# -Convex Geometries are -free.

## Abstract

Let be a finite set and a collection of subsets of . Then is an alignment of if and only if is closed under taking intersections and contains both and the empty set. If is an alignment of , then the elements of are called convex sets and the pair is called an aligned space. If , then the convex hull of is the smallest convex set that contains . Suppose . Then is an extreme point for if . The collection of all extreme points of is denoted by . A convex geometry on a finite set is an aligned space with the additional property that every convex set is the convex hull of its extreme points. Let be a connected graph and a set of vertices of . A subgraph of containing is a minimal -tree if is a tree and if every vertex of is a cut-vertex of the subgraph induced by . The monophonic interval of is the collection of all vertices of that belong to some minimal -tree. A set of vertices in a graph is -convex if it contains the monophonic interval of every -set of vertices is . A set of vertices of a graph is -convex if for every pair of vertices in , the vertices on every induced path of length at least 3 are contained in . A set is -convex if it is both - and - convex. We show that if the -convex sets form a convex geometry, then is -free.

Key Words: minimal trees,
monophonic intervals of sets, -monophonic convexity, convex geometries

AMS subject classification: 05C75, 05C12, 05C17

## 1 Introduction

Let and be graphs. Then is an induced subgraph of if is a subgraph of and for every , if and only if . We say a graph is -free if it does not contain as an induced subgraph. Suppose is a collection of graphs. Then is -free if is -free for every . If is a path or cycle that is a subgraph of , then has a chord if it is not an induced subgraph of , i.e., has two vertices that are adjacent in but not in . An induced cycle of length at least is called a hole.

Let be a finite set and a collection of subsets of . Then is an alignment of if and only if is closed under taking intersections and contains both and the empty set. If is an alignment of , then the elements of are called convex sets and the pair is called an aligned space. If , then the convex hull of is the smallest convex set that contains . Suppose . Then is an extreme point for if . The collection of all extreme points of is denoted by . A convex geometry on a finite set is an aligned space with the additional property that every convex set is the convex hull of its extreme points. This property is referred to as the Minkowski-Krein-Milman () property. For a more extensive overview of other abstract convex structures see [13]. Convexities associated with the vertex set of a graph are discussed for example in [3]. Their study is of interest in Computational Geometry and has applications in Game Theory [2].

Convexities on the vertex set of a graph are usually defined in terms of some type of ‘intervals’. Suppose is a connected graph and two vertices of . Then a geodesic is a shortest path in . Such geodesics are necessarily induced paths. However, not all induced paths are geodesics. The -interval (respectively, -interval) between a pair of vertices in a graph is the collection of all vertices that lie on some geodesic (respectively, induced path) in and is denoted by (respectively, ).

A subset of vertices of a graph is said to be -convex (-convex) if it contains the -interval (-interval) between every pair of vertices in . It is not difficult to see that the collection of all -convex (-convex) sets is an alignment of . A vertex is an extreme point for a -convex (or -convex) set if and only if is simplicial in the subgraph induced by , i.e., every two neighbours of in are adjacent. Of course the convex hull of the extreme points of a convex set is contained in , but equality holds only in special cases. In [6] those graphs for which the -convex sets form a convex geometry are characterized as the chordal -fan-free graphs(see Fig. 1). These are precisely the chordal, distance-hereditary graphs (see [1, 7]). In the same paper it is shown that the chordal graphs are precisely those graphs for which the -convex sets form a convex geometry.

For what follows we use to denote an induced path of order . A vertex is simplicial in a set of vertices if and only if it is not the centre vertex of an induced in . Jamison and Olariu [8] relaxed this condition. They defined a vertex to be semisimplicial in if and only if it is not a centre vertex of an induced in .

