Decomposable Convexities in Graphs and Hypergraphs

Given a connected hypergraphHwith vertex set V, a convexity space onH is a subset CC of the powerset of V that contains∅, V, and the singletons; furthermore,CC is closed under intersection and every set inCC is connected inH. e members ofCC are called convex sets. e convex hull of a subset X of V is the smallest convex set containing X. By a cluster ofH we mean any nonempty subset of V in which every two vertices are separated by no convex set. We say that a convexity space onH is decomposable if it satis�es the following three axioms: (i) the maximal clusters ofH form an acyclic hypergraph, (ii) every maximal cluster ofH is a convex set, and (iii) for every nonempty vertex set X, a vertex does not belong to the convex hull of X if and only if it is separated from X by a convex cluster. We prove that a decomposable convexity spaceCC onH is fully speci�ed by the maximal clusters ofH in that (1) there is a closed formula which expresses the convex hull of a set in terms of certain convex clusters ofH and (2)CC is a convex geometry if and only if the subspaces of CC induced by maximal clusters ofH are all convex geometries. Finally, we prove the decomposability of some known convexities in graphs and hypergraphs taken from the literature (such as “monophonic” and “canonical” convexities in hypergraphs and “all-paths” convexity in graphs).


Introduction
A (�nite) convexity space [1] over a �nite nonempty set  is a subset  of the powerset of  that contains ∅ and  and is closed under intersection.e members of  are called convex sets.e convex hull of a subset  of , denoted by ⟨, is the smallest convex set containing .It is well known that (i)   ⟨, (ii) if    then ⟨  ⟨, and (iii) ⟨⟨  ⟨.A convexity space  over  is a point-convexity space [2] if, for every element  of , the singleton { is a convex set.Abstract convexity theory has been applied to graphs [3][4][5][6] and hypergraphs [4,7].A convexity space on a connected hypergraph is a point-convexity space  such that every convex set is connected.If  is a convex set, the subspace of  induced by  is the convexity space   {       on the (connected) subhypergraph induced by .
Most convexities in graphs and hypergraphs have been stated in terms of "feasible" paths [8]; accordingly, a vertex set  is convex if  contains all vertices on every feasible path joining two vertices in .For example, geodetic convexity, monophonic convexity, and all-paths convexity are obtained using shortest paths, chordless paths, or all paths, respectively.
We note that in graphs monophonic convexity (convexity, for short) and all-paths convexity (ap-convexity, for short) enjoy similar properties.For example, Duchet [5] proved that "a vertex  does not belong to the -convex hull of a set  if and only if  is separated from  by a clique, " and known results on ap-convexity [3,9] imply that "a vertex  does not belong to the ap-convex hull of a set  if and only if  is separated from  by a cut-vertex." is similarity also applies to the algorithms to compute -convex hulls [7,[9][10][11] and ap-convex hulls [9].
In this paper, we show that -convexity and ap-convexity belong to a wide class of convexity spaces on hypergraphs, which we call "decomposable" and de�ne by means of three axioms.To this end, we need to introduce the notion of a "cluster." Let  be a convexity space on a connected hypergraph ℋ with vertex set .A cluster of ℋ is a nonempty subset of  in which every two vertices are separated by no convex set.e cluster hypergraph of ℋ is the set of maximal clusters of ℋ.A convexity space on ℋ is decomposable if it satis�es the following three axioms.Axiom 1. e cluster hypergraph of ℋ is acyclic.Axiom 2. Every edge of the cluster hypergraph of ℋ is a convex set.Axiom 3.For every nonempty vertex set , a vertex does not belong to the convex hull of  if and only if it is separated from  by a convex cluster.
Our main result is that if  is a decomposable convexity space on ℋ, then  enjoys the following two properties.
(D1) there is a closed formula which expresses the convex hull of a set in terms of certain convex clusters of ℋ.
(D2)  is a convex geometry if and only if the subspaces of  induced by maximal clusters of ℋ are all convex geometries.
e rest of the paper is organized as follows.In Section 2 we recall more-or-less standard de�nitions and some preliminary results on acyclic hypergraphs, which will be used in the sequel.In Section 3 we state general properties of the cluster hypergraph.In Section 4 we prove some properties of convexity spaces that satisfy Axioms 1 and 2. In Section 5 we prove that decomposable convexity spaces enjoy properties (D1) and (D2).In Section 6, we give three examples of decomposable convexity spaces taken from the literature.Finally, Section 7 contains a closing note.

