# On the Line Graph for Zero-Divisors ofC(X)

Authors: Ghada AlAfifi; Emad Abu Osba
Journal: International Journal of Combinatorics
 (2013)

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## Abstract

Let beX a completely regular Hausdorff space and let C(X) be the ring of all continuous real valued functions defined on X. In this paper, the line graph for the zero-divisor graph of C(X) is studied. It is shown that this graph is connected with diameter less than or equal to 3 and girth 3. It is shown that this graph is always triangulated and hypertriangulated. It is characterized when the graph is complemented. It is proved that the radius of this graph is 2 if and only if X has isolated points; otherwise, the radius is 3. Bounds for the dominating number and clique number are also found in terms of the density number of X.


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## Body

## 1. Introduction

LetX be a completely regular Hausdorff space and C(X) the ring of all continuous real valued functions defined on X.For eachf∈C(X), let Z(f)={x∈X:f(x)=0}, coz(f)=X∖Z(f), Supp(f)=ClXcoz(f), and Ann(f)={g∈C(X):fg=0}.For all notations and undefined terms concerning the ringC(X), the reader may consult [1].If|X|=1, then C(X) is a field isomorphic to ℝ. So we will assume that |X|>1.LetR be a commutative ring. Z(R) is the set of zero-divisors of R, and Z*(R)=Z(R)∖{0}. The zero-divisor graph of R, Γ(Z*(R)), usually written as Γ(R), is the graph in which each element of Z*(R) is a vertex, and two distinct vertices f and g are adjacent if and only if fg=0. For further details about this graph, see [2] and the survey [3] for a list of references.The line graph of a graphG, denoted by L(G), is a graph whose vertices are the edges of G and two vertices of L(G) are adjacent wherever the corresponding edges of G are incident to a common vertex; see [4]. In this case, if a,b are adjacent vertices in G, then [a,b] is a vertex in L(G). For any undefined terms in graph theory, the reader may consult [5].The zero-divisor graph forC(X) was introduced and studied in [6]. A more general study for reduced rings was done in [7].In this paper we will study the line graph for the zero-divisor graphΓ(C(X)).An elementf∈Z*(C(X)) if and only if IntXZ(f)≠ϕ. Let f,g∈Z*(C(X)). Then [f,g] is a vertex in L(Γ(C(X))) if fg=0. Since Γ(C(X)) is an undirected graph, then [f,g]=[g,f]. We will study when L(Γ(C(X))) is connected and calculate its diameter and girth. We will show that L(Γ(C(X))) is always triangulated and hypertriangulated and characterize when L(Γ(C(X))) is complemented. We will find the radius and give bounds for the dominating and clique numbers.


## 2. Connectedness

LetG be a graph and let u and v be two distinct vertices in G. The distance  d(u,v) between u and v is the length of the shortest path joining them in G; if no such path exists, we set d(u,v)=∞. The associate number  e(u) of a vertex u of a graph G is defined to be e(u)=max{d(u,v):u≠v}. A vertex v is center in G if e(v)≤e(u) for any vertex u∈V(G). The radius of G is defined to be ρ(G)=min{e(v):v∈V(G)} and the diameter of G is diam(G)=max{e(v):v∈V(G)}. The graph G is connected if any two of its vertices are linked by a path in G; otherwise G is disconnected. In this section, we will show that L(Γ(C(X))) is connected, and we will also calculate its diameter and radius.It was shown in [6] that Γ(C(X)) is connected with diam(Γ(C(X)))≤3. We now show that a similar result also holds for L(Γ(C(X))).Theorem 1.
If|X|>1 holds, then L(Γ(C(X))) is connected with
(1)diam(L(Γ(C(X))))≤3.Proof.
IfX={a,b}, [f1,f2] and [g1,g2] are nonadjacent vertices in L(Γ(C(X))), then fi≠gj for i,j=1,2. We may assume that coz(f1)=coz(g1)={a} and coz(f2)=coz(g2)={b}.
Thus we have the path[f1,f2]_[f1,g2]_[g1,g2]. So L(Γ(C(X))) is connected with diam(L(Γ(C(X))))=2.
