End tangles in S k

To complete our proof as outlined, let any finite setZ ⊆V be given. For everyz∈ZrKchoose (A_z, B_z)∈τwithz∈A_zrB_z: this exists, becausez /∈K. Since (K, V)∈τ, byLemma 5.1.4we have (A⁰, B⁰)∈τ for

A⁰:=K∪ [

z∈ZrK

Az and B⁰:=V ∩ \

z∈ZrK

Bz.

As desired, (A⁰, B⁰)Z= (A, B)Z(which is (Z, Z∩K), since (A, B) = (V, K)):

every z ∈ZrK lies in some Az and outside that Bz, so z ∈A⁰rB⁰, while everyz∈Z∩K lies inK⊆A⁰ and also, by definition ofK, in everyBz (and hence inB⁰), since (Az, Bz)∈τ.

This proof ofTheorem 35concludes our exposition of material from [11].

We conclude the section with the remark that an end tangleτ of an infinite graphGis defined by anℵ0-block if and only if it is defined by some setX ⊆V: for ifτis defined by someX ⊆V, thenXmust be infinite, as otherwise{X, V} ∈ S would witness thatX does not define τ. But ifX is infinite then so isK:=

T{B |(A, B)∈τ} ⊇X, and we can follow the proof ofTheorem 35 to show that Kis anℵ₀-block definingτ.

Example 5.1.5. Let Gbe as in Example 5.1.1, that is, a single rayv₀v₁. . . with endω. The same argument as inExample 5.1.1 shows thatτ∩S_k is not closed in S_k for k > 2. However, ω does induce a closed tangle in S₁: the set τ∩S₁={(∅, V)}is closed inS₁.

Example 5.1.6. LetGbe the infinite grid, ω the unique end ofGandτ the tangle induced byω inS=S(G). SinceGis locally finite it does not contain an ℵ0-block. Thereforeτ cannot be defined by an ℵ0-block and is thus not closed in S by Theorem 35. However, for every k∈ N, it is easy to see that τ∩S_k is closed inS_k: indeed, for fixedk∈N, the size ofA is bounded in terms ofk for every (A, B) ∈τ ∩S_k. Thus, for every (A⁰, B⁰) ∈S_krτ, any Z ⊆V(G) with|Z∩A⁰|sufficiently large witnesses that (A⁰, B⁰) does not lie in the closure of τ: if Z is large enough that |Z∩A⁰|>|A|for every (A, B)∈τ∩S_k, then no (A, B)∈τ∩S_kwill agree with (A⁰, B⁰) onZ.

Example 5.1.6shows that the end tangleτ of the infinite grid is not defined by a setX ⊆V in the sense of (1). Moreover, even for any fixedk∈N>5, we cannot find a setX ⊆V withX ⊆B for all (A, B)∈τ∩S_k: for everyx∈X the separation ({x} ∪N(x), V r{x}) lies inτ∩S_k. Thus, even the tangles thatτ induces inS_k are not defined by any subset ofV. However, they come reasonably close to it: for X =V and any (A, B)∈τ, the majority of X lies in B, that is, we have |A∩X| <|B∩X|. In fact, for any fixed k ∈N, any finite set X ⊆V that is at least twice as large as max{|A| |(A, B)∈τ∩S_k} has the property that|A∩X|<|B∩X|for every (A, B)∈τ∩S_k. Therefore, even though no τ∩S_k is defined by a X ⊆V, we can for each k ∈N find a (finite)X ⊆V which ‘defines’τ∩S_k by simple majority.

To make the above observation formal, for an end tangleτ ofG, let us call a setX ⊆V adecider setforτ (resp., forτ∩S_k) if we have|A∩X|<|B∩X| for every (A, B)∈τ (resp., (A, B)∈τ∩S_k). Thus, if we have a decider set for an end tangleτ, given a separation (A, B)∈S, the decider set tells us which of (A, B) and (B, A) lies inτ by a simple majority vote. ByTheorem 35, every end tangle that is closed in S has a decider set, since theℵ₀-block defining it is such a decider set. In analogy with this, we shall show in the remainder of this section that for an end tangleτ its induced tangleτ∩S_kis closed inS_k if and only ifτ∩S_khas a finite decider set. Such a finite decider set can be thought of as a local encoding of the tangle, or a local witness to the tangle being closed in S_k.

