Maximal graphical realization of a topology

Given a topological space, the graphical realizations of it with as many edges as possible, called maximal graphical realizations, are studied here. Every ﬁ nite topological space admits a maximal graphical realization. However, there are graphs which are not maximal graphical realizations of any topology. A tree of odd order is never a maximal graphical realization of a topological space. Maximal graphical realization of a topology is a cycle if and only if it is C 3 . It is shown that chain topologies admit unique maximal graphical realizations. A lower bound for the size of a maximal graphical realization is also obtained.


Introduction
Acharya [1] initiated the study of set-valuations by introducing the notions of set-indexer and topological set-indexer for a given graph.Subsequently, several authors [4], [5], [7], [8], [11], [15] studied set-valuations of graphs and obtained many significant results.In [12,16], the authors investigated topological set-indexers and derived the topological number of certain graphs.The additional study [13] established topological set-gracefulness of certain stars, paths and related graphs.Further, the topological set-gracefulness of subgraphs, especially spanning subgraphs, of a topologically set-graceful graph has been examined in [6].Later in [14], the authors concentrated on topologies formed by the vertex labels of a topological set-indexer of a given graph.The graph then is called a graphical realization of the topology and the topological space is said to be a graphically realizable space.
This paper sheds more light on graphical realizations by exploring graphical realizations with maximum size for a given topological space.Such a graphical realization is called maximal graphical realization and every topology with finitely many open sets has a maximal graphical realization.It is proved that star graph is a maximal graphical realization of a finite topological space if and only if the space is quasi-discrete.Further, no cycle of order greater than three is a maximal graphical realization of a topology.It is shown that chain topologies admit unique maximal graphical realizations and conjectured that the converse is also true.Investigations are also carried out on the lower bound for the size of a maximal graphical realization.

Preliminaries
In this section we include certain definitions and known results needed for the subsequent development of the study.For a nonempty set X, the set of all subsets of X is denoted by 2 X .By A c , we mean, the complement of a set A. We always denote a graph under consideration by G and its vertex and edge sets by V and E respectively.By we mean the subgraph of G induced by the vertices v 1 , . . ., v n .The order and size of a graph G is denoted by o(G) and s(G) respectively.When it is said that two graphs are different we mean they are non-isomorphic.All graphs considered in this paper are simple.Definition 2.1.[2] Let G = (V, E) be a given graph and X be a nonempty set.Then a mapping f : V → 2 X , or f : E → 2 X , or f : V ∪ E → 2 X is called a set-assignment or set-valuation of the vertices or edges or both.Definition 2.2.[2] Let G be a given graph and X be a nonempty set.Then a set-valuation f : In this case, X is called an indexing set of G. Clearly a graph can have many indexing sets and the minimum of the cardinalities of the indexing sets is said to be the set-indexing number of G, denoted by γ(G).The set-indexing number of K 1 is defined to be zero.
where d e is the ceiling function.
and the corresponding set-indexer is called a set-graceful labeling of G. Definition 2.7.[2] A set-indexer f of a graph G with indexing set X is said to be a topological set-indexer (t-set-indexer) if f (V ) is a topology on X and X is called the topological indexing set(t-indexing set) of G.The minimum number among the cardinalities of such topological indexing sets is said to be the topological number (t-number) of G, denoted by τ (G) and the corresponding t-set indexer is called the optimal t-set-indexer of G.A graph G is said to be topologically set-graceful or t-set-graceful if γ(G) = τ (G).
Theorem 2.8.[2] Every graph with at least two vertices has a t-setindexer.