Assume that|X|>2 holds, and [f1,f2] and [g1,g2] are nonadjacent vertices in L(Γ(C(X))). We have 3 cases:
Case I.
f
i
g
j
=
0 for some i,j∈{1,2}. In this case, [f1,f2]_[fi,gj]_[g1,g2] is a path in L(Γ(C(X))).
Case II.
f
i
g
j
≠
0 for all i,j∈{1,2}, but d(fi,gj)=2 for some i,j∈{1,2}. In this case, there exists h∈Z*(C(X))∖{fi,gj} such that fi_h_gj is a path in Γ(C(X)). So [f1,f2]_[fi,h]_[h,gj]_[g1,g2] is a path in L(Γ(C(X))). Note that if there exist h,k∈Z*(C(X)) such that [f1,f2]_[h,k]_[g1,g2], then hk=0 and h=fi for some i∈{1,2} and k=gj for some j∈{1,2} or h=gi for some i∈{1,2} and k=fj for some j∈{1,2}, which contradicts the fact that figj≠0 for all i,j∈{1,2}.
Case III.  d(fi,gj)=3 for all i,j∈{1,2}. This case will not occur since otherwise, if d(f1,gj)=3 holds for j=1,2 and f1f2=0, then coz(f2)⊆Z(f1), and so coz(f2)⊆IntXZ(f1). Thus coz(f2)∩IntXZ(gj)=ϕ for j=1,2; hence, coz(f2)⊆Supp(gj), j=1,2, which implies that coz(f2)∩coz(gj)≠ϕ for j=1,2. Moreover, IntXZ(gj)⊆IntXZ(f2) for j=1,2, and so ϕ≠IntXZ(gj)=IntXZ(gj)∩IntXZ(f2) for j=1,2. Hence d(f2,gj)=2 for j=1,2; see Lemma 1.2 in [6], a contradiction.
Sincediam(Γ(C(X)))≤3, we conclude that L(Γ(C(X))) is connected with diam(L(Γ(C(X))))≤3.Using Theorem1 and Lemma  1.2 in [6], we can conclude the following.Corollary 2.
Assume that|X|>2 holds. Assume further that [f1,f2] and [g1,g2] are distinct elements in L(Γ(C(X))). Then(1)
d([f1,f2],[g1,g2])=1 if and only if fi=gj for some i,j∈{1,2};(2)
d([f1,f2],[g1,g2])=2 if and only if fi≠gj for all i,j∈{1,2}, and Z(fi)∪Z(gj)=X for some i,j∈{1,2};(3)
d([f1,f2],[g1,g2])=3 if and only if fi≠gj for all i,j∈{1,2}, and Z(fi)∪Z(gj)≠X for all i,j∈{1,2}.In fact, we can be more precise concerning the diameter ofL(Γ(C(X))).Theorem 3.
Assume|X|>1 holds. Then L(Γ(C(X))) is connected with diam(L(Γ(C(X))))=2 if |X|=2 or 3; otherwise diam(L(Γ(C(X))))=3.Proof.
Assume|X|=2 or 3. Let [f1,f2] and [g1,g2] be nonadjacent vertices in L(Γ(C(X))). Then |coz(fi)|=1 for at least one i∈{1,2}. We may assume |coz(f1)|=1. But Z(g1)∪Z(g2)=X; hence, coz(f1)⊂Z(gj) for some j∈{1,2}; that is, f1gj=0 for some j∈{1,2}, which implies that d([f1,f2],[g1,g2])=2 by the above corollary. Since this is true for every two nonadjacent vertices in L(Γ(C(X))), then diam(L(Γ(C(X))))=2 whenever |X|=2 or 3.
Assume that|X|≥4 holds. Pick four distinct points x1, x2, x3, and x4 in X. Since X is a completely regular Hausdorff space, there exist four mutually disjoint open sets Ui, where i∈{1,2,3,4} and xi∈Ui. Consider the functions hi∈Z*(C(X)) such that hi(xi)=1, and hi(X∖Ui)=0 for each i∈{1,2,3,4}. Clearly hihj=0, whenever i≠j. Consider the functions k1=h1+h2, k2=h3+h4, k3=h1+h3 and k4=h2+h4. Obviously ki∈Z*(C(X)) for each i∈{1,2,3,4} and k1k2=k3k4=0; hence, [k1,k2], [k3,k4] are two distinct vertices in L(Γ(C(X))) with ki≠kj whenever i≠j and kikj≠0 for every i∈{1,2} and every j∈{3,4}. By the above corollary, d([k1,k2],[k3,k4])=3. Hence diam(L(Γ(C(X))))=3.Now, we calculate the radius ofL(Γ(C(X))). It was shown in [6] that if f∈Z*(C(X)), then
(2)e(f)={2ifcoz(f)isasingleton3otherwise.A similar result holds forL(Γ(C(X))).Theorem 4.
