0.5 Acknowledgements and basis of this thesis
1.0.6 A tree-decomposition distinguishing the topological ends . 31
In this section, we prove Theorem 1.0.1 already mentioned in the Introduction.
A key lemma in the proof of Theorem 1.0.1 is the following.
Lemma 1.0.45. Let G be a graph with a finite nonempty set W of vertices.
ThenGhas a star decomposition (S, Qs|s∈V(S))of finite adhesion such that each topological end lives in someQswhere sis a leaf.
Moreover, only the central partQc contains vertices ofW, and for each leaf s, there lives an topological end inQs, andQs\Qc is connected.
Proof that Lemma 1.0.45 implies Theorem 1.0.1. We shall recursively construct a sequence Tn = (Tn, Ptn|t ∈ V(Tn)) of tree-decomposition of G of finite ad-hesion as follows. We starting by picking a vertex v of G arbitrarily and we obtainT1 by applying Lemma 1.0.45 with W ={v}. Assume that we already constructedTn. For each leafsofTn, we denote byWsthe set of those vertices in Qs also contained in some other part of Tn. Note thatWs is contained in the part adjacent toQs and thus is finite. By Lemma 1.0.45, we obtain a star decompositionTsofG[Qs] such that now∈Wsis contained in a leaf part ofTs and such that each topological end living inQslives in a leaf of Ts. We obtain Tn+1fromTnby replacing each leaf partQsbyTs, which is well-defined as the setWsis contained in a unique part ofTs.
By r, we denote the center of T1. For each j < m < n, the balls of radius j aroundr inTm andTn are the same. Thus we takeT to be the tree whose nodes are those that are eventually a node ofTn. For eacht∈V(T), the parts Ptn are the same fornlarger than the distance betweentandr, and we takePt
to be the limit of thePtn.
It is easily proved by induction that each vertex inWsforsa leaf ofTn has distance at leastn−1 fromv in G. Thus for each j < n the ball of radius j aroundv in Gis included in the union over all partsPtn where t is in the ball of radiusjaroundrinTn. Hence (T, Pt|t∈V(T)) is a tree-decomposition, and it has finite adhesion by construction.
It remains to show that the ends ofT define precisely the topological ends ofG, which is done in the following four sublemmas.
Sublemma 1.0.46. Each topological endω of Glives in an end ofT.
Proof. There is a unique leaf s of Tn such that ω lives in Psn. Letsn be the predecessor ofsinTn. Thenωlives in the end ofTto whichs1s2. . .belongs.
Sublemma 1.0.47. In each endτ ofT, there lives a vertex end of G.
Proof. For a directed pathsP, we shall denote by←P−the directed path with the inverse ordering of that ofP.
Lets1s2... be the ray inT starting atrthat belongs toτ. By construction, the setsWsi are disjoint and finite. For eachw∈Wsi, we pick a pathPw from wto v. SinceWsi−1 separates wfrom v, there is a firstw0 ∈Wsi−1 appearing on Pw. Now we apply the Infinity Lemma in the form of [52, Section 8] on the graph whose vertex set is the disjoint union of the setsWsi, and we put in all the edges ww0. Thus this graph has a ray w1w2... where wi ∈Wsi. Then K=v←−−Pw1w1←−−Pw2w2←−−Pw3...is an infinite walk with the property that the distance betweenvand a vertexkonKis at leastnifkappears after←−−Pwn. In particular, Ktraverses each vertex only finitely many times. ThusKis a connected locally finite graph, and thus contains a rayR. SinceRmeets each of the setsWsi, the end to whichRbelongs lives in τ, as desired.
Sublemma 1.0.48. No two distinct vertex ends ω1 and ω2 of G live in the same endτ ofT.
Proof. Suppose for a contradiction, there are suchω1,ω2andτ. LetUbe a finite separator separatingω1from ω2 and letnbe the maximal distances betweenv and a vertex inU. Then there is a leafsof Tn+1 such that τ lives inQs. Let Ci be the component of G−U in which ωi lives. Since Ws separatesU from Qs\Ws, it must be that the connected setQs\Wsis contained in a component ofG−U. Asωi lives inQs\Wsby assumption, it must be thatQs\Ws⊆Ci. HenceC1andC2 intersect, which is the desired contradiction.
