Cycles through all finite vertex sets in infinite graphs

European Journal of Combinatorics 65 (2017) 259–275

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Cycles through all finite vertex sets in infinite graphs✩ André Kündgen a , Binlong Li b , Carsten Thomassen c a

Department of Mathematics, California State University San Marcos, San Marcos, CA 92096, United States Department of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, PR China c Department of Applied Mathematics and Computer Science, Technical University of Denmark, DK-2800 Lyngby, Denmark b

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Article history: Received 28 February 2017 Accepted 23 June 2017

a b s t r a c t A closed curve in the Freudenthal compactification |G| of an infinite locally finite graph G is called a Hamiltonian curve if it meets every vertex of G exactly once (and hence it meets every end at least once). We prove that |G| has a Hamiltonian curve if and only if every finite vertex set of G is contained in a cycle of G. We apply this to extend a number of results and conjectures on finite graphs to Hamiltonian curves in infinite locally finite graphs. For example, Barnette’s conjecture (that every finite planar cubic 3-connected bipartite graph is Hamiltonian) is equivalent to the statement that every one-ended planar cubic 3-connected bipartite graph has a Hamiltonian curve. It is also equivalent to the statement that every planar cubic 3-connected bipartite graph with a nowhere-zero 3flow (with no restriction on the number of ends) has a Hamiltonian curve. However, there are 7-ended planar cubic 3-connected bipartite graphs that do not have a Hamiltonian curve. © 2017 Elsevier Ltd. All rights reserved.

1. Introduction Diestel [6] launched the ambitious project of extending results on finite Hamiltonian cycles to Hamiltonian circles, that is simple closed curves traversing each vertex and each end precisely once ✩ This work was done when Li and Kündgen visited the Technical University of Denmark during the Academic year 2016– 2017. Li was supported by National Natural Science Foundation of China (Nos. 11601429 & 11671320). Kündgen and Thomassen were supported by the ERC Advanced Grant GRACOL, Project No. 320812. E-mail addresses: [email protected] (A. Kündgen), [email protected] (B. Li), [email protected] (C. Thomassen). http://dx.doi.org/10.1016/j.ejc.2017.06.006 0195-6698/© 2017 Elsevier Ltd. All rights reserved.

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in the Freudenthal compactification |G| of a locally finite graph G. Diestel specifically suggested to find Hamiltonian circles in the square of a graph in order to obtain a unification of Fleischner’s theorem [11] and its counterpart for one-ended graphs by Thomassen [28]. This was done by Georgakopoulos [13]. Georgakopoulos [14] (see also [6]) then conjectured that every 4-connected line graph has a Hamiltonian circle. Bruhn (see [6]) conjectured that Tutte’s theorem on Hamiltonian cycles in 4connected planar graphs can be extended to Hamiltonian circles. These conjectures are open but significant progress has been made by Lehner [24] on the former and by Bruhn and Yu [3] on the latter. Several other results in this area can be found in [5,17,21–23]. In the present paper we consider another natural way of extending results on finite Hamiltonian cycles to infinite locally finite graphs, namely the following property: for every finite vertex set S, the graph has a cycle containing S. We prove that this property is related to the Hamiltonian circle problem in that the property holds if and only if the Freudenthal compactification |G| has a Hamiltonian curve, that is, a simple closed curve traversing each vertex precisely once (and hence each end at least once). The most difficult part here is the implication from infinite to finite graphs. For a finite vertex set S we let GS be the graph obtained from G[S ], the subgraph induced by S, by adding an edge between each pair of vertices that are joined by a path internally disjoint from S. If G has a cycle whose vertex set contains S, then clearly GS has a Hamiltonian cycle. The converse is not true for a fixed S. However, our result implies that, if GS has a Hamiltonian cycle for each finite vertex set S, then also G has a cycle containing S, for each finite S. As the proof goes via Hamiltonian curves, one may see this as a way of using infinite substructures to find finite substructures. As pointed out by Georgakopoulos [13], we can often go in the other direction and obtain Hamiltonian curves by appropriate ad hoc compactness arguments. Our characterization tells more precisely when that can be done, and we illustrate this by extending various results and problems on finite graphs. A particularly interesting example is Barnette’s conjecture that every finite planar bipartite cubic 3-connected graph has a Hamiltonian cycle. An infinite planar bipartite cubic 3connected graph need not have a Hamiltonian curve. Indeed, there are 7-ended counterexamples. Instead we show that Barnette’s conjecture for finite graphs is equivalent with the following: Every infinite planar cubic 3-connected graph having a nowhere-zero 3-flow has a Hamiltonian curve. Using this we also conclude that Barnette’s conjecture for finite graphs is also equivalent with the following: Every infinite planar cubic 3-connected bipartite one-ended graph has a Hamiltonian curve. 2. Notation and terminology All graphs are locally finite in this paper, that is, all vertices have finite degree. We follow Diestel [8] in our basic terminology for infinite graphs G. Specifically, a ray is a one-way infinite path, and two such rays in a graph G are equivalent if for every finite vertex set S both rays have a tail in the same component of G − S. An end α of G is an equivalence class of rays, and this can be viewed as a particular ‘‘point at infinity’’. For a finite vertex set S we now denote the unique component of G − S containing a tail of every ray in α by C (S , α ). We also let Ω (G) be the set of ends of G, and Ω (S , α ) be the set of all ends β with C (S , β ) = C (S , α ). To define a topology on G we associate each edge uv with a homeomorphic image of [0, 1] where 0,1 map to u, v (and where different edges may only intersect at common endpoints). Basic open neighborhoods of points that are vertices or contained in edges are defined in the usual way, that is we start with the basic open neighborhoods in the topology of the 1-complex. For an end α we proceed as follows: For every finite vertex set S we let the basic neighborhood ˆ C (S , α ) consist of C (S , α ), Ω (S , α ) and the edges between C (S , α ) and S (except their endvertices in S). It is not hard to verify that a connected locally finite graph G together with its ends Ω (G) is a compact Hausdorff space in this topology called the Freudenthal compactification |G| of G (see [8] Proposition 8.6.1). The closure of C (S , α ) in this topology is C (S , α ) ∪ Ω (S , α ). Let x = (xi )∞ i=1 be a vertex sequence of G. We say that a set A contains almost all terms of x if there exists j such that xi ∈ A for all i ≥ j. So the sequence x converges to an end α ∈ Ω (G) if every C (S , α ) for S finite contains almost all terms of x. Thus the closure of a vertex set X is X ∪ Ω (X ), where Ω (X ) denotes the set of ends of G such that there is a sequence in X converging to the end. Conversely, α ∈ Ω (G) − Ω (X ) if and only if there is a finite vertex set S such that C (S , α ) is disjoint from X .

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A (closed) curve of |G| is the image of a continuous map from the unit circle S 1 to |G|. The map itself is called a parametrization of the curve. A circle in |G| is a curve that is homeomorphic to the unit circle S 1 in R2 . An arc of |G| is a homeomorphic image of the unit interval [0, 1] in |G|. For any set of vertices X we let an X -curve be a closed curve in |G| such that every vertex in X is used precisely once. Thus a V (G)-curve, which we will also call a Hamiltonian curve, meets every vertex exactly once, but may meet an end arbitrarily often. A Hamiltonian circle in |G| is a circle which meets every vertex (and every end) exactly once. An X -curve is strict if it does not use vertices outside of X . Example 1. Observe that the one-ended locally finite graph G obtained from a grid with vertex-set

{(x, y) : x, y ∈ Z} by removing every edge of the form (x, 0)(x, 1) and (x, 0)(x, −1) when x is an even integer, has a Hamiltonian curve, but does not have a Hamiltonian circle. 3. Hamiltonian curves The main purpose of this paper is to better understand the property that every finite vertex set of a locally finite graph G is contained in a cycle, where a cycle is a finite connected subgraph of G such that every vertex has degree 2. Our main result relates this property to topological properties of the Freudenthal compactification. Theorem 1. The following are equivalent for any locally finite graph G. (i) For every finite vertex set S, G has a cycle containing S. (ii) For every finite vertex set S, there is an S-curve in |G|. (iii) |G| has a Hamiltonian curve. While (i) H⇒ (ii) follows from the observation that every cycle through S is an S-curve, the converse is less obvious. We will prove (ii) H⇒ (iii) in Section 5 and (iii) H⇒ (i) in Section 6. Our next result gives additional characterizations that are a bit more technical, but useful for applications. Recall that a 2-factor is a 2-regular spanning subgraph. Theorem 2. The following are equivalent for any locally finite graph G. (i) (ii) (iii) (iv)

|G| has a Hamiltonian curve. G has a 2-factor F such that every finite cut intersects F a positive even number of times. For every finite vertex set S with at least three vertices, GS has a Hamiltonian cycle. For every finite vertex set S, there is a vertex set S ′ containing S such that GS ′ has a Hamiltonian cycle.

