OUT-GRAPHIC TOPOLOGY ON DIRECTED GRAPHS

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INTRODUCTION
Graph theory is one of discrete mathematics structures.It is simple and can represent a lot of mathematical combinations.In 1736, L. Euler introduce graph theory for the first time for solving the problem of the Königsberg seven bridges [5].Later, graph theory becomes a fundamental mathematical tool for a large of domain as chemistry, marketing and computers network.
Making a topology on the graphs enriches its structures and gives a more applications as in economy domain, the traffick flow study [2,8,10], medical application and blood circulation [6,9,11,12].
A topology is called an Alexandroff topology if any intersection of open sets is an open set [3,13].The graphic topology is an Alexandroff topology, and so the study of open sets, closed sets, homeomorphism, compactedness can be done through minimal basis.
In 2013, Jafarian Amiri et al. [7] introduced an Alexandroff topology on the set of vertices of a simple graph .This topology is known as graphic topology.In [14], the authors investigated the graphic topology and solved partially an open problem mentioned in [7] (Problem 2 page 658).
In the paper [15] in 2022, Zoman et al. define the Z -graphic topology Z G which conserves the connectivity of the simple graph.Graphic topology was also be defined on fuzzy graph by Alzubaidi et al. in [4].
In the paper [1], Kacimi et al. defined two topologies on the edges set of directed graph called compatible and incompatible edges topologies.In this paper,we introduce the out-graphic topology on directed graph on the vertices set.Our motivation is the investigation of some properties of directed graphs by their corresponding topology.In Section 2, we give some preliminaries about directed graph theory and topology.Also, we introduce a subbasis for the out-graphic topology and give some examples.In section 3, we prove some elementary results.Section 4 is devoted to more properties of the out-graphic topology.In Section 5, functions between digraphs are investigated.Finally, Connectivity or disconnectivity of the out-graphic topology of some graphs are studied.

PRELIMINARIES
In this paper, we will define and study the out-graphic topology on directed graphs.Recall that a directed graph G (or digraph) is a given nonempty set V and a set of ordered pairs E, subset of V × V .This means, a graph is called a directed graph if each edge e ∈ E has a direction.When e = (x, y) ∈ E, x is called the tail of the edge e and y the head of the edge.We also say e is an edge from x to y and we write x → y.
∈ E and there is no multiple edges from x to y, for any two vertices x, y of G.
Definition 2.1.A digraph G = (V, E) is called complete if it is simple and for any distinct x, y ∈ V , there exist a unique edge from x to y and a unique edge from y to x.
Definition 2.2.A digraph G = (V, E) is said an oriented graph if for all x, y ∈ V , at most one of (x, y) and (y, x) is in E. If G is an oriented graph such that for all x, y ∈ V , we have and the int-neighborhood set of x by (2) We remark that That is, x does not dominates any y ∈ V .
In this digraph, the vertex x 1 is an isolated vertex.
In order to define the out-graphic topology for V , we suppose that the digraph G = (V, E) is simple and without isolated vertices.So, for all Let us consider (4) Now, if y ∈ V and since G without isolated vertices, we get I y = / 0. There exists z ∈ I y and so, by (3) there exists z ∈ V such that y ∈ O z .Therefore V ⊂ ∪ x∈V O x and then Hence S out G is a subbasis for a topology of V called out-graphic topology and denoted by T out G .
Example 3.For the graph G given by Then, the basis of the topology T out G is We get For a vertex x of G, the out-degree of x is defined as the cardinal of the out-neighborhood O x and the int-degree of x is the cardinal of the int-neighborhood I x .
The minimum out-degree and the minimum int-degree of a digraph G = (V, E) are given by However, the maximum out-degree and maximum int-degree of G are respectively is finite.That is, I x is a finite set for all x ∈ V .
We remark that any finite digraph is locally finite.
Definition 2.6.Let (V, T ) be a topological space and X ⊂ V .
(iii) The closure of X in V is the smallest closed set of V containing X.It will be denoted by X.
, where x i ∈ V and e i an edge from x i to x i+1 , (ii) Two vertices x and y are said connected if there is a directed path from x to y and a directed path from y to x.
(iii) The directed graph G is called strongly connected if any two distinct vertices are connected.
In this paper, a digraph means a simple locally finite digraph without isolated vertex.