Preliminaries
Let  be a �nite set.A hypergraph [12] on  is a nonempty set of subsets of  whose union recovers .e elements of  are the vertices of ℋ and the sets in ℋ are the (hyper) edges of ℋ.If   , then we call ℋ  {} the empty hypergraph.A nonempty hypergraph is trivial if it has exactly one edge.A hypergraph is simple if no edge is a subset of another edge.A partial hypergraph of hypergraph ℋ is a nonempty subset of ℋ.A cover of ℋ is a simple hypergraph  on  such that each edge of ℋ is contained in some edge of .
A vertex is a leaf if it belongs to exactly one edge.Two vertices are adjacent if they belong together to some edge.A vertex  is adjacent to a vertex set  if  contains a vertex  such that  and  are adjacent.
A clique is a nonempty vertex set in which every two vertices are adjacent.A partial edge is a nonempty vertex set included in some edge.A hypergraph is conformal if every clique is a partial edge.e clique hypergraph of a hypergraph ℋ is the hypergraph whose edges are precisely the maximal cliques of ℋ.
A (vertex) path is a sequence of distinct vertices ( 1 ,  2 , … ,   ),   1, such that if   1 then  ℎ and  ℎ+1 are adjacent for ℎ  1, … ,   1; the path is said to join  1 and   from  1 to   and to pass through any set  such that   { 2 , … ,  1 } ≠ .A hypergraph is connected if every two vertices are joined by a path.e connected components of a hypergraph are its maximal partial hypergraphs that are connected.
Let  be a nonempty subset of ; the subhypergraph of ℋinduced by  is the hypergraph on , denoted by ℋ(), whose edges are the nonempty intersections of  with edges of Let  be a nonempty proper subset of ; by ℋ   we denote the subhypergraph of ℋ induced by   .If ℋ ′ is a connected component of ℋ  , the neighborhood of ℋ ′ , denoted by (ℋ ′ ), is the set of vertices in  that are adjacent to the vertex set of ℋ ′ .
A vertex  is separated from a set  by  if      and  is disjoint from the vertex set of the connected component of ℋ   containing .
In the next two subsections we deal with "separators" and "acyclic hypergraphs."  [13] of ℋ if there exists a vertex pair for which  is a minimal separator.