If[f1,f2] is a vertex in L(Γ(C(X))), then
(3)e([f1,f2])={2ifcoz(fi)isasingletonforsomei∈{1,2}3otherwise.Proof.
It is clear thate([f1,f2])≠1 holds for any vertex [f1,f2] in L(Γ(C(X))).
Assumecoz(f1)={a}. Let [g1,g2] be a vertex in L(Γ(C(X))) that is not adjacent to [f1,f2]. Then fi≠gj for all i,j∈{1,2}. But X=Z(g1)∪Z(g2) and so a∈Z(gj) for some j∈{1,2}. Thus f1gj=0 and [f1,f2]_[f1,gj]_[g1,g2] is a path in L(Γ(C(X))), and d([f1,f2],[g1,g2])=2. Hence e([f1,f2])=2. Now assume that a1,a2∈coz(f1) and a3,a4∈coz(f2). Let U1,U2,U3,U4 be mutually disjoint open sets in X such that ai∈Ui for each i and let Vi be an open set in X such that ai∈Vi⊆ClXVi⊆Ui for each i. Define h1,h2∈C(X) such that h1(a1)=1, h1(X∖V1)=0, h2(a3)=1, and h2(X∖V3)=0. Let g1=h1∨h2. Then g1≠fi and g1fi≠0 for each i∈{1,2}. Similarly define k1,k2∈C(X) such that k1(a2)=1, k1(X∖V2)=0, k2(a4)=1, and k2(X∖V4)=0. Let g2=k1∨k2. Then g2≠fi and g2fi≠0 for each i∈{1,2}. Moreover, g1g2=0, so g1,g2∈Z*(C(X)) and it follows by Corollary 2 that d([f1,f2],[g1,g2])=3, and since diam(L(Γ(C(X))))≤3, we have e([f1,f2])=3.Corollary 5.
If|X|>1 holds, then ρ(L(Γ(C(X))))=2 if and only if X has isolated points. Otherwise ρ(L(Γ(C(X)))=3.


## 3. Cycles

Acycle in a graph G is a closed walk such that no vertex, except the initial and the final vertex, appears more than once, while the girth of G, which is denoted by gr(G), is the length of the shortest cycle in G. If a graph G has no cycles, then its girth gr(G)=∞. If u and v are two vertices in G, by c(u,v), we mean the length of the smallest cycle containing u and v, and  we write c(u,v)=∞ if there is no cycle containing u and v. An edge which joins two vertices of a cycle but is not itself an edge of the cycle is a chord of that cycle. A graph G is chordal if every cycle of length greater than three has a chord. A graph G is called triangulated if each vertex in G is a vertex of a triangle. A graph G is called hypertriangulated if each edge in G is an edge of a triangle.In this section, we will calculate the girth ofL(Γ(C(X))), find the shortest cycle containing two vertices, and show that L(Γ(C(X))) is never chordal. We will also show that L(Γ(C(X))) is always triangulated and hypertriangulated.It was shown in [6] that if X has at least 3 points, then gr(Γ(C(X)))=3. A similar but yet stronger case holds for L(Γ(C(X))).Theorem 6.
If|X|>1 holds, then L(Γ(C(X))) is triangulated and hypertriangulated.Proof.
Let[f,g] be any vertex in L(Γ(C(X))).
Then[f,g]_[2f,g]_[3f,g]_[f,g] is a triangle in L(Γ(C(X))). Thus L(Γ(C(X))) is triangulated. Now let [f,g]_[f,h] be any edge in L(Γ(C(X))). Then [f,g]_[f,h]_[f,rh]_[f,g] is a cycle in L(Γ(C(X))) for some r∈ℝ∖{0,1} such that g≠rh, and so [f,g]_[f,h] is an edge in a triangle. Thus L(Γ(C(X))) is hypertriangulated.Corollary 7.
