Let Gbe a graph and (A, B) a separation of G. Ashape for (A, B) is a word v1x1v2x2. . . xn−1vn withvi∈A∩Bandxi∈ {l, r}such that no vertex appears twice. We call the vi the vertices of the shape. Every ray R induces a shape
σ =σR(A, B) on every separation (A, B) of finite order in the following way:
Let <R be the natural order on V(R) induced by the ray, where v <R w if w lies in the unique infinite component of R−v. The vertices of σ are those vertices of Rthat lie inA∩B and they appear inσ in the order given by<R. Forvi, vi+1 the pathviRvi+1 has edges only inAor only inB but not in both.
In the first case we putl betweenvi and vi+1 and in the second case we put r betweenvi andvi+1.
Let (A1, B1),(A2, B2) be separations withA1∩B2=∅and thus also A1⊆ A2 and B2 ⊆ B1. Let σi be a nonempty shape for (Ai, Bi). The word τ = v1x1v2. . . xn−1vn is an allowed shape linking σ1 to σ2 withvertices v1. . . vn if the following holds.
• vis a vertex ofτ if and only if it is a vertex ofσ1 orσ2,
• ifv appears beforewin σi, thenv appears beforewin τ,
• v1is the initial vertex of σ1 andvn is the terminal vertex ofσ2,
• xi∈ {l, m, r},
• the subwordvlw appears inτ if and only if it appears inσ1,
• the subwordvrwappears inτ if and only if it appears inσ2,
• vi6=vj fori6=j.
Each ray R defines a word τ =τR[(A1, B1),(A2, B2)] = v1x1v2. . . xn−1vn
with verticesviandxi∈ {l, m, r}as follows. The vertices ofτare those vertices ofR that lie inA1∩B1 orA2∩B2and they appear in τ in the order given by
<R. Forvi, vi+1 the pathviRvi+1 has edges either only inA1, only inA2∩B1, or only inB2. In the first case we setxi =landτcontains the subwordvilvi+1. In the second case we set xi =mand τ contains the subword vimvi+1. In the third case we set xi=randτ contains the subword virvi+1.
For a rayR to induce an allowed shape τR[(A1, B1),(A2, B2)] we need at least thatRstarts inA2. However, each ray inω has a tail such that whenever it meets an Ai it also starts in thatAi. Let us call such rayslefty. A 2-ray is lefty if both its rays are.
Remark 21. Let (A1, B1), and (A2, B2) be two separations of finite order with A1 ⊆ A2, and B2 ⊆ B1. For every lefty ray R meeting A1, the word τR[(A1, B1),(A2, B2)] is an allowed shape linking σR(A1, B1) and σR(A2, B2).
From now on let us fix a locally finite graphGwith a thin endωof vertex-degreek. And let ((Ai, Bi))i∈Nbe a sequence capturingω such that each mem-ber has orderk.
A 2-shape for a separation (A, B) is a pair of shapes for (A, B). Every 2-ray induces a 2-shape coordinatewise in the obvious way. Similarly, anallowed 2-shape is a pair of allowed shapes.
Clearly, there is a global constant c1 ∈ N depending only on k such that there are at most c1 distinct 2-shapes for each separation (Ai, Bi). Similarly, there is a global constant c2∈Ndepending only onksuch that for all i, j∈N there are at mostc2distinct allowed 2-shapes linking a 2-shape for (Ai, Bi) with a 2-shape for (Aj, Bj).
For most of the remainder of this subsection we assume that for everyi∈N there is a setDi consisting of at least c1·c2·i edge-disjoint 2-rays inG. Our aim will be to show that in these circumstances there must be infinitely many edge-disjoint 2-rays.
By taking a tailor if necessary, we may assume that every 2-ray in eachDi
is lefty.
Lemma 22. There is an infinite setJ ⊆Nand, for eachi∈N, a tailorDi0 of Di of cardinalityc2·isuch that for all i∈Nandj∈J all2-rays inD0i induce the same2-shape σ[i, j]on(Aj, Bj).
Proof. We recursively build infinite sets Ji⊆Nand tailorsDi0 ofDi such that for all k≤iand j ∈Ji all 2-rays in D0k induce the same 2-shape on (Aj, Bj).
For alli≥1, we shall ensure thatJi is an infinite subset ofJi−1 and that the i−1 smallest members ofJi andJi−1 are the same. We shall takeJ to be the intersection of all theJi.