Proof. The equivalence of (i) and (ii) follows from Theorem 2.5 in [7], but we include the proof for completeness. To see that (i) H⇒ (ii) observe that the vertices and edges in a Hamiltonian curve C of |G| clearly form a 2-factor F . If [A, B] is a finite cut, then the Jumping Arc Lemma (see [8]) implies that every arc with one endpoint each in A and B must contain an edge of [A, B]. Since any a ∈ A and b ∈ B are connected by an arc in C , it follows that [A, B] ∩ E(F ) is nonempty. Each time C meets [A, B] it crosses between A and B (and vice versa), so since C is closed and [A, B] is finite, this implies that C meets [A, B] an even number of times. To see that (ii) H⇒ (iii) observe that (ii) implies that G is connected, and since G is locally finite, G − S has finitely many components C1 , . . . , Cm . Contracting each Ci to a vertex ci we obtain a finite (multi)graph G′ . We let F ′ denote the subgraph of G′ corresponding to F . Each cut in G′ corresponds to a cut in G with the same number of edges, and by (ii) every such cut has a positive even number of edges from F . Thus F ′ is a connected spanning subgraph of G′ in which every vertex has even degree. Thus F ′ has an Euler tour v0 , v1 , . . . , vn = v0 . Every vertex in S appears only once in this tour. Every vertex ci is immediately preceded and succeeded by a vertex in S, and in GS these vertices must be adjacent. Thus suppressing the vertices of the form ci from the tour we obtain a Hamiltonian cycle in GS .

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Clearly (iii) H⇒ (iv). To prove that (iv) H⇒ (i) it suffices to prove that (iv) implies Theorem 1(ii). For this, let S be any finite vertex set, and let S ′ be as in Theorem 2(iv). We replace every edge uv ∈ E(GS ′ ) \ E(G) in a Hamiltonian cycle of GS ′ by a u, v -path in G that is internally disjoint from S ′ . This yields an S-curve, so that (iv) H⇒ (i). □ In the next section we shall use Theorem 2(iii) to generalize some Hamiltonian results from finite graphs to infinite graphs. Theorem 2 also implies the following characterization for Hamiltonian circles. Theorem 3. Let G be a locally finite graph. Then every Hamiltonian circle in |G| corresponds to a 2-factor F of G such that (i) every finite cut intersects F a positive even number of times, and (ii) for all distinct e1 , e2 ∈ F there is a finite cut [A, B] in G such that F ∩ [A, B] = {e1 , e2 }. Conversely, if a 2-factor F satisfies (i) and (ii), then the closure of F is a Hamiltonian circle. Proof. For the forward direction let C be a Hamiltonian circle. From Theorem 2 we see that the vertices and edges of C form a 2-factor F satisfying (i). To see that (ii) is satisfied, consider the two components C1 and C2 of C −{e1 , e2 }, and let A = V (C1 ) and B = V (C2 ). If [A, B] is infinite, then it contains an infinite matching with edges ai bi where i is a natural number, ai ∈ A, and bi ∈ B. By taking subsequences we may assume that a1 , a2 , . . . and b1 , b2 , . . . are convergent (since |G| is compact). The sequences must converge to ends α ∈ C1 and β ∈ C2 respectively, since these arcs are compact. Since ai bi is an edge, it follows that α = β , so that this end occurs twice on C , a contradiction. For the backwards direction observe that the closure of F is a Hamiltonian curve C because of condition (i). It remains to see that C has no repeated end α . Suppose (reductio ad absurdum) that x1 , x2 are different preimages of α in a parametrization σ : S 1 → C . Let e1 , e2 be edges on C whose preimages are in separate components of S 1 − {x1 , x2 }, and let [A, B] be the finite cut that meets C only in {e1 , e2 }. As all ends of G[A] are distinct from all ends of G[B], but α is contained in both we have obtained a contradiction. □ 4. From Hamiltonian cycles to Hamiltonian curves Before we prove Theorem 1 we mention some applications in the same spirit as Diestel [6] except that we focus on curves rather than circles. 4.1. Degree conditions Dirac’s theorem that every finite graph G with minimum degree at least |V (G)|/2 is Hamiltonian, has been the inspiration for many results on Hamiltonian cycles in finite graphs. While most of these conditions cannot be formulated for infinite locally finite graphs, Diestel [8] mentions a degree condition by Asratian and Khachatrian [20] that does: d(u) + d(w ) ≥ |N(u) ∪ N(v ) ∪ N(w )|, for every induced path uvw. A close inspection of the proof shows that if G is a (finite or infinite) graph satisfying that condition, then it also satisfies the following: If C = v1 v2 . . . vk v1 is a cycle in G and v is a vertex in G − V (C ) joined to a vertex in C , then G has a cycle of the form (i) v1 v2 . . . vi vvi+1 . . . vk v1 or (ii) v1 v2 . . . vi v uvi+1 . . . vk v1 (u ̸ ∈ V (C )) or (iii) v1 v2 . . . vi−1 vj−1 vj−2 . . . vi vvj vj+1 . . . vk v1 where j > i. Clearly G satisfies the condition of Theorem 1(i) and has therefore a Hamiltonian curve. Diestel [6] conjectures that, if G satisfies the condition by Asratian and Khachatrian [20], then |G| has a Hamiltonian circle. It is not hard to prove this under the stronger condition that every cycle can be extended as in (i) or (ii).

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4.2. Prisms and forbidden induced subgraphs It is easy to see that every finite subset of an infinite 2-connected graph is contained in a finite 2-connected subgraph, for example by using an ear decomposition. This observation combined with the implication (i) H⇒ (iii) in Theorem 1 makes it possible to extend various Hamiltonian results on finite graphs. One such example is the result of Duffus et al. [9] that every 2-connected claw-free and net-free graph is Hamiltonian. Here a claw is an induced K1,3 subgraph and a net is an induced K3 with a pendant edge at each vertex. For definitions and other similar results, see [10]. We immediately obtain that a locally finite graph satisfying these conditions must contain a Hamiltonian curve. The Cartesian product of a graph G and a complete graph with two vertices is called the prism of G. Theorem 4 (Paulraja [26]). If G is a finite cubic 3-connected graph, then the prism of G has a Hamiltonian cycle. Although this is a result about 3-connected graphs, the proof is based on certain spanning 2connected subgraphs, as the proof in [26] shows that, for any bipartite 2-connected graph H of maximum degree 3, the prism of H has a Hamiltonian cycle. As observed in [31], a bipartite spanning subgraph H of a (2k − 1)-edge-connected graph G with as many edges as possible is k-edge-connected. Therefore, the graph G in Theorem 4 contains a spanning bipartite 2-connected subgraph H of maximum degree 3. It is easy to extend this to the case where G is infinite, cubic and 3-connected. Every finite subset of the resulting graph H is contained in a 2-connected finite subgraph of H. Using Theorem 1, we conclude Theorem 5. Let G be an infinite cubic 3-connected graph. Then the prism of G has a Hamiltonian curve. For higher connectivity we have to argue slightly differently because an infinite k-connected graph need not contain a 3-edge-connected finite subgraph. (To see this, take the k-regular tree and add, for each distance class from a fixed vertex, a Hamiltonian path in that distance class.) We illustrate this by Theorems 6 and 9. The point here is that the arguments are unified. As mentioned earlier, the analogous Hamiltonian circle problems are much harder. A graph is cyclically 4-edge-connected if the deletion of any set of fewer than 4 edges leaves a graph with at most one component containing a cycle. For a cubic graph this is equivalent to the graph being 3-connected and, moreover, the only cuts with precisely 3 edges are those incident with a vertex. Recall the following three conjectures. Conjecture 1 (Matthews, Sumner [25]). Every finite 4-connected claw-free graph is Hamiltonian. Conjecture 2 (Thomassen [29]). Every finite 4-connected line graph is Hamiltonian. Conjecture 3 (Ash and Jackson [1]). Every finite cubic cyclically 4-edge-connected graph has a cycle incident with all edges of the graph. As every line graph is claw-free the Matthews–Sumner Conjecture is stronger than that of Thomassen. As every graph in Conjecture 3 has a 4-connected line graph, the conjecture of Thomassen is stronger than that of Ash and Jackson. However, Fleischner and Jackson [12] proved that the two latter conjectures are equivalent, and Ryjáček [27] proved that the two former conjectures are equivalent. Theorem 6. If Conjectures 1–3 are true, then the statements (i), (ii), (iii), (iv) below hold. (i) Every 4-connected infinite locally finite claw-free graph has a Hamiltonian curve. (ii) Every 4-connected line graph of an infinite locally finite graph has a Hamiltonian curve. (iii) If G is an infinite cubic cyclically 4-edge-connected graph G, then the line graph of G has a Hamiltonian curve. (iv) For every finite edge set E in an infinite cubic cyclically 4-edge-connected graph, the graph has a cycle incident with all edges of E. Conversely, if one of the statements (i), (ii), (iii), (iv) is true, then each of (i), (ii), (iii), (iv) is true, and Conjectures 1–3 are true.