If|X|>1 holds, then gr(L(Γ(C(X))))=3.Example 8.
LetX={a,b,c} with the discrete topology. Then Γ(C(X)) is not triangulated nor hypertriangulated. Let
(4)f(x)={0x=a,1otherwise.
Iff_g_h_f is a triangle, then g=0 or h=0; hence, f is not a vertex in any triangle in Γ(C(X)). For any k∈Z*(C(X)) such that fk=0, the edge [f,k] is not an edge in any triangle in Γ(C(X)), but [f,k] as a vertex in L(Γ(C(X))) is a vertex in the triangle [f,k]_[f,2k]_[f,3k]_[f,k] and the edge [f,k]_[f,h] is an edge in the triangle [f,k]_[f,h]_[f,rh]_[f,k] for some r∈ℝ.We now find the length of the shortest cycle containing any two distinct vertices inL(Γ(C(X))).Theorem 9.
Let[f1,f2] and [g1,g2] be two distinct vertices in L(Γ(C(X))). Then(1)
c([f1,f2],[g1,g2])=3 if and only if fi=gj  for some i,j∈{1,2};(2)
c([f1,f2],[g1,g2])=4 if and only if fi≠gj  for all i,j∈{1,2} and for some i∈{1,2}, figj=0 for all j∈{1,2} or f1gi=0 and f2gj=0 where {i,j}={1,2};(3)
c([f1,f2],[g1,g2])=5 if and only if fi≠gj  for all i,j∈{1,2} and for only one i∈{1,2}, figj=0 for only one j∈{1,2};(4)
c([f1,f2],[g1,g2])=6 if and only if fi≠gj  for all i,j∈{1,2} and figj≠0 for all i,j∈{1,2}.Proof.
(1) Assumec([f1,f2],[g1,g2])=3 holds. It is clear that if there is a triangle containing [f1,f2] and [g1,g2], then fi=gj for some i,j∈{1,2}.
For the converse assume thatf1=g1. Then there exists r∈ℝ∖{0} such that rg2∉{f1,f2,g2}, and so
(5)[f1,f2]_[f1,g2]_[f1,rg2]_[f1,f2]
is a cycle of length 3 in L(Γ(C(X))) containing [f1,f2] and [g1,g2]. Thus c([f1,f2],[g1,g2])=3.
(2) Assume thatc([f1,f2],[g1,g2])=4 holds. Then it follows by (1) that fi≠gj holds for all i,j∈{1,2}. Hence we have the cycle
(6)[f1,f2]_[a,b]_[g1,g2]_[c,d]_[f1,f2],
where a,c∈{f1,f2} and b,d∈{g1,g2}. Assume that a=f1, c=f2, b=g1, and d=g2, which implies that f1g1=0 and f2g2=0 hold. Now if a=c=f1, b=g1, and d=g2, then f1g1=0 and f1g2=0. Finally, if a=c=f1, b=d=g2, then [a,b]=[c,d], a contradiction.
For the converse, assumefi≠gj for all i,j∈{1,2}. If f1gj=0 holds for all j∈{1,2}, then
(7)[f1,f2]_[f1,g2]_[g1,g2]_[g1,f1]_[f1,f2]
is a cycle of length 4 in L(Γ(C(X))) containing [f1,f2] and [g1,g2], while if f1g1=0 and f2g2=0 hold, then
(8)[f1,f2]_[f1,g1]_[g1,g2]_[g2,f2]_[f1,f2]
is a cycle of length 4 in L(Γ(C(X))) containing [f1,f2] and [g1,g2]. Hence in both cases we have c([f1,f2],[g1,g2])=4.
(3) Assume thatc([f1,f2],[g1,g2])=5 holds. Then fi≠gj for all i,j∈{1,2}. If figj≠0 for all i,j∈{1,2}, and we have the cycle
(9)[f1,f2]_[k1,l1]_[g1,g2]_[k2,l2]_[k3,l3]_[f1,f2]
of length 5 in L(Γ(C(X))), then k1∈{f1,f2} and l1∈{g1,g2}. But k1l1=0, contradicting the assumption. Similarly we will have a contradiction, if we have the cycle
(10)[f1,f2]_[k1,l1]_[k2,l2]_[g1,g2]_[k3,l3]_[f1,f2].