Acyclic Hypergraphs.
A hypergraph is acyclic if there exists a running-intersection ordering ( 1 ,  2 , … ,   ) of its edges, that is, if   1 then, for each   1 there exists   <  such that  1  ⋯   1            . ( Several equivalent de�nitions of acyclicity exist [14].A test for acyclicity is given by the following algorithm which reduces a hypergraph ℋ to the empty hypergraph {} if and only if ℋ is acyclic [14].
Graham Reduction.Repeatedly apply the following two operations until neither can be longer applied: (vertex deletion): remove a vertex if it is a leaf.
(edge deletion): remove an edge if it is contained in another edge.
A more efficient algorithm to test acyclicity was given in [15].e following two facts state well known properties of MVSs of an acyclic hypergraph [14,16] and they will be used in the sequel.Fact 1.Let ℋ be an acyclic, simple, and connected hypergraph.e MVSs of ℋ are exactly the partial edges of ℋ that are removed during the Graham reduction of ℋ. Fact 2. Let ℋ be an acyclic, simple, and connected hypergraph.If  is an MVS of ℋ, then there exist two edges  and  of ℋ such that      and, for every      and     ,  is the only minimal separator for { .
For our purposes, we need a modi�cation of Graham reduction when there is a set  of sacred vertices that may not be deleted by vertex deletion.Let GR(ℋ  denote the result of applying Graham reduction to a hypergraph ℋ with the vertex deletion rule modi�ed to disallow the removal of any vertex in , the application of rules proceeding until such time as no more applications are possible [16].We call GR(ℋ  the Graham reduction of ℋ with sacred set .It is well known [16] that GR(ℋ  is de�ned uniquely, that is, if a deletion rule is applicable to a vertex or edge, applying a rule elsewhere does not make the �rst inapplicable; so, the reduction procedure proceeds to do everything that is ever possible, independent of the order of reductions chosen.In other words, the Graham reduction process enjoys the socalled Church-Rosser property [16], which makes it a pruning process according to the terminology used in [17].It is also well known [16] that if ℋ is an acyclic, simple and connected hypergraph, then (i) GR(ℋ  is an acyclic, simple, and connected hypergraph; (ii) the edges of GR(ℋ  are the maximal edges of an induced subhypergraph of ℋ whose vertex set contains ; (iii)  is a subset of the vertex set of GR(ℋ ; (iv) the MVSs of ℋ are the MVSs of GR(ℋ  plus the partial edges ℋ removed during the Graham reduction process.
e following lemma contains one more property of GR(ℋ  which will be used later on. Lemma 1 (see [18]).Let ℋ be an acyclic hypergraph on ,  a nonempty proper subset of , and  a vertex in   .e following holds: (i) a vertex  of ℋ belongs to the vertex set of GR(ℋ  if and only if  is separated from  by no partial edge of ℋ; (ii) if  is not a vertex of GR(ℋ , then there exists an edge of GR(ℋ  that separates  from  in ℋ; (iii) if  is an edge of GR(ℋ , then  is the union of    with the neighborhoods of the connected components of ℋ −  whose vertex sets are not disjoint from .Corollary 2. Let ℋ be an acyclic hypergraph on ,  a nonempty proper subset of , and  a vertex in   .If A is an edge of GR(ℋ ,  is not a vertex of GR(ℋ , and   { is a partial edge of ℋ, then A is contained in every partial edge of ℋ that separates  from  in ℋ. Proof.First of all, observe that the connected component of ℋ− that contains  has an empty intersection with .Let  be a partial edge of ℋ that separates  from  in ℋ, and let  be the vertex set of a connected component ℋ ′ of ℋ− such that   .en  must contain the neighborhood of ℋ ′ for, otherwise,  and each vertex in  would be in the same connected component of ℋ − .Moreover,      for, otherwise,  and some vertex in    would be in the same connected component of ℋ − .From part (iii) of Lemma 1, it follows that   .

The Cluster Hypergraph
Let  be a convexity space on a connected hypergraph ℋ. Recall from the Introduction that a cluster of ℋ is a nonempty vertex set that is separated by no convex set of ℋ, and the cluster hypergraph of ℋ is the hypergraph, henceforth denoted by , whose edges are precisely the maximal clusters of ℋ.In this section, we shall state some general properties of the cluster hypergraph.eorem 3. Let ℋ be a connected hypergraph and  a convexity space on ℋ. e cluster hypergraph  of ℋ is a cover of the clique hypergraph of ℋ and is a conformal hypergraph.
Proof.Since every clique of ℋ is a cluster, every maximal clique of ℋ is a partial edge of , that is,  is a cover of the clique hypergraph of ℋ.Moreover, since two vertices  and  of  are adjacent (in ) if and only if {  is a cluster of ℋ, one has that every clique of  is a cluster of ℋ and, hence, a partial edge of .
By eorem 3,  is a cover of ℋ so that every separator of  is also a separator of ℋ.We now prove that the converse also holds for convex separators of ℋ. Lemma 4. Let ℋ be a connected hypergraph,  a convexity space on ℋ,  the cluster hypergraph of ℋ, and  a convex set of ℋ.Two vertices that are separated by  in ℋ are also separated by  in .
Proof.Let  and  be two vertices that are separated by  in ℋ.Since  is a convex set, the vertex pair {  is not a cluster of ℋ and, hence,  and  are not adjacent in .Suppose, by contradiction, that  and  are not separated by  in .en, there exists in  a path ( 1    2  …     ,   , such that  ℎ ∉  for all ℎ, 1 ≤ ℎ ≤ .Since each vertex pair { ℎ   ℎ+1  is a partial edge of , the set { ℎ   ℎ+1  is a cluster of ℋ and, since  is a convex set,  ℎ and  ℎ+1 cannot be separated by  in ℋ so that in ℋ there exists a path  ℎ from  ℎ to  ℎ+1 such that no vertex on  ℎ belongs to .Combining the paths  1   2  …   −1 , we obtain a path  in ℋ from  to  and no vertex on  belongs to , which contradicts the hypothesis that  and  are separated by  in ℋ. e next lemma is useful for the sequel.

