the latter [5] laidthe foundations for extending the theory of separation systems to a broad class of infiniteseparation systems.The theory of abstract separation systems introduced in [2] aims to generalize thenotion of tangles, a notion originally invented and studied by Robertson and Seymourin [7] to capture highly connected objects or regions in graphs, from graphs to othertypes of highly cohesive regions in graphs, matroids, or other combinatorial structures.The fundamental idea of Robertson and Seymour in [7] was to describe dense objectsin graphs not directly, say by specifying a set of vertices, but indirectly, by having eachlow-order separation of the graph point towards that object. In contrast to specifyinga list of vertices, this indirect approach allows one to capture objects or regions in thegraph that are highly-connected in a global sense but not locally. A typical example forsuch a region is a large grid in a graph: since every vertex of a grid has low degree, thegrid cannot be said to be locally highly connected. However, a large grid still constitutesa dense and highly cohesive structure in a graph, as witnessed by the fact that a largegrid forces high tree-width.For tangles in graphs, Robertson and Seymour [7] proved two fundamental types oftheorems: a tree-of-tangles theorem , which shows how to find a tree-like shape in thegraph which displays all the different tangles in that graph, and a tangle-tree dualitytheorem , which shows that if a graph has no tangles (of a particular order), then theentire graph can be cut by low-order separations into a tree-like structure witnessing thisabsence of tangles. Both of these types of theorems can be established in the framework1 a r X i v : . [ m a t h . C O ] S e p f abstract separation systems. In fact, for the special case of separations of graphs,Robertson and Seymour’s original version of these theorem can be obtained from thegeneralized abstract theorems [4, 6].In [5] the foundations were laid for extending the tree-of-tangles theorem and thetangle-tree duality theorem to infinite separation systems: [5] introduced, and studied,separation systems that are profinite – i.e. which are determined by the finite separationsystems they induce. For this class of separation systems it is then possible to establisha tree-of-tangles theorem and a tangle-tree-duality theorem by applying and lifting theirfinite versions using compactness arguments [1].The central object in both the tree-of-tangles theorem as well as the tangle-tree du-ality theorem for abstract separation systems, apart from the tangles themselves, is thestructure of a tree set , a nested separation system without any trivial elements: thetree-of-tangles theorem finds a tree set which distinguishes a given set of tangles, andthe tangle-tree duality theorem finds a tree set witnessing that there are no tangles. Theunderstanding of extensions of these theorems from finite to profinite separation systemsthus necessitate a thorough understanding of the properties of profinite tree sets.A recurring question for profinite separation systems is the following: if every inducedfinite subsystem has a certain property, does this property carry over to the profinitesystem – and conversely, if the profinite separation system has a certain structure, canits induced finite subsystems be assumed to have that structure, too? In [5] affirmativeanswers to both parts of this question were given for two of the most basic properties ofseparation systems: nestedness and regularity. A profinite separation system is nestedas soon as all its finite subsystems are; and every nested profinite separation system canbe obtained from suitable finite nested systems. For the second part a straightforwardcompactness argument is used to show that, in fact, all relevant separations of almost allfinite subsystems are nested. The same assertion, and indeed the same general argument,holds for regularity, too [5, Proposition 5.6].In this paper we give a positive answer to the above question for the structural propertyof being a tree set, that is, being nested and containing no trivial separations (see [2, 5]for formal definitions). Concretely, we show the following: Theorem 1.1. (i) Every inverse limit of finite tree sets is a tree set.(ii) Every profinite tree set is an inverse limit of finite tree sets.The first part of Theorem 1.1 is similar to the first part of Proposition [5, Proposi-tion 5.6]. However, the second part of Theorem 1.1 is much more difficult to show. Thereason for this is that the compactness arguments used for nestedness and regularity donot work for tree sets: as [5, Example 5.7] shows it is possible that a separation is non-trivial in the profinite tree set, but that each of its induced finite separations is trivialin the respective finite subsystem. This makes it impossible to obtain the profinite treeset from finite tree sets by simply passing to suitable finite subsystems, as in the proofof [5, Proposition 5.6]. 2n order to prove Theorem 1.1 we will formulate a way of breaking up a given profinitetree τ set into finitely many parts in such a way that these parts form a finite tree set. Thetechnical details of this are somewhat involved and are laid out in Section 3.2. Followingthat, in Section 3.3, we will show that a carefully selected family of these finite tree setscan be used to re-obtain the profinite tree set τ . To get this re-assembly of τ to work weshall need to assume some that τ has certain structural properties; thus, to finish ourproof of Theorem 1.1, we then need to verify that all profinite tree sets indeed have thesestructural properties. In doing so we will also obtain a characterization of the profinitetree sets in purely combinatorial terms.Finally, in Section 4, we apply the knowledge gained in the previous sections to extendcertain representation theorems to profinite tree sets. In those theorems we seek torepresent a tree set τ as a separation system of bipartitions of a suitable groundset.This is easy to do in principle (see [3]), but becomes a challenging problem when onewants to minimize the groundset used for the representation. For definitions and a basic discussion of abstract separation systems as well as of treesets we refer the reader to [2] and [3]. Additionally, we shall use the following terms.We call a separation co-small if its inverse is small. If an oriented separation → r istrivial and this is witnessed by some separation s , we also call the orientations → s , ← s of s witnesses of the triviality of → r . If σ is a splitting star of some separation system → S ,and σ has size at least three, we call σ a branching star of → S and its elements branchingpoints of → S .We shall be using the following two lemmas from [2]: Lemma 2.1 (Extension Lemma) . [2, Lemma 4.1] Let S be a set of unoriented separa-tions, and let P be a consistent partial orientation of S . (i) P extends to a consistent orientation O of S if and only if no element of P isco-trivial in S . (ii) If → p is maximal in P , then O in (i) can be chosen with → p maximal in O if andonly if → p is nontrivial in → S . (iii) If S is nested, then the orientation O in (ii) is unique. Nested separation systems without degenerate or trivial elements are known as treesets . A subset σ ⊆ → S splits a nested separation system → S if → S has a consistent orientation O such that O ⊆ (cid:100) σ (cid:101) and σ is precisely the set of maximal elements of O . Conversely,a consistent orientation O of → S splits (at σ ) if it is contained in the down-closure of theset σ of its maximal elements.The consistent orientations of a finite nested separation system can be recovered fromits splitting subsets by taking their down-closures, but infinite separation systems canhave consistent orientations without any maximal elements.3 emma 2.2. [2, Lemmas 4.4, 4.5] The splitting subsets of a nested separation system → S without degenerate elements are proper stars. Their elements are neither trivial norco-trivial in → S . If S has a degenerate element s , then { → s } is the unique splitting subsetof → S . The splitting stars of the edge tree set → E ( T ) of a tree T , for example, are the sets (cid:126)F t of edges at a node t all oriented towards t . By this correspondence, the nodes of T canbe recovered from its edge tree set; if T is finite, its nodes correspond bijectively to itsconsistent orientations.Given two separation systems R, S , a map f : → R → → S is a homomorphism of separationsystems if it commutes with their involutions and respects the ordering on → R . Formally,we say that f commutes with the involutions of → R and → S if (cid:0) f ( → r ) (cid:1) ∗ = f ( ← r ) for all → r ∈ → R .It respects the ordering on → R if f ( → r ) ≤ f ( → r ) whenever → r ≤ → r . Note that the conditionfor f to be order-respecting is not ‘if and only if’: we allow that f ( → r ) ≤ f ( → r ) also forincomparable → r , → r ∈ R . Furthermore f need not be injective. It can therefore happenthat → r ≤ → r with → r (cid:54) = → r but f ( → r ) = f ( → r ), so f need not preserve strict inequality.A bijective homomorphism of separation systems whose inverse is also a homomorphismis an isomorphism .We shall now prove two handy lemmas which provide sufficient conditions for a homo-morphism of separation systems to be an isomorphism. These lemmas will be tailoredtowards their intended applications in Section 3 but may be of some use in general.As all trivial separations are small every regular nested separation system is a treeset. These two properties, regularity and nestedness, are preserved by homomorphismsof separations systems, albeit in different directions: the image of nested separations isnested, and the pre-image of regular separations is regular. Lemma 2.3.
Let f : → R → → S be a homomorphism of separation systems. If S is regularthen so is R ; and if R is nested then so is its image in S .Proof. First suppose some → r ∈ → R is small, that is → r ≤ ← r . But then f ( → r ) ≤ f ( ← r ) = (cid:0) f ( → r ) (cid:1) ∗ , so → S contains a small element. Therefore if S is regular then R must be too.Now consider two unoriented separations s, s (cid:48) ∈ S . If there are r, r (cid:48) ∈ R with s = f ( r )and s (cid:48) = f ( r (cid:48) ) and R is nested, then, say, → r ≤ → r (cid:48) and thus → s := f ( → r ) ≤ f ( → r (cid:48) ) =: → s (cid:48) .Hence if R is nested its image in S is nested too.Lemma 2.3 makes it possible to show that a homomorphism f : → R → → S of separationsystems is an isomorphism of tree sets without knowing beforehand that either R or S is a tree set: Lemma 2.4.
Let f : → R → → S be a bijective homomorphism of separation systems. If R is nested and S is regular then f is an isomorphism of tree sets.Proof. From Lemma 2.3 it follows that both R and S are regular and nested, whichmeans they are regular tree sets. Therefore all we need to show is that the inverse of f
4s order-preserving, i.e. that → r ≤ → r whenever f ( → r ) ≤ f ( → r ). Let → r , → r ∈ → R with f ( → r ) ≤ f ( → r ) be given. Since R is nested r and r have comparable orientations.If → r ≥ → r then f ( → r ) = f ( → r ), implying → r = → r and hence the claim.If → r ≤ ← r then f ( → r ) ≤ f ( → r ) , f ( ← r ), contradicting the fact that S is a regular treeset.Finally if → r ≥ ← r then f ( ← r ) ≤ f ( → r ), contradicting the fact that S is regular.Hence → r ≤ → r , as desired.In our applications we sometimes already know that S is a tree set, but not that S is regular. The proof of Lemma 2.4 still goes through though if we know that thepre-images of small separations are small: Lemma 2.5.