Theorem 2.10.[12] If G 0 is a spanning sub graph of G, then τ (G 0 ) ≤ τ (G).Definition 2.11.[10] A topological space (X, τ ) is said to be finite if the set X is finite.Definition 2.12.[14] Let X be a nonempty set.A topology τ on X is said to be graphically realizable if there exists a graph G = (V, E) and a set-indexer f : V ∪ E → 2 X such that f (V ) = τ .In this case G is said to be a graphical realization of τ .Also the topological space (X, τ ) is called a graphically realizable space.By G τ , we denote the collection of all graphs which realizes (X, τ ).That is, G τ is the set of all graphical realizations of the topological space (X, τ ).Remark 2.13.In [14], it has been shown that any finite topological space (X, τ ); |τ | = n is graphically realizable by K 1,n−1 .It can be achieved by labeling the central vertex of K 1,n−1 with ∅ and the remaining vertices with the n − 1 nonempty elements of τ in any order.Here onwards we refer to it, the star realization of (X, τ ).
Theorem 2.14.[14] Every topology with finitely many open sets is graphically realizable.
Theorem 2.15.[14] A graph G of order n (> 2) realizes only chain topologies if and only if Theorem 2.16.[14] Every graph of order n realizes every chain topology with n open sets.Definition 2.17.[14] A topological space (X, τ ) realized by a graph G is called an optimal space of G if τ (G) = |X|.In this case G is said to be an optimal graphical realization of τ .The collection of all optimal graphical realizations of a topological space (X, τ ) is denoted by Theorem 2.19.[14] Every optimal topological space is T 0 .Theorem 2.20.[13] every open set is closed and vice versa.

Maximal Graphical Realization of a Topology
Definition 3.1.A graph G is said to be a maximal graphical realization of a topological space (X, τ ) if 1. G is a graphical realization of (X, τ ) and 2. s(G) ≥ s(H) whenever H is a graphical realization of (X, τ ) Evidently, a topological space may possess many maximal graphical realizations.Obviously, spaces realized by the complete graphs have unique maximal graphical realization.
The collection of all maximal graphical realizations of a given topological space (X, τ ) is denoted by M τ .Clearly, M τ ⊆ G τ .
Remark 3.2.In contrast with the optimal graphical realization, every finite topological space has a maximal graphical realization.Hence M τ 6 = ∅, even if O τ = ∅.On the other hand, every element of O τ need not be in M τ .For example, consider the topological space (X, τ ); X = {a, b, c}, τ = {X, ∅, {a}, {b}, {a, b}, {a, c}}.Two different elements in O τ are given below: Proof 1.Let H = (U, F ) be a maximal graphical realization of (X, τ ) with t-set-indexer g.Then g(U .
∈ E, then since no more edges can be drawn in G keeping f fixed, there must be an edge Let G be a graph.Any graph H obtained from G by joining atleast one pair of nonadjacent vertices is called an extension of G and we say G is extendable to H.
Following is a useful consequence of theorem 3.3.
Corollary 3.5.Let (X, τ ) be a finite topological space.Any graphical realization of (X, τ ) is extendable to a maximal graphical realization of (X, τ ) with the same vertex labels.
Remark 3.11.There are graphs which are not maximal graphical realizations of any topology (X, τ ).For example, are not maximal graphical realizations of any topology with n open sets (by theorem 2.15).Theorem 3.12.Let G be a maximal graphical realization of a discrete space (X, τ ).Then G is set-graceful as well as t-set-graceful.
Proof 4. For any graphical realization H of the discrete space (X, τ ) we must have n = o(H) = 2 |X| and s(H) ≤ n − 1.By the star realization of (X, τ ) we have Remark 3.13.Since there are many graphs of odd order which are both set-graceful and t-set-graceful, the converse of the above theorem is not true.For, such a graph cannot even be a graphical realization of a discrete space.K 3 is one such graph.
The following is a simple result.
Theorem 3.14.Let G be a set-graceful graph with indexing set X and f be a t-set-indexer of G with the same indexing set X. Then G is a maximal graphical realization of (X, f (V )).
Lemma 3.15.Let (X, τ ) be an optimal space of a graph G.
Proof 5. Let A be the collection of all symmetric differences of the open sets A 1 , . . ., A n of τ ; |τ | = n.Without loss of generality we may assume that A n−1 = X and Theorem 3.16.Let (X, τ ) be a topological space.Then the star realization of (X, τ ) is maximal and optimal if and only if (X, τ ) is discrete |X| by theorem 2.20 and hence (X, τ ) is discrete.Now suppose n ≥ 5 so that |X| ≥ 3. To prove that (X, τ ) is discrete we need only to show that n = 2 |X| .