Axioms 1 and 2
In this section we state some properties of convexity spaces that satisfy Axioms 1 and 2. Recall that a convexity space on ℋ satis�es Axiom 1 if the cluster hypergraph of ℋ is acyclic, and satis�es Axiom 2 if every maximal cluster of ℋ is a convex set.First of all, observe that the following condition is equivalent to Axiom 2. Fact 3. Let ℋ be a connected hypergraph.A convexity space  on ℋ satis�es Axiom 2 if and only if the convex hull of every cluster is a cluster.

Proof of ("only if ").
Let  be any cluster of ℋ and  a maximal cluster containing .By Axiom 2,  is a convex set.Since   , one has ⟨  ⟨   and, hence, ⟨ is a cluster.

Proof of ("if ").
Let  be any maximal cluster of ℋ.We need to prove that   ⟨.Of course,   ⟨.By hypothesis, ⟨ is a cluster.Let  be a maximal cluster of ℋ that contains ⟨.en, one has   ⟨  .Since  is a maximal cluster, one has    so that   ⟨.
e following are two properties of convexity spaces that satisfy both Axioms 1 and 2. eorem 8. Let ℋ be a connected hypergraph,  a convexity space on ℋ and  the cluster hypergraph of ℋ.If  satis�es Axioms 1 and 2, then every MVS of  is a convex set.
Proof.Since  satis�es Axiom 1, is an acyclic hypergraph.Let  be any MVS of .By Fact 2,      for some two edges  and  of .Since  satis�es Axiom 2,  and  are convex sets of ℋ so that  is a convex set of ℋ. Proof.We �rst prove that (i) every MVCS of ℋ is an MVS of  and, then, that (ii) every MVS of  is an MVCS of .

Proof of (i).
Let  be an MVCS of ℋ.By eorem 6, there exists an MVS  of  such that   .By eorem 8,    so that   .

Decomposable Convexity Spaces
Recall that a convexity space  on ℋ is decomposable if  satis�es Axioms 1 and 2 and, for every subset  of , the convex hull of  equals the set of vertices that are separated from  by no convex cluster of ℋ.
Example 10 (continued).e convexity space  is not decomposable since the convex hull of the set {    is the whole vertex set  but the vertex  is separated from {    by the convex cluster {.
In the next two subsections, we shall prove the properties (D1 and D2) mentioned in the introduction.

Property (D1).
Given any convexity space  on ℋ that satis�es Axioms 1 and 2, we �rst provide a closed formula for the set of vertices that are separated from a set  by no convex cluster of ℋ.It will follow that if  is decomposable, then the same formula is an expression of the convex hull of .

Lemma 11.
Let ℋ be a hypergraph on ,  a convexity space on ℋ that satis�es Axioms 1 and 2,  the cluster hypergraph of ℋ, and  a nonempty subset of .e set of vertices that are separated from  by no convex cluster of ℋ is given by the union of the convex hulls of edges of GR( .
Proof.Let  *  ⋃ GR( .We now prove that    * if and only if  is separated from  by some convex cluster of ℋ.

Proof of ("if ").
Let  be a vertex of ℋ that is separated from  by a convex cluster of ℋ.Let  such a convex cluster.Since  is a convex set,  separates  from  in  by Lemma 5.Moreover, since  is a cluster of ℋ,  is a partial edge of .In order to prove that    * , we �rst show that  is not a vertex of GR(  and, then,    for every edge  of GR( .
Since in  the vertex  is separated from  by the partial edge  of ,  is not a vertex of GR(  by part (i) of Lemma 1.Consider now any edge  of GR( .If    then, since  is a subset of the vertex set of GR( , one has that   .Assume that    and suppose, by contradiction, that   .By Corollary 2, one has that    and, hence,     .To sum up,    by assumption, and    since  separates  from , which is in contradiction with the inclusion   .