Let f : → R → → S be a bijective homomorphism of separation systems. If R is nested, S is a tree set, and → r ∈ → R is small whenever f ( → r ) is small, then f is anisomorphism of tree sets.Proof. It suffices to show that the inverse of f is order-preserving. Let → r , → r ∈ → R with f ( → r ) ≤ f ( → r ) be given. If f ( → r ) = f ( → r ) or f ( ← r ) = f ( → r ) we have → r ≤ → r byassumption. Therefore we may assume that f ( → r ) (cid:8) f ( → r ) and hence r (cid:54) = r . Since R is nested r and r have comparable orientations.If → r ≥ → r then f ( → r ) = f ( → r ) contradicting f ( → r ) (cid:8) f ( → r ).If → r (cid:8) ← r then f ( → r ) (cid:8) f ( → r ) , f ( ← r ), contradicting the fact that S contains notrivial element.Finally if → r ≥ ← r then f ( ← r ) (cid:8) f ( → r ) , f ( ← r ), again contradicting the fact that S contains no trivial element.Hence → r ≤ → r , as desired.Finally, we introduce the following non-standard notation: for separations → r and → s insome separation system we write → r (cid:8) → s if → r ≤ → s and r (cid:54) = s . Note that this is not thesame as → r < → s : if → r < → s , that is, if → r ≤ → s and → r and → s differ as oriented separations,then r and s could still be the same unoriented separation if → r = ← s . On the other hand,if → r (cid:8) → s , then r and s must be distinct as unoriented separations. In nearly all placesin this paper in which we consider separations → r and → s with → r ≤ → s , and → r and → s aredifferent oriented separations, we shall also want r and s to be different as unorientedseparations and hence will write → r (cid:8) → s . To avoid confusion we shall, from now on,never use the symbol ‘ < ’ again. We refer the reader to [5] for an introduction to inverse limits of sets, inverse systems ofseparation systems, and profinite separation systems. In particular we assume familiaritywith Section 3, Section 4, and Section 5 up to Example 5.7 of [5]; we shall follow theterms and notation given there. 5et S = ( → S p | p ∈ P ) be an inverse system of finite separation systems and → S = lim ←− S its inverse limit. The separations in → S are of the form → s = ( → s p | p ∈ P ) with → s p ∈ → S p .To enhance readability, when no confusion is possible, we will simply write → s ∈ → S forseparations in → S and implicitly assume that → s = ( → s p | p ∈ P ). Thus, if no context isgiven, → s p will always be the projection of → s to → S p .We are interested in the relations between properties of the → S p and the properties of → S :which structural properties of separation systems can be ‘projected downwards’ from → S ,and which can be ‘lifted upwards’ from the → S p to → S ? In [5] this question was answeredfor the two properties of being nested, and being regular: both of these properties ‘liftup’ in the sense that if all → S p are nested (resp. regular), then → S is nested (resp. regular),too. Moreover, both of these properties also ‘project downwards’: if → S is nested (resp.regular), then the → S p can be assumed to be nested (resp. regular), too. More precisely:every nested (resp. regular) profinite separation system is the inverse limit of nested(resp. regular) finite separation systems.Indeed, the following was shown in [5]: Proposition 3.1 ([5]) . (i) Every inverse limit of finite regular separation systems is regular.(ii) Every profinite regular separation system is an inverse limit of finite regular sepa-ration systems.The same assertion holds with ‘regular’ being replaced by ‘nested’ and follows from [5,Lemma 5.4].The proof of Proposition 3.1 is straightforward: (i) follows directly from the definitionof an inverse limit of separation systems. For (ii), one observes that if every → S p in aninverse system S = ( → S p | p ∈ P ) contains a small separation, then these small separationlift to a common element of → S = lim ←− S , which will be small, too. Therefore if → S isregular then S must already contain a sub-system of regular finite separation systemswhose inverse limit is → S .However, already in [5] it was observed that the same is not true for trivial separations:it is possible that some → s = ( → s p | p ∈ P ) ∈ → S is non-trivial in → S , but its projections → s p ∈ → S p are trivial in → S p for all p ∈ P . See [5, Example 5.7] for an example of this behaviour.The problem with these so-called finitely trivial separations is that the witnesses of thetriviality of → s p in → S p may not lift to a witness of the triviality of → s in → S .The aim of this section is to overcome the difficulties laid out above and establish thefollowing theorem: Theorem 1.1. (i) Every inverse limit of finite tree sets is a tree set.(ii) Every profinite tree set is an inverse limit of finite tree sets.As [5, Example 5.7] shows, it is not possible to prove (ii) of Theorem 1.1 with adirect compactness argument. Therefore a new approach is needed in order to establish6heorem 1.1. Before we get to this we shall briefly deal with the much easier specialcase of regular tree sets:
Theorem 3.2. (i) Every inverse limit of regular finite tree sets is a regular tree set.(ii) Every regular profinite tree set is an inverse limit of regular finite tree sets.
Proof.
Assertion (i) follows directly from [5, Lemma 5.4], [5, Proposition 5.6(i)] andthe fact that every regular nested separation system is a regular tree set as all trivialseparations are small.For (ii) let → S be a regular tree set with → S = lim ←− S for an inverse system S = ( → S p | p ∈ P ). We may assume without loss of generality that S is surjective; then every → S p is nested. Furthermore as seen in the proof of [5, Proposition 5.6(ii)] there exists some p ∈ P for which → S p is regular. Now ( → S p | p ≥ p ) is the desired inverse system of finiteregular tree sets.Let us now return to Theorem 1.1. Its first part is straightforward to prove: Proof of Theorem 1.1 (i). Let S = ( → S p | p ∈ P ) be an inverse system of finite treesets and → S = lim ←− S . Suppose some → s = ( → s p | p ∈ P ) ∈ → S is trivial in → S with witness r .Since P is a directed set there is a p ∈ P such that → s p (cid:54) = → r p and → s p (cid:54) = ← r p . But → s (cid:8) → r , ← r in → S implies that → s p (cid:8) → r p , ← r p in S p , contrary to the assumption that S p is a tree set.Moreover if every S p is nested then so is → S by [5, Lemma 5.4].Therefore → S is a tree set. (cid:3) We will postpone the proof of (ii) until the end of Section 3.3. Our approach shallbe to decompose a given tree set τ into finitely many parts which together form a finitetree set. The family of all these ‘quotients’ of τ should then form an inverse systemwhose inverse limit is precisely τ . However the exact definition of these decompositionsis fairly technical, and we introduce it in the next section. This section lays the technical foundations for the proof of Theorem 1.1. Given a treeset τ our aim is to find a way of defining finite quotients of τ that form an inverse systemof finite tree sets whose inverse limit is isomorphic to τ . The latter part of this will bedone in the next two sections, while in this section we define these ‘finite quotients’ andanalyse their properties.To this end for any finite set of stars in τ we define an equivalence relation on τ whichessentially breaks up τ into finitely many chunks. After proving a few basic facts aboutthis equivalence relation we find certain conditions that ensure that the equivalenceclasses of τ form a finite tree set, as needed for the proof of Theorem 1.1(ii).Following this main part of the section we analyse these equivalence relations a bitmore and prove a few key lemmas. 7et τ be a tree set. A selection of τ is a non-empty finite set D ⊆ τ of orientedseparations with | σ ∩ D | (cid:54) = 1 for every splitting star σ of τ .Let us show that any selection D of τ divides τ into different sections between thestars that meet D . We make this precise by defining an equivalence relation ∼ D on τ .Recall that, for → r , → s ∈ τ , we write → r (cid:8) → s as shorthand notation for → r ≤ → s and r (cid:54) = s .For a separation → s ∈ τ and a selection D let D + ( → s ) := (cid:110) → d ∈ D | → d (cid:8) → s (cid:111) and D − ( → s ) := D + ( ← s ) . Two separations → s , → r are D - equivalent for a selection D , denoted as → s ∼ D → r , if D + ( → s ) = D + ( → r ) and D − ( → s ) = D − ( → r ). This is an equivalence relation with finitely many classes.We write [ → s ] D for the equivalence class of → s ∈ τ under ∼ D . A separation → d ∈ D distinguishes → r and → s if → d ∈ (cid:0) D + ( → r ) (cid:1) (cid:52) (cid:0) D + ( → s ) (cid:1) or → d ∈ (cid:0) D − ( → r ) (cid:1) (cid:52) (cid:0) D − ( → s ) (cid:1) . Thus → r and → s are D -equivalent if and only if no → d ∈ D distinguishes them.For a selection D of τ and separations → r ≤ → s it follows from the definitions that D + ( → r ) ⊆ D + ( → s ) and D − ( → r ) ⊇ D − ( → s ). Furthermore D + ( → s ) ∩ D − ( → s ) = ∅ for all → s ∈ τ as any element of this intersection would be trivial with witness → s . This impliesthat → s ∼ D ← s if and only if D + ( → s ) = D − ( → s ) = ∅ . But D + ( → s ) ∪ D − ( → s ) is never emptyand in fact contains an element of every splitting star that meets D , so → s (cid:54)∼ D ← s forevery → s ∈ τ .The next lemma shows a few basic properties of ∼ D . The first of these is especiallyimportant, as it will enable us to turn the equivalence classes of ∼ D on τ into a separa-tion system. Lemma 3.3.