In contrast, it is easy to see that every end tangle has an infinite decider set:

Proposition 5.1.7. For any endω of Gthe end tangleτ induced byω has an infinite decider set.

Proof. For any rayR∈ωits vertex set V(R) is a decider set forτ.

Since no end tangle ofS can have a finite decider set the existence of decider sets for end tangles is thus only interesting for finite decider sets of the tangle’s restrictions to someS_k.

We shall complement this local witness of a given end tangle being closed with a more global type of witness. For this we need to introduce some notation.

The(vertex) degreedeg(ω) of an endω ofGis the largest size of a family of pairwise disjointω-rays³. A vertexv∈V dominatesan endωif it sends infinitely

3Here our notation deviates from that in [10], whered(ω) is used for the degree ofω.

many disjoint paths to some (equivalently: to each) ray in ω. We write dom(ω) for the number of vertices ofGwhich dominateω. An endω isundominated if dom(ω) = 0; it isfinitely dominatedif finitely many (including zero) vertices of G dominateω; and, finally, ω is infinitely dominated if dom(ω) = ∞. We will show that the category that an end ω belongs to depends just on these parameters deg(ω) and dom(ω). Concretely, we will show the following:

Theorem 36. Letτthe end tangle induced by an endωofG. Then the following statements hold:

(i) τ is closed inS if and only ifdom(ω) =∞.

(ii) τ is not closed inS but τ∩S_k is closed inS_k for everyk∈Nif and only ifdeg(ω) =∞anddom(ω)<∞.

(iii) τ∩S_kis not closed inS_kfor somek∈Nif and only ifdeg(ω) + dom(ω)<∞.

Theorem 36will be a consequence ofTheorem 35and the following theorem, which characterizes for whichk∈Na given end tangle is closed inS_kand makes the connection to finite decider sets:

Theorem 37. Let τ be the end tangle induced by an end ω of Gand letk∈N. Then the following are equivalent:

(i) τ∩S_k is closed inS_k; (ii) deg(ω) + dom(ω)>k;

(iii) τ∩S_k has a finite decider set;

(iv) τ∩S_k has a decider set of size exactly k.

Let us first proveTheorem 36fromTheorem 35andTheorem 37:

Proof of Theorem 36. We will show (i) using Theorem 35 and (ii) with The-orem 37. The third statement is then an immediate consequence of the first two and the fact that an end tangle which is closed inS is also closed inS_k.

To see that (i) holds, let us first suppose that τ is closed in S. Then, byTheorem 35,τ is defined by anℵ0-blockK⊆V. It is easy to see that every vertex inKdominatesω, hence dom(ω) =∞asK is infinite.

For the converse, suppose that ω is infinitely dominated, and let us show thatτis closed inS. For every separation (A, B)∈τ every vertex dominatingω must lie inB. Therefore K:=T{B|(A, B)∈τ}is infinite and thus, as seen in the proof ofTheorem 35, anℵ0-block definingτ, which is hence closed inS.

Let us now show (ii). By (i),τ is not closed in S if and only if dom(ω)<∞. On the other hand, byTheorem 37,τ∩S_k is closed inS_kfor allk if and only if deg(ω) + dom(ω)>kfor allk∈N, that is, if at least one summand is infinite.

Thus (ii) holds.

Claim (iii) now follows either from claim (i) and (ii), or directly from The-orem 37.

We will conclude this section by provingTheorem 37. For this we shall need two lemmas. The first lemma can be seen as an analogue of Menger’s Theorem between a vertex set and an end. Given a set X ⊆V and an end ω, we say that F⊆V separatesX fromω if everyω-ray which meetsX also meetsF.