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Proof. Assume Conjecture 1 is true. Let G be a 4-connected infinite claw-free graph G. Let S ⊂ V (G) be a finite set. If necessary, we extend S to a bigger set (also denoted S) such that G[S ] is connected and |S | ≥ 5. Now GS is 4-connected and claw-free (as a claw in GS would yield a claw in G), and hence GS is Hamiltonian. By Theorem 2, G has a Hamiltonian curve. This proves (i). Clearly, (i) implies (ii), and (ii) implies (iii). Also, (iv) follows from (iii) combined with Theorem 1 because a cycle C in the line graph L(G) corresponds to a closed walk in G with no edge-repetition but incident with all edges in G that are vertices in C , as observed by Harary and Nash–Williams [19]. As G is cubic, a closed walk with no repeated edges is a cycle. For the converse, it suffices to prove that (iv) implies Conjecture 3 (because (i) H⇒ (ii) H⇒ (iii) H⇒ (iv ) ). Let G be a finite cubic cyclically 4-edge-connected graph. We shall prove that G has a cycle incident with all edges of the graph. Let x, y be two adjacent vertices of G. Let H be an infinite cubic cyclically 4-edge-connected graph containing disjoint paths u1 u2 . . . u5 and v1 v2 . . . v5 and edges u1 v1 , u2 v2 , . . . , u5 v5 . (It is easy to see that such an H exists.) For each even i delete ui , vi and replace them by a copy of G − x − y such that the two neighbors of x, respectively y, in G are joined to ui−1 , vi−1 , respectively ui+1 , vi+1 . It is easy to see that the resulting graph G′ is still cyclically 4-edge-connected. Then a cycle in G′ satisfying the conclusion of (iv) with E being sufficiently large contains one or two paths in a copy of G − x − y which can be used to obtain a cycle in G that satisfies the conclusion of Conjecture 3. □ It is tempting to conjecture that the assumptions in (i), (ii), (iii) imply a Hamiltonian circle. However, that would imply not only the long-standing Conjectures 1–3 but also the recent conjecture of Georgakopoulos [14] that the line graph of a 4-edge-connected graph has a Hamiltonian circle. 4.3. Planar graphs Tutte [32] proved Theorem 7. Every finite 4-connected planar graph is Hamiltonian. In fact, Tutte proved a stronger statement which enabled Harant and Senitsch [18] to derive the following extension: Theorem 8. Let G be a planar graph and X be a subset of V (G) with at least two nonadjacent vertices. If no two vertices of X are separated by fewer than 4 vertices, then X is contained in a cycle of G. Theorems 8 and 1 now imply Theorem 9. Every 4-connected locally finite planar graph has a Hamiltonian curve. Proof. Let G be 4-connected locally finite planar graph and let S be a finite subset of V (G). It suffices to prove that S is contained in a cycle. For every two vertices u, v in S, we choose four internally disjoint paths between u and v . They exist by Menger’s theorem. Let G1 be the finite subgraph of G obtained as the union of all the chosen paths. Then G1 satisfies the assumption of Theorem 8, and hence S is contained in a cycle in G1 . □ We now turn to a challenging case of the Hamiltonian curve problem. The remaining part of this section is about cubic planar graphs. Tait’s conjecture says that every finite planar cubic 3-connected graph has a Hamiltonian cycle. Barnette’s conjecture (which is a special case of Tait’s conjecture) says that every finite planar cubic 3-connected bipartite graph has a Hamiltonian cycle. Tait’s conjecture was disproved by Tutte, see [2,15], while Barnette’s conjecture is still open. Recall that a nowhere-zero 3-flow in a graph (finite or infinite) is an orientation of its edges in which each edge is assigned a flow value 1 or 2 such that every finite cut [A, B] is balanced, that is the total flow from A to B is the same as the total flow from B to A. Note that, in a cubic graph, the edges with flow 2 form a perfect matching, and the edges with flow 1 form a 2-factor where all vertices have outdegree 0 or 2. This partition into outdegree 0 and outdegree 2 vertices shows that a cubic graph with a nowhere-zero 3-flow must be bipartite. Conversely, every cubic bipartite graph has a perfect

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Fig. 1. Replacing an active vertex v .

matching. (Indeed, it is easy to see that the statement that every finite (infinite) cubic bipartite graph has a perfect matching is equivalent to the statement that every finite (infinite) bipartite graph of maximum degree 3 has a matching that covers every vertex of degree 3. While the finite version of the former statement immediately follows from Hall’s theorem [16], the infinite version of the latter follows from its finite version by a standard compactness argument.) If the graph is finite then each perfect matching can play the role of the set of edges with flow value 2 in a nowhere-zero 3flow, but this need not be true for infinite graphs. (For example, consider the two-way infinite ladder obtained by taking two copies of a two-way infinite path P and joining each vertex to its copy. A perfect matching M in one copy of P together with the matching corresponding to E(P) − M in the other copy cannot be extended to a nowhere-zero 3-flow in this way.) This leads to two possible extensions of Barnette’s conjecture for infinite graphs: Strong version of Barnette’s conjecture: Every planar cubic 3-connected bipartite graph has a Hamiltonian curve. Weak version of Barnette’s conjecture: Every planar cubic 3-connected graph with a nowhere-zero 3-flow has a Hamiltonian curve. We start by giving a general procedure for constructing a cubic bipartite infinite planar graph G∞ with |A| ends from a cubic finite planar graph G0 with a distinguished independent set of vertices A in G0 , which we call active. We will also have the property that G∞ has a Hamiltonian curve if and only if G0 has a Hamiltonian cycle. When G0 is bipartite A could be any nonempty independent set, but when G0 is nonbipartite, we let A be an independent set such that G − A is bipartite. (Such an A exists because G0 is 3-colorable, by Brooks’ theorem, see [2].) We choose an orientation of the edges of G0 such that all vertices not in A have indegree or outdegree 0. Now obtain a new oriented graph G1 by replacing each active vertex v as suggested by Fig. 1. That is, we replace v by a 6-cycle of inactive vertices v1 , v2 , . . . , v6 so that each edge is oriented towards the vertex of even index, and a new active vertex v ′ inside the 6-cycle. For each of the neighbors u1 , u2 , u3 of v create exactly one of the two edges ui vj with j ∈ {2i − 1, 2i}, where we let j be even if the head of ui v is at v (and odd otherwise) and orient it so the head is at vj if j is even, and the tail is at vj otherwise. We make v ′ incident with each of the three remaining vertices of the 6-cycle, and we orient its incident edges so that every inactive vertex has indegree 0 or 3. Figure 1 also shows the edge-orientations and the vertex-2-coloring of the inactive vertices. Then G1 is 3-connected, and G1 is Hamiltonian if and only if G0 is. Repeating this process with each active vertex we obtain a graph G∞ in the limit that is cubic 3-connected and planar and has as many ends as G0 has active vertices. In particular, the limit G∞ can be chosen to be one-ended if G0 is bipartite. Moreover G∞ is bipartite since in the orientation each vertex has indegree 0 or 3. Observe that the following three statements are equivalent: (i) G0 has a Hamiltonian cycle. (ii) G∞ has a Hamiltonian curve. (iii) G∞ has a Hamiltonian circle. Indeed (ii) and (iii) are equivalent since every Hamiltonian curve of G∞ is a Hamiltonian circle: For each end α of G∞ we can find a nested sequence of basic neighborhoods ˆ C (Sn , α ) such that the edge cuts [C (Sn , α ), Sn ] are of size 3, and so the Jumping Arc Lemma implies that there are at most 3 pairwise edge-disjoint arcs at α . Clearly, (ii) implies (i). Moreover (i) implies (ii) by Theorem 1 since every finite vertex-set S in G∞ is contained in V (Gi ) for some i. A Hamiltonian cycle in G0 can easily be modified to a Hamiltonian cycle in Gi which can be modified to a cycle in G∞ containing S.