LetJ0=Nand letD00 be the empty set. Now, for somei≥1, assume that sets Jk andD0k have been defined for all k < i. By replacing 2-rays inDi by their tails, if necessary, we may assume that each 2-ray in Di avoidsA`, where
` is the (i−1)st smallest value ofJi−1. AsDi containsc1·c2·imany 2-rays, for each j ∈ Ji−1 there is a set Sj ⊆Di of size at least c2·i such that each 2-ray in Sj induces the same 2-shape on (Aj, Bj). As there are only finitely many possible choices forSj, there is an infinite subsetJi ofJi−1on whichSj
is constant. For D0i we pick this value of Sj. Since each d ∈ Di0 induces the empty 2-shape on each (Ak, Bk) withk≤`we may assume that the first i−1 elements ofJi−1 are also included inJi.
It is immediate that the setJ =T
i∈NJi and theD0i have the desired prop-erty.
Lemma 23. There are two strictly increasing sequences (ni)i∈N and (ji)i∈N
withni∈Nandji∈J for alli∈Nsuch thatσ[ni, ji] =σ[ni+1, ji]andσ[ni, ji] is not empty.
Proof. LetH be the graph onNwith an edgevw ∈E(H) if and only if there are infinitely many elements j∈J such thatσ[v, j] =σ[w, j].
As there are at mostc1 distinct 2-shapes for any separator (Ai, Bi), there is no independent set of size c1+ 1 in H and thus no infinite one. Thus, by Ramsey’s theorem, there is an infinite clique in H. We may assume without loss of generality thatH itself is a clique by moving to a subsequence of theD0i if necessary. With this assumption we simply pickni=i.
Now we pick theji recursively. Assume thatji has been chosen. As i and i+ 1 are adjacent in H, there are infinitely many indicies ` ∈ N such that
σ[i, `] =σ[i+ 1, `]. In particular, there is such an ` > ji such thatσ[i+ 1, `] is not empty. We pickji+1 to be one of those`.
Clearly, (ji)i∈N is an increasing sequence andσ[i, ji] =σ[i+ 1, ji] as well as σ[i, ji] is non-empty for alli∈N, which completes the proof.
By moving to a subsequence of (Di0) and ((Aj, Bj)), if necessary, we may assume by Lemma 22 and Lemma 23 that for alli, j∈Nalld∈D0i induce the same 2-shape σ[i, j] on (Aj, Bj), and thatσ[i, i] =σ[i+ 1, i], and thatσ[i, i] is non-empty.
Lemma 24. For alli∈Nthere isDi00⊆D0i such that|D00i|=i, and alld∈D00i induce the same allowed 2-shape τ[i]that links σ[i, i] andσ[i, i+ 1].
Proof. Note that it is in this proof that we need all the 2-rays inDi00to be lefty as they need to induce an allowed 2-shape that linksσ[i, i] andσ[i, i+ 1] as soon as they contain a vertex from Ai. As |Di0| ≥i·c2 and as there are at most c2
many distinct allowed 2-shapes that link σ[i, i] andσ[i, i+ 1] there isDi00⊆D0i with|D00i|=isuch that alld∈Di00induce the same allowed 2-shape.
We enumerate the elements ofDj00as follows: dj1, dj2, . . . , djj. Let (sji, tji) be a representation ofdji. LetSji =sji∩Aj+1∩Bj, and letSi =S
j≥iSij. Similarly, letTij=tji ∩Aj+1∩Bj, and letTi=S
j≥iTij.
Clearly,SiandTi are vertex-disjoint and any two graphs inS
i∈N{Si,Ti}are edge-disjoint. We shall find a rayRi in each of theSi and a rayR0i in each of theTi. The infinitely many pairs (Ri, R0i) will then be edge-disjoint 2-rays, as desired.
Lemma 25. Each vertexv of Si has degree at most 2. If v has degree 1 it is contained inAi∩Bi.
Proof. Clearly, each vertex v ofSi that does not lie in any separator Aj∩Bj
has degree 2, as it is contained in precisely one Sij, and all the leaves ofSij lie inAj∩Bj andAj+1∩Bj+1 asdji is lefty. Indeed, inSij it is an inner vertex of a path and thus has degree 2 in there. Ifvlies inAi∩Biit has degree at most 2, as it is only a vertex ofSij for one value ofj, namelyj=i.