Let τ be a tree set, D a selection and → r , → s , → t ∈ τ . (i) If → r ∼ D → s then ← r ∼ D ← s . (ii) If → r ≤ → s ≤ → t and → r ∼ D → t then → r ∼ D → s ∼ D → t . (iii) If there is no → d ∈ D with → d (cid:8) → s , and → r ≤ → s , then → r ∼ D → s . (iv) If → r ≤ → s ≤ ← t and → s ∼ D → t then → r ∼ D → s . (v) If → r ≥ → s ≥ ← t and → s ∼ D → t then → r ∼ D → s .Proof. (i) This follows from D + ( ← s ) = D − ( → s ) = D − ( → r ) = D + ( ← r )8nd D − ( ← s ) = D + ( → s ) = D + ( → r ) = D − ( ← r ) . (ii) By the observation above D + ( → r ) ⊆ D + ( → s ) ⊆ D + ( → t ) = D + ( → r )and similarly D − ( → r ) = D − ( → s ), hence → r ∼ D → s .(iii) By assumption D + ( → r ) = D + ( → s ) = ∅ . Furthermore D − ( → s ) ⊆ D − ( → r ). Supposethere is a → d ∈ D with → d (cid:8) ← r but not → d (cid:8) ← s . Then ← d (cid:8) → s as → d (cid:54) (cid:8) → s by assumption.Let σ be the splitting star containing → d and → e ∈ σ ∩ D with → d (cid:54) = → e . Then → e ≤ ← d (cid:8) → s by the star property, contradicting the assumption that there is no such → e ∈ D .(iv) As D + ( → s ) ⊆ D + ( ← t ) = D + ( ← s ) there can be no → d ∈ D with → d (cid:8) → s as it would betrivial with witness s . Thus → s ∼ D → r by (iii).(v) This follows from (i) and (iv).We now define precisely in which way we want to turn the equivalence classes of τ into a separation system.Let D be a selection of a tree set τ . Then write τ /D := ( { [ → s ] D | → s ∈ τ } , ≤ , ∗ )with ([ → s ] D ) ∗ := [ ← s ] D and [ → s ] D ≤ [ → r ] D if there are → s (cid:48) ∈ [ → s ] D and → r (cid:48) ∈ [ → r ] D with → s (cid:48) ≤ → r (cid:48) .If for some selection D the relation ≤ of τ /D is a partial order then τ /D is a separationsystem by Lemma 3.3(i). In that case τ /D would even be nested because τ is.Our aim is to ensure that τ /D is a tree set. For this we first find sufficient conditionsfor ≤ to be a partial order, and then show that these conditions are strong enough toensure that τ /D does not contain any trivial elements.The relation ≤ on τ /D is reflexive by definition, thus we need to show that it istransitive and anti-symmetric. For the latter no further assumptions are needed, so webegin by proving the anti-symmetry. Lemma 3.4.
Let τ be a tree set and D a selection. If → r ≤ → x and → s ≥ → y for → r , → s , → x , → y ∈ τ with → r ∼ D → s and → x ∼ D → y then also → r ∼ D → x .Proof. We have D + ( → r ) ⊆ D + ( → x ) = D + ( → x ) ⊆ D + ( → s ) = D + ( → r )and similarly D − ( → r ) = D − ( → x ).This shows that ≤ on τ /D is antisymmetric. To prove transitivity we need furtherassumptions, as the following example demonstrates.9 xample 3.5. Let T be the following graph.1 2 3 4 56The edge tree set τ ( T ) (as defined in [3]) is regular and we have (6 , ≤ (3 ,
2) and(4 , ≤ (3 , D = { (1 , , (3 , , (3 , , (5 , } the edges (2 ,
3) and(3 ,
4) get identified in τ ( T ) /D , implying[(6 , D ≤ [(3 , D = [(4 , D ≤ [(3 , D in τ ( T ) /D . But (6 , (cid:54)≤ (3 , ≤ is not transitive on τ ( T ) /D .This example exploits the fact that there is a branching star (i.e., a splitting star ofsize at least three) that does not meet D between two splitting stars that do meet D . Inorder to prevent this counterexample to transitivity one could ask that D meets everybranching star that lies between two elements of D . But this alone is not enough toensure that ≤ on τ /D is transitive: if we replace the separation (6 ,
3) in Example 3.5above with a chain of order type ω the resulting tree set would not have any branchingstars, but the transitivity of ≤ would still fail for the same reason. Therefore we alsoneed an assumption on τ that ensures that whenever there is a three-star as in theexample above we can also find a branching star, which would then be subject to thecondition on D .Recall that → b ∈ τ is a branching point of τ if → b lies in a splitting star of size at leastthree.Call a selection D of a tree set τ branch-closed if → b ∈ D for every branching point → b of τ for which there are → d , → d ∈ D with → d ≤ → b ≤ ← d . Furthermore τ is chain-complete if every non-empty chain C ⊆ τ has a supremum in τ .Let us see an example of a chain-complete tree set: Example 3.6.
Let X be a non-empty set of positive real numbers and let τ ( X ) := (cid:32) (cid:91) x ∈ X { x, − x } , (cid:52) , ∗ (cid:33) , where x ∗ := − x , and x (cid:52) y if and only if x ≤ y as real numbers and x and y have thesame sign. Then τ ( X ) is a tree set. Moreover, τ ( X ) is chain-complete if and only if X is compact as a subset of R .We claim that the two conditions that τ is chain-complete and D branch-closed areenough to ensure that ≤ on τ /D is transitive and hence a partial order. Before weprove this claim we need to establish some basic properties of chain-complete tree sets,beginning with the fact that every chain has not only a supremum but an infimum too: This differs from the common definition of a chain-complete poset, which usually omits the ‘non-empty’, as it does not imply that τ has a smallest element. emma 3.7. Let τ be a chain-complete tree set and C a chain. Then C has an infimumin τ .Proof. Consider the chain C (cid:48) := (cid:110) ← t | → t ∈ C (cid:111) and let ← s be its supremum in τ . Then → s is the infimum of C .Recall that an orientation O of τ is splitting if every element of O lies below somemaximal element of O .The usual way to find a branching star in a tree set τ is to define a consistent orienta-tion with three or more maximal elements and then show that it is splitting. The firstpart can be done with the Extension Lemma (Lemma 2.1). For the latter part the fol-lowing lemma provides a sufficient condition for a consistent orientation to be splitting.It turns out that having two maximal elements is already enough, if τ is chain-complete: Lemma 3.8.
Let τ be a chain-complete tree set and O a consistent orientation of τ withtwo or more maximal elements. Then O is splitting.Proof. Let → r , → s be two maximal elements of O and let → t ∈ O be any separation. If → t lies below → r or → s there is nothing to show. If not consider the up-closure C := (cid:98) → t (cid:99) ⊆ O of → t in O . As O is consistent C is a chain, which has a supremum → m ∈ τ by assumption. ← r and ← s are upper bounds for C , so ← m ≥ → r , → s . But → r and → s are maximal in O ,implying ← m / ∈ O and hence → m ∈ O .With Lemma 3.8 we can now show that if we have a three-star in a chain-completetree set we can find a branching star ‘in the same location’: Proposition 3.9.
Let τ be a chain-complete tree set and σ a star with exactly threeelements. Then there is a unique branching star σ (cid:48) of τ such that every element of σ lies below a different element of σ (cid:48) .Proof. Let σ = { → r , → s , → t } and R = (cid:110) → x ∈ τ | → r ≤ → x ≤ ← s and → r ≤ → x ≤ ← t (cid:111) . Then R is a chain and by assumption → r (cid:48) = sup R exists. As ← s and ← t are lower boundsfor R we have → r (cid:48) ∈ R . Similarly define S = (cid:110) → x ∈ τ | → s ≤ → x ≤ ← r and → s ≤ → x ≤ ← t (cid:111) and T = (cid:110) → x ∈ τ | → t ≤ → x ≤ ← r and → t ≤ → x ≤ ← s (cid:111) as well as → s (cid:48) = sup S and → t (cid:48) = sup T . Since R, S and T are disjoint { → r (cid:48) , → s (cid:48) , → t (cid:48) } forms athree-star. Observe that there is no → x ∈ τ that lies strictly between → r (cid:48) and ← s (cid:48) : for → x ∈ τ with → r (cid:48) ≤ → x ≤ ← s (cid:48) then neither → x ≤ → t nor → x ≥ → t since that would make r or s trivial,11espectively. Thus either → x ≤ ← t or → x ≥ → t , giving → x ∈ R or ← x ∈ S , respectively, andthus equality with → r (cid:48) or ← s (cid:48) by their maximality.Similarly, there are no separations strictly between → s (cid:48) and ← t (cid:48) , or between → t (cid:48) and ← r (cid:48) .Applying Lemma 2.1 to { → r (cid:48) , → s (cid:48) , → t (cid:48) } yields a consistent orientation O in which → r (cid:48) ismaximal. By the above observation → s (cid:48) and → t (cid:48) are maximal in O too. It follows fromLemma 3.8 that O is splitting, so the set σ (cid:48) of its maximal elements is the desiredbranching star.The uniqueness follows from the fact that if σ and σ are two distinct splitting stars,there is a → s ∈ σ which is an upper bound for all elements of σ but one. Hence if threeseparations lie below different elements of σ , at least two of them will lie below thesame element of σ .This proposition is useful as it often allows us to work with splitting stars withoutloss of generality in the context of selections. We will also use it in the next section,especially the uniqueness part which is not important in this section.We now show in three steps that ≤ on τ /D is transitive for branch-closed D andchain-complete τ . First we show that if a counterexample to the transitivity exists itmust be a three-star with one element equivalent to the inverse of the second, as inExample 3.5. Then we apply Proposition 3.9 to this three-star to obtain a branchingstar, of which we show that it is still a counterexample to the transitivity. Finally wederive a contradiction to the assumption that D is branch-closed. Lemma 3.10.
Let τ be a tree set and D a selection. If ≤ on τ /D is not transitive thenthere is a three-star { → r , → s , ← s } such that → s ∼ D → s but neither → r ∼ D → s nor → r ∼ D ← s .Proof. Suppose there are [ → x ] D , [ → y ] D , [ → z ] D ∈ τ /D such that [ → x ] D ≤ [ → y ] D and [ → y ] D ≤ [ → z ] D but [ → x ] D (cid:54)≤ [ → z ] D . Pick → r ∈ [ → x ] D , → s , → s ∈ [ → y ] D and → t ∈ [ → z ] D with → r ≤ → s and → s ≤ → t .At most one of → r and → t can be D -equivalent to ← s by assumption. Suppose that → r (cid:54)∼ D ← s (the case → t (cid:54)∼ D ← s is symmetrical).Because τ is a tree set s and s have comparable orientations. If → s ≤ → s then → r ≤ → t , contradicting [ → x ] D (cid:54)≤ [ → z ] D . By Lemma 3.3(iv) and (v) → s (cid:54)≤ ← s and → s (cid:54)≥ ← s as → r (cid:54)∼ D → s , ← s . Hence → s ≥ → s . Furthermore → r (cid:54)≤ → s as → r (cid:54)≤ → t , and → r (cid:54)≥ → s byLemma 3.3(i), so { → r , ← s , → s } must be a three-star.This completes the first of the three steps. In the next step we show that if a coun-terexample to the transitivity of ≤ exists there is a counterexample which is a branchingstar. Lemma 3.11.