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Fig. 2. Nonhamiltonian planar graph with 7 active vertices.

It follows immediately that every counterexample G0 to Tait’s conjecture can be used to find a counterexample G∞ to the strong infinite version of Barnette’s conjecture as follows. The counterexamples to Tait’s conjecture found independently by Barnette, Bosak and Lederberg shown in Fig. 2 see also [15] page 361, has a set A of 7 independent vertices (marked as squares) whose deletion leaves a bipartition indicated by the black and white vertices in Fig. 2. So, the strong version of Barnette’s conjecture fails for infinite graphs with 7 ends. It may hold for graphs with fewer ends, in particular one-ended graphs, see Theorem 11 below. We now show that the weak version is the right extension of Barnette’s conjecture to infinite graphs. Theorem 10. Barnette’s conjecture (for finite graphs) is equivalent to the following conjecture on infinite graphs: Every infinite planar cubic 3-connected graph with a nowhere-zero 3-flow has a Hamiltonian curve. Proof. We will first use a counterexample G0 to Barnette’s conjecture to create a one-ended counterexample to the weak infinite version. As G0 is finite and bipartite it has a nowhere-zero 3flow, and we let exactly one vertex v be active. We will again replace v by a 6-cycle and an active vertex v ′ to obtain a new counterexample G1 . If uv was the edge of flow value 2, then we create the edge uv1 and orient it in the same way. The edges v x, v y with flow value 1 will be similarly replaced by v3 x and v5 y. v ′ will be adjacent to v2 , v4 , v6 . If we assign a flow value of 2 to uv1 , v2 v3 , v4 v, v5 v6 then there is a unique way to orient the remaining nine edges so that we have a nowhere-zero 3flow for G1 . Repeating this process with each active vertex we obtain a graph G in the limit that is cubic 3-connected and planar and has exactly one end. Moreover G has a nowhere-zero 3-flow since every finite cut in G corresponds to a finite cut in some Gi , and it cannot have a Hamiltonian curve since contracting all the new vertices that were created to the original active vertex would yield a Hamiltonian cycle in G0 . Assume next that Barnette’s conjecture is true, and let G be an infinite planar cubic 3-connected graph with a nowhere-zero 3-flow. Let S ′ be a finite vertex set, and fix a plane drawing of G. By Theorem 2 it suffices to prove that, for some finite vertex set S containing S ′ , GS is Hamiltonian. First we enlarge S ′ to a finite set S ′′ so that G[S ′′ ], the subgraph of G induced by S ′′ , is a subdivision of a 3-connected graph. Next we enlarge S ′′ to S so that each vertex not in S has at most one neighbor in S. We claim that G[S ] can be extended to a finite planar, cubic, 3-connected, bipartite graph G′ . Once this claim has been proved, it will complete the proof because G′ is Hamiltonian, by Barnette’s conjecture. To prove this claim we first reverse the direction of each edge of flow 2 in G and ignore the flow values. The resulting oriented graph has the property that each finite cut is balanced modulo 3. In particular, each vertex has outdegree 0 or 3, and hence each cycle is alternating. We first replace each component H of G − S by a cycle C such that we get a finite cubic plane graph G1 . It is easy to see that G1 is 3-connected. We subdivide each edge of the cycle C at least once and give

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each edge an orientation. It is an easy exercise to prove that this can be done such that each cycle in the resulting graph is alternating. For each vertex a of degree 2 we add a new vertex a′ of degree 1 and give the edge aa′ an orientation so that a has outdegree 0 or 3. The new vertices joined to vertices in the subdivision of C are added inside the cycle C . If two new vertices a′ , b′ are consecutive in the cyclic ordering inside C (inherited by the cyclic ordering on C of the neighbors), and the edges a′ a, b′ b have the same orientation, we identify a′ , b′ and add a new vertex c of degree 1 joined to the identified vertex and we give it an orientation such that the neighbor of c has outdegree 0 or 3. We proceed in this way until we reach a situation where the directions to the vertices of degree 1 alternate. We then add an edge between each degree 1 vertex and the degree 1 vertices immediately preceding and succeeding it to give every vertex degree 3. The new edges form an even cycle C ′ , and these edges can now be oriented so that each vertex has outdegree 0 or 3. (If C ′ has length 2, we delete C ′ and replace its two incident edges by a single edge.) The only problem is that we may reach a situation with precisely two vertices x, y of degree 1 such that, say, both have outdegree 1. Then we identify x, y into a vertex z of outdegree 2. We now complete the proof by showing that this vertex z cannot appear. Recall that z corresponds to some component H of G − S. We have now replaced H by a finite graph H ′ . The cut between G[S ] and H ′ is also the cut between G[S ] and H and is therefore balanced modulo 3. Now contract G[S ] into a single vertex s and consider the graph H ′′ consisting of s and H ′ , and the cut between s and H ′ . In H ′′ all vertices, except z, are balanced modulo 3. But then also z is balanced modulo 3, a contradiction which shows that the vertex z will never appear in this construction. □ Theorem 11. Barnette’s conjecture (for finite graphs) is equivalent to the following conjecture on infinite graphs: Every infinite one-ended planar cubic 3-connected bipartite graph has a Hamiltonian curve. Proof. If there is a counterexample to Barnette’s conjecture for finite graphs, then the construction preceding Theorem 10 gives a one-ended counterexample to the strong version of Barnette’s conjecture. On the other hand, if G is an infinite one-ended planar cubic 3-connected bipartite graph with bipartition B1 , B2 , then G has a perfect matching M (as explained after Theorem 9). We direct all edges in M from B2 to B1 and give them flow 2. All other edges are directed from B1 to B2 and are given flow 1. As every finite cut separates a finite vertex set from an infinite vertex set, it follows that each cut is balanced, that is, G has a nowhere-zero 3-flow. Now Theorem 11 follows from Theorem 10. □ The 7-ended counterexample to the strong version of Barnette’s conjecture suggests that the end structure is crucial for Barnette’s conjecture. To make this more precise, let the bipartite index bi(G) be the smallest number of edges you must delete in order to make the graph bipartite. The bipartite index was used in [30] for configurations in graphs of large connectivity. Here we use it to refine Barnette’s conjecture. Let cT be the smallest bipartite index of a counterexample to Tait’s conjecture. It is easy to see that, for any cubic graph G, the bipartite index of G is the smallest number of vertices you must delete in order to make the graph bipartite. (Clearly, G has a set U of at most bi(G) vertices such that G − U is bipartite. Conversely, if U is a set of vertices such that G − U is bipartite, that is, G − U has a proper vertex-2-coloring, then this coloring can be extended to a non-proper vertex-2-coloring of G such that each vertex in U has at most one neighbor in the same color class. Hence bi(G) ≤ |U |.) (This observation shows that, for a cubic non-bipartite graph G that does not contain K4 , the bipartite index equals the chromatic surplus which in [4] is defined as the number of vertices in a smallest color class when G is properly vertex-colored with χ (G) colors. But, we shall not need this.) Thus cT ≤ 7 as shown by the counterexample of Barnette, Bosak and Lederberg. Barnette’s conjecture is equivalent to the statement that cT ≥ 2: a cubic graph cannot have bipartite index 1, as in that case the number of edges in the bipartition would need to be both 1 and 2 modulo 3. We now show that the smallest number of ends in a counterexample to the strong version of Barnette’s conjecture equals cT , provided Barnette’s conjecture is true. Theorem 12. Assume that Barnette’s conjecture for finite graphs is true. Then the following statements hold (i) There exists a counterexample with precisely cT ends to the strong version of Barnette’s conjecture. (ii) Every planar cubic 3-connected bipartite graph with fewer than cT ends has a Hamiltonian curve.