Lemma 5.1.8. Letω be an undominated end ofGandX ⊆V a finite set. The largest size of a family of disjointω-rays which start inX is equal to the smallest size of a setT ⊆V separating X fromω.

Proof. Let T be a set separatingX from ω of minimal size. Clearly, a family of disjoint ω-rays which all start in X cannot be larger than T, as every ray in that family must meet T. So let us show that we can find a family of |T| disjointω-rays starting inX.

Observe that, as ω is undominated, for every vertex v ∈ V we can find a finite set Tv ⊆ V r{v} which separates v from ω. Thus, for every finite set Y ⊆V we can find a finite set inV rY separatingY fromω: for instance, the setS

{TvrY |v∈Y}.

Pick a sequence of finite setsTn ⊆V inductively by setting T0 := T and picking as T_n a set of minimal size with the property thatT_n−1∩T_n=∅and thatT_nseparatesT_n−1fromω; these sets exist by the above observation. LetC_n be the component of G−T_n that containsω. ClearlyT_n⊆C_n−1.

We claim thatC :=T

n∈NC_n =∅. To see this, consider any v ∈C and a shortestv–X path inG. This path must pass throughT_n for everyn∈N, which is impossible since the separatorsTn are pairwise disjoint. ThereforeC must be empty.

By the minimality of eachTn, the sets Tn are of non-decreasing size, and furthermore Menger’s Theorem yields a family of |Tn| many disjoint paths between Tn and Tn+1 for each n ∈ N, as well as |T0| many disjoint paths betweenXandT0. By concatenating these paths we obtain a family of|T0|=|T| many rays starting inX. To finish the proof we just need to show that these rays belong to ω. To see this, let ω⁰ be another end of G, and T⁰ a finite set separatingω andω⁰. SinceC=∅we have Cn∩T⁰=∅ for sufficiently largen, which shows that the rays constructed do not belong toω⁰ and hence concludes the proof.

An immediate consequence ofLemma 5.1.8is that, for an undominated endω, every finite setX ⊆V can be separated fromω by at most deg(ω) many vertices.

In fact, we can state a slightly more general corollary:

Corollary 5.1.9. Letω be a finitely dominated end ofGandX⊆V a finite set.

ThenX can be separated fromω by someT ⊆V with|T|6deg(ω) + dom(ω). Proof. LetD be the set of vertices dominatingω and consider the graphG⁰:=

G−D and the set X⁰ := XrD. ByLemma 5.1.8 there is a set T⁰ ⊆V(G⁰) of size at most deg(ω) separatingX⁰ fromω inG⁰. SetT :=T⁰∪D. Then T separatesX fromω inGand has size|T|=|T⁰|+|D|6deg(ω) + dom(ω).

The second lemma we shall need for our proof ofTheorem 37roughly states that, for an end of high degree, we can find a large family of disjoint ω-rays whose set of starting vertices is highly connected inG, even after removing the tails of these rays:

Lemma 5.1.10. Letω be an end of Gandk6deg(ω) + dom(ω). Then there are a set X ⊆V of k vertices and a set Rof disjoint ω-rays with the follow-ing properties: every vertex in X is either the start-vertex of a ray in R, or dominates ω and does not lie on any R ∈ R, and furthermore for any two sets A, B ⊆ X there are min(|A|,|B|) many disjoint A–B-paths in G whose internal vertices meet no ray in Rand no vertex ofX.

Proof. Pick a setDof vertices dominatingω and a setRof disjointω-rays not meetingD such that|D|+|R|=k; we shall find suitable tails of the rays inR such that their starting vertices together withD are the desired setX.

Using the fact that the vertices in D dominate ω and that the rays inR belong toω, we can pick for each pairx1, x2of elements ofD∪ Ranx1–x2-path in Gin such a way that these paths are pairwise disjoint with the exception of possibly having a common end-vertex in D. Let P be the set of these paths.