Let τ be a chain-complete tree set and D a selection. If ≤ on τ /D is nottransitive then there is a three-star { → r , → s , ← s } of branching points such that → s ∼ D → s but neither → r ∼ D → s nor → r ∼ D ← s .Proof. By Lemma 3.10 there is a three-star { → x , → y , ← y } with → y ∼ D → y and → x (cid:54)∼ D → y , ← y . An application of Proposition 3.9 yields a branching star σ with a three-star { → r , → s , ← s } ⊆ σ for which → x ≤ → r and → y ≤ → s ≤ → s ≤ → y . From Lemma 3.3(ii)12t follows that → y ∼ D → s ∼ D → s ∼ D → y . Furthermore Lemma 3.3(i) and (iv) implythat → r (cid:54)∼ D → s and → r (cid:54)∼ D ← s , as otherwise → x ∼ D → y or → x ∼ D ← y contrary toassumption. Thus { → r , → s , ← s } is the desired three-star.For the third step we need to show that there are elements of D that allow us to applythe branch-closedness of D to derive a contradiction. Lemma 3.12.
Let τ be a chain-complete tree set and D a selection. Let { → r , → s , ← s } bea three-star in τ with → s ∼ D → s but neither → r ∼ D → s nor → r ∼ D ← s . Then there are → d , → d ∈ D with → d (cid:8) → s and ← d (cid:8) ← s .Proof. Let → d ∈ D distinguish → r and → s ; we will show that → d (cid:8) → s . This → d cannotlie in D + ( → r ) \ D + ( → s ) as then it would also distinguish → s and → s . Furthermore D − ( → s ) = D − ( → s ) ⊆ D − ( → r ) by the star property, so → d cannot lie in D − ( → s ) \ D − ( → r )either. If → d ∈ D − ( → r ) \ D − ( → s ) then any → e ∈ σ ∩ D with → e (cid:54) = → d would distinguish → s and → s , where σ is the splitting star containing → d . Therefore → d ∈ D + ( → s ) \ D + ( → r ), soin particular → d (cid:8) → s .Repeating this argument for a → d ∈ D that distinguishes → r and ← s shows ← d (cid:8) ← s and hence the claim.Finally we put the above lemmas together to prove that ≤ in τ /D is transitive. Lemma 3.13.
Let τ be a chain-complete tree set and D a branch-closed selection. Then ≤ on τ /D is transitive.Proof. Suppose ≤ is not transitive. Then by Lemma 3.11 there is a three-star { → r , → s , ← s } of branching points such that → s ∼ D → s but neither → r ∼ D → s nor → r ∼ D ← s . ApplyingLemma 3.12 to this star yields → d , → d ∈ D with → d (cid:8) → s ≤ → s (cid:8) → d . As D is branch-closed and → s a branching point this implies → s ∈ D ; but then → s ∈ D + ( → s ) \ D + ( → s ),contrary to the assumption that → s ∼ D → s .Hence ≤ is transitive as claimed.Therefore ≤ on τ /D is a partial order, so τ /D is a nested separation system for chain-complete τ and branch-closed D . To prove that τ /D is a tree set it is thus left to showthat it does not contain any trivial elements.The next example shows that τ /D may well contain a trivial element even in caseswhere ≤ is a partial order. However, this too exploits that D is not branch-closed, andwe will subsequently prove that τ /D is indeed a tree set for branch-closed D . Example 3.14.
Let T be the following graph.1 2 3 456 7813he edge tree set τ ( T ) is regular. For the selection D = { (2 , , (6 , , (3 , , (8 , } wehave (1 , ∼ D (4 ,
3) and thus [(1 , D ≤ [(2 , D , [(3 , D . As D distinguishes (1 , ,
3) and from (3 ,
2) this means that [(1 , D is trivial in τ ( T ) /D .The proof that τ /D has no trivial elements if τ is chain-complete and D is branch-closed will again be carried out in multiple steps. First we show that the configurationfrom Example 3.14 is the only possible type of counterexample. Following that we provethat if this counterexample occurs there are elements of D we can use to apply thebranch-closedness of D with. Lemma 3.15.
Let τ be a chain-complete tree set and D a branch-closed selection. If τ /D contains a trivial element then there are → r , → s , → x ∈ τ with → r ≤ → x ≤ ← s and → r ∼ D → s but neither → r ∼ D → x nor → r ∼ D ← x .Proof. If τ /D contains a trivial element then there are → r , → x ∈ τ with → r ≤ → x and[ → r ] D (cid:8) [ → x ] D , [ ← x ] D in τ /D . Then there are → s ∈ [ → r ] D , → y ∈ [ → x ] D with → s ≤ ← y .As neither → r ∼ D → x nor → r ∼ D ← x by assumption Lemma 3.3 (iv) and (v) imply → x (cid:54)≥ ← y and → x (cid:54)≤ ← y . Furthermore if → x ≤ → y then → r ≤ → x ≤ → y ≤ ← s , so we are done.This leaves the case → x ≥ → y . If → r ≤ ← y then { → r , → y , ← x } is a three-star as in Lemma 3.11and 3.12, which we know is impossible as shown in the proof of Lemma 3.13 if D is branch-closed. By Lemma 3.3(ii) → r ≥ → y would imply → r ∼ D → x , so this is also impossible. Hencethe only relation r and y can have is → r ≤ → y , and then → r ≤ → y ≤ ← s as desired.The next step is to find → d , → d ∈ D with certain relations to the separations fromLemma 3.15, which we can later apply the assumption that D is branch-closed to so asto obtain a contradiction. Lemma 3.16.
Let τ be a tree set and D a selection. Let → r , → s , → x ∈ τ with → r ≤ → x ≤ ← s and → r ∼ D → s but neither → r ∼ D → x nor → r ∼ D ← x . Then there are → d , → d ∈ D with → d (cid:8) ← r , ← s , → x , → d (cid:8) ← r , ← s , ← x . Proof. By → r ≤ ← s and → r ∼ D → s we have D + ( → r ) = D + ( → s ) ⊆ D − ( → r ) and hence D + ( → r ) = D + ( → s ) = ∅ . Therefore either → d (cid:8) ← r , ← s or ← d (cid:8) ← r , ← s for all → d ∈ D . Thus if → d ∈ D distinguishes → r and → x then → d (cid:8) ← r , → x is the only possibility, and if → d ∈ D distinguishes → s and ← x then → d (cid:8) ← s , ← x is the only possibility. The claim now follows from theassumption that → r ∼ D → s but neither → r ∼ D → x nor → s ∼ D ← x .Finally we combine the above lemmas and use Proposition 3.9 to prove that τ /D hasno trivial elements. Lemma 3.17.
Let τ be a chain-complete tree set and D a branch-closed selection. Then τ /D contains no trivial element.Proof. Suppose τ /D contains a trivial element. From Lemma 3.15 and 3.16 it followsthat there are → r , → s , → x ∈ τ with → r ≤ → x ≤ ← s and → r ∼ D → s but neither → r ∼ D → x nor → r ∼ D ← x , as well as → d , → d ∈ D with → d (cid:8) ← r , ← s , → x , → d (cid:8) ← r , ← s , ← x . { → x , → d , ← s } then yields a branching star σ andsome → b ∈ σ with → x ≤ → b . Then → d ≤ → b ≤ ← d by → d ≤ → x and the star propertyand hence → b ∈ D as D is branch-closed. But → r ≤ → b (cid:8) ← s by the star property, so → b distinguishes → r and → s , contradicting → r ∼ D → s .Therefore τ /D cannot contain a trivial element.We have assembled all the parts necessary to show that τ /D is a finite tree set: Proposition 3.18.
Let τ be a chain-complete tree set and D a branch-closed selection.Then τ /D is a finite tree set.Proof. As D is finite there are only finitely many subsets of D and hence only finitelymany equivalence classes of ∼ D , so τ /D is finite. The relation ≤ on τ /D is reflexive bydefinition, anti-symmetric by Lemma 3.4 and transitive by Lemma 3.13 and thus a partialorder. The involution ([ → s ] D ) ∗ = [ ← s ] D is order-reversing: if [ → s ] D ≤ [ → r ] D with → s ≤ → r then ← s ≥ ← r and thus [ ← s ] D ≥ [ ← r ] D . Therefore τ /D is a separation system. Any twounoriented separations { [ → s ] D , [ ← s ] D } , { [ → r ] D , [ ← r ] D } in τ /D have comparable orientations,because their representatives s and r are nested. Finally Lemma 3.17 shows that τ /D has no trivial elements and is thus a finite tree set.With this we have accomplished the main goal of this section. In the next two sectionswe will define a suitable directed set D of selections of τ and show τ ∼ = lim ←− ( τ /D | D ∈ D ).To help with this in the remainder of this section we establish a few independent factsabout the behaviour of τ /D for later use. We show that the relation of → r and → s in τ can sometimes be recovered from the relation of [ → r ] D and [ → s ] D in τ /D , and that theequivalence classes of ∼ D in τ are chain-complete. The latter will be crucial in the sur-jectivity proof in the next sections.If τ /D is a tree set this implies that r and s in τ have to have the same relation as[ → r ] D and [ → s ] D in τ /D , at least if those are different classes: Lemma 3.19.
Let τ be a tree set, → r , → s ∈ τ and D a selection for which τ /D is a treeset. If [ → r ] D (cid:8) [ → s ] D then → r (cid:8) → s .Proof. Any other relation between r and s implies either [ → r ] D = [ → s ] D or that one of[ → r ] D and [ ← s ] D would be trivial in τ /D .For the study of τ /D it is essential to know the behaviour of chains of τ with regardto ∼ D . It turns out that the equivalence classes of τ are chain-complete themselves if τ is; we don’t even need the assumption that D is branch-closed for this: Proposition 3.20.