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Proof. We have earlier constructed a counterexample with precisely 7 ends to the strong version of Barnette’s conjecture. A cT -ended example is constructed exactly in the same way. Suppose now (reductio ad absurdum) that there is a counterexample G to the strong version of Barnette’s conjecture with fewer than cT ends. We shall construct a counterexample to Tait’s conjecture with bipartite index < cT and thereby reach a contradiction. We do this as in the proof of Theorem 10: We consider a finite vertex set S such that G[S ] is 2-connected and G − S has only infinite components. The number of these is less than cT . Consider one of these components, say H. As in the proof of Theorem 10 we replace H by a finite bipartite graph H ′ such that the union of G[S ] and all these H ′ is finite, bipartite, non-Hamiltonian, 3-connected and cubic, except that some (maybe each) H ′ may contain a vertex z of degree 2, in which case we replace the two incident edges by a single edge joining two vertices in the same part of the bipartition. In this way we obtain a counterexample to Tait’s conjecture with bipartite index < cT because the graph under consideration has less than cT ends. □ 5. Proof of Theorem 1 (ii) H⇒ (iii) If C is an X -curve with a fixed parametrization, then the vertices of X occur on C in a cyclic order called the cyclic order of X induced by C . In general we say that a cyclic order O (on a set X ) and an S-curve C (with S ⊆ X ) are compatible on S if the cyclic order on S induced by C is the restriction of O to S. If x, y, z are three elements of a cyclically ordered set X , we use x → y → z to denote that x, y, z occur in this order. We call a cyclic order O on a set X ⊆ V (G) embeddable if for all finite sets S ⊆ X and T ⊆ V (G) \ X , there is an S-curve C in |G| − T that is compatible with O on S. The following lemma will be used in the proof of the remaining two implications of Theorem 1 and will be proved in Section 7. Lemma 1. O is an embeddable cyclic ordering of a set of vertices X in a locally finite graph G if and only if there is a strict X -curve C such that the cyclic order of X induced by C is O. To show that Lemma 1 implies (ii) H⇒ (iii) it suffices to find an embeddable ordering O of X = V (G). By (ii), G is connected, and since G is locally finite, it follows that X = V (G) = {v1 , v2 , . . .} is countable. Let Xi = {v1 , . . . , vi } and let Ci be an Xi -curve in |G|. We will construct a cyclic order O on X , by first constructing cyclic orders Oi on Xi such that for i < j, Oi is the restriction of Oj to Xi . Moreover for all j i ≥ j we find an Xi -curve Ci whose order induced on Xj is Oj . 1 2 Let Ci = Ci = Ci , and let O1 and O2 be the trivial cyclic order on X1 and X2 , respectively. For j ≥ 3, j we define Ci and Oj recursively. Since Xj is finite there are only finitely many cyclic orders on Xj , but j−1 there are an infinite number of Xj -curves that induce Oj−1 on Xj−1 , namely Ci for all i ≥ j. Thus we j

can find a cyclic order Oj on Xj that is induced by infinitely many of these Xj -curves. Let (Ci )∞ i=j be the j−1 ∞ )i=j

subsequence of (Ci

that induces Oj on Xj .

Now we define the limit of (Oi )∞ i=1 as a cyclic order O on V (G) such that for any vi , vj , vk ∈ V (G), vi → vj → vk in O if and only if vi → vj → vk in Omax{i,j,k} . Clearly Oi is the restriction of O on Xi for all i ≥ 1. To see that O is embeddable, consider finite sets S ⊆ X and T ⊆ V (G) − X . Thus T = ∅, and j we can choose j large enough so that S ⊆ Xj . Thus Cj is the desired S-curve. 6. Proof of Theorem 1 (iii) H⇒ (i) We prove the following slightly stronger statement (Theorem 13) that immediately implies our result, as well as the much easier fact that every finite vertex set that lies on a circle in a locally finite graph also lies on a cycle. Theorem 13. Let A be a finite vertex set of a locally finite graph G. If |G| has a strict B-curve for some B with A ⊆ B ⊆ V (G), then G has a cycle containing A.

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Let C be the class of all such strict curves. We define a partial order on C by saying that C1 < C2 if and only if V (C1 ) ⊂ V (C2 ). Our goal will be to use a minimal element C1 of this partial order to find the desired cycle. If Λ is a linearly ordered set, then we say that (Ci )i∈Λ is a decreasing generalized sequence in C if Ci ∈ C for all i ∈ Λ and V (Ci ) ⊆ (Cj ) for all i, j ∈ Λ with i > j. Claim 1. If (Ci )i∈Λ is a decreasing generalized sequence in C for some linearly ordered set Λ, then there is ⋂ a strict X -curve with X = {V (Ci ) : i ∈ Λ}. Proof. As V (G) is countable, (Ci )i∈Λ has a decreasing subsequence whose intersection is X (because, for every vertex not in X , there is an element of (Ci )i∈Λ not containing that ⋂∞vertex). So we may assume that Λ is the set of natural numbers, that is (Ci )i∈Λ = (Ci )∞ . Let X = i=1 i=1 V (Ci ) = {v1 , v2 , . . .} and Xj = {v1 , v2 , . . . , vj }. Either X is countably infinite, or X = Xj for some j. We first construct a cyclic order O on X by constructing an order Oj for all j such that Xj is defined. We will also construct infinite sequences C j of curves in C such that the order on Xj induced by its curves is Oj . 2 ∞ 2 Let Ci1 = Ci2 = Ci , C 1 = (Ci1 )∞ i=1 , C = (Ci )i=1 , and let O1 and O2 be the trivial cyclic order on X1 = {v1 } and X2 = {v1 , v2 }, respectively. Now for j ≥ 3, since Xj is finite there are only finitely many cyclic orders on Xj . Thus we can find a cyclic order Oj on Xj that is induced by an infinite number of j j−1 with all curves that induce Oj . curves in C j−1 . Let C j = (Ci )∞ i=1 be the subsequence of C So for every j ≥ 1, we have a cyclic order Oj on Xj and an infinite sequence C j of curves that induces Oj . Moreover, if k ≥ j, then Oj is the restriction of Ok on Xj . If X = Xj for some j, then we let O = Oj . Otherwise X is infinite and we define the limit of (Oi )∞ i=1 as a cyclic order O on X such that for any vi , vj , vk ∈ X , vi → vj → vk in O if and only if vi → vj → vk in Omax{i,j,k} . Clearly Oi is the restriction of O on Xi for all i ≥ 1. To finish the proof of Claim 1 using Lemma 1 it remains to see that O is embeddable. Consider finite sets S ⊆ X and T ⊆ V (G) \ X . Since S is finite we can choose j large enough that S ⊆ Xj . Since every j curve Ci ∈ C j induces the cyclic order Oj on Xj , which is the restriction of O on Xj , it follows that each j j Ci is an S-curve that is compatible with O on S. Since T is finite we can choose i large enough that Ci contains no vertex in T . □ Recall that every C ∈ C is a strict B-curve for some B with A ⊆ B ⊆ V (G). Thus C is the image of some parametrization σ : S 1 → |G|, and we let V (C ) = B = V (G) ∩ σ (S 1 ). Since every v ∈ V (C ) has a unique preimage in S 1 , we may speak of its location on C without ambiguity. Since σ is continuous we can also define Ω (C ) = Ω (G) ∩ σ (S 1 ). For each end α ∈ Ω (C ) there may be more than one point in S 1 mapping to α , and we call these images the copies of α in C . We generally use α1 , α2 , . . . to denote different copies of α (although they, strictly speaking, are equal to α ), and we let kC (α ) denote the number of copies of α in C . We say that a point in C is either a vertex or a copy of an end. Claim 1 shows that in the partial order on C every decreasing sequence has a lower bound. Thus by Zorn’s Lemma [33] C has at least one minimal element, which we will call C1 . If V (C1 ) is finite, then C1 is the desired cycle containing A. Thus we may assume that V (C1 ) is infinite, so that the set Ω (C1 ) is nonempty. If x, y are points in a curve C ∈ C with a fixed parametrization σ : S 1 → C and x0 , y0 ∈ S1 are preimages of x, y, then the image of the arc from x0 to y0 in S 1 (where we fix an orientation of − → ← − S 1 throughout), denoted by C [x, y], is called a closed interval of C ; we use C [y, x] to denote the

− →

← −

same interval with the opposite orientation. The corresponding open intervals C (x, y) and C (x, y) − → ← − are obtained from C [x, y] and C [x, y] respectively by removing the points x, y. If C is an A-curve, then a maximal closed interval of C that is internally disjoint from A is an A-interval of C . So C has |A| A-intervals. Let α be an end of G and α1 , α2 be two distinct copies of α in C . We call {α1 , α2 } an end pair of C .