Hence, we may assume thatv∈Aj∩Bjfor somej > i. Thus,σ[j, j] contains v andl:σ[j, j] :rcontains precisely one of the four following subwords:
lvl, lvr, rvl, rvr
(Here we use the notationp:qto denote the concatenation of the wordpwith the wordq.) In the first caseτ[j−1] containsmvmas a subword and τ[j] has nomadjacent tov. ThenSij−1 contains precisely 2 edges adjacent tov andSji has no such edge. The fourth case is the first one withl andrandj andj−1 interchanged.
In the second and third cases, each ofτ[j−1] andτ[j] has precisely onem adjacent tov. So bothSij−1and Sij contain precisely 1 edge adjacent tov.
Asv appears only as a vertex ofSi`for`=j or`=j−1, the degree ofvin Si is 2.
Lemma 26. There are an odd number of vertices inSi of degree1.
Proof. By Lemma 25 we have that each vertex of degree 1 lies inAi∩Bi. Letv be a vertex inAi∩Bi. Then,σ[i, i] containsvandl:σ[i, i] :rcontains precisely one of the four following subwords:
lvl, lvr, rvl, rvr
In the first and fourth case v has even degree. It has degree 1 otherwise. As l : σ[i, i] :rstarts with l and ends withr, the word lvr appear precisely once more than the word rvl. Indeed, between two occurrences of lvr there must be one of rvl and vice versa. Thus, there are an odd number of vertices with degree 1 in Si.
Lemma 27. Si includes a ray.
Proof. By Lemma 25 every vertex of Si has degree at most 2 and thus every component ofSi has at most two vertices of degree 1. By Lemma 26Si has a component C that contains an odd number of vertices with degree 1. Thus C has precisely one vertex of degree 1 and all its other vertices have degree 2, thus C is a ray.
Corollary 28. Gcontains infinitely many edge-disjoint 2-rays.
Proof. By symmetry, Lemma 27 is also true withTi in place ofSi. ThusSi∪ Ti includes a 2-rayXi. TheXiare edge-disjoint by construction.
Recall that Lemma 15 states that a countable graph with a thin endω and arbitrarily many edge-disjoint double rays all whose subrays converge toω, also has infinitely many edge-disjoint double rays. We are now in a position to prove this lemma.
Proof of Lemma 15. By Lemma 20 it suffices to show that G contains a sub-graph H with a single end which is thin such thatH has infinitely many edge-disjoint 2-rays. By Corollary 17,Ghas a subgraphH with a single end which is thin such that H has arbitrarily many edge-disjoint 2-rays. But then by the argument aboveHcontains infinitely many edge-disjoint 2-rays, as required.
With these tools at hand, the remaining proof of Theorem 1 is easy. Let us collect the results proved so far to show that each graph with arbitrarily many edge-disjoint double rays also has infinitely many edge-disjoint double rays.
Proof of Theorem 1. LetGbe a graph that has a setDiofiedge-disjoint double rays for eachi∈N. Clearly,Ghas infinitely many edge-disjoint double rays if its subgraphS
i∈NDi does, and thus we may assume without loss of generality that G=S
i∈NDi. In particular,Gis countable.
By Corollary 10 we may assume that each connected component ofG in-cludes only finitely many ends. As each component inin-cludes a double ray we may assume that G has only finitely many components. Thus, there is one
component containing arbitrarily many edge-disjoint double rays, and thus we may assume that Gis connected.
By Corollary 8 we may assume that all ends ofGare thin. Thus, as men-tioned at the start of Section 4, there is a pair of ends (ω, ω0) ofG(not neces-sarily distinct) such thatGcontains arbitrarily many edge-disjoint double rays each of which converges precisely to ω and ω0. This completes the proof as, by Lemma 13 Ghas infinitely many edge-disjoint double rays if ω and ω0 are distinct and by Lemma 15 G has infinitely many edge-disjoint double rays if ω=ω0.
6 Outlook and open problems
We will say that a graphH isedge-ubiquitousif every graph having arbitrarily many edge-disjointH also has infinitely many edge-disjointH.
Thus Theorem 1 can be stated as follows: the double ray is edge-ubiquitous.
Andreae’s Theorem implies that the ray is edge-ubiquitous. And clearly, every finite graph is edge-ubiquitous.