Sublemma 1.0.49. No vertexudominates a vertex end ω living in some end ofT.
Proof. Suppose for a contradictionudoes. Letnbe the distance betweenuand vinG. Then there is a leafsofTn+1such thatωlives inQs. Thus the finite set Wsseparatesufromω, contradicting the assumption thatudominatesω.
Sublemma 1.0.46, Sublemma 1.0.47, Sublemma 1.0.48 and Sublemma 1.0.49 imply that the ends ofT define precisely the topological ends ofG, as desired.
Remark 1.0.50. Let (T,≤) be the tree order on T as in the proof of Theo-rem 1.0.1 where the root r is the smallest element. We remark that we con-structed(T,≤)such that(T, Pt|t∈V(T))has the following additional property:
For each edgetuwith t≤u, the vertex setS
w≥uV(Pw)\V(Pt)is connected.
Moreover, we construct (T, Pt|t∈V(T))such that ifst and tuare edges of T withs≤t≤u, thenV(Ps)∩V(Pt)andV(Pt)∩V(Pu)are disjoint.
In order to prove Lemma 1.0.45, we need the following.
Lemma 1.0.51.LetGbe a connected graph andW ⊆V(G)finite and nonempty.
Then there is a setX of disjoint edge setsX of finite boundary such that every vertex end not dominated by somew∈W lives in some X ∈ X and no edge e in anyX∈ X is incident with a vertex ofW.
Proof that Lemma 1.0.51 implies Lemma 1.0.45. We may assume thatGin Lemma 1.0.45 is connected. LetC=V(E\S
X)∪S
X∈X∂(X). ForX ∈ X letQX consist of sets of the form∂(X)∪Q, whereQis a component ofG−∂(X) withQ⊆V(X).
LetQbe the union over X of the setsQX. LetRbe the set of those H in Q such that some topological end lives inV(H). Note that each topological end lives in someR∈ R and thatW does not intersect any such R. We obtainC0 fromC by adding the vertex sets of allH ∈ Q \ R. We considerS=R ∪ {C0} as a star with centerC0. It is straightforward to verify that (S, s|s∈V(S)) is a star decomposition with the desired properties.
The rest of this section is devoted to the proof of Lemma 1.0.51. We shall need the following lemma.
Lemma 1.0.52. LetGbe a connected graph andW ⊆V(G)finite. There is a nested setN of nonempty separations of finite order such that every vertex end not dominated by some w ∈ W lives in some X ∈ N and no edge e in some X∈N is incident with a vertex ofW.
Moreover, if X, Y ∈ N are distinct with X ⊆ Y, then the order of Y is strictly larger than the order ofX.
Proof. We obtain GW from G by first deleting W and then adding a copy of Kω, the complete graph on countably many vertices, which we join completely to the neighbourhood of W. Applying Corollary 1.0.44 to GW, we obtain a nested set N0 of separations of finite order such that any two vertex ends of GW are distinguished efficiently by a separation in N0. Let τ be the vertex end to which the rays of the newly added copy ofKω belong. Let N00 consist of those separations inN0 that distinguish τ efficiently from some other vertex end. As the separations in N00 distinguish efficiency, no X ∈ N00 contains an edge incident with a vertex of the newly added copy ofKω.
Given k ∈ N, a k-sequence (Xα|α ∈ γ) (for N00) is an ordinal indexed sequence of elements ofN00of order at mostksuch that ifα < β, thenXα⊆Xβ. We obtainN000 fromN00 by addingS
α∈γXαfor allk-sequences (Xα) for all k.
Clearly,N00⊆N000 andN000is nested. Givenk∈N, the setNk consists of those X ∈ N000 of order at most k, and Nk0 consists of the inclusion-wise maximal elements ofNk.
We let N = S
k∈NNk0. By construction, each X ∈ N contains no edge incident with a vertex of the newly added copy of Kω, and thus it can be considered as an edge set ofG, whose boundary is the same as the boundary in GW. We claim that N has all the properties stated in Lemma 1.0.52: By construction, eachX ∈ N is nonempty. Since N ⊆N000, the set N is nested.