Dragan, Nicolai and Brandstädt [5] introduced another convexity notion that relies on induced paths. The -interval between a pair of vertices in a graph , denoted by , is the collection of all vertices of that belong to an induced path of length at least . Let be a graph with vertex set . A set is -convex if and only if for every pair of vertices of the vertices of the -interval between and belong to . As in the other cases the collection of all -convex sets is an alignment. Note that an -convex set is not necessarily connected. It is shown in [5] that the extreme points of an -convex set are precisely the semisimplicial vertices of . Moreover, those graphs for which the -convex sets form a convex geometry are characterized in [5] as the (house, hole, domino, )-free graphs (see Fig. 1).

More recently a graph convexity that generalizes -convexity was introduced (see [11]). The Steiner interval of a set of vertices in a connected graph , denoted by , is the union of all vertices of that lie on some Steiner tree for , i.e., a connected subgraph that contains and has the minimum number of edges among all such subgraphs. Steiner intervals have been studied for example in [9, 12]. A set of vertices in a graph is -Steiner convex (-convex) if the Steiner interval of every collection of vertices of is contained in . Thus is -convex if and only if it is -convex. The collection of -convex sets forms an aligned space. We call an extreme point of a -convex set a -Steiner simplicial vertex, abbreviated vertex.

The extreme points of -convex sets , i.e., the vertices are characterized in [4] as those vertices that are not a centre vertex of an induced claw, paw or , in see Fig. 1. Thus a vertex is semisimplicial. Apart from the -convexity, for a fixed , other graph convexities that (i) depend on more than one value of and (ii) combine the convexity and the geodesic counterpart of the -convexity were introduced and studied in [10]. In particular characterizations of convex geometries for several of these graph convexities are given.

The notion of an induced path between a pair of vertices can be extended to three or more vertices. This gives rise to graph convexities that extend the -convexity. Let be a set of at least two vertices in a connected graph . A subgraph containing is a minimal -tree if is a tree and if every vertex is a cut-vertex of . Thus if , then a minimal -tree is just an induced path. Moreover, every Steiner tree for a set of vertices is a minimal -tree. The collection of all vertices that belong to some minimal -tree is called the monophonic interval of and is denoted by . A set of vertices is -monophonic convex, abbreviated as -convex, if it contains the monophonic interval of every subset of vertices of . Thus a set of vertices in is a monophonic convex set if and only if it is a -convex set. By combining the - convexity with the -convexity introduced in [5], we obtain a graph convexity that extends the graph convexity studied in [10]. More specifically we define a set of vertices in a connected graph to be -convex if is both - and -convex. In this paper we show that if the -convex alignment forms a convex geometry then is -free. We use the fact that these graphs are -free for several other graphs . In particular is easily seen to be house, hole, and domino free. Moreover the graphs of Fig. 2 are forbidden. A graph is a replicated twin if it is isomorphic to any one of the four graphs shown in Fig. 2(a), where any subset of the dashed edges may belong to . The collection of the four replicated twin graphs is denoted by . A graph is a tailed twin if it is isomorphic to one of the two graphs shown in Fig. 2(b) where again any subset of the dotted edges may be chosen to belong to . We denote the collection of tailed twin ’s by .

## 2 -Convex Geometries are -Free

Recall that the graphs for which the -convex sets form a convex geometry are characterized in [5] as the (house, hole, domino, )-free graphs. The proof of this characterization depends on the following useful result also proven in [5]:

###### Theorem 1.

If is a (house, hole, domino, )-free graph, then every vertex of is either semisimplicial or lies on an induced path of length at least between two semisimplicial vertices.

In [5] several ‘local’ convexities related to the -convexity were studied. For a set of vertices in a graph , is where is the collection of all vertices adjacent with some vertex of . A set of vertices in a graph is connected if is connected. The following useful result was established in [5].

###### Theorem 2.

A graph G is (house, hole, domino)-free if and only if is -convex for all connected sets of vertices of .

###### Theorem 3.

If is a graph such that is a convex geometry, then is -free.

###### Proof.

Observe first that is (house, hole, domino, , )-free. Suppose is a house, hole, domino, replicated twin or a tailed twin . Then has at most one vertex. Suppose is a graph that contains as an induced subgraph. Then the set of extreme points of the convex hull of is contained in the collection of vertices of . So the convex hull of the extreme points of the -convex hull of is empty or consists of a single vertex. So in this case the -convex alignment of does not form a convex geometry.