We could ask which other graphs are edge-ubiquitous. It follows from our result that the 2-ray is edge-ubiquitous. Let G be a graph in which there are arbitrarily many edge-disjoint 2-rays. Letv∗Gbe the graph obtained fromG by adding a vertexv adjacent to all vertices ofG. Thenv∗G has arbitrarily many edge-disjoint double rays, and thus infinitely many edge-disjoint double rays. Each of these double rays uses v at most once and thus includes a 2-ray ofG.
The vertex-disjoint union of k rays is called a k-ray. The k-ray is edge-ubiquitous. This can be proved with an argument similar to that for Theorem 1:
LetGbe a graph with arbitrarily many edge-disjointk-rays. The same argument as in Corollaries 10 and 8 shows that we may assume that Ghas only finitely many ends, each of which is thin. By removing a finite set of vertices if necessary we may assume that each component ofGhas at most one end, which is thin.
Now we can find numberskC indexed by the componentsC ofGand summing to k such that each component C has arbitrarily many edge-disjointkC-rays.
Hence, we may assume thatGhas only a single end, which is thin. By Lemma 16 we may assume that Gis locally finite.
In this case, we use an argument as in Subsection 5.3. It is necessary to use k-shapes instead of 2-shapes but other than that we can use the same combina-torial principle. IfC1 andC2are finite sets, a (C1, C2)-shapingis a pair (c1, c2) wherec1is a partial colouring ofNwith colours fromC1 which is defined at all but finitely many numbers and c2 is a colouring ofN(2) with colours fromC2
(in our argument above, C1 would be the set of all k-shapes andC2 would be the set of all allowedk-shapes for all pairs of k-shapes).
Lemma 29. LetD1, D2, . . .be a sequence of sets of(C1, C2)-shapings whereDi
has size i. Then there are strictly increasing sequences i1, i2, . . . andj1, j2, . . . and subsets Sn⊆Din with|Sn| ≥nsuch that
• for anyn∈Nall the values ofc1(jn)for the shapings(c1, c2)∈Sn−1∪Sn
are equal (in particular, they are all defined).
• for anyn∈N, all the values ofc2(jn, jn+1)for the shapings (c1, c2)∈Sn
are equal.
Lemma 29 can be proved by the same method with which we constructed the sets D00i from the setsDi. The advantage of Lemma 29 is that it can not only be applied to 2-rays but also to more complicated graphs likek-rays.
Atalon is a tree with a single vertex of degree 3 where all the other vertices have degree 2. An argument as in Subsection 5.2 can be used to deduce that talons are edge-ubiquitous from the fact that 3-rays are. However, we do not know whether the graph in Figure 2 is edge-ubiquitous.
Figure 2: A graph obtained from 2 disjoint double rays, joined by a single edge.
Is this graph edge-ubiquitous?
We finish with the following open problem.
Problem 30. Is the directed analogue of Theorem 1 true? More precisely: Is it true that if a directed graph has arbitrarily many edge-disjoint directed double rays, then it has infinitely many edge-disjoint directed double rays?
It should be noted that if true the directed analogue would be a common generalization of Theorem 1 and the fact that double rays are ubiquitous with respect to the subgraph relation.
7 Acknowledgement
We appreciate the helpful and accurate comments of a referee.
References
[1] Th. Andreae. ¨Uber maximale Systeme von kantendisjunkten unendlichen Wegen in Graphen. Math. Nachr., 101:219–228, 1981.
[2] Thomas Andreae. On disjoint configurations in infinite graphs. Journal of Graph Theory, 39(4):222–229, 2002.
[3] Thomas Andreae. Classes of locally finite ubiquitous graphs. Journal of Combinatorial Theory, Series B, 103(2):274 – 290, 2013.
[4] R. Diestel. Graph Theory. Springer, 4th edition, 2010.
[5] R. Halin. ¨Uber unendliche Wege in Graphen. Math. Annalen, 157:125–137, 1964.
[6] R. Halin. Die Maximalzahl fremder zweiseitig unendlicher Wege in Graphen.
Math. Nachr., 44:119–127, 1970.
[7] H.A. Jung. Wurzelb¨aume und unendliche Wege in Graphen. Math. Nachr., 41:1–22, 1969.
[8] John Lake. A problem concerning infinite graphs. Discrete Mathematics, 14(4):343 – 345, 1976.
[9] D.R Woodall. A note on a problem of Halin’s. Journal of Combinatorial Theory, Series B, 21(2):132 – 134, 1976.