Thus we must have figj=0 for only one i∈{1,2} and only one j∈{1,2}.
For the converse, assumefi≠gj for all i,j∈{1,2} and for only one i∈{1,2}, figj=0 for only one j∈{1,2}, say f1g1=0. By (1) and (2), there is no cycle of length 3 or 4 containing both [f1,f2] and [g1,g2]. There exists r∈ℝ∖{0} such that rg1∉{f1,f2,g1}, and so the cycle
(11)[f1,f2]_[f1,g1]_[g1,g2]_[g2,rg1]_[rg1,f1]_[f1,f2]
is of length 5 in L(Γ(C(X))) containing [f1,f2] and [g1,g2]. Hence we have c([f1,f2],[g1,g2])=5.
(4) Ifc([f1,f2],[g1,g2])=6 holds, then by (1), (2), and (3) fi≠gj holds for all i,j∈{1,2} and figj≠0 for all i,j∈{1,2}.
For the converse, assume thatfi≠gj for all i,j∈{1,2} and figj≠0 for all i,j∈{1,2}. By the previous steps c([f1,f2],[g1,g2])>5. It follows by Corollary 2 that d([f1,f2],[g1,g2])=3, and so there exists h∈Z*(C(X))∖{f1,f2,g1,g2} such that [f1,f2]_[f1,h]_[h,g1]_[g1,g2] is a path in L(Γ(C(X))). There exists r∈ℝ∖{0} such that rh∉{f2,g2,h}. The cycle
(12)[f1,f2]_[f1,h]_[h,g1]_[g1,g2]_[g1,rh]_[rh,f1]_[f1,f2]
is of length 6 in L(Γ(C(X))) containing [f1,f2] and [g1,g2]. Hence we have c([f1,f2],[g1,g2])=6.Theorem 10.
The graphL(Γ(C(X))) is never chordal.Proof.
This is true since forf∈Z*(C(X)) there exists g∈Z*(C(X)), with fg=0, and hence [f,g]_[g,2f]_[2f,2g]_[2g,f]_[f,g] is a cycle of length 4 in L(Γ(C(X))), where no chord can be added.


## 4. Dominating Sets

A proper prime ideal in a ringR that contains no smaller prime ideal is called a minimal prime ideal, and the set of all minimal prime ideals in R will be denoted by Min(R). For any subset S of a ring R, we define the hull of S to be h(S)={P∈Min(R):S⊂P}.LetX be a topological space. The density of X, denoted by d(X), is the smallest cardinal number of the form |A|, where A is a dense subset of X. The weight of X, denoted by ϖ(X), is the smallest cardinality of the form |ß|, where ß is a base for X. The cellularity of the space X is c(X)=sup{|F|:F is a family of pairwise disjoint nonempty open subsets of X}. A zero-set Z in X is said to be a middle zero-set if there exist two proper zero-sets E and F such that Z=E∩F and E∪F=X. A space X is middle P-space if every nonempty middle zero-set in X has nonempty interior. Let ℵ0 denote the infinite countable cardinal number, and let ℵ1 denote the first infinite uncountable cardinal number. If ℵn is an infinite cardinal number, then let ℵn+1=2ℵn.A simple graphG in which all the vertices of G are pairwise adjacent is called complete graph. A complete graph on n vertices is denoted by Kn. A maximal complete subgraph of a graph G is called a clique. The clique number of G is ω(G)=sup{|H|:H is a complete subgraph of G}. In a graph G, a dominating set is a set of vertices A such that every vertex outside A is adjacent to at least one vertex in A. The dominating number of a graph G, denoted by dt(G), is the smallest number of the form |A|, where A is a dominating set. For distinct vertices a and b in a graph G, we say that a and b are orthogonal, written as a⊥b if a and b are adjacent and there is no vertex c of G which is adjacent to both a and b; see [8]. A graph G is called complemented, if for each vertex a of G, there is a vertex b of G (called a complement of a) such that a⊥b.In this section we will estimate the dominating number and the clique number forL(Γ(C(X))) and characterize when L(Γ(C(X))) is complemented.Note that ifD is a dominating set for L(Γ(C(X))), then for each vertex [f,g] in L(Γ(C(X))), we have [f,g]∈D or [f,h]∈D or [g,k]∈D for some h,k∈Z*(C(X)).Lemma 11.