− → − → ← − by C [α2 , α1 ] are also curves. We call the curve C ′ a shortening of C at (α1 , α2 ), and C ′′ the flip of C at (α1 , α2 ). Let x1 , x2 , y1 , y2 be four points in C . The two pairs {x1 , x2 } and {y1 , y2 } are crossed if exactly one of x1 → y1 → x2 and x1 → y2 → x2 holds in C ; otherwise they are parallel. The pair {x1 , x2 } is

Suppose {α1 , α2 } is an end pair of C . Then C ′ = C [α2 , α1 ] and C ′′ consisting of C (α2 , α1 ) followed

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self-similar if x1 , x2 are contained in a common A-interval. Two pairs {x1 , x2 } and {y1 , y2 } are similar if x1 , y1 are contained in a common A-interval and x2 , y2 are contained in a common A-interval. We call a self-similar end pair a 1-end shortcut; a pair of crossed similar end pairs is a 2-end shortcut; and a triple of end pairs in which two of them are parallel and similar, and both of them are crossed with the third one, is a 3-end shortcut. Claim 2. C1 has no 1-, 2-, or 3-end shortcut.

− →

Proof. Suppose first that C1 has a 1-end shortcut. Let {α1 , α2 } be an end pair such that C1 (α1 , α2 ) is contained in an A-interval. Then the shortening of C1 at (α1 , α2 ) is an A-curve with vertex set smaller than C1 (in the partial ordering of C ), a contradiction. Suppose now that C1 has a 2-end shortcut. Let {α1 , α2 } and {β1 , β2 } be two crossed similar end pairs. Let C ′ be the flip of C1 at (β1 , β2 ). Then C ′ is a curve with 1-end shortcut {α1 , α2 }. Shortening C ′ yields a curve in C with vertex set smaller than C1 , a contradiction. Finally we suppose that C1 has a 3-end shortcut. Let {α1 , α2 } and {β1 , β2 } be two parallel similar end pairs, both of which are crossed with {γ1 , γ2 }. Let C ′ be the flip of C1 at (γ1 , γ2 ). Then C ′ is a curve in which {α1 , α2 } and {β1 , β2 } form a 2-end shortcut. Thus there is a curve in C with vertex set smaller than C1 , a contradiction. □ Since C1 has no self-similar end pair by Claim 2, we can see that every end of G has at most |A| copies in C1 . An end α is crossed in C ∈ C if there is an end pair {α1 , α2 } (representing α ) that is crossed with at least one end pair in C (representing an end distinct from α ). An end α is polygonal in C ∈ C if kC (α ) ≥ 3. Claim 3. C1 has a finite number of polygonal ends and a finite number of crossed ends. Proof. Let I1 , I2 , I3 be three distinct A-intervals of C1 . We claim that there is at most one end of G that has copies in all three intervals. Suppose not. Let α, β be such two ends, and αi , βi ∈ Ii for i = 1, 2, 3. Then for two of the three intervals, say I1 , I2 , the copies of α and β appear in the same order. Thus {α1 , α(2 } and ) {β1 , β2 } form a 2-end shortcut, a contradiction to Claim 2. This implies that there are at | A| most 3 polygonal ends in C1 .

Let I1 , I2 be two A-intervals of C1 . We claim that there are at most two end pairs in I1 ∪ I2 that are crossed with some other end pairs. Suppose otherwise we let {α1 , α2 }, {β1 , β2 }, {γ1 , γ2 } be three such end pairs, where αi , βi , γi ∈ Ii for i = 1, 2. Now the three end pairs are parallel; for otherwise there will be a 2-end shortcut. We suppose without loss of generality that α1 → β1 → γ1 , and thus γ2 → β2 → α2 in C1 . Let {δ1 , δ2 } be an end pair crossed with {β1 , β2 }. Since {δ1 , δ2 } is not self-similar or similar to {β1 , β2 }, it will cross either {α1 , α2 } or {γ1 , γ2 }. In either case, we have a 3-end shortcut, a contradiction. This implies that there are at most 2

( | A| ) 2

crossed ends in C1 .

□

A path connecting two vertices x, y of a curve that is internally vertex-disjoint with the curve is an x, y-bridge of the curve. Let P be an x, y-bridge of C ∈ C . If {x, y} is self-similar and the interval − → C [x, y] that is vertex-disjoint with A contains a copy of an end α , then P is called a 1-copy shortcut of C (associated with α ). If {x, y} is similar and crossed with an end pair {α1 , α2 }, then P is a 2-copy shortcut of C (associated with α ). If {x, y} is crossed with an end pair {α1 , α2 }, and similar and parallel with an end pair {α2 , α3 } (where α1 = α2 = α3 = α ), then P is a 3-copy shortcut of C (associated with α ). (See Fig. 3 where A = {a1 , . . . , a6 } and α = α1 = α2 = α3 .) − → If C ∈ C and P is a 1-copy shortcut of C from x to y, then C ′ = C [y, x] ∪ P is also a curve in C . We call C ′ the jump of C at (x, y). Claim 4. Let C ∈ C and α ∈ Ω (G) such that kC (α ) is finite. We can transform C to a curve C ′ ∈ C with kC ′ (α ) = 0 by a finite number of flips and jumps. Proof. Suppose that kC (α ) > 0. We show that C has a 1-, 2-, or 3-copy shortcut associated with α .

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Fig. 3. A 1-copy, 2-copy and 3-copy shortcut P associated with α .

− → − → C [a, b] containing α , then since C [b, a] is a closed set − → not containing α , there must be a basic open set ˆ C (S , α ) that is disjoint from C [b, a]. Set X1 = − → − → C (S , α ) ∩ V ( C [a, α )), and X2 = C (S , α ) ∩ V ( C (α, b]). Since C (S , α ) is connected there is a path P connecting two vertices x ∈ X1 and y ∈ X2 that is internally disjoint from X1 ∪ X2 . Thus P is a 1-copy shortcut of C associated with α . Now we assume that there are at least two A-intervals containing α . Let I = {Ii : i = 1, . . . , k} be − → the set of A-intervals that contain α , where Ii = C (ai , bi ) and αi ∈ Ii . We assume that the intervals ⋃ appear in this order along C . Let ˆ C (S , α ) be a neighborhood of α that is disjoint from C \ I . Let − → Xi = C (S , α ) ∩ V ( C (αi , αi+1 )) for 1 ≤ i ≤ k, where the subscripts are taken modulo k. Since C (S , α ) is connected, there is an x, y-path P that is internally disjoint from all sets Xi such that x ∈ Xi , y ∈ Xj for some i ̸ = j. If x, y are in the same A-interval, then P is a 1-copy shortcut. Suppose now that x, y are in distinct A-intervals. Since every A-interval in I has a copy of α , {x, y} is similar with an end pair of α . Since every two sets Xi , Xj are separated by some copies of α , {x, y} is crossed with an end pair of α . This implies that P is either a 2-copy shortcut or a 3-copy shortcut. Let P be a 1-, 2- or 3-copy short cut from x to y associated with α . If P is a 1-copy shortcut, then the jump of C at (x, y) is a curve C ′ with kC ′ (α ) < kC (α ). If P is a 2-copy shortcut, where {α1 , α2 } is an end pair similar and crossed with {x, y}, then P is a 1-copy shortcut of the flip of C at (α1 , α2 ). (The If there is only one A-interval I =

1-copy shortcut in Fig. 3 is obtained from the 2-copy shortcut in this way.) Thus we have a curve C ′ with kC ′ (α ) < kC (α ). If P is a 3-copy shortcut, where {α2 , α3 } is an end pair similar and parallel with {x, y} and {α1 , α2 } crossed with {x, y}, then P is a 2-copy shortcut of the flip of C at (α1 , α3 ). (Again see Fig. 3.) Thus we have a curve C ′ with kC ′ (α ) < kC (α ). Since kC (α ) is finite, there is a finite sequence of flips and jumps, that yields a curve not containing any copy of α . □

We say an end is good for a curve C if it is neither polygonal nor crossed in C . By Claim 3, there are only finitely many ends that are not good for C1 . By using Claim 4 repeatedly, we can obtain a curve C2 from C1 by a finite number of flips and jumps such that C2 contains only ends that are good in C1 . The next claim will now show that every end in C2 is good for C2 . Claim 5. Let C be a curve and C ′ be a curve obtained from C by a flip or jump. For any α ∈ Ω (G), if α is good for C , then α is also good for C ′ . Proof. Since α is good in C , kC (α ) ≤ 2, and if kC (α ) = 2, then its unique end pair is parallel with all other end pairs. Since the flip and jump cannot increase copies of any end, kC ′ (α ) ≤ 2. So α is not polygonal in C ′ . If kC ′ (α ) = 0 or 1, then α is good for C ′ . So we assume that kC ′ (α ) = kC (α ) = 2. Let {α1 , α2 } be the unique end pair of α , both in C and C ′ . Since {α1 , α2 } is good for C , for any end

− →

− →

pair {β1 , β2 } with α ̸ = β , either β1 , β2 ∈ C (α1 , α2 ) or β1 , β2 ∈ C (α2 , α1 ). If C ′ is a flip of C , then

− →

− →′

− →

− →′

V ( C (α1 , α2 )) = V ( C (α1 , α2 )) and V ( C (α2 , α1 )) = V ( C (α2 , α1 )). This implies {α1 , α2 } is also good for C ′ .