Let τ be a chain-complete tree set, D a selection, and → t ∈ τ . Then [ → t ] D is chain-complete.In particular [ → t ] D has a maximal (and a minimal) element. roof. Let C be a chain in the equivalence class [ → t ] D with supremum → s in τ . The claimis trivial if → s ∈ C . Thus we may assume that → s / ∈ C . Then ← s cannot lie in a splittingstar σ of τ , as in that case some other element → s (cid:48) of σ would be an upper bound of C with → s (cid:48) (cid:8) → s .Pick some → r ∈ C ; we will verify that → r ∼ D → s . As → r ≤ → s we have D + ( → r ) ⊆ D + ( → s )and D − ( → s ) ⊆ D − ( → r ).Consider → d ∈ D − ( → r ). As all elements of C are D -equivalent ← d is an upper boundfor C and hence → s ≤ ← d . On the one hand → d (cid:54) = ← s as ← s does not lie in a splittingstar, on the other hand → d (cid:54) = → s as then → r would be trivial. Therefore → s (cid:8) ← d and thus D − ( → r ) ⊆ D − ( → s ).Now consider → d ∈ D + ( → s ). This → d cannot be an upper bound for C , hence either → d ∈ D + ( → r ) or → d ≤ ← r . In the latter case ← d is an upper bound for C implying → d (cid:8) → s ≤ ← d and thus that → d would be trivial. Therefore D + ( → s ) ⊆ D + ( → r ) and hence → r ∼ D → s .For a subset B ⊆ τ and a selection D write [ B ] D := { [ → b ] D | → b ∈ B } ⊆ τ /D . Adirect consequence of Lemma 3.3(ii) and Proposition 3.20 is that for a chain C ⊆ τ thesupremum of [ C ] D is the class of the supremum of C in τ : Corollary 3.21.
Let τ be a chain-complete tree set, D a branch-closed selection, C achain and → s the supremum of C in τ . Then [ → s ] D = max [ C ] D in τ /D .Proof. The relation ∼ D has finitely many equivalence classes, so Lemma 3.3(ii) impliesthat some final segment of C is completely contained in some class [ → t ] D of τ . Theset [ C ] D in τ /D is again a chain, and as [ → t ] D contains a final segment of C it is themaximum of [ C ] D in τ /D . Proposition 3.20 now implies → s ∈ [ → t ] D = max[ C ] D .Infinite splitting stars of τ play an important role in the upcoming Section 3.3. Wenow analyze their behaviour with regard to ∼ D . This turns out to be quite simple: if asplitting star σ meets D , then all elements of σ ∩ D are pairwise non-equivalent, and allelements of σ \ D get identified: Lemma 3.22.
Let τ be a chain-complete tree set, D a branch-closed selection and σ asplitting star that meets D . Then → r ∼ D → s for distinct → r , → s ∈ σ if and only if → r , → s / ∈ D .In particular if σ is infinite there is exactly one equivalence class of ∼ D containinginfinitely many elements of σ , and every other equivalence class contains at most oneelement of σ .Proof. Let → r , → s ∈ σ be two distinct separations. For the forward direction suppose that → r ∈ D . Then → r ∈ D − ( → s ) but → r / ∈ D − ( → r ), so → r (cid:54)∼ D → s .For the backward direction assume that → r , → s / ∈ D . Then D + ( → s ) = ∅ as otherwise → s ∈ D by the assumptions that D is branch-closed and σ meets D . Similarly D + ( → r ) = ∅ .Moreover D − ( → r ) \ { → s } = D − ( → s ) ∪ D + ( → s ) as → r , → s lie in a splitting star, so D − ( → r ) = D − ( → s ) by → s / ∈ D .To apply Lemma 3.22 in practice it is useful to have a sufficient condition for σ tomeet D . The following lemma accomplishes this by showing that a splitting star σ of τ must meet D as soon as it meets at least three equivalence classes of ∼ D :16 emma 3.23. Let τ be a chain-complete tree set, D a branch-closed selection, and σ asplitting star which meets at least three equivalence classes of ∼ D . Then σ meets D .Proof. Suppose that σ ∩ D = ∅ . Then there is → t ∈ σ with D + ( → t ) (cid:54) = ∅ . Consider → r , → s ∈ σ with → r (cid:54)∼ D → t and → s (cid:54)∼ D → t . We will show that → r ∼ D → s , contradicting theassumption that σ meets three equivalence classes. First note that D + ( → r ) = D + ( → s ) = ∅ by the assumptions that D is branch-closed and → t / ∈ D . Moreover D − ( → r ) \ { → s } = D − ( → s ) ∪ D + ( → s ) as → r , → s lie in a splitting star, so D − ( → r ) = D − ( → s ) by → s / ∈ D . Hence → r ∼ D → s . In this section we prove the second assertion of Theorem 1.1:
Theorem 1.1. (i) Every inverse limit of finite tree sets is a tree set.(ii) Every profinite tree set is an inverse limit of finite tree sets.For this we find a set of properties such that we can obtain every tree set with theseproperties as an inverse limit of finite tree sets, making use of the ‘quotients’ definedin the previous section. Following that we show that every profinite tree set has theseproperties. In doing this we also obtain a characterization of the profinite tree sets inpurely combinatorial terms.The general strategy is as follows. For a tree set τ we define a suitable directed set D ofselections such that τ /D is a finite tree set for each D ∈ D . We will use Proposition 3.18for this, so τ needs to be chain-complete and the selections in D must be branch-closed.For → s ∈ τ and selections D ⊆ D (cid:48) we then have [ → s ] D (cid:48) ⊆ [ → s ] D , so by taking theseinclusions as the bonding maps ( τ /D | D ∈ D ) is an inverse system of finite tree sets. Itthen remains to prove that τ and τ (cid:48) := lim ←− ( τ /D | D ∈ D ) are isomorphic; for this wedefine the map ϕ : τ → τ (cid:48) as ϕ ( → s ) = ([ → s ] D | D ∈ D ) . By the observation above ϕ ( → s ) is indeed always a compatible choice (and thus an el-ement of τ (cid:48) ), and ϕ is a homomorphism of separation systems by Lemma 3.3(i) andthe definition of ≤ in τ /D . We shall verify the assumptions of Lemma 2.5: that ϕ is abijection and that pre-images of small separations are small. The latter will be done bysimple case-checking.The map ϕ is injective if and only if for all distinct → r , → s ∈ τ the set D contains aselection D with [ → r ] D (cid:54) = [ → s ] D . An almost sufficient condition for this is that there is asplitting star of τ between any two given separations. To make this formal we say thata tree set τ is splittable if for every → r , → s ∈ τ with → r (cid:8) → s there is a splitting star σ of τ with → r (cid:48) , ← s (cid:48) ∈ σ such that → r ≤ → r (cid:48) (cid:8) → s (cid:48) ≤ → s . Then ϕ is injective if τ is splittable and theset (cid:83) D contains all elements of τ that lie in non-singleton splitting stars . In fact we will use a slightly smaller but still sufficient set D of selections: (cid:83) D will miss only a singleseparation of every infinite splitting star, which we will deal with separately. ϕ we will rely on Corollary 3.21. This is the most technicalpart of the proof. Essentially, if τ (cid:48) contains some separation → t ∗ which is not in ϕ ( τ ), wewill try to ‘sandwich’ → t ∗ by taking maximal chains C and C (cid:48) in τ whose images under ϕ lie below and above → t ∗ respectively. Then → t ∗ will lie between the images under ϕ of thesupremum → s of C and the infimum → s (cid:48) of C (cid:48) . However, no element of τ will lie inbetween → s and → s (cid:48) , enabling us to show that for sufficiently large selections D ∈ D there can beno compatible choice of elements of τ /D inbetween [ → s ] D and [ → s (cid:48) ] D apart from ϕ ( → s ) and ϕ ( → s (cid:48) ) themselves. This approach works as long as → t ∗ is not maximal or minimal in τ (cid:48) ; inthose cases we need to be more careful, and we will adress this below.What remains is the choice of the selections in D . Proposition 3.18 demands thatall D ∈ D are branch-closed. For the injectivity of ϕ we want (cid:83) D to contain all non-singleton splitting stars of τ . Lastly D needs to be a directed set. However if τ containstwo non-singleton splitting stars with infinitely many branching points inbetween themthen clearly we cannot achieve all three of these simultaneously. We therefore need toassume that τ contains only finitely many branching points inbetween any two non-singleton splitting stars.To make this formal, for unoriented separations s, s (cid:48) ∈ τ let C ( s, s (cid:48) ) denote the setof all branching points → b of τ for which there are orientations → s , → s (cid:48) with → s ≤ → d ≤ → s (cid:48) or → s (cid:48) ≤ → d ≤ → s . Then C ( s, s (cid:48) ) is always the disjoint union of two chains, and if C ( s, s (cid:48) )meets a splitting star σ in an element other than s or s (cid:48) then it meets σ in exactly twoelements. Thus for → s ≤ → s (cid:48) , if there is a → d ∈ C ( s, s (cid:48) ) with → s ≤ → d (cid:8) → s (cid:48) , then C ( s, s (cid:48) )contains a ← d (cid:48) with → s ≤ → d (cid:8) → d (cid:48) ≤ → s (cid:48) . Furthermore a selection D of τ is branch-closed ifand only if C ( s, s (cid:48) ) ⊆ D for all → s , → s (cid:48) ∈ D .If we assume that C ( s, s (cid:48) ) is finite for all regular s, s (cid:48) in τ then no two non-singletonsplitting stars can have infinitely many branching points between them as each of thesestars must contain a regular separation. Under this assumption the set of all branch-closed selections is a directed set.However the set of all branch-closed selections is a bit too large to ensure the surjec-tivity of ϕ ; consider for example an infinite splitting star σ of τ . As seen in Lemma 3.22for every branch-closed selection D that meets σ the set σ \ D lies inside one equivalenceclass of ∼ D . The family → t ∗ := ([ σ \ D ] D | D ∩ σ (cid:54) = ∅ ) is a compatible choice whichunfortunately is not in the image of ϕ : for any → s ∈ σ there is some selection D thatcontains → s and hence shows that ϕ ( → s ) (cid:54) = → t ∗ . Therefore we need to ‘reserve’ some → s ∈ σ as the → s ∈ τ with ϕ ( → s ) = → t (cid:48) , which we can achieve by putting only those branch-closedselections D of τ into D that do not contain → s . Since → t ∗ is small we need to pick a smallseparation → s ∈ σ for this. The assumption on τ needed for this to work is thus thatevery infinite splitting star contains a small separation. The fact that (cid:83) D now missesthis selected → s does not interfere with the injectivity of ϕ : since → s is small and henceminimal in τ it is not needed to distinguish any two separations of τ and is also neverrequired to be in a selection to make it branch-closed.We will now show that under the assumptions outlined above we do indeed get anisomorphism between τ and τ (cid:48) . Proposition 3.24.