PRELIMINARY RESULTS
Theorem 3.1.Let G be a digraph.Then (V, T out G ) is an Alexandroff space.
Proof.In order to prove that T out G is an Alexandroff topology, it is sufficient to prove that any intersection of elements in the subbasis S out G is an open set.Let U a subset of V and consider this means U ⊂ I z .Since G is locally finite, we deduce that U is a finite set and so Let G be a digraph, then the out-graphic topology T out G of V has a minimal basis The first characterisation of the minimal basis is the following.
Theorem 3.2.Let G be a digraph and let x ∈ V .Then, V x = ∩ y∈I x O y and V x is a finite set.
Proof.Since G is without isolated vertices, is the smallest open set containing x, we get Conversely, since S out G is a subbasis for the topology and V x is the smallest open set containing x, there exists For all y ∈ U, x ∈ O y .Therefore, for all y ∈ U, y ∈ I x .Then, U ⊂ I x and so, The result follows.
Corollary 3.1.Let G be a digraph and let x, y ∈ V two distinct vertices. Proof.
Let G be a digraph and let x, y ∈ V .Then, y ∈ V x if, and only if, Proof.
(i) Suppose that there exists z ∈ V x ∩I x .Since z ∈ V x , we get from Proposition 3.1 But z ∈ I x and so z ∈ I z : this is impossible since the graph is simple.We deduce that Using the above result, we remark that {x} ⊂

SOME PROPERTIES OF GRAPHIC TOPOLOGY
A topological space V is called compact if for any family {A i } i∈I of open sets satisfying there exists a finite set J ⊂ I such that For diagraphs, we have the following result.
Proposition 4.1.Let G = (V, E) be a digraph.Then, (V, T out G ) is a compact topological space if and only if V is finite.
Proof.First, suppose that V is a compact topological space.Consider the minimal basis U G given by (5).U G is an open cover of V , so there exists a finite subcover M of U G .Since it is minimal as basis, M = U G .Therefore, V is finite from (5).
Conversely, if V is finite, from any open cover we have a finite subcover.Hence, the result follows.Proof.The idea is to prove that for all x ∈ U − , we have x ∈ V x ⊂ U − .Let x ∈ U − and y ∈ V x , the smallest open set containing x. From Proposition 3.1, I x ⊂ I y .We set for the graphic topology on G.
Proof.We will prove that Consider an element z ∈ V x ∩ F − .We have and so x ∈ F − and the result follows.
Proposition 4.4.Let G = (V, E) be a finite digraph.Then, the following results hold.
From Proposition 3.2, we have get V x ⊂ {x} and so V x = {x} and the result follows.E 2 ) be two digraphs.We say that G 1 and G 2 are isomorphic if there exists a bijection φ :

ON FUNCTIONS BETWEEN DIGRAPHS
Further more, the two spaces V 1 and V 1 are said homeomorphic if there exists a continuous bijective φ : Without loss of generality we can suppose that U is in the subbasis S G 2 and so U = O y , for some y ∈ V 2 .
Let x = φ −1 (y), we have In general, the converse is not true.Let us consider the following two graphs: They have the same out-graphic topology (the discrete topology) but they are not isomorphic.
We have also the following characterisation of homeomorphic graphic topology spaces.
a function.Then the following properties are equivalent.
(i) φ is an homeomorphism between the topological spaces Proof.First, suppose that φ is an homeomorphism and let x, y ∈ V 1 .If I x ⊂ I y , from the Proposition 5.1 we get I φ (x) ⊂ I φ (y) .Now, we suppose that I φ (x) ⊂ I φ (y) .By using the Proposition 5.1 for the continuous function φ −1 , we get I x ⊂ I y .
Conversely, suppose that the result (ii) is true.
It is clear that the function φ is continuous by using Proposition 5.1.
We want to prove that φ −1 is also continuous.Let x , y ∈ V 2 such that I x ⊂ I y .
We have and so Again, from Proposition 5.1, the function φ −1 is continuous.