Suppose n 6 = 2 |X| .Then by lemma 3.15, we have Conversely let (X, τ ) be a discrete space so that |τ | = 2 |X| .Then the star realization K 1,|τ |−1 has 2 |X| − 1 edges so that it is a maximal graphical realization of (X, τ ).Now the optimality of K 1,|τ |−1 follows from theorem 2.20. 2 Theorem 3.17.Let f be a t-set-indexer of the graph G = K n \ E(K 1,3 ) with indexing set X. Then the removal of atmost one set from f (V ) results in a chain topology on X.
is a chain topology, then there is nothing to be proved.If f (V ) is not a chain topology, then there exists -a contradiction.
Let C and D be any two sets in Suppose, on the contrary C ∪D / ∈ {C, D}.Let u, v, w and x be the vertices of Theorem 3.18.Let f be a t-set-indexer of the graph G = K n \ E(K 3 ); n ≥ 4 with indexing set X. Then the removal of atmost one set from f (V ) results in a chain topology.
is a chain topology, then there is nothing to be proved.If f (V ) is not a chain topology, then there exists A, B ∈ f (V ) such that A ∪ B 6 = A and A∪B 6 = B. Since f (V ) is a topology on X, there exists four distinct vertices ) and H = (U, F ) be the graph K n \ E(K 3 ).Then for every t-set indexer f of G, there exists a t-set indexer g of H satisfying f (V ) = g(U ) and conversely.
Since f is a t-set-indexer of G, by theorem 3.17, the removal of atmost one set from f (V ) results in a chain topology.If f (V ) itself is a chain topology, then we can define a set-indexer say g 1 of H by g 1 (u i ) = f (v i ) for i = 1, . . ., n.
Clearly, g 1 (U ) = f (V ) and then g 1 is a t-set-indexer of H. Otherwise, there exists A ∈ f (V ) such that f (V ) \ A is a chain topology.
Clearly there exists B(6 = A) in f (V ) such that A∪B 6 = A, B. Now consider a set-valuation, say g 2 of H defined by g 2 Hence atleast 3 edges should be absent in the subgraph of H induced by the vertices with labels A, B, A ∪ B and Therefore, the edge labels of H under g 2 are distinct so that g 2 is a t-set-indexer of H.
Conversely, let g be a t-set-indexer of H. Then by theorem 3.18, the removal of atmost one set from g(U ) results in a chain topology.If g(U ) itself is a chain topology, then we can define a set-indexer say Otherwise, there exists C ∈ g(U ) such that g(U ) \ C is a chain topology.Evidently, there exists Hence atleast 3 edges should be absent in the subgraph of G induced by the vertices with labels C, D, C ∪ D and Consequently, no more edges can be drawn in K 1,|τ |−1 such that the resulting graph realizes (X, τ ) keeping g fixed.Then by theorem 3.3 we have K A consequence of the above theorem is, Corollary 3.28.No tree of odd order is a maximal graphical realization of a topological space.
Theorem 3.29.Maximal graphical realization of a topology is a cycle if and only if it is C 3 .