The “Moreover”-part is clear by construction. Thus it remains to show that each
vertex endωofGnot dominated by some vertex inW lives in some element of N.
LetR be a ray belonging to ω. Since ω is not dominated by any vertex in W, for eachw∈W there is a finite vertex setSw separating a subrayRw ofR fromw. ThenS =S
w∈WSw\W separatesR0=T
w∈WRw from W in Gbut also in GW. Letω0 be the vertex end ofGW to which R0 belongs. Note that S witnesses thatω0 6=τ. Thus there is someX ∈N000 in which ω0 lives. Letk be the order ofX. By Zorn’s lemma,N000 contains an inclusion-wise maximal element X0 of order at most k includingX. By constructionX0 is in Nk0 and includes a subray ofR0. Thusω lives inX0, which completes the proof.
Next we show how Lemma 1.0.52 implies Lemma 1.0.51. A good candidate for X in Lemma 1.0.51 might be the inclusion-wise maximal elements of N. However, there might be an infinite strictly increasing sequence of members in N, whose orders are also strictly increasing, so that we cannot expect that the union over the members of this sequence has finite order, and hence cannot be inN. Thus we have to make a more sophisticated choice forX than just taking the maximal members ofN.
Lemma 1.0.52 implies Lemma 1.0.51. Let N be as in Lemma 1.0.52. Let X ∈ N be such that there is another Y ∈N with X ⊆ Y, then the order of Y is strictly larger than the order of X. We denote the set of such Y of minimal order byD(X). LetH be the digraph with vertex setN where we put in the directed edgeXY if Y ∈D(X). Aconnected component of H, is a connected component of the underlying graph ofH.
Sublemma 1.0.53. Let X0, Y0 ∈N. Then X0 ⊆Y0 if and only if there is a directed path fromX0 toY0. Moreover, ifX, Y ∈N are not joined by a directed path, then they are disjoint.
Proof. Clearly, if there is a directed path from X0 to Y0, then X0 ⊆Y0. Con-versely, let X0, Y0 ∈ N with X0 ⊆ Y0. Let (Xn) be a sequence of distinct separations in N such that X0 ⊆ X1 ⊆ ... ⊆ Xn ⊆ Y0. By Lemma 1.0.52, n ≤ |∂(Y0)| − |∂(X0)|+ 1. Thus there is a maximal such chain (Zn), which satisfiesZ1=X0 andZn =Y0 andXi+1∈D(Xi) for allibetween 1 andn−1.
HenceZ1...Zn is a path fromX0 toY0.
To see that “Moreover”-part, letX, Y ∈N. AsGis connected, there is an edgeeincident with some vertex inW. Sinceeis not inX∪Y andX andY are nested,X andY must be disjoint if they are not joined by a directed path.
Sublemma 1.0.54. Each vertex v ofH has out-degree at most 1
Proof. Suppose for a contradiction v has out-degree at least 2. Then there are distinct X, Y ∈ D(v) so that neither X ⊆ Y nor Y ⊆ X. Thus X and Y are disjoint by Sublemma 1.0.53. Since v ⊆ X ∩Y, this is the desired contradiction.
Sublemma 1.0.55. Any undirected pathP joining two verticesv and w con-tains a vertex usuch that vP u andwP u are directed paths which are directed towardsu.
Proof. It suffices to show that P contains at most one vertex of out-degree 0 onP. If it contained two such vertices then between them would be a vertex of out-degree 2, which is impossible by Sublemma 1.0.54.
We define X as the union of sets XC, one for each component C of H. TheXC are defined as follows: IfC has a vertex vC of out-degree 0, then by Sublemma 1.0.55Ccannot contain a second such vertex and for any other vertex v in C, there is a directed path from v to vC directed towards vC. Hence vC
includes any otherv∈V(C). We letXC={vC}.
Otherwise,C includes a ray X1X2. . . asC cannot contain a directed cycle by Sublemma 1.0.53. In this case, we takeXC to be the set consisting of the Yi = Xi\Xi−1 for each i∈ N, where Y1 =X1. Note that the order of Yi is bounded by the sum of the orders ofXi andXi−1, and thus finite.