Now for each ray in Rpick a tail of that ray which avoids all the paths in P. LetR⁰ be the set of these tails andX the union of their starting vertices andD. We claim thatX andR⁰ are as desired.

To see this, let us show that for any setsA, B⊆X we can find min(|A|,|B|) many disjointA–B-paths inGwhose internal vertices avoidD as well asV(R⁰) for every R⁰ ∈ R⁰. Clearly it suffices to show this for disjoint sets A, B of equal size. So let A, B ⊆ X be two disjoint sets with n := |A| = |B| and let R_A,B be the set of all rays in R that contain a vertex from A orB. For each pair (a, b) ∈ A×B there is a unique path P ∈ P such that each of its end-vertices either isaorb (ifa∈D orb∈D) or lies on a ray in R_A,B which containsaorb; letPa,b be thea–b-path obtained fromP by extending it, for each of its end-vertices that is not eitheraorb, along the corresponding ray inR up to aorb. LetPA,B be the set of all these pathsPa,b. Note that the internal vertices of each pathPa,b∈ PA,B meet none of the rays inR⁰ or vertices inX. We claim thatA andB cannot be separated by fewer thann=|A|vertices in

G⁰ := [

Pa,b∈PA,B

Pa,b;

the claim will then follow from Menger’s Theorem. So suppose that some setT ⊆V(G⁰) of size less thannis given. Letxandy be the number of vertices inA andB, respectively, whose ray inRA,B does not meet T. There are xy many paths inPA,Bbetween these vertices in AandB. Since these paths are disjoint outside their corresponding ray segments, each vertex ofT can lie on at most one of them. Thus ifxy>nthere must be aT-avoiding path in PA,B

whose end-vertices’ rays inRA,B also do not meetT.

Sincex+y>n+ 1 we havexy>x(n+ 1−x). The right-hand side of this inequality, as a function ofxwith domain [n−1], is minimized by takingx= 1, wherefore it evaluates ton. Thusxy>n, which shows thatTdoes not separateA andB in G⁰.

We can thus apply Menger’s Theorem to obtainndisjointA–B-paths inG⁰, which are the desired disjoint paths inGwhose internal vertices avoid the rays inR⁰ and vertices inX: the only vertices that are contained both in V(G⁰) as well as in eitherX or a ray fromR⁰ are vertices fromAorB, which cannot be internal vertices of then=|A|=|B|disjointA–B-paths.

We are now ready to proveTheorem 37:

Theorem 37. Let τ be the end tangle induced by an end ω of Gand letk∈N. Then the following are equivalent:

(i) τ∩S_k is closed inS_k; (ii) deg(ω) + dom(ω)>k;

(iii) τ∩S_k has a finite decider set;

(iv) τ∩S_k has a decider set of size exactly k.

Proof. We will show (i)⇒(ii)⇒(iv)⇒(iii)⇒(i).

To see that (i) ⇒(ii), let us first suppose that deg(ω) + dom(ω)< k and show thatτ∩S_k is not closed inS_k. LetD be the set of dominating vertices of ω. Since|D|= dom(ω)< k the separation (V, D) lies in S_k. By definition ofτ we have (D, V)∈τand (V, D)∈/τ. Thus it suffices to show that (V, D) lies in the closure ofτ∩S_k inS_k. This will be the case if for every finite setX ⊆V there is a separation (A, B)∈τ∩S_k with (A, B)X = (V, D)X.

So letX be a finite subset ofV. ByCorollary 5.1.9some setT of size at most deg(ω) + dom(ω) separatesXfromω. LetCbe the component ofGcontainingω. We define the separation (A, B) by setting A := V rC and B := T ∪C. Then (A, B) is a separation ofGwith

|(A, B)|=|T|6deg(ω) + dom(ω)< k,

so (A, B)∈S_k. In fact (A, B) lies inτ∩S_k asω lives inB. Furthermore we haveX ⊆AandD⊆B since no vertex dominatingω can be separated fromω byT. Therefore (A, B)X = (X, X∩D) = (V, D)X, showing that (V, D) lies in the closure ofτ∩S_k inS_k.