Linkages in Large Graphs of Bounded Tree-Width
Jan-Oliver Fröhlich
∗Ken-ichi Kawarabayashi
†‡Theodor Müller
§Julian Pott
¶Paul Wollan
kAbstract
We show that all sufficiently large (2k+ 3)-connected graphs of bounded tree-width arek-linked. Thomassen has conjectured that all sufficiently large(2k+ 2)-connected graphs arek-linked.
1 Introduction
Given an integerk ≥1, a graph Gisk-linked if for any choice of2k distinct verticess1, . . . , sk andt1, . . . , tk of Gthere are disjoint paths P1, . . . , Pk in G such that the end vertices of Pi are si and ti for i = 1, . . . , k. Menger’s theorem implies that every k-linked graph isk-connected.
One can conversely ask how much connectivity (as a function of k) is required to conclude that a graph is k-linked. Larman and Mani [12] and Jung [8] gave the first proofs that a sufficiently highly connected graph is also k-linked. The bound was steadily improved until Bollobás and Thomason [3] gave the first linear bound on the necessary connectivity, showing that every 22k-connected graph is k-linked. The current best bound shows that 10k-connected graphs are also k-linked [18].
What is the best possible functionf(k)one could hope for which implies anf(k)-connected graph must also bek-linked? Thomassen [20] conjectured
∗Fachbereich Mathematik, Universität Hamburg, Hamburg, Germany.
jan-oliver.froehlich@math.uni-hamburg.de
†National Institute of Informatics, Tokyo, Japan.
‡JST, ERATO, Kawarabayashi Large Graph Project.
§Fachbereich Mathematik, Universität Hamburg, Hamburg, Germany.
¶Fachbereich Mathematik, Universität Hamburg, Hamburg, Germany.
kDepartment of Computer Science, University of Rome "La Sapienza", Rome, Italy.
wollan@di.uniroma1.it. Research supported by the European Research Council under the European Unions Seventh Framework Programme (FP7/2007 - 2013)/ERC Grant Agree-ment no. 279558.
that(2k+2)-connected graphs arek-linked. However, this was quickly proven to not be the case by Jørgensen with the following example [21]. Consider the graph obtained fromK3k−1 obtained by deleting the edges of a matching of sizek. This graph is(3k−3)-connected but is notk-linked. Thus, the best possible function f(k) one could hope for to implyk-linked would be3k−2.
However, all known examples of graphs which are roughly3k-connected but notk-linked are similarly of bounded size, and it is possible that Thomassen’s conjectured bound is correct if one assumes that the graph has sufficiently many vertices.
In this paper, we show Thomassen’s conjectured bound is almost correct with the additional assumption that the graph is large and has bounded tree-width. This is the main result of this article.
Theorem 1.1. For all integers k and w there exists an integer N such that a graph Gis k-linked if
κ(G)≥2k+ 3, tw(G)< w, and |G| ≥N.
where κ is the connectivity of the graph andtw is the tree-width.
The tree-width of the graph is a parameter commonly arising in the theory of graph minors; we will delay giving the definition until Section 2 where we give a more in depth discussion of how tree-width arises naturally in tackling the problem. The value 2k + 2 would be best possible; see Section 8 for examples of arbitrarily large graphs which are (2k+ 1)-connected but not k-linked.
Our work builds on the theory of graph minors in large, highly connected graphs begun by Böhme, Kawarabayashi, Maharry and Mohar [1]. Recall that a graph Gcontains Kt as a minor if Kt can be obtained from a subgraph ofGby repeatedly contracting edges. Böhme et al. showed that there exists an absolute constantc such that every sufficiently largect-connected graph contains Kt as a minor. This statement is not true without the assumption that the graph be sufficiently large, as there are examples of small graphs which are(t√
logt)-connected but still have noKt minor [11, 19]. In the case where we restrict our attention to small values of t, one is able to get an explicit characterisation of the larget-connected graphs which do not contain Kt as a minor.
Theorem 1.2 (Kawarabayashi et al. [10]). There exists a constantN such that every 6-connected graphGon N vertices either contains K6 as a minor or there exists a vertexv∈V(G) such thatG−v is planar.
Jorgensen [7] conjectures that Theorem 1.2 holds for all graphs without the additional restriction to graphs on a large number of vertices. In 2010, Norine and Thomas [17] announced that Theorem 1.2 could be generalised to arbitrary values of t to either find a Kt minor in a sufficiently large t-connected graph or alternatively, find a small set of vertices whose deletion leaves the graph planar. They have indicated that their methodology could be used to show a similar bound of 2k+ 3on the connectivity which ensures a large graph is k-linked.