Since noY ∈N contains an edge incident with somew ∈W, the same is true for anyY ∈ X. Any two distinct X, Y ∈ X are disjoint: IfX and Y are in the same XC, this is clear by construction. Otherwise it follows from the definition ofYi and Sublemma 1.0.53. Thus it remains to prove the following:
Sublemma 1.0.56. Each vertex end ω not dominated by some vertex of W lives in someX∈ X.
Proof. By Lemma 1.0.52, there is some Z ∈N in whichω lives. Let C be the component ofH containingZ. IfXC ={vC}, then Z ⊆vC. Otherwise let the Xi and theYi be as in the construction of XC. IfZ =Xj for somej. Then we pickj minimal such thatω lives in Xj. Since ω does not live inXj−1, it must live inYj, as desired.
Thus we may assume thatZis not equal to anyXj. LetP be a path joining Z and X1 =Y1. By Sublemma 1.0.55, P contains a vertex u such that ZP u and X1P u are directed paths which are directed towards u. If u= X1, then Z ⊆ Y1, and we are done. OtherwiseX1P u is a subpath of the rayX1X2. . . since the out-degree is at most 1 so thatu=Xj for some j.
We pick P such that the j with u= Xj is minimal and have the aim to prove that then Z ⊆Yj. Since Z ⊆Xj, it remains to show that Z and Xj−1
are disjoint. Suppose for a contradiction, there is a directed pathQjoining Z andXj−1. IfQ is directed towards Z, thenZ =Xm for somem, contrary to our assumption. ThusQis directed towardsXj−1. But thenZQXj−1P X1 has a smallerj-value, which contradicts the minimality ofP. Hence there cannot be such a Q, and thus Z and Xj−1 are disjoint by Sublemma 1.0.53. Having shown thatZ⊆Yj, we finish the proof by concluding that thenω also lives in Yj.
Finally we deduce Corollary 1.0.2.
Proof that Theorem 1.0.1 implies Corollary 1.0.2. By Theorem 1.0.1, G has a tree-decomposition (T, Pt|t∈V(T)) of finite adhesion such that the ends ofT define precisely the topological ends ofT, and we choose this tree-decomposition as in Remark 1.0.50. In particular, we can pick a rootrofT such that for each edgetuwitht≤u, the vertex setS
w≥uV(Pw)\V(Pt) is connected.
Thus for each such edge tu, there is a finite connected subgraph Su of G[S
w≥uV(Pw)] that contains V(Pt)∩V(Pu). Let Qt be a maximal subfor-est of the union of theSu, where the union ranges over all upper neighboursu of t. We recursively build a maximal subset U of V(T) such that if a, b ∈U, thenQa andQb are vertex-disjoint. In this construction, we first add the nodes ofT with smaller distance from the root. This ensures by the “Moreover”-part of Remark 1.0.50 thatU contains infinitely many nodes of each ray of T.
LetS0 be the union of thoseQt witht ∈U. We obtainS by extendingS0 to a spanning tree of G, and rooting it at some v ∈ V(S) arbitrarily. By the Star-Comb-Lemma [52, Section 8], each spanning tree of G contains for each topological endω a ray belonging toω.
Thus it remains to show that S does not contain two disjoint raysR1 and R2 that both belong to the same topological end ω of G. Suppose there are such R1, R2 and ω. Let t1t2. . . be the ray of T in which ω lives. Let n be so large that both R1 and R2 meet Ptn. Then for each m ≥ n, the set Stm
contains a path joiningR1 andR2. Thus the set Qtm−1 contains such a path.
SinceQtm−1⊆S for infinitely manym, the treeScontains a cycle, which is the desired contradiction.
Chapter 2
Canonical
tree-decompositions
In Section 2.1, we show that every finite graph has a canonical tree-decomposition distinguishing all robust blocks efficiently. We improve these tree-decompositions in Section 2.2 so that they additionally distinguish all the maximal tangles. In Section 2.2 we investigate further properties of these tree-decompositions.
In Section 2.4, we use our techniques to give a simpler proof of the tangle-tree theorem of Robertson and Seymour. In Section 2.5, we study the block number.