If is a set of vertices of a graph , then .

To show that contains no as an induced subgraph we prove a series of lemmas.

###### Lemma 1.

Suppose is a graph for which is a convex geometry. Then for every , .

###### Proof.

By the above observation is (house, hole, domino, , )-free. If then . So we may assume . If , there is a vertex that lies on an induced path between two vertices of . Among all such induced paths of length at least containing , let be one with a minimum number of edges. Suppose is a path. Clearly ; otherwise, . Let . (Suppose .) Then is not adjacent with two non-adjacent vertices of any induced path; otherwise, lies on an induced path.

Case 1 Suppose and lie on a common induced path . We may assume precedes on such a path. Moreover, we may assume that all internal vertices of are not on . For if , , then either or contains , say the former. Since is an induced path, so is . Hence . Thus must have length at least ; otherwise is adjacent with a pair of nonadjacent vertices of , implying that contains an induced path passing through , contrary to assumption. But then we have a contradiction to our choice of .

Let and . Then is connected for . By Theorem 2, is -convex. Since and both belong to , every vertex of must be adjacent with an internal vertex of . This is true in particular for . Since followed by Q and then is an path that contains it cannot be induced. Some vertex of or a vertex of must be adjacent with an internal vertex of ; say the former occurs. Let be the first vertex of that is adjacent with an internal vertex of . Let be the first vertex on that is adjacent with a vertex of (possibly is ). Let be the last vertex of adjacent with . Then the path is an induced path and thus does not contain . So is an internal vertex of or of ; suppose the former. Since is connected, is -convex by Theorem 2. Since and as has length at least , must contain every vertex of . Thus . Hence is adjacent with a vertex of . Since is adjacent with an internal vertex of , is not adjacent with any vertex of nor . Since is an induced path, the only vertex of to which can be adjacent is . So follows on . Since and belong to and as is an induced path containing , it follows that must be adjacent with ; otherwise, we have a contradiction to our choice of . Let be the last vertex on to which is adjacent. Then ; otherwise is an induced cycle of length at least . Let be the first internal vertex of to which is adjacent. (By an earlier observation exists.) Then ; otherwise, lies on an induced path. Also ; otherwise, is a house. If , let be the neighbour of on . Then and is an induced path between two vertices of having length and containing , contrary to our choice of . So . So is a cycle of length at least . Since , is an induced -cycle. Let be the first vertex after z on to which is adjacent (perhaps ). Then is an induced cycle and hence has length or . This cycle together with the -cycles produces either a house or a domino both of which are forbidden. So we may assume that is an induced path between vertices and of that do not belong to the same induced path. Indeed we may assume if and are any non-adjacent vertices that lie on the same induced path, then

Case 2 Suppose and lie on two internally disjoint paths and , respectively. We may assume ; otherwise, we are in Case 1.

We show first that no internal vertex of belongs to or . Suppose some internal vertex of or , say belongs to or . However, no internal vertex of belongs to ; otherwise, either the situation arises that was considered in Case 1 or there is an induced path containing . So we may assume that an internal vertex of lies on . Let be the last such vertex. Then contains and is an induced path between two vertices of that is shorter than . So has length ; otherwise we have a contradiction to our choice of . So must be the path . Since has length at least and by our choice of one of the neighbours of on must be . So one of the configurations shown in Fig. 3 must occur where solid lines are edges and dashed lines represent subpaths of and . We may assume that the configuration in (a) occurs. The argument for the configuration in (b) is similar.