Let τ be a chain-complete splittable tree set with no infinite regular plitting star, in which C ( s, s (cid:48) ) is finite for all regular s, s (cid:48) in τ . For every infinitesplitting star σ let ν ( σ ) be a small separation in σ , and let D be the set of all branch-closed selections D of τ with ν ( σ ) / ∈ D for every infinite splitting star σ of τ .Then ( τ /D | D ∈ D ) is an inverse system of finite tree sets and the map ϕ : τ → lim ←− ( τ /D | D ∈ D ) with ϕ ( → s ) = ([ → s ] D | D ∈ D ) is an isomorphism of tree sets.Proof. By Proposition 3.18 each τ /D is a finite tree set. The set D ordered by inclusionis a directed set: to see this, let D, D (cid:48) ∈ D and set E := D ∪ D (cid:48) ∪ (cid:91) → s , → s (cid:48) ∈ D ∪ D (cid:48) C ( s, s (cid:48) ) . Then E ∈ D with D, D (cid:48) ⊆ E : E is finite by the assumption that C ( s, s (cid:48) ) is finite for allregular s, s (cid:48) , and E is branch-closed by construction. It contains no ν ( σ ) of any infinitesplitting star σ , as neither D nor D (cid:48) does and thus ν ( σ ) / ∈ C ( s, s (cid:48) ) for all → s , → s (cid:48) ∈ D ∪ D (cid:48) .Furthermore E meets no splitting star in exactly one element: since D and D (cid:48) areselections, neither of them meets any splitting star in exactly one separation; and for C ( s, s (cid:48) ) with → s , → s (cid:48) ∈ D ∪ D (cid:48) , if C ( s, s (cid:48) ) contains a element of some splitting star, andthat star does not already meet D or D (cid:48) , then C ( s, s (cid:48) ) contains at least two elements ofthat star. Thus E is a selection with E ∈ D , showing that D is indeed a directed set.Therefore ( τ /D | D ∈ D ) is an inverse system with the surjective bonding maps f DD (cid:48) : τ /D → τ /D (cid:48) defined as f ([ → s ] D ) = [ → s ] D (cid:48) for D (cid:48) ⊂ D . Note that these bonding maps are well-defined by the definition of ∼ D , andare homomorphisms of tree sets by Lemma 3.3(i). Thus τ (cid:48) := lim ←− ( τ /D | D ∈ D ) is atree set by Theorem 1.1(i).The map ϕ is a homomorphism of tree sets by Lemma 3.3(i) and the definition of ≤ in τ /D : if → r ≤ → s then [ → r ] D ≤ [ → s ] D for all D ∈ D and hence ϕ ( → r ) ≤ ϕ ( → s ). The claimthus follows from Lemma 2.5 if we can show that ϕ is a bijection, and that pre-imagesof small separations are small.For the injectivity let → r , → s ∈ τ be two distinct separations. Then ϕ ( → r ) (cid:54) = ϕ ( → s ) followsfrom the assumption that τ is splittable, unless one of → r and → s is ν ( σ ) for some infinitesplitting star σ . Suppose → r = ν ( σ ). If s does not meet σ the injectivity again followsfrom τ being splittable; if on the other hand s meets σ then any D ∈ D that meets s witnesses ϕ ( → r ) (cid:54) = ϕ ( → s ) by Proposition 3.22.To show that pre-images of small separations are small we will show that the imagesof non-small separations are non-small. So consider a non-small → s ∈ τ ; we will finda D ∈ D for which [ → s ] D is non-small, witnessing that ϕ ( → s ) is not small. If there are → r , → t ∈ τ with → r (cid:8) → s (cid:8) → t , then by Lemma 3.19 we can obtain a suitable selection D ∈ D by applying the splittability of τ to the pair → r , → s and → s , → t , and then taking as D those two-stars and all branching points between them. We may therefore assume thatthere is no → r ∈ τ with → r (cid:8) → s (the other case is symmetrical). If → s lies in a splitting star19 of τ , then σ is not a singleton star, and any non-singleton finite subset of σ containing → s but not ν ( σ ) is a selection D ∈ D for which [ → s ] D is regular. So suppose that → s does not lie in a splitting star. Then → s cannot be co-small, and there is a separation → t ∈ τ with → s (cid:8) → t and t regular. Consider C ( s, t ), which is finite by assumption since s and t are both regular. If C ( s, t ) is empty then by applying the splittability of τ to → s , → t we obtain a two-element subset of a splitting star which is a selection D ∈ D with[ → s ] D regular. So suppose that C ( s, t ) is non-empty. From the assumption that thereis no → r ∈ τ with → r (cid:8) → s we know that ← s cannot be a branching point of τ and hence ← s / ∈ C ( s, t ). Moreover → s / ∈ C ( s, t ) since we assumed that → s does not lie in any splittingstar of τ . Thus there exists a → d ∈ C ( s, t ) with → s (cid:8) → d ; pick a minimal such → d . Let σ bethe branching star containing → d , and let ← d (cid:48) be any element of σ other than → d or ν ( σ ) (if σ is infinite). Note that, if σ is infinite, then → d cannot be ν ( σ ) since then → s (cid:8) → d wouldbe small. Thus D := { → d , ← d (cid:48) } is a selection in D , and the minimality of → d implies that[ → s ] D cannot be small.For the surjectivity of ϕ let → t ∗ = ( → t ∗ D | D ∈ D ) ∈ τ (cid:48) = lim ←− ( τ /D | D ∈ D ) and assumefor a contradiction that there is no → s ∈ τ with ϕ ( → s ) = → t ∗ . This implies that → t ∗ D is aninfinite equivalence class for each D ∈ D : for if → t ∗ D were finite for some D ∈ D , then theinjectivity of ϕ and the fact that D is a directed set would enable us to find a D (cid:48) ⊆ D in D for which → t ∗ D (cid:48) is a singleton, in which case → t ∗ would be the image of the single elementof → t ∗ D (cid:48) .Set X := (cid:110) → s ∈ τ | ϕ ( → s ) ≤ → t ∗ (cid:111) , Y := (cid:110) → s ∈ τ | ϕ ( → s ) ≥ → t ∗ (cid:111) . For every s ∈ τ exactly one of → s , ← s lies in X ∪ Y . Therefore at most one of X and Y isempty. We distinguish two cases. Case 1:
Both X and Y are non-empty.Let C be a maximal chain in X with supremum → s in τ and C (cid:48) a maximal chain in Y with infimum → s (cid:48) . Then ϕ ( C ) and ϕ ( C (cid:48) ) are chains too. From Corollary 3.21 it followsthat ϕ ( → s ) and ϕ ( → s (cid:48) ) are respectively the supremum of ϕ ( C ) and the infimum of ϕ ( C (cid:48) ) in τ (cid:48) , hence ϕ ( → s ) ≤ → t ∗ ≤ ϕ ( → s (cid:48) ). In fact we even have ϕ ( → s ) (cid:8) → t ∗ (cid:8) ϕ ( → s (cid:48) ) by the assumptionthat → t (cid:48) does not lie in the image of ϕ . Pick D ∈ D large enough that [ → s ] D ≤ → t ∗ D ≤ [ → s (cid:48) ] D with → t ∗ D (cid:54) = [ → s ] D and → t ∗ D (cid:54) = [ → s (cid:48) ] D . As τ /D is a tree set this implies [ → s ] D (cid:8) → t ∗ D (cid:8) [ → s (cid:48) ] D .Now consider some → t ∈ τ with [ → t ] D = → t ∗ D . By Lemma 3.19 we have → s (cid:8) → t (cid:8) → s (cid:48) , so ← t ∈ X ∪ Y would imply that one of → s and → s (cid:48) is co-trivial. Therefore → t ∈ X ∪ Y , whichcontradicts the maximality of either C or C (cid:48) . This concludes Case 1. Case 2:
One of X and Y is empty.We may assume that Y is empty (the case that X is empty is analogous).Let C be a maximal chain in X with supremum → s . By Corollary 3.21 ϕ ( → s ) is thesupremum of C in τ (cid:48) and hence ϕ ( → s ) ≤ → t ∗ . Moreover ϕ ( ← s ) (cid:54) = → t ∗ and therefore ϕ ( → s ) (cid:8) → t ∗ .Let D (cid:48) be the set of all D ∈ D with [ → s ] D (cid:8) → t ∗ D . This is a cofinal set in D . For each D ∈ D (cid:48) let → M ( D ) be the set of minimal elements of the equivalence class → t ∗ D , which isnon-empty by Proposition 3.20. Consider a D ∈ D (cid:48) and → r ∈ → M ( D ). Then → s (cid:8) → r , andby the maximality of C and the minimality of → r there can be no → t ∈ τ with → s (cid:8) → t (cid:8) → r .Lemma 3.8 thus implies that → s and ← r lie in a common splitting star σ of τ . As σ is the20nique splitting star of τ containing → s , and both D ∈ D (cid:48) and → r ∈ → M ( D ) were arbitrary,this shows ← M ( D ) := { ← r | → r ∈ → M ( D ) } ⊆ σ for every D ∈ D (cid:48) . We will show that → M ( D )and thus σ is infinite for every D ∈ D (cid:48) and deduce ← t ∗ = ϕ ( ν ( σ )).Suppose for a contradiction that there is a D ∈ D (cid:48) with → M ( D ) finite and let D (cid:48) ∈ D (cid:48) with D (cid:48) ⊇ D be such that [ → r ] D (cid:48) (cid:54) = → t ∗ D (cid:48) for each → r ∈ → M . Pick a → u ∈ → M ( D (cid:48) ). Then ← u ∈ ← M ( D (cid:48) ) ⊆ σ . By compatibility [ → u ] D = → t ∗ D and hence → r ≤ → u with → r (cid:54) = → u for some → r ∈ → M ( D ), contradicting ← r , ← u ∈ σ . Therefore → M ( D ) ⊆ σ is infinite for every D ∈ D (cid:48) .Pick a ← r ∈ σ with ← r (cid:54) = → s and fix some D ∈ D (cid:48) with [ ← r ] D (cid:54) = [ → s ] D and [ ← r ] D (cid:54) = → t ∗ D . ThenLemma 3.22 and Lemma 3.23 imply that ← t ∗ D (cid:48) = [ ν ( σ )] D (cid:48) for every D (cid:48) ∈ D (cid:48) with D (cid:48) ≥ D as ν ( σ ) / ∈ D (cid:48) by the definition of D . As the set of all D (cid:48) ∈ D (cid:48) with D (cid:48) ≥ D is cofinal in D this shows ← t ∗ = ϕ ( ν ( σ )). This concludes Case 2.We have shown that ϕ is a bijection. The claim now follows from Lemma 2.5.In order to establish Theorem 1.1 we now show that all profinite tree sets meet theassumptions of Proposition 3.24, i.e. that every profinite tree set is chain-complete,splittable, contains no infinite regular splitting star, and has finite C ( s, s (cid:48) ) for all regularseparations s and s (cid:48) .The following lemma from [5] implies that every profinite separation system is chain-complete: Lemma 3.25 ([5]) . Let S = ( → S p | p ∈ P ) be an inverse system of finite separationsystems and → S = lim ←− S . If C ⊆ → S is a non-empty chain, then C has a supremum andan infimum in → S . Both these lie in the closure of C in → S . Let us now show that profinite tree sets are splittable, using the knowledge that theyare chain-complete:
Proposition 3.26.