GRAPHIC TOPOLOGY AND CONNECTEDNESS
Definition 6.1.Let (X, T ) a topological space.We say that X is connected if whenever X = U ∪ V and U ∩ V = / 0, we have U = / 0 or V = / 0. That is, X can not be the union of two disjoint proper open sets.Definition 6.2.A digraph G = (V, E) is called strongly connected if for all x, y ∈ V there exist a path from x to y and a path from y to x.
For a digraph G = (V, E), we define the connected components as follows.
(ii) V i ∩V j = / 0, for all i = j; (iii) For i = 1, 2, • • • , for all x, y ∈ V i , there exist a path from x to y and a path from y to x.
(iv) For all x ∈ V i , y ∈ V j and i = j, there is no pair of paths: one from x to y and one from y to x.
Then, each subset V i is called connected component of the digraph G.
As a particular cases, a strongly connected digraph has one connected component.Also, a finite digraph has a finite connected components.
When the graph G = (V, E) is undirected, and it is not connected, the connected components are open sets for the graphic topology T G and so, (V, T G ) is a disconnected topological space.
But if G = (V, E) is not strongly connected digraph, the topology T out G can be connected as in the following example.and so, T out G is connected.
In the following example, the graph is strongly connected but T out G is not connected.We have V 1 = / 0 , V 2 = / 0 and V = V 1 V 2 since S out G is a subbasis for the topology T out G .The result follows from the fact that V 1 ∩V 2 ⊂ U 1 ∩U 2 = / 0.
In fact, we have the following result.Proof.Since G is a cycle (x 1 , • • • , x n ).We have V x i = {x i }, for all i = 1, . . ., n and so T out G is the discrete topology.

Definition 2 . 3 .
Let G = (V, E) be a simple digraph.The complement of G is the digraph G = (V, E) defined by (x, y) ∈ E if, and only if (x, y) / ∈ E. Example 1.The complement of the following graph is given by For x ∈ V , we define the out-neighborhood set of x as (1) where V x is the intersection of all open sets containing x, i.e. the smallest open set containing the vertex x.
Finally, we have the following result.Proposition 3.3.Let G be a directed graph and x ∈ V .We have y ∈ {x} if and only if I y ⊂ I x .This means, {x} = {y ∈ V ; I y ⊂ I x }.Proof.y ∈ {x} if and only if, for all open set O containing y, O ∩ {x} = / 0. But, this is equivalent to V y ∩ {x} = / 0. So, y ∈ {x} if and only if x ∈ V y and the result follows by Proposition 3.1.

Example 6 .Example 7 .
This is an example of not strongly connected graph with disconnected out-{c} d {a, c} {b} {c, d} and so, T out G is not connected.In this example, we have a strongly connected graph with connected out-{a , b, c} {a , b, c} {a} a {a} {a} {a , b, c} b {a, b , c} {a, b , c} {b} b {b} {b} {a, b , c} c {a, b, c } {a, b, c } {c} c {c} {c} {a, b, c } then, the topological space (V, T out G ) is connected.Theorem 6.1.Let G = (V, E) be a bipartite digraph, then T out G is disconnected.Proof.Suppose that V = U 1 ∪U 2 , with U 1 ∩U 2 = / 0 and xy ∈ E and (x, y) / ∈ A × B ⇒ (x, y) ∈ B × A. Set V 1 = x∈A O x ⊂ U 2 and V 2 = x∈B O x ⊂ U 1 .

Proposition 6 . 1 .Proposition 6 . 2 .
For a strongly connected digraph G = (V, E), the space (V, T out G ) is disconnected.If G = (x 1 , • • • , x n ) is a cycle of order n ≥ 3 (strongly connected), then T out G is disconnected.