Let us now show that (ii)⇒(iv). So let us assume that deg(ω) + dom(ω)>k. Then by Lemma 5.1.10 we find a set X ⊆ V of size k and a family R of ω -rays such that every vertex of X either dominatesω or is the start-vertex of a ray in R, and such that for any A, B ⊆X we can find min(|A|,|B|) many disjoint A–B-paths in Gwhose internal vertices meet neither X nor any ray in R.

We claim thatX is the desired decider set forτ∩S_k. To see this, let (A, B) be any separation in τ∩S_k; we need to show that|A∩X|<|B∩X|. Let us writeXArB := (ArB)∩X and XBrA := (BrA)∩X as well as X_A∩B :=

(A∩B)∩X. It then suffices to prove |XArB|<|XBrA|.

So suppose to the contrary that|XArB|>|XBrA|. Note first that no vertex inXArBdominatesωas witnessed by the finite-order separation (A, B)∈τ⊆Sk. Therefore, for every vertex inXArB, we have a ray inRstarting at that vertex.

Each of those disjoint rays must pass through the separatorA∩B, and none of them hits XA∩B. Furthermore by Lemma 5.1.10there are |XBrA| many disjoint X_ArB–X_BrA-paths whose internal vertices avoid R and X. These paths, too, must pass the separatorA∩B without meetingX_A∩B or any of the rays above. Thus we have

|A∩B|>|X_A∩B|+|XArB|+|XBrA|=|X|=k,

a contradiction since (A, B)∈Sk and hence|A∩B|< k. Therefore we must have |XArB|<|XBrA|, which immediately implies|A∩X|<|B∩X|.

Finally, let us show that (iii)⇒(i). So let X ⊆V(G) be a finite decider set for τ∩S_k. We need to show that no (A, B) ∈S_krτ lies in the closure of τ∩S_k. For this let (A, B)∈S_krτ be given; thenX witnesses that (A, B) does not lie in the closure ofτ. To see this, let any (C, D)∈τ be given. SinceX is a decider set for τ we have |C∩X| < |D∩X|, and since (A, B) ∈/ τ we have|A∩X|>|B∩X|. Therefore (A, B) and (C, D) do not agree onX, which thus witnesses that (A, B) does not lie in the closure ofτ∩S_k inSk.

Note that in our proof above that (iii) implies (i) we did not make use of the assumption that the tangle τ was induced by an end of G: indeed, every orientation ofSk that has a finite decider set is closed inS_k.

For an end tangleτ that is closed in S we can say slightly more about its decider sets in S_k: for everyk∈N the restrictionτ∩S_k has a decider set of size exactlyk which is a (< k)-inseparable set. Finding these (< k)-inseparable decider sets is straightforward: such an end tangleτis defined by anℵ0-blockK byTheorem 35, and every subsetX⊆K of sizekis a (< k)-inseparable decider set for τ∩S_k. However, having a (< k)-inseparable decider set forτ∩S_k for allk∈Nis not a characterizing property for the closed end tangles ofG: Example 5.1.11. Forn∈NletKⁿ be the complete graph onnvertices. LetG be the graph obtained from a ray R = v₁v₂. . . by replacing each vertex v_n with the complete graph Kⁿ, making each vertex from the Kⁿ replacing v_n adjacent to all vertices from theKⁿ⁺¹ replacing v_n+1. Then Ghas a unique endω; letτ be the end tangle induced by ω. Sinceω is undominatedτ is not closed inS =S(G) byTheorem 36. However, for everyk∈N, the tangleτ∩S_k has a (< k)-inseparable decider set of sizek: the cliqueK^k which replaced the vertexvk ofRis such a (< k)-inseparable decider set.

Im Dokument Tangles and where to find them (Seite 105-112)