LetD be a dominating set in L(Γ(C(X))). Then |D|≥ℵ1.Proof.
Note first that|Z*(C(X))|≥ℵ1, since if f∈Z*(C(X)), then rf∈Z*(C(X)) for each r∈ℝ∖{0}. Also note that if f∈Z*(C(X)), then |Ann(f)|≥ℵ1, since if g∈Ann(f)∖{0}, then rg∈Ann(f)∖{0} for each r∈ℝ∖{0}.
Now, if for eachf∈Z*(C(X)), there exists g∈Ann(f)∖{0} such that [f,g]∈D, then |D|≥|Z*(C(X))|≥ℵ1, while if there exists f∈Z*(C(X)) such that [f,g]∉D for each g∈Ann(f)∖{0}, then since D is a dominating set, we must have [g,hg]∈D, for each g∈Ann(f)∖{0} and some hg∈Ann(g)∖{0}. Thus |D|≥|Ann(f)|≥ℵ1.Theorem 12.
If|X|>1 holds, then d(X)≤dt(L(Γ(C(X)))).Proof.
LetD be any dominating set for L(Γ(C(X))).
For every vertex[f,g]∈D, choose xf∈IntXZ(f), yf∈coz(f), zg∈IntXZ(g) and wg∈coz(g). Let A={xf,yf,zg,wg:[f,g]∈D}.
Claim.
A is dense in X.
It is sufficient to prove that for any vertexf in Γ(C(X)), A∩IntXZ(f)≠ϕ. Let [f,g] be a vertex in L(Γ(C(X))). Then we have 3 cases.
Case I. Consider [f,g]∈D. Then clearly, A∩IntXZ(f)≠ϕ.
Case II. Consider [f,g]∉D, but [f,h]∈D for some h∈Z*(C(X)). Then again A∩IntXZ(f)≠ϕ.
Case III. Consider [f,g]∉D, but [g,h]∈D for some h∈Z*(C(X)). Then coz(g)⊆IntXZ(f), since fg=0. But A∩coz(g)≠ϕ; hence, A∩IntXZ(f)≠ϕ.
ThusA is dense in X, and so d(X)≤|A|≤4|D|=|D|, because |D|≥ℵ1. Since this is true for any dominating set, we have d(X)≤dt(L(Γ(C(X)))).Theorem 13.
If|X|>1 holds, then ϖ(X)≤dt(L(Γ(C(X)))).Proof.
LetD be any dominating set for L(Γ(C(X))).
LetB={coz(f), IntXZ(f), coz(g), IntXZ(g):[f,g]∈D}. Let U be any non-empty proper open set in X, and let a∈U, b∉U. There exists an open sets K and W in βX such that U=K∩X and a∈W⊆ClβXW⊆K. Define h∈C(βX) such that h(βX∖K)=1 and h(ClβXW)=0. This implies that a∈W⊆ClβXW⊆Z(h)⊆K, and so a∈W∩X⊆Z(h)∩X⊆K∩X=U and b∈coz(h). Let g=h∣X. Then a∈W∩X⊆Z(h)∩X=Z(g). Define f∈C(X) such that f(a)=1 and f(X∖W∩X)=0. Then clearly f,g∈Z*(C(X)) and fg=0.
We have 3 cases.
Case I. Consider [f,g]∈D. Then clearly, a∈coz(f)⊆U.
Case II. Consider [f,g]∉D, but [f,h]∈D for some h∈Z*(C(X)). Then again a∈coz(f)⊆U.
Case III. Consider [f,g]∉D, but [g,h]∈D for some h∈Z*(C(X)). Then a∈IntXZ(g)⊆U.
ThusB is a base for X, and so ϖ(X)≤|B|≤4|D|=|D|. Since this is true for any dominating set, we have ϖ(X)≤dt(L(Γ(C(X)))).Using the argument in the last proof, one can deduce easily the following corollary.Corollary 14.
If|X|>1 holds, and D is a dominating set in L(Γ(C(X))), then the set {coz(f), IntXZ(f), coz(g), IntXZ(g):[f,g]∈D} is a base for open sets in X.We now relate dominating sets inΓ(C(X)) with dominating sets in L(Γ(C(X))).Theorem 15.