Proof of ("only if ").
Assume that    * .We now prove that there exists a convex cluster that separates  from  in ℋ. Recall that the vertex set of GR(  is a subset of  * and  is a subset of  * for every edge  of GR( .erefore, since    * ,  is not a vertex of GR(  and    for every edge  of GR( .Since  is not a vertex of GR( , by part (ii) of Lemma 1 there exists an edge  of GR(  that separates  from  in  and, since    * and    * , one has that    and, hence,  separates  from  in .Since  is a cover of ℋ,  is separated from  by the convex cluster  in ℋ.
At this point, we are in a position to prove property (D1) of decomposable convexity spaces.eorem 12. Let ℋ be a hypergraph on ,  a decomposable convexity space on ℋ,  the cluster hypergraph of ℋ, and  a subset of .e convex hull of  is given by the union of the convex hulls of the edges of GR( , that is, Assume that  is a convex set.en   ⟨ so that, by ( 2), one has that   ⋃ GR(  and   ⟨ for every edge  of GR( , which are exactly conditions (a) and (b).

Property (D2) .
We begin with a few basic de�nitions and preliminary results.A vertex  in a convex set  is an extreme point of  if    is a convex set.A convexity space is a convex geometry if it enjoys the Minkowsky-Krein-Milman property: every convex set is the convex hull of the set of its extreme points.
In order to prove property (D2), we make use of the following two characterizations of convex geometries.
Lemma 14 (e.g., see [17]).A convexity space  is a convex geometry if and only if, for every set  in ,   , there exists    such that    belongs to .e next characterization of convex geometries is based on the notion of a �descending path, � which is de�ned as follows.Let  and  be two convex sets with   .A descending path [17] from  to  in the lattice of  is an ordering ( 1  …     of the elements of    such that, for each , 1 ≤  ≤ , the set    1   2  …     is convex.Lemma 15 (see [4]).A convexity space  is a convex geometry if and only if, for every set  in ,   , there exists a descending path from  to  in the lattice of .

Corollary 16.
Let  be a convex geometry and let  and  be two convex sets.If  is a subset of ,   , then there exists a descending path from  to  in the lattice of the induced subspace (.
At this point, we are in a position to prove property (D2) of decomposable convexity spaces.eorem 17.Let ℋ be a connected hypergraph,  a decomposable convexity space on ℋ, and  the cluster hypergraph of ℋ. e convexity space  is a convex geometry if and only if the subspaces of  induced by edges of  are all convex geometries.Proof.Proof of ("only if ").Let  be any edge of .Since  is decomposable,  is a convex set by Axiom 2. Let ( be the subspace of  induced by .Consider any convex set  in (,   .By Corollary 16, there exists a descending path from  to  in the lattice of (, which by Lemma 15 proves that ( is a convex geometry.

Proof of ("if ").
Let  be any convex set with   .By Lemma 14, we need to prove that there exists a vertex   such that    is convex.By eorem 13,  equals the vertex set of GR(  and every edge of GR(  is a convex set.Distinguish two cases depending on whether or not GR(  is a partial hypergraph of  (i.e., every edge of GR(  is an edge of ).
Case 1. GR(  is a partial hypergraph of .Since   , all the edges of  that are not edges of GR(  have been deleted during the Graham reduction of  with sacred set .Let  be the edge of  that has been deleted; then, its residual part, say , was found to be contained in an edge of GR( .en,  is an MVS of  (see the properties of the Graham reduction).Since  is decomposable,  is a convex set by Axiom 2, and  is a convex set by eorem 8.
Case 2. GR(  is not a partial hypergraph of .en there exists an edge  of GR(  that is strictly contained in an edge of .Let  be an edge of  that contains .Since  is decomposable,  is a convex set by Axiom 2, and  is a convex set by condition (b) of eorem 13.In both cases, by hypothesis, ( is a convex geometry.erefore, since  is a proper subset of , by Lemma 14 there exists      such that the set    is convex.We will show that    is a convex set, which by Lemma 14 proves that  is a convex geometry.Since     , by the Church-Rosser property of the Graham reduction procedure, we can assume that  is the last vertex that is deleted during the Graham reduction of  with sacred set .erefore, in Case 1 the hypergraph GR(  can be viewed as being obtained from GR(  by adding the edge   , and in Case 2 the hypergraph GR(  can be viewed as being obtained from GR(  by deleting the edge  and adding the edge   .Since  is a convex set, by condition (a) of eorem 13 the vertex set of GR(  equals , so that the vertex set of GR(    equals   , which proves that    satis�es condition (a) of eorem 13.Morever, an edge of GR(    is either    or an edge of GR(  and, since    is a convex set and every edge of GR(  is a convex set by condition (b) of eorem 13, one has that every edge of GR(    is a convex set, which proves that    satis�es condition (b) of eorem 13.So, by eorem 13,    is convex.
It is worth observing that, by eorem 13, every descending path in a decomposable convex geometry  on ℋ is a Graham elimination ordering of vertices of the cluster hypergraph  (or, equivalently, is a perfect elimination ordering of the chordal graph on the vertex set of ℋ where two vertices are adjacent if and only if they form a cluster of ℋ).