Since is induced, is not adjacent to or and is not adjacent with . Let and be the neighbours of on and , respectively. If is adjacent with a vertex of then is an induced path of length containing and whose end vertices lie on the same induced path. By Case 1, this situation cannot occur. So the only vertex of to which can be adjacent is . Similarly, the only vertex of to which can be adjacent is . Using a similar argument and the fact that , we see that is not adjacent with any vertex of . Moreover, is not adjacent with any vertex of ; otherwise, lies on an induced path. The path obtained by taking followed by and then is an path that contains . Hence this path is not induced. Suppose first that . So some vertex of is adjacent with some vertex of . Since is not adjacent with any vertex of , some vertex of is adjacent with some vertex of . Let be a vertex closest to on that is adjacent with a vertex of and let be such a neighbour of closest to on . Observe that and ; otherwise, the cycle is an induced cycle of length at least . Let be the vertex closest to on that is adjacent with a vertex of (possibly ). Let be the neighbour of on closest to . By the above observation . The cycle is induced and has length at least unless and . So is adjacent with both and . Observe that is either adjacent with both and or neither of these two vertices; otherwise, is a house. We show next that no vertex of is adjacent with . Suppose is a vertex on closest to that is adjacent with First observe that for if , then is a house. So must be the neighbour of on ; otherwise, has a hole. However then is a domino. So is not adjacent with any vertex of . Let . Then is a cycle of length at least and hence has chords. Now is not adjacent with any vertex of other than ; otherwise, lies on an induced path of length between two vertices of that belong to the same induced path, a case already dealt with. Since , . Suppose is adjacent with an internal vertex of . Let be such a vertex closest to . So . Since contains no holes, is adjacent with . But then is a domino. So the neighbour of on is incident with a chord of . Since has no holes . But then is a domino. Suppose now that . Then . Let . Then is a cycle of length at least and hence has no chords. Since neither nor are incident with chords of , . If is a house. Note that is not adjacent with an internal vertex vertex of ; otherwise, if is such a neighbour of , then is an induced path of length between two vertices of that lie on the same induced path, a case already considered. Let be the neighbour of on and the neighbour of on . Then either or is an edge; otherwise, has a hole. But then or is a domino. So no internal vertex of belongs to or to .

Let . Let and be the neighbours of on and , respectively and and the neighbours of on and , respectively. Let and . Since is connected for , it follows from Theorem 2 that is -convex. Since for , every vertex of is adjacent with a vertex of for . In particular is adjacent with a vertex of for . However, is not adjacent with a pair of nonadjacent vertices of nor a pair of nonadjacent vertices of . So without loss of generality we may assume that is adjacent with a vertex of and a vertex of . Also is not adjacent with either or ; otherwise, lies on an induced path.

If is adjacent with two non-adjacent vertices of (or if is adjacent with two nonadjacent vertices of ), then ( and , respectively) and (or , respectively) is an induced path between two vertices of that is shorter than and contains . By our choice of this can only happen if has length .

We consider two subcases that depend on the length of .

Subcase 2.1 Suppose has length .

Then or is , say . The case where can be argued similarly. From the above, we may assume that is adjacent with an internal vertex of and an internal vertex of . The only vertex of that can be adjacent with is ; otherwise, lies on an induced path. So . Now it follows that is not adjacent with a vertex of . Thus is a house unless . If , then ; otherwise, or is a house. So is a tailed twin which is forbidden. So this subcase cannot occur.

Subcase 2.2 Suppose has length at least .

By an earlier observation, is not adjacent with a pair of non-adjacent vertices of and is not adjacent with a pair of non-adjacent vertices of . By assumption, is adjacent with an internal vertex of and an internal vertex of . Suppose . So is not adjacent with a vertex of nor a vertex of .

Fact 1 No vertex of is adjacent with a vertex of and no vertex of is adjacent with a vertex of .

Proof of Fact 1. Suppose some vertex of is adjacent with a vertex of . Let be the largest integer less than such that is adjacent with a vertex of . Let be a neighbour of on closest to on this path. Then is a cycle of length at least . If , then has length at least and three consecutive vertices of are not incident with a chord of the cycle. This implies that has a hole; which is forbidden. So . Clearly . Let . Then is a cycle of length at least . Thus contains an induced cycle of length at least that contains the edge . Since contains no holes, has length or . Since neither nor is adjacent with a vertex of nor a vertex of and as , it is not difficult to see that the vertices of and induce a house or a domino. So no vertex of is adjacent with a vertex of . By an identical argument we can show that no vertex of is adjacent with a vertex of .