Profinite tree sets are splittable.Proof.
Let τ = lim ←− ( S p | p ∈ P ) be a profinite tree set and → r , → s ∈ τ with → r (cid:8) → s . Fix p ∈ P such that → r p (cid:8) → s p and set X := (cid:8) → x ∈ τ | → r ≤ → x ≤ → s and → x p = → r p (cid:9) . Then X is a chain with → r ∈ X which by Lemma 3.25 has a supremum → r (cid:48) = (max X q | q ∈ P ). Now set Y := (cid:8) → y ∈ τ | → r ≤ → y ≤ → s with → r p ≤ → y p and → r p (cid:54) = → y p (cid:9) . This is a chain with → s ∈ Y and infimum → s (cid:48) = (min Y q | q ∈ P ). By definition we have → r ≤ → r (cid:48) (cid:8) → s (cid:48) ≤ → s . Furthermore no → t ∈ τ lies strictly between → r (cid:48) and → s (cid:48) as such a → t wouldlie in X ∪ Y and thus contradict the definition of either → r (cid:48) or → s (cid:48) . By Lemma 2.1 there isa consistent orientation O of τ extending { → r (cid:48) , ← s (cid:48) } in which → r (cid:48) and therefore ← s (cid:48) is maximal.Lemma 3.8 says that O is splitting, so → r (cid:48) and ← s (cid:48) lie in a common splitting star.The next assumption made by Proposition 3.24 is that every infinite splitting star ofthe tree set at hand contains a small separation. This is the same as asking that thetree set contains no regular infinite splitting star. In fact we can show slightly more forprofinite tree sets: 21 roposition 3.27. Let τ be a profinite tree set. Then every infinite star which ismaximal by inclusion contains a small separation.Proof. Suppose σ ⊆ τ is an infinite maximal star, and τ = lim ←− ( S p | p ∈ P ). Let σ p be the projection of σ to S p . For every p ∈ P there must be some → s p ∈ S p which isthe image of infinitely many elements of σ . As σ is a star such a → s p has to be small.For p ∈ P let σ (cid:48) p be the set of all → s p ∈ S p which are the projection of infinitely manyelements of σ . Then ( σ (cid:48) p | p ∈ P ) is an inverse system of finite sets with a non-emptyinverse limit, and its elements are also elements of τ . Let → s ∈ lim ←− ( σ (cid:48) p | p ∈ P ) be suchan element. As every → s p is small so is → s . Moreover → s p ≤ ← r p for all p ∈ P and → r ∈ σ ,so → s ∈ σ by maximality.As every splitting star of a tree set is also an inclusion-maximal star, Proposition 3.27clearly implies that profinite tree sets contain no regular infinite splitting stars.Let us call a tree set star-finite if it contains no infinite star. Then Proposition 3.27implies that all regular profinite tree sets are star-finite: Corollary 3.28.
Regular profinite tree sets are star-finite.Proof.
If a profinite tree set contains an infinite star by Proposition 3.27 it also containsa small separation. Hence regular profinite tree sets do not contain infinite stars.A tree set τ that contains no infinite star clearly contains no regular infinite splittingstar either. Furthermore if C ( s, s (cid:48) ) was infinite for any s, s (cid:48) ∈ τ then the set of allseparations in τ \ C ( s, s (cid:48) ) that belong to a branching star of τ which meets C ( s, s (cid:48) ) is aninfinite star in τ . Therefore Lemma 3.25 together with Proposition 3.24, 3.26 and 3.27implies the following characterization of the regular profinite tree sets: Theorem 3.29.
A regular tree set τ is profinite if and only if it is chain-complete,splittable and star-finite. (cid:3) To complete the proof of Theorem 1.1 it remains to show that that C ( s, s (cid:48) ) is finitefor all regular s, s (cid:48) in a profinite tree set τ . We do this in three steps. First we showthat every infinite chain has some limit element. Then we show that if → m ∈ τ is thesupremum of a chain of branching points it must be co-small; and finally we infer that C ( s, s (cid:48) ) can only be infinite if one of s and s (cid:48) is non-regular.The first step is more about posets and chain-complete tree sets than about profinitetree sets: Lemma 3.30.
Let τ be a chain-complete tree set and C an infinite chain in τ . Thenthere is a sub-chain C (cid:48) ⊆ C that does not contain both its infimum and its supremumin τ .Proof. We may assume that C contains its infimum and supremum in τ as otherwise C (cid:48) := C is as desired.Let us define C ≤ → s := (cid:8) → r ∈ C | → r ≤ → s (cid:9) → s ∈ C and define L := (cid:8) → s ∈ C | C ≤ → s is finite (cid:9) . Then L is a non-empty sub-chain of C . Let → l be the supremum of L in τ ; if → l / ∈ L then L is as desired. If on the other hand → l ∈ L then L is finite, so R := C \ L isinfinite. Let → r be the infimum of R in τ . If → r ∈ R then C ≤ → r is infinite, so there isa → t ∈ C ≤ → r \ ( L ∪ { → r } ). But this contradicts the fact that → r is the infimum of R .Therefore → r / ∈ R and R is the desired sub-chain.Now we prove that the supremum of a chain of branching points is co-small. Theproof of this is somewhat analogous to the proof of Proposition 3.27: Lemma 3.31.
Let τ = lim ←− ( S p | p ∈ P ) be a profinite tree set, C a chain of infinitelymany branching points and → m the supremum of C in τ . If → m / ∈ C then → m is co-small.Proof. As co-small separations are maximal in τ we may assume without loss of gener-ality that C is a chain of order type ω .Let C = { → s n | n ∈ N } with → s n (cid:8) → s n +1 for all n ∈ N . For every n ∈ N pick an element → t n of the branching star containing → s n with → t n (cid:8) → s n +1 . Let → m = ( → m p | p ∈ P ) bethe supremum of C . Then by Lemma 3.25 for any fixed p ∈ P there is an n ∈ N with → s np = → m p ; let k ( p ) ∈ N be the minimal such index and write T p := (cid:110) → t np | n ≥ k ( p ) (cid:111) . Observe that if n ≥ k ( p ) for some p ∈ P then → t n ≤ → m, ← s n by definition and hence → t np ≤ → m p as well as → t np ≤ ← s np = ← m p , and as a consequence also → t np ≤ ← t np .Moreover k ( p ) ≤ k ( q ) for all p, q ∈ P with p ≤ q , so ( T p | p ∈ P ) is an inverse systemwhose inverse limit is a subset of τ . Let → t ∈ lim ←− ( T p | p ∈ P ). By the above observationwe have → t ≤ → m as well as → t ≤ ← m , and thus t = m , since otherwise m would witnessthat → t is trivial. Moreover → t is small by the above observation. Therefore one of → m and ← m is small; but the first of these is impossible since then then every → s n would be trvial.Therefore ← m is small, that is, → m is co-small.With these two lemmas we can now show that C ( s, s (cid:48) ) is finite for all regular s, s (cid:48) inprofinite tree sets: Proposition 3.32.
Let τ be a profinite tree set and s, s (cid:48) ∈ τ two regular unorientedseparations. Then C ( s, s (cid:48) ) is finite.Proof. Suppose that C ( s, s (cid:48) ) is infinite. Then C ( s, s (cid:48) ) is the disjoint union of two infinitechains. Let C be one of them. By Lemma 3.30 there is a sub-chain C (cid:48) of C that doesnot contain both its infimum and supremum in τ ; suppose that C (cid:48) does not contain itssupremum (the other case is symmetrical). Let → m be the supremum of C (cid:48) . Lemma 3.31implies that → m is co-small. But one of → s , ← s , → s (cid:48) or ← s (cid:48) is an upper bound for C (cid:48) . As co-smallelements are maximal in τ it follows that m = s or m = s (cid:48) , a contradiction.Lemma 3.25 and Proposition 3.24, 3.26, 3.27 and 3.32 combine into the followingtheorem characterizing the profinite tree sets:23 heorem 3.33. A tree set τ is profinite if and only if it is chain-complete and splittable,contains no regular infinite splitting star, and has the property that C ( s, s (cid:48) ) is finite forall regular s, s (cid:48) . (cid:3) Moreover we can now prove Theorem 1.1(ii), that is, that every profinite tree set isan inverse limit of finite tree sets:
Proof of Theorem 1.1 (ii). Let τ be a profinite tree set. From Theorem 3.33 it followsthat τ meets the assumptions of Proposition 3.24, which together with Proposition 3.18implies that τ is an inverse limit of finite tree sets. (cid:3) Therefore the profinite tree sets are indeed precisely those tree sets that are an in-verse limit of finite tree sets.Let us conclude this section by proving by example that the four properties in The-orem 3.33 are independent of each other, in the sense that none of them is implied bythe others:
Example 3.34.