𝑚𝑚-Convexity in Hypergraphs.
Let ℋ be a connected hypergraph with vertex set .A path  in ℋ is chordless if, for every two nonconsecutive vertices  and  on ,  and  are not adjacent.A subset  of  is -convex if either    or  contains all vertices on every chordless path joining two vertices in  [4,5].Equivalently, a subset  of  is -convex if either    or    or the neighborhood of every connected component of ℋ −  is a clique of ℋ [5,7].Note that every clique of ℋ is an -convex set.By ⟨  we denote the -convex hull of .Efficient algorithms for computing -convex hulls in graphs were given in [9][10][11].A similar algorithm for computing -convex hulls in hypergraphs was given in [7]; it makes use of the acyclic hypergraph whose edges are the maximal sets that are not separable by cliques of ℋ [19].is hypergraph is called the prime hypergraph of ℋ [7].Let  ℋ denote the prime hypergraph of ℋ.We can summarize the algorithm in [7] into the following formula: where In order to prove the decomposability of -convexity, we need the following lemma.Lemma 18.Let ℋ be a connected hypergraph, and let  be the m-convexity space on ℋ. e cluster hypergraph of ℋ equals the prime hypergraph of ℋ.
Proof.It is sufficient to prove that two vertices of ℋ are separated by an -convex set of ℋ if and only if they are separated by a clique of ℋ.
Proof of (if).e statement follows from the fact that every clique of ℋ is an -convex set of ℋ.
Proof of (only if).We need to prove that if two vertices  and  of ℋ are separated by an -convex set of ℋ, then they are separated by a clique of ℋ.Let  be an convex set separating  and .us,  and  belong to two distinct connected components of ℋ − .Let ℋ ′ be the connected component of ℋ −  containing , and let  be the neighborhood of ℋ ′ .en, ℋ ′ is also a connected component of ℋ −  and the vertex  belongs to another connected component of ℋ − .erefore,  and  are separated by  and, since  is an -convex set of ℋ and  is the neighborhood of a connected component of ℋ − ,  is a clique of ℋ, which proves the statement.eorem 19.Let ℋ be a connected hypergraph.e convexity space on ℋ is decomposable.
Proof.Let  be the -convexity space on ℋ.By Lemma 18, the cluster hypergraph of ℋ is acyclic so that, by Fact 4, every-edge of the cluster hypergraph of ℋ is an -convex set.erefore,  satis�es Axioms 1 and 2.Moreover, by Lemmas 11 and 18 and by Fact 5, the right-hand side of (5) equals the set of vertices that are separated from  by no -convex cluster of ℋ.Finally, by the equality in (5),  also satis�es Axiom 3, which proves the statement.