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− →

Suppose now that C ′ is a jump of C at (x, y). If C (x, y) contains α1 or α2 , then kC ′ (α ) ≤ 1, contrary

− →

− →

to our assumption. So we assume that x, y ∈ C (α1 , α2 ) or x, y ∈ C (α2 , α1 ). Then clearly {α1 , α2 } is not crossed with any end pair in C ′ . Thus α is also good for C ′ . □

− →

− →

Claim 6. Let I1 = C2 [a1 , b1 ] and I2 = C2 [a2 , b2 ] be two A-intervals of C2 such that there is an end pair {α1 , α2 } between I1 and I2 , that is α1 ∈ I1 and α2 ∈ I2 . Then there are two sub-intervals

− →

− →

I1′ = C2 [a′1 , b′1 ] ⊂ I1 and I2′ = C2 [a′2 , b′2 ] ⊂ I2 with a′1 , a′2 , b′1 , b′2 ∈ V (G) such that every end pair between I1 and I2 is contained in I1′ ∪ I2′ and all other end pairs are contained in C2 \ (I1′ ∪ I2′ ). Proof. Let ΩD be the set of ends of G that have copies in both I1 and I2 , and let Ω1 and Ω2 be the set of copies of ends of ΩD in I1 and I2 , respectively. If there is a convergent sequence (α1i )∞ i=1 in Ω1 , then i ∞ i ∞ (α i ) ∞ i=1 converges to an end α of G. By the continuity of the curve, (α2 )i=1 is also convergent, and (α1 )i=1

and (α2i )∞ i=1 converge to two copies of the same end α . This implies that Ω1 and Ω2 are closed in C2 .

− →

− →

Let α1 , β1 ∈ Ω1 be the first and last copies of ends of ΩD that appear in C2 [a1 , b1 ]. Thus C2 [α1 , β1 ] contains no copies of ends in C2 \ (I1 ∪ I2 ). Let ˆ C (Sα , α ) be a neighborhood of α with a1 , a2 , b1 , b2 ∈ Sα and disjoint from C2 \ (I1 ∪ I2 ), and let ˆ C (Sβ , β ) be a neighborhood of β with a1 , a2 , b1 , b2 ∈ Sβ and

− →

disjoint from C2 \ (I1 ∪ I2 ). Let a′1 ∈ Sα be the last vertex to appear in C2 [a1 , α1 ] and b′1 ∈ Sβ be the

− → − → − → − → − → ′ implying that C2 [a1 , α1 ] and C2 [β1 , b′1 ] contain no copies of ends incident with C2 \ (I1 ∪ I2 ). So we − → let I1′ = C2 [a′1 , b′1 ]. − → In the same way, we can find I2′ = C2 [a′2 , b′2 ] ⊆ I2 that contains no copies of ends incident with C2 \ (I1 ∪ I2 ). □

first vertex to appear in C2 [β1 , b1 ]. Note that V ( C2 (a′1 , α1 ]) ⊆ C (Sα , α ), and V ( C2 [β1 , b′1 )) ⊆ C (Sβ , β ),

Lemma 2. Let G be a locally finite graph, and X , U1 , U2 be disjoint vertex-sets with |U1 | = |U2 | = k. If there are k vertex-disjoint U1 , U2 -arcs in |G| − X − Ω (X ), then there are k vertex-disjoint U1 , U2 -paths in G − X. Proof. Observe that X ∪ Ω (X ) is closed. Thus for every end α of G with α ̸ ∈ Ω (X ), there is a finite vertex set S such that C (S , α ) is disjoint from X . It follows that every end α ̸ ∈ Ω (X ) corresponds to an end of G′ = G − X . Thus every arc in |G| − X − Ω (X ) corresponds to an arc in |G′ | with the same vertex-set, and hence there are k vertex-disjoint U1 , U2 -arcs in |G′ |. If there are no k vertex-disjoint U1 , U2 -paths in G′ , then it follows from Menger’s Theorem that there is a set V of less than k vertices such that there is no U1 , U2 -path in G′ − V . But, it is easy to see that every u1 , u2 -arc P with ui ∈ Ui intersects V , a contradiction. □

− →

− →

Suppose that C2 has an end pair between two A-intervals I1 = C2 [a1 , b1 ] and I2 = C2 [a2 , b2 ]. − → − → Let I1′ = C2 [a′1 , b′1 ] and I2′ = C2 [a′2 , b′2 ] be as in Claim 6 and let X = V (C2 \ (I1′ ∪ I2′ )). Then

− →

− →

Y = C2 [b′1 , a′2 ] ∪ C2 [b′2 , a′1 ] is a closed set containing X and thus Ω (X ), so that I1′ and I2′ yield vertex-disjoint {a′1 , b′2 }, {a′2 , b′1 }-arcs in G − X − Ω (X ). By Lemma 2, there are two vertex-disjoint {a′1 , b′2 }, {a′2 , b′1 }-paths P1 , P2 in G − X . Thus (C2 \ (I1′ ∪ I2′ )) ∪ (P1 ∪ P2 ) is a curve in C with no end pair in the intervals that replaced I1 ∪ I2 . Repeating this process finitely many times, we obtain a curve C3 that contains no end pairs, i.e., C3 is a circle containing A. − → Suppose that C3 has an end contained in an A-interval I = C [a, b], and let X = V (C2 \ I). By the same reasoning as above I yields an arc in G − X − Ω (X ), so that by Lemma 2 with k = 1, there is − → an a, b-path P in G − X . Thus P ∪ C [b, a] is a circle in C that contains no end in I. Repeating this for every A-interval that contains an end, we obtain a circle C4 that is incident with no end, i.e., C4 is a cycle containing A. 7. Proof of Lemma 1 The backwards implication is trivial, as the curve C shows that O is embeddable. So suppose that O is an embeddable ordering of X . If X is finite, then letting S = X and T = NG (X ) − X yields a cycle C