Let R be a one-way infinite ray. Then the edge tree set τ ( R ) is a regulartree set which is splittable and star-finite but not chain-complete. Example 3.35.
Let τ := ([ − , − ∪ [1 , , (cid:52) , ∗ ) , where x ∗ := − x , and x (cid:52) y if and only if x ≤ y as real numbers and x and y have thesame sign. Then τ is chain-complete and star-finite, but τ is not splittable since theonly splitting stars are {− } and { } . Example 3.36.
Let S be an infinite graph-theoretical star. Then the edge tree set τ ( S )is a tree set which is chain-complete and splittable with finite C ( s, s (cid:48) ) for all s, s (cid:48) ∈ τ ( S )but which contains a regular infinite splitting star. Example 3.37.
Let B be the tree set with ground set → m, ← m and → s n , ← s n , → t n and ← t n forevery n ∈ N , with the following relations:1. → m ≥ → s n , → t n and ← m ≤ ← s n , ← t n for all n ∈ N ,2. → s i ≤ → s j and ← s i ≥ ← s j if and only if i ≤ j ,3. → t i ≤ ← t j if and only if i (cid:54) = j ,4. → s i ≤ ← t j and ← s i ≥ → t j if and only if i ≤ j ,5. → s i ≥ → t j and ← s i ≤ ← t j if and only if i ≥ j and i (cid:54) = j .Then B is a tree set which is chain-complete and splittable with no infinite splittingstar, but C ( s , m ) is infinite despite s and m being regular.24 Representing profinite tree sets
In [3] Diestel explored the various ways in which a tree set can be represented by a nestedsystem of bipartitions of some groundset. Typically the groundset used to represent atree set τ is the set O ( τ ) of consistent orientations of τ , or a suitable subset of O ( τ ).The representation comes in the form of a map ϕ : τ → B ( O ( τ )), where B ( X ) denotesthe set of (non-trivial) oriented bipartitions of X , such that ϕ is an isomorphism of treesets between τ and its image. As B ( X ) is a regular separation system only regular treesets can be represented by such bipartitions.Diestel proved that the set of directed consistent orientations of a tree set can be usedto represent that tree set, where an orientation O is directed if O is a directed set. Atree set is ever-branching if it contains no inclusion-maximal proper star of order 2. Theorem 4.1 ([3]) . Let τ be an ever-branching regular tree set, ˜ O = ˜ O ( τ ) the set of alldirected consistent orientations of τ , and ϕ : τ → B ( ˜ O ) the map ϕ ( → s ) := (cid:16) ˜ O ( ← s ) , ˜ O ( → s ) (cid:17) , where ˜ O ( → s ) := { O ∈ ˜ O | → s ∈ O } . Then ϕ is an isomorphism of tree sets between τ andits image in B ( ˜ O ) . In the remainder of this section we will show that every profinite tree set without asplitting two-star fulfills the assumptions of Theorem 4.1, i.e. that it is ever-branching.Following that we shall use our insights from Section 3.3 to show that every regularprofinite tree set can be represented by the bipartitions of its closed consistent orienta-tions.The first part of this is straightforward:
Lemma 4.2.
A splittable tree set is ever-branching if and only if it has no splittingtwo-star.Proof.
For the forward direction note that every splitting two-star of a tree set witnessesthat that tree set is not ever-branching.For the backward direction, let τ be a splittable tree set with no splitting two star and { → r , → s } a two-star in τ with r (cid:54) = s . As τ is splittable there is a splitting star σ of τ thatcontains separations → r (cid:48) and → s (cid:48) with → r ≤ → r (cid:48) (cid:8) ← s (cid:48) ≤ ← s . By assumption σ is not a two-star.Hence replacing → r (cid:48) and → s (cid:48) in σ with → r and → s yields a proper star which includes { → r , → s } as a proper subset, showing that τ is ever-branching.As every profinite tree set is splittable by Theorem 3.33 this shows that every regularprofinite tree set without a splitting two-star can be represented as in Theorem 4.1.In fact for profinite tree sets the directed orientations have a very simple description: Lemma 4.3.
In a chain-complete tree set every directed consistent orientation has agreatest element. roof. Let τ be a chain-complete tree set, O a consistent directed orientation of τ , and C a maximal chain in O . The claim follows from the chain-completeness of τ if O is thedown-closure of C . Suppose there is a → s ∈ O with → s (cid:54)≤ → t for every → t ∈ C . As C ismaximal and O is consistent this means → s (cid:8) ← t for all → t ∈ C . Fix some → t ∈ C . As O is directed there exists → r ∈ O with → s , → t ≤ → r . Then → r / ∈ C by the choice of → s . Inparticular r (cid:54) = t . Consider some → u ∈ C with → t ≤ → u . If → r ≤ → u then → s would be trivialwith witness r ; if → r ≤ ← u then → t would be trivial with witness r ; and finally → r ≥ ← u would contradict the consistency of O . Therefore u and r must be related as → r ≥ → u .As this holds for every → u ∈ C with → t ≤ → u we know that → r is an upper bound for C ,contradicting the maximality of C .Clearly the converse of Lemma 4.3 holds as well: every orientation of a tree set thathas a greatest element is directed as witnessed by that element. Therefore we haveestablished the following theorem, which is essentially a re-formulated special case ofTheorem 4.1: Theorem 4.4.
Let τ be a regular profinite tree set with no splitting two-star, O (cid:48) = O (cid:48) ( τ ) the set of all consistent orientations of τ that have a greatest element, and ϕ : τ → B ( O (cid:48) ) the map ϕ ( → s ) := (cid:0) O (cid:48) ( ← s ) , O (cid:48) ( → s ) (cid:1) , where O (cid:48) ( → s ) := { O ∈ O (cid:48) | → s ∈ O } . Then ϕ is an isomorphism of tree sets between τ and its image in B ( O (cid:48) ) . For regular profinite tree sets that do contain splitting two-stars we can still prove asuccinct representation theorem. To do this we use the set O ( τ ) of all splitting consistentorientations as a groundset rather than all directed orientations. For any splitting two-star the corresponding consistent orientation is then contained in O ( τ ) and can be usedto distinguish the two elements of that star. Theorem 4.5.
Let τ be a regular profinite tree set, O = O ( τ ) the set of all splittingconsistent orientations of τ , and ϕ : τ → B ( O ) the map ϕ ( → s ) := (cid:0) O ( ← s ) , O ( → s ) (cid:1) , where O ( → s ) := (cid:8) O ∈ O | → s ∈ O (cid:9) . Then ϕ is an isomorphism of tree sets between τ andits image in B ( O ) .Proof. We check the assumptions of Lemma 2.4.To see that O ( → s ) is non-emtpy for → s ∈ τ , let C be a maximal chain in τ containing → s . By Lemma 2.1 C extends to a consistent orientation O of τ . As τ is chain-completeby Theorem 3.29 and C was chosen maximal C must have a greatest element → m ∈ C .This → m is in fact even the greatest element of O : for any → r ∈ O either → r ≤ → m or ← r ≥ → m by consistency, but the latter case would contradict the maximality of C . Therefore O is splitting and hence O ∈ O ( → s ).Additionally O = O ( → s ) ˙ ∪O ( ← s ) for each → s ∈ τ , so ϕ is indeed a map into B ( O ).26learly ϕ commutes with the involution. It is also order-preserving: for → r , → s ∈ τ with → r ≤ → s we have O ( → s ) ⊆ O ( → r ) by consistency and hence ϕ ( → r ) ≤ ϕ ( → s ). This shows that ϕ is a homomorphism.It remains to prove that ϕ is injective. For this consider → r , → s ∈ τ . If r = s theneither → r = → s , in which case there is nothing to show, or → r = ← s , in which case everyorientation in O ( → r ) does not contain → s and hence witnesses ϕ ( → r ) (cid:54) = ϕ ( → s ). Thus wemay assume that r (cid:54) = s . If → r and → s point away from each other then every orientation in O ( → r ) does not contain → s by consistency and thus witnesses ϕ ( → r ) (cid:54) = ϕ ( → s ). If → r and → s point towards each other then every orientation in O ( ← r ) contains → s but not → r . Finallyif → r and → s are comparable, say → r ≤ → s , then by the splittability of τ there is a splittingstar σ of τ such that → r and ← s lie below different elements of σ . The orientation of τ induced by σ then contains → r but not → s , witnessing that ϕ ( → r ) (cid:54) = ϕ ( → s ) and concludingthe proof.We conclude with two remarks on Theorem 4.5. First, by [5, Theorem 7.4] the splittingorientations of a regular tree set τ are precisely those consistent orientations of τ thatare closed as a set in the inverse limit topology of τ ; see [5, Section 7] for more. Thus inTheorem 4.5 we could equivalently have used the set of all closed consistent orientationsas a groundset.Finally, one can prove Theorem 4.5 with a slightly smaller groundset, namely withoutusing splitting orientations that have three or more maximal elements: throughout theproof we have only used orientations with a greatest element, with the exception of thevery last step. Given → r , → s ∈ τ with → r (cid:8) → s we find a splitting star σ with → r and ← s below distinct elements of σ . If σ is a two-star then its corresponding orientation of τ contains → r but not → s ; and if σ has size three or greater it contains a separation → t with → r , ← s ≤ → t , in which case any orientation with a greatest element containing ← t alsocontains → r and ← s . References [1] N. Bowler, R. Diestel, and F. Mazoit. Tangle-tree duality in infinite graphs andmatroids. In preparation.[2] R. Diestel. Abstract separation systems.
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