𝑐𝑐-Convexity in Hypergraphs.
Let ℋ be a connected hypergraph with vertex set .A subset  of  is -convex if either    or    or the neighborhood of every connected component of ℋ −  is a partial edge of ℋ [7].Note that every partial edge of ℋ is a -convex set.By ⟨  we denote the -convex hull of .
An efficient algorithm for computing -convex hulls in hypergraphs was given in [12]; it makes use of the acyclic hypergraph whose edges are the maximal sets that are not separable by partial edges of ℋ [20].e hypergraph is called the compact hypergraph of ℋ [7,18,20].Let  ℋ denote the compact hypergraph of ℋ.We can summarize the algorithm in [7] into the following formula: In order to prove the decomposability of c-convexity, we need the following lemma whose proof is analogous to the proof of Lemma 18.
Lemma 20.Let ℋ be a connected hypergraph and let  be the -convexity space on ℋ. e cluster hypergraph of ℋ equals the compact hypergraph of ℋ. eorem 21.Let ℋ be a connected hypergraph.e convexity space on ℋ is decomposable.
Proof.Let  be the -convexity space on ℋ.By Lemma 20, the cluster hypergraph of ℋ is acyclic so that, by Fact 6, every-edge of the cluster hypergraph of ℋ is a -convex set.erefore,  satis�es Axioms 1 and 2.Moreover, by Lemmas 11 and 20 and by Fact 7, the right-hand side of ( 7) equals the set of vertices that are separated from  by no convex cluster of ℋ.Finally, by the equality in (7),  also satis�es Axiom 3, which proves the statement.

ap-Convexity in Graphs.
Let ℋ be a connected graph with vertex set .A subset  of  is ap-convex if either    or  contains all vertices on every path joining two vertices in  [3,9].By ⟨  we denote the ap-convex hull of .
Let ℋ be the block hypergraph of ℋ; that is, ℋ is the acyclic hypergraph whose edges are the maximal sets that are not separable by cut vertices of ℋ. From results in [3,9], it follows that the ap-convex hull of a subset  of  is given by the following formula: where ( is the edge of ℋ that contains .At this point, it is a mere exercise to prove that the cluster hypergraph of ℋ equals the block hypergraph of ℋ and that the ap-convexity space on ℋ is decomposable.

Closing Note
Given a convexity space  on a connected hypergraph ℋ, we introduced the notions of a cluster of ℋ and of the cluster hypergraph of ℋ. �sing them, we de�ne decomposability of  by means of three axioms and proved that decomposability implies the existence of a closed formula (see (2) above) which expresses the convex hull of a vertex set  in terms of the convex hulls of edges of GR(, , where  is the cluster hypergraph of ℋ.Moreover, we proved that a decomposable convexity space  is a convex geometry if and only if the subspaces of  induced by the edges of  are all convex geometries.Finally, we showed that both convexity and -convexity in hypergraphs are decomposable, and that ap-convexity in graphs is decomposable.An open problem is the search for other decomposable known F 3 hypergraph convexities.We conjecture that not only apconvexity is decomposable in hypergraphs too, but also every convexity that is coarser than -convexity (such as "simplepath convexity" [4]) is decomposable.On the other hand, it is easy to see that geodetic convexity (-convexity) is not decomposable in the general case as is proven by the following example.However, since -convexity and -convexity are equivalent in distance-hereditary graphs [4], one has that convexity is decomposable in distance-hereditary graphs.
Example 22. Consider the (hyper)graph ℋ shown in Figure 3 and the -convexity space on ℋ. Recall from the introduction that a vertex set  is -convex if  contains all vertices on every shortest path joining two vertices in .
It is easy to see that the sets , ,  and , ,  are the only minimal vertex -convex separators of ℋ.Moreover, the cluster hypergraph of ℋ is   , , , , , , , , , , , , , , , .Since  is not an acyclic hypergraph, the -convexity space on ℋ is not decomposable.

Figure 1
Figure 1 illustrates part (iii) of Lemma 1 for the two cases in which      and     .