If|X|>1 holds, and D is a dominating set in L(Γ(C(X))), then the set D′={f,g:[f,g]∈D} is a dominating set for Γ(C(X)).Proof.
Letf∈Z*(C(X))∖D′. Then [f,g]∉D for all g∈Ann(f)∖{0}. Since D dominates L(Γ(C(X))), then for each g∈Ann(f)∖{0}, there exists hg∈Ann(g)∖{0} such that [g,hg]∈D. Thus g∈D′ and f is adjacent to g. Hence D′ dominates Γ(C(X)).Theorem 16.
LetD be a dominating set in Γ(C(X)) and let D′={[f,g]:f∈D and g∈Ann(f)∖{0}}. Then D′ dominates L(Γ(C(X))).Proof.
Let[f,g] be a vertex in L(Γ(C(X))) that is not in D′. Then f,g∉D. Since D dominates Γ(C(X)), there exists h∈D such that fh=0. Thus [f,h]∈D′ and [f,g] is adjacent to [f,h]. Hence D′ dominates L(Γ(C(X))).LetD be a dominating set in Γ(C(X)) such that dt(Γ(C(X)))=|D|, and let D′={[f,g]:f∈D and g∈Ann(f)∖{0}}. It was shown in [6] that d(X)≤dt(Γ(C(X)))≤ϖ(X), and so |D|≤ϖ(X) and |D′|≤∑f∈D|Ann(f)|≤|D||C(X)|, since |Ann(f)|≤|C(X)|. Thus |D′|≤ϖ(X)|C(X)|.Theorem 17.
LetX be an infinite space with d(X)=ℵn.
Thenℵn≤dt(L(Γ(C(X))))≤ℵn+1.Proof.
We have shown in Theorem12 that ℵn=d(X)≤dt(L(Γ(C(X)))), and by the remark after Theorem 16, dt(L(Γ(C(X))))≤ϖ(X)|C(X)|. But it is known that |C(X)|≤2d(X)=2ℵn, see [9], and ϖ(X)≤2d(X)=2ℵn, see [10], and so dt(L(Γ(C(X))))≤ϖ(X)|C(X)|≤2ℵn2ℵn=2ℵn=ℵn+1.Corollary 18.
LetX be an infinite space with d(X)=ℵn. If d(X)<ϖ(X), then ϖ(X)=dt(L(Γ(C(X))))=ℵn+1.Proof.
Sinceℵn=d(X)<ϖ(X), then using Theorem 13, we have ℵn+1=2ℵn≤ϖ(X)≤dt(L(Γ(C(X))))≤ℵn+1.Corollary 19.
IfX is an infinite space, then d(X)≤ϖ(X)≤dt(L(Γ(C(X))))≤2d(X).Example 20.
Using Lemma11 and Theorem 16, we get that dt(L(Γ(C(ℝ))))=ℵ1, since d(ℝ)=ℵ0. Similarly dt(L(Γ(C(ℕ))))=ℵ1.Example 21.
ℕ is dense in βℕ and so d(βℕ)=ℵ0. But note that ℵ0=d(βℕ)<ℵ1=ϖ(βℕ). So dt(L(Γ(C(βℕ))))=ℵ1.We now investigate cliques and the clique number inL(Γ(C(X))).Theorem 22.
IfK is a finite clique in L(Γ(C(X))), then |K|=3.Proof.
IfK={[f,g],[g,h]} is a clique, then there exists r∈ℝ∖{0,1} such that f≠rh. Thus [g,rh] is adjacent to both [f,g] and [g,h], a contradiction. So |K|≥3. If K={[f,g]·[g,h],[f,h]}, then clearly K is a complete subgraph of L(Γ(C(X))). If [i,j] is a vertex in L(Γ(C(X))) that is adjacent to all vertices in K, then [f,g]_[i,j] implies that i=f or i=g. If i=f, then j=h or j=g, since [i,j] is adjacent to [g,h]. So [i,j]=[f,g] or [i,j]=[f,h] and in both cases we have [i,j]∈K. A similar result will also be obtained if i=g. Thus K is a clique with |K|=3. Now assume that K is a clique in L(Γ(C(X))) with |K|>3 and let [f,g]∈K and [h,k]∈K∖{[f,g]} with h≠g and k≠g. So we may assume that k=f since [h,k] is adjacent to [f,g]. Let [i,j]∈K∖{[f,g],[f,h]} with i≠f and j≠f. Then we may assume that i=g and j=h, since [i,j] is adjacent to both [f,g] and [f,h]. Thus {[f,g]·[g,h],[f,h]}⊂K. But as shown earlier any vertex adjacent to all vertices in {[f,g]·[g,h],[f,h]} must be one of them, contradicting the fact that |K|>3. Thus i=f or j=f. Hence {[f,g]:g∈Ann(f)∖{0}}⊆K, and therefore K is an infinite set.The following proposition was proved in [6].Proposition 23.