Fact 2 No vertex of is adjacent with any vertex of .

Proof of Fact 2. Let be the first vertex of that is adjacent with some vertex of . Let be a neighbour of on that is closest to . Then the path is an induced path. So is -convex and hence contains all induced paths of length at least . Since , and since both and contain an internal vertex adjacent with , both and have length at least . So contains all the vertices of and and hence and . So also contains . Thus every vertex of is adjacent with a vertex of or with a vertex of . But by assumption is adjacent with an internal vertex of both and . So is adjacent with a pair of non-adjacent vertices of or a pair of non-adjacent vertices of , neither of which is possible.

From Facts and , it follows that no vertex of the path is adjacent with a vertex of the path . Hence the subgraph induced by the path is an induced path that contains ; contrary to the assumption that . This completes the proof of Case 2.

Case 3 Suppose that belongs to an induced path and to an induced path where and intersect at vertices other than and . We may assume that and do not both belong to nor both to ; otherwise, Case 1 occurs. Let be the last vertex prior to on that is also a vertex of (perhaps ). Let be the first vertex after on that belongs to . So . Let be the last vertex prior to on that also belongs to and the first vertex after on that also belongs to . So .

Subcase 3.1 Suppose contains both and . (Note may precede on .) In this case we can apply the argument used in Case 2 with and replaced by and and and replaced by and . Hence this subcase cannot occur.

Subcase 3.2 Suppose does not contain both and . Then and either lie on or on . We will assume the former case occurs. The arguments for the latter case are similar. We may assume precedes on . The case where precedes on is similar. First suppose that has length . Then is the only interior vertex of and is adjacent with two nonadjacent vertices of . Let be the first vertex on that is adjacent with , and the last vertex of adjacent with . Since , . If precedes on , then the path obtained by taking followed by and then is an induced path that contains both and . Thus we can apply the argument used in Case 1 to this path to obtain a contradiction. If follows on , then we can use the path and the path and apply the argument used in Case 2 with and instead of and , respectively.

We now assume that has length at least . Since is connected it follows, from Theorem 2, that is -convex. Since contains both and it must contain every internal vertex of . So each internal vertex of is adjacent with an internal vertex of . If no internal vertex of is adjacent with a vertex of or , then we can replace in with to obtain an induced path that contains both and . By applying the argument used in Case 1 to this path we obtain a contradiction. Let and be the neighbours of that precede and succeed on . Let be the neighbour of on .

Suppose first that some internal vertex of is adjacent with some vertex of . If , then is also adjacent with some internal vertex of . So ; otherwise, is adjacent with two nonadjacent vertices of which leads to a situation where the arguments of either Case 1 or Case 2 apply. If has length at least , then it follows, from Theorem 2, that is adjacent with every vertex of including ; this is not possible as and are nonadjacent vertices on the induced path . So and and is the only vertex of to which is adjacent. Suppose contains . If contains at least four vertices, then the subgraph induced by and the vertices of must contain a hole, house or domino. (We use the fact that cannot be adjacent to nonadjacent vertices of ; otherwise, one can again argue that Case 1 or Case 2 occurs.) Suppose now that . Let be the neighbour of on . Then is a tailed twin since is nonadjacent with and .

Suppose thus that does not belong to . Then we may assume that is the first internal vertex on that is adjacent with . Let be the neighbour of on . By the above we know that . If , then is a house which is forbidden. So assume . Let be the neighbour of on . Since and has no holes, . But then is a domino, which is forbidden. So is the only internal vertex of that is adjacent with vertices of . Let be the neighbour of on and let and be the neighbours of on and , respectively. One can argue as in the previous situation that the only internal vertex of that is possibly adjacent with a vertex of is .

Now let be the first vertex on that is adjacent with (possibly ) and let be the last vertex on to which is adjacent (possibly ). If belongs to , then the path obtained by taking followed by and then is an induced path containing both and . By Case 1 this produces a contradiction. Suppose thus that belongs to