Corollary 9 .
Let ℋ be a connected hypergraph,  a convexity space on ℋ, and  the cluster hypergraph of ℋ.If  satis�es Axioms 1 and 2, then the MVCSs of ℋ are exactly the MVSs of .
then  is said to separate every two vertices that are in distinct connected components of ℋ  .Note that  separates two vertices  and  if and only if   {, }   and every path joining  and  passes through .A separator  separates a subset  of  if  separates two distinct vertices in .A set  is a minimal separator for a vertex pair {, } if  and  are separated by  and are separated by no proper subset of .A set  is a minimal vertex separator (MVS) Let  be the vertex set of a connected component of ℋ and  the vertex set of a connected component of   such that     .By eorem 3,  is a cover of ℋ and, since     , one has   .Suppose, by contradiction, that   , and let    and     .Since neither  nor  are in , , and  are separated by  in ℋ.By Lemma 4,  and  are separated by  in , which contradicts the fact that  and  are both in .enextresultrelates the MVSs of the cluster hypergraph of ℋ with "minimal vertex convex separators" of ℋ, which we de�ne as follows.A convex set  of ℋ is a minimal convex separator of ℋ for a vertex pair {  if  and  are separated by  and are separated by no convex proper subset of .A set  is a minimal vertex convex separator (MVCS) of ℋ if there exists a vertex pair for which  is a minimal convex separator.Let ℋ be a connected hypergraph,  a convexity space on ℋ, and  the cluster hypergraph of ℋ.Every MVCS of ℋ is the convex hull of an MVS of .Proof.Let  be an MVCS of ℋ, and let {  be a vertex pair for which  is a minimal convex separator of ℋ.Since  is a convex set,  separates  and  in  by Lemma 4; therefore,  must contain a minimal separator of  for { .Let  be such an MVS of .Since   , one has ⟨  ⟨   and, hence, neither  nor  belongs to ⟨.erefore, since  separates  and  in  and since   ⟨, also the set ⟨ separates  and  in  and, since  is a cover of ℋ, the set ⟨ separates  and  in ℋ.Finally, since ⟨   and  is a minimal convex separator of ℋ for { , one has ⟨  .Consider the (hyper)graph ℋ with vertex set   {     that is shown in Figure2, and the convexity space  that only contains , , the �ve singletons, and the set {  .e MVCSs of ℋ are the set { (which is a minimal convex separator for the vertex pairs { , { , and { ) and the set {   (which is a minimal convex separator for the vertex pair { ).e cluster hypergraph of ℋ is   {{   {   { .e MVSs of  are the set { (which is a minimal separator for the vertex pairs { , { , and { ) and the set {  (which is a minimal separator for the vertex pair { ).e MVCS {   of ℋ is the convex hull of the MVS {  of , and the MVCS { of ℋ is the convex hull of the MVS { of .
Lemma 5. Let ℋ be a connected hypergraph,  a convexity space on ℋ,  the cluster hypergraph of ℋ, and  a convex set.If the vertex sets of a connected component of ℋ −  and ).Let  be an MVS of .Since  is acyclic, by Fact 2 there exists a vertex pair {  for which  is the only minimal separator in .erefore,  and  are not adjacent in  so that {  is not a cluster of ℋ.Let  be a minimal convex separator of ℋ for { .By Lemma 4,  separates  and  in .Since  is the only minimal vertex separator for {  in , one has that   .On the other hand, since  is convex by eorem 8 and since  is a cover of ℋ,  is also a convex separator of ℋ for { .Finally, since  is a minimal convex separator of ℋ for { , one has   .
Example 10.Consider again the (hyper)graph ℋ of Figure2and the convexity space  on ℋ that only contains ∅, , the singletons and the four sets {  , {  , {  and { .e MVCSs of ℋ are { and {  and the cluster hypergraph is   {{   {   { .Since  is an acyclic hypergraph and its edges are all convex sets,  satis�es Axioms 1 and 2. Note that the MVCSs of ℋ are exactly the MVSs of .
) Let ℋ be a connected hypergraph on ,  a decomposable convexity space on ℋ, and  the cluster hypergraph of ℋ.A nonempty subset  of  is convex if and only if Proof of ("if ").By condition (a),  equals the vertex set of GR(  and, by condition (b), ⟨   for every edge of GR( .By (2), one has First of all, recall that for any subset  of , one has that  is a subset of the vertex set of GR( .
(a)  equals the vertex set of GR( , and (b) every edge of GR(  is a convex set.
Fact 6.If  is an edge of  ℋ, then   ℋ,    and, hence, ⟨   .It follows that every edge of  ℋ is a -convex set.Fact 7. If  is cluster of ℋ (i.e.,  is a partial edge of  ℋ), then GR(  ℋ,    and, hence, ⟨    (.