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on X whose induced cyclic order is O, as desired. So we may assume that X = {v1 , v2 , . . .} is infinite, and we let Xi = {v1 , v2 , . . . , vi }, and Oi be the ordering on Xi induced by O. To construct a strict X -curve in |G|, that is, the image of a σ : S 1 → |G|, we will give three maps σV , σE , σΩ , whose images will be the vertices, edges and ends of the curve, respectively. We will use (a, b) ([a, b] etc.), to denote a cyclic interval in S 1 . So for any a, b ∈ S 1 , the intervals (a, b) and [b, a] form a partition of S 1 . To describe σV we pick distinct points qi ∈ S 1 for which we define σV (qi ) = vi . We shall pick qi such that the order of Qi = {q1 , . . . , qi } on S 1 is the same as that of Xi in Oi . To start let q1 , q2 be opposite points in S 1 . Suppose we already picked Qj−1 and consider vj . Note that Xj is finite, and vj has a predecessor vi and a successor vk in the cyclic order Oj . Let qj be the midpoint of the interval (qi , qk ) of S 1 . Since this interval contains no point whose image is a vertex of Xj−1 it follows that S 1 induces the cyclic order Oj on Qj . Continuing in this fashion we obtain a map σV : Q → X such that S 1 induces the cyclic order O on Q = {q1 , q2 , . . .}. This implies that σV preserves the order of O. The next two claims will help us to construct the map σE . An open interval (a, b) that is disjoint from Q is called a gap. Claim 7. If (a, b) is a gap, then there are qr , qt ∈ Q such that (qr , qt ) is a gap containing (a, b). Proof. Let (a′ , b′ ) be the maximal gap containing (a, b). Let l be the length of the gap (a′ , b′ ) and let ε be any positive real number that is much smaller than l. Since (a′ , b′ ) is a maximal gap, both (a′ − ε, a′ ] and [b′ , b′ + ε ) contain a point in Q . Let qr be the point in Q ∩ ((a′ − ε, a′ ] ∪ [b′ , b′ + ε )) with the smallest subscript. Without loss of generality we assume that qr ∈ (a′ − ε, a′ ]. Let qt be the point in Q ∩ ((qr , a′ ] ∪ [b′ , b′ + ε )) with the smallest subscript. Note that qr is the predecessor of qt in Ot . We claim that qt ∈ [b′ , b′ + ε ). To prove this claim, let qs be the successor of qt in Ot . Then s < t and the point qt is the midpoint of (qr , qs ). If qt ∈ (qr , a′ ], then qs ∈ (qr , a′ + ε ), implying that qs ∈ (qr , a′ ], contradicting the choice of qt . Thus qt ∈ [b′ , b′ +ε ) which proves the claim. We also claim that there is no vertex in Q ∩ (qr , qt ). For suppose (reductio ad absurdum) that qs is a point in Q ∩ (qr , qt ) with the smallest subscript. Then s > r , t and qr , qt are the predecessor and successor of qs in Os , respectively. So qs is the midpoint of (qr , qt ), implying that qs ∈ (a′ , b′ ), a contradiction. Thus (qr , qt ) = (a′ , b′ ) is a gap containing (a, b). □ Claim 8. Every vertex x ∈ X has a predecessor x− and a successor x+ in O, with x− , x+ ∈ NG (x). Proof. Let S = (NG (x) ∩ X ) ∪{x} and T = NG (x) \ X . Then S , T are finite sets, since G is locally finite. Let C be an S-curve guaranteed by embeddability. We consider the successor x+ of x in C . (The argument for the predecessor is similar.) Clearly x+ ∈ NG (x). Since C contains no vertex of T , it follows that x+ ∈ S. For every other x′ ∈ S − {x, x+ } we get x → x+ → x′ in the order induced by C , and thus in O. We claim that x+ is the successor of x in O. Suppose (reductio ad absurdum) that there is a vertex y with x → y → x+ in O. Let S ′ = (NG (x) ∩ X ) ∪ {x, y} and T ′ = NG (x) \ X . Then S ′ , T ′ are finite sets, and we can let C ′ be an S ′ -curve guaranteed by embeddability. We consider the successor x′ of x on C ′ . Since C ′ contains no vertex of T ′ , x′ ∈ NG (x) ∩ X , and we conclude that x → x′ → y in the order induced by C ′ , and thus in O. So, x → x′ → y → x+ , which contradicts x → x+ → x′ . □ Now we can construct the map σE . If I = (a, b) is a maximal gap of S 1 , then by Claim 7 σV (a), σV (b) are successive vertices in O, and by Claim 8 they are adjacent in G. Let I be the union of all gaps of S 1 and define σE : I → E(G) such that each maximal gap (a, b) is mapped homeomorphically onto the edge σV (a)σV (b). ∞ 1 Claim 9. Let (pi )∞ i=1 be a sequence in Q that converges to a point in p ∈ S \ (Q ∪ I ). Then (σV (pi ))i=1 converges to an end α ∈ |G|.

Proof. Let xi = σV (pi ), and let S ⊆ V (G) be an arbitrary finite set. We claim that there is a component of G − S containing all but finitely many terms of (xi )∞ i=1 . Once this sub-claim has been established, it follows that no two subsequences of (xi )∞ i=1 can converge to distinct ends, and hence Claim 9 follows. To prove this sub-claim, let Q ′ = {q ∈ Q : σV (q) ∈ S }. Then Q ′ is finite and divides S 1 into a finite

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′ number of intervals. Since (pi )∞ i=1 converges to p ̸ ∈ Q , one of these intervals, say I, contains all but ∞ finitely many terms of (pi )i=1 . We claim that for any two points pj , pk ∈ I ∩ Q , xj and xk belong to the same component of G − S. Since S ′ = (S ∩ X ) ∪ {xj , xk } and T ′ = S − X are finite and O is embeddable, there is an S ′ -curve that avoids T ′ and has the same order on S ′ as O. Since O has the same order on X as σV , it follows that there is no element of S ∩ X between xj and xk in O. Thus the S ′ -curve contains an xj , xk -arc that avoids S ∩ X , as well as T ′ = S − X . It follows that there is a component of G − S containing almost all vertices in (xi )∞ i=1 . This proves Claim 9. □

Let p be a point in S 1 \ (Q ∪ I ). By Claim 7, p is the limit of a sequence (pi )∞ i=1 in Q . By Claim 9 ′ ∞ ) (σV (pi ))∞ converges to an end α of G. Suppose that (p is another sequence in Q that also converges i i=1 i=1 to p in S 1 . Consider the sequence p = (p1 , p′1 , p2 , p′2 , . . .). Note that p converges to p, so by Claim 9 we ′ ∞ can see that (σV (pi ))∞ i=1 and (σV (pi ))i=1 both converge to α . Thus we can define σΩ (p) = α . We define the map σ : S 1 → |G| by setting σ (p) = σV (p) if p ∈ Q , σ (p) = σE (p) if p ∈ I and σ (p) = σΩ (p) otherwise. To show that the image of σ is the desired X -curve it remains to prove that σ is continuous. Claim 10. The map σ is continuous. Proof. Clearly σ is continuous at all points in Q ∪ I . Now we consider a point p ∈ S 1 \ (Q ∪ I ). 1 Let p = (pi )∞ i=1 be a sequence in S that converges to p. Let α = σ (p) and let S ⊆ V (G) be an arbitrary finite set. We shall prove that the neighborhood ˆ C (S , α ) contains almost all terms of (σ (pi ))∞ i=1 . (This will prove Claim 10.) Let Q ′ = {q ∈ S 1 : σ (q) ∈ S }. Then Q ′ is finite and divides S 1 into a finite number of intervals. (If necessary we can enlarge S so that we may assume |Q ′ | ≥ 2.) Since p ̸ ∈ Q ∪ I , p is contained in one of these intervals, say I = (r , t), where r , t ∈ Q ′ . Let r + , t − be the successor of r and predecessor of t in O, respectively, and let I ′ = [r + , t − ]. Since ((r , r + ) ∪ (t − , t)) ⊆ I , we conclude that p ∈ I ′ − Q . Thus I ′ contains almost all terms of p. Let X ′ = σ (I ′ ) ∩ V (G). Like in the proof of Claim 9, it follows that all vertices in X ′ are contained in a component H of G − S. Let q be an arbitrary point in I ′ . If q ∈ Q , then H contains σ (q). If q ∈ I , then q ∈ (a, b) with a, b ∈ Q ∩ I ′ . Thus σ (q) is an inner point of the edge σ (a)σ (b) and σ (a), σ (b) ∈ X ′ . This implies that σ (q) is contained in H. If q ∈ S 1 \ (Q ∪ I ), then for any sequence q = (σ (qi ))∞ i=1 in Q that converges to q, X ′ contains almost all terms of (σ (qi ))∞ i=1 , implying that H = C (S , σ (q)). By letting q = p, we conclude that H = C (S , σ (p)). This implies that ˆ C (S , σ (p)) contains almost all terms of (σ (pi ))∞ i=1 . Thus σ is continuous at p. □ This completes the proof of Lemma 1. Acknowledgments The authors thank Karl Heuer and the reviewers for their careful reading and helpful suggestions on this manuscript. References [1] P. Ash, B. Jackson, Dominating cycles in bipartite graphs, in: J.A. Bondy, U.S.R. Murty (Eds.), Progress in Graph Theory, Academic Press, New York, 1984, pp. pp. 81—87. [2] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, The MacMillan Press Ltd., London, 1976. [3] H. Bruhn, X. Yu, Hamilton cycles in planar locally finite graphs, SIAM J. Discrete Math. 22 (2008) 1381–1392. [4] S.A. Burr, J.W. Grossman, Ramsey numbers of graphs with long tails, Discrete Math. 41 (1982) 223–227. [5] Q. Cui, J. Wang, X. Yu, Hamilton circles in infinite planar graphs, J. Combin. Theory Ser. B 99 (2009) 110–138. [6] R. Diestel, Locally finite graphs with ends: A topological approach, II. Applications, Discrete Math. 310 (2010) 2750–2765. [7] R. Diestel, Locally finite graphs with ends: A topological approach, I. Basic theory, Discrete Math. 311 (2011) 1423–1447. [8] R. Diestel, Graph Theory, fifth ed., Springer, New York, 2016. [9] D. Duffus, M. Jacboson, R.J. Gould, Forbidden subgraphs and the Hamiltonian theme, in: The Theory and Applications of Graphs, John Wiley and Sons, New York, 1981, pp. pp. 297–316.

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