The graphΓ(C(X)) is triangulated if and only if X has no isolated points, while Γ(C(X)) is hypertriangulated if and only if X is a connected middle P-space.Corollary 24.
(1) Every vertex[f,g] in L(Γ(C(X))) belongs to a finite clique if and only if X is a connected middle P-space.
(2) For everyf∈Z*(C(X)) there exists g∈Z*(C(X)) such that [f,g] is a vertex in a finite clique if and only if X has no isolated points.Proof.
The result follows by the above theorem together with Proposition23.Using Corollary24 together with Proposition 2.2 and Corollary  2.5 in [6], we get the following.Theorem 25.
The following are equivalent.(1)
Γ(C(X)) is complemented.(2)
For allf∈Z*(C(X), there exists g∈Z*(C(X)) such that Z(f)∪Z(g)=X and IntXZ(f)∩IntXZ(g)=ϕ.(3)
For allf∈Z*(C(X)), there exists g∈Z*(C(X)) such that h(g)=h(Ann(f)).(4)
For allf∈Z*(C(X), there exists g∈Z*(C(X)) such that [f,g] is not a vertex in any finite clique in L(Γ(C(X))).(5)
Min(C(X)) is compact.Corollary 26.
IfK is an infinite clique in L(Γ(C(X))), then K={[f,g]:g∈Ann(f)∖{0}} for some f∈Z*(C(X)).Thus iff∈Z*(C(X)) such that |Ann(f)| is maximal, then ω(L(Γ(C(X))))=|Ann(f)|≤|C(X)|≤2d(X).LetF={Oi:i∈Λ} be a family of pairwise disjoint non-empty open subsets of X with maximum cardinality. For every i∈Λ there exists 0≠fi∈C(X) such that Supp(fi)⊆Oi.LetΩ be the power set of F. For each β∈Ω∖{F,ϕ}, define fβ:X→ℝ as follows:
(13)fβ(x)={∑fi(x)x∈ClX(⋃Oi∈βOi),0x∈X∖⋃Oi∈βOi.Lemma 27.
The functionsfβ defined above are well-defined continuous functions.Proof.
The function is well defined, since for eachi∈Λ, fi=0 on the boundary of Oi. Since the functions fi and 0 are continuous on the closed sets ClX(⋃Oi∈βOi) and X∖⋃Oi∈βOi, then it follows that the function is continuous.Theorem 28.
If|X|>1 holds, then 2c(X)≤ω(L(Γ(C(X))))≤2d(X).Proof.
LetF={Oi:i∈Λ} be a family of pairwise disjoint non-empty open subsets of X with |F|=c(X), and for all Oi∈Flet fi be a nonzero continuous function with Supp(fi)⊂Oi. Let Ω be the power set of F∖{O1}. Then for every β∈Ω∖{ϕ} the function gβ defined by
(14)gβ(x)={fi(x)x∈Oi∈β,0otherwise
is a well-defined continuous function; see Lemma 27. Clearly gβ∈Z*(C(X)) and f1gβ=0 for every β∈Ω∖{ϕ}; hence, the induced subgraph H of {[f1,gβ]:β∈Ω∖{ϕ}} in L(Γ(C(X))) is a complete subgraph of L(Γ(C(X))). Since this is true for every family of pairwise disjoint non-empty open subsets of X, then 2c(X)≤ω(L(Γ(C(X)))). The second inequality comes from the fact |C(X)|≤2d(X) and the previous corollary; that is, 2c(X)≤ω(L(Γ(C(X))))≤2d(X).Example 29.
IfX is a discrete space with |X|=ℵm, then ℵm+