2 Outline
In this section, we motivate our choice to restrict our attention to graphs of bounded tree-width and give an outline of the proof of Theorem 1.1.
We first introduce the basic definitions of tree-width. A tree-decompos-ition of a graphGis a pair(T,X) whereT is a tree andX ={Xt ⊆V(G) : t ∈ V(T)} is a collection of subsets of V(G) indexed by the vertices of T. Moreover,X satisfies the following properties.
1. S
t∈V(T)Xt =V(G),
2. for all e ∈ E(G), there exists t∈ V(T) such that both ends of e are contained inXt, and
3. for allv∈V(G), the subset {t∈V(T) :v∈Xt}induces a connected subtree of T.
The sets inX are sometimes called thebags of the decomposition. The width of the decomposition is maxt∈V(T)|Xt| −1, and the tree-width ofGis the minimum width of a tree-decomposition.
Robertson and Seymour showed that if a 2k-connected graph contains K3k as a minor, then it is k-linked [15]. Thus, when one considers (2k+ 3)-connected graphs which are notk-linked, one can further restrict attention to graphs which exclude a fixed clique minor. This allows one to apply the excluded minor structure theorem of Robertson and Seymour [16]. The structure theorem can be further strengthened if one assumes the graph has large tree-width [5]. This motivates one to analyse separately the case when the tree-width is large or bounded. The proofs of the main results in [1] and [10] similarly split the analysis into cases based on either large or bounded tree-width.
We continue with an outline of how the proof of Theorem 1.1 proceeds.
Assume Theorem 1.1 is false, and letGbe a(2k+ 3)-connected graph which
is not k-linked. Fix a set {s1, . . . , sk, t1, . . . , tk}such that there do not exist disjoint paths P1, . . . , Pk where the ends of Pi are si and ti for all i. Fix a tree-decomposition(T,X)of Gof minimal width w.
We first exclude the possibility thatT has a high degree vertex. Assume t is a vertex of T of large degree. By Property 3 in the definition of a tree-decomposition, if we delete the setXt of vertices fromG, the resulting graph must have at least degT(t) distinct connected components. By the connectivity of G, each component contains 2k+ 3internally disjoint paths from a vertexvto2k+3distinct vertices inXt. If the degree oftis sufficiently large, we conclude that the graph Gcontains a subdivision of Ka,2k+3 for some large value a. We now prove that that if a graph contains such a large complete bipartite subdivision and is 2k-connected, then it must be k-linked (Lemma 7.1).
We conclude that the tree T does not have a high degree vertex, and consequently contains a long path. It follows that the graph Ghas a long path decomposition, that is, a tree-decomposition where the tree is a path.
As the bags of the decomposition are linearly ordered by their position on the path, we simply give the path decomposition as a linearly ordered set of bags (B1, . . . , Bt) for some large value t. At this point in the argument, the path-decomposition (B1, . . . , Bt) may not have bounded width, but it will have the property that |Bi ∩Bj| is bounded, and this will suffice for the argument to proceed. Section 3 examines this path decomposition in detail and presents a series of refinements allowing us to assume the path decomposition satisfies a set of desirable properties. For example, we are able to assume that |Bi∩Bi+1| is the same for alli, 1≤i < t. Moreover, there exist a setP of|B1∩B2|disjoint paths starting inB1 and ending inBt. We call these paths the foundational linkage and they play an important role in the proof. A further property of the path decomposition which we prove in Section 3 is that for each i, 1 < i < t, if there is a bridge connecting two foundational paths inP inBi, then for allj,1< j < t, there exists a bridge connecting the same foundational paths in Bj. This allows us to define an auxiliary graph H with vertex setP and two vertices of P adjacent inH if there exists a bridge connecting them in some Bi 1< i < t.
Return to the linkage problem at hand; we have 2k terminalss1, . . . , sk
and t1, . . . , tk which we would like to link appropriately, andB1, . . . , Bt is our path decomposition with the foundational linkage running through it.
Let the setBi∩Bi+1be labeled Si. As our path decomposition developed in the previous paragraph is very long, we can assume there exists some long subsection Bi, Bi+1, . . . , Bi+a such that no vertex of s1, . . . , sk, t1, . . . , tk is contained in Si+a
i Bi −(Si−1∪Si+a) for some large value a. By Menger’s