Proof 14.Let (X, τ ); |τ | = n be a given topological space and suppose C n ∈ M τ where n ≥ 3.By the star realization of (X, τ ) we have K 1,n−1 ∈ G τ and let g be the corresponding t-set-indexer.Let V = {v 0 , v 1 , . . ., v n−1 } with d(v 0 ) = n − 1 be the vertex set of K 1,n−1 .Then we have g(v 0 ) = ∅ and g(v k ) = X for some v k 6 = v 0 .Since C n ∈ M τ , by corollary 3.5, we can extend K 1,n−1 to a maximal graphical realization G = (V, E) of (X, τ ) just by joining two distinct pendant vertices v i and v j , keeping all the vertex labels fixed.Let f be the corresponding t-set-indexer which is an extension of g.It is claimed that either v i = v k or v j = v k .Suppose v k / ∈ {v i , v j } and let f (v i ) = A and f (v j ) = B. Obviously A and B are nonempty and B 6 = X \ A. Since (v k , v i ), (v k , v j ) / ∈ E and G ∈ M τ , both X \ A and X \ B are edge labels of G under f .But f | K 1,n−1 = g and g corresponds to the star realization of K 1,n−1 .Consequently, there are vertices v a , v b ∈ V \{v i , v j , v k , v 0 } such that f (v a ) = X \A and f ( Because of obvious reasons, v l / ∈ {v 0 , v i , v j , v k , v a , v b }.But then f (v 0 , v l ) = A ⊕ B = f (v i , v j ) -a contradiction.Hence, we must have v k = v i or v k = v j as claimed.Without loss of generality assume that v k = v i so that f (v i ) = A = X.Then f (v i , v j ) = X \ B and X \ B / ∈ f (V ).Suppose G contains a fourth vertex v x other than v 0 , v i and v j .Let f (v x ) = D. Clearly, D / ∈ {X, B, ∅, X \ B}.Since (v i , v x ) / ∈ E and G ∈ M τ , there exists a vertex v y ∈ V \ {v 0 , v i , v j , v x } such that f (v y ) = X \ D.
where '⊕' denotes the binary operation of taking the symmetric difference of the sets in 2 X 2. the restriction maps f | V and f | E are both injective.

Figure 1 :Figure 2 :
Figure 1: Optimal graphical realization Let A be any edge label of H. Then there are vertices a and b in H such that A = g(a) ⊕ g(b).Since g(a), g(b) ∈ f (V ), there are vertices c, d in G such that f (c) = g(a) and f (d) = g(b)
a contradiction.Thus, we have {C, D, C ∪ D, C ∩ D} = {A, B, A ∪ B, A ∩ B}.Since A / ∈ {C, D} we must have A ∈ {C ∪ D, C ∩ D}.If A = C ∪ D, then B ∈ {C, D, C ∩ D} so that B ⊆ A, which is not possible.Therefore, we must have A = C ∩ D. Then B ∈ {C, D, C ∪ D} so that A ⊆ B, which is also not true.Hence we must have C ∪ D = C or C ∪ D = D as claimed and f (V ) \ A is a chain topology.2

2 Theorem 3 .
so that atleast three edges are absent in G[u, v, w, x].Consequently, we must have G[u, v, w, x] ⊆ K 1,3 or G[u, v, w, x] ⊆ K 3 ∪ K 1 .Now we claim that {u, v, w, x} 6 = {v 1 , v 2 , v 3 , v 4 }.Otherwise, s(G[u, v, w, x]) ≥ 4, a contradiction.Thus, we have {C, D, C ∪ D, C ∩ D} = {A, B, A ∪ B, A ∩ B}.Since A / ∈ {C, D} we must have A ∈ {C ∪ D, C ∩ D}.If A = C ∪ D, then B ∈ {C, D, C ∩ D} so that B ⊆ A, which is not possible.Therefore, we must have A = C ∩ D. Then B ∈ {C, D, C ∪ D} so that A ⊆ B, which is also not true.Hence, we must have C ∪ D = C or D, as claimed and f (V ) \ A is a chain topology.19.Let G = (V, E) be the graph K n \E(K 1,3 2 and hence for any A ∈ τ , we have A c ∈ τ .Clearly, n is even and every open set is closed also.Let x, y ∈ X with x 6 = y.By theorem 2.19, every optimal space is T 0 and hence there exists an open set U containing x but not y.Then the disjoint open sets U and X \ U are such that x ∈ U , y / ∈ U , x / ∈ X \ U and y ∈ X \ U .This implies that (X, τ ) is a finite T 1 space and hence discrete.Then |τ | = 2 |X| , a contradiction.Therefore, |A| ≥ n + 1.But all the elements of A except ∅ are edge labels of H. Consequently, s(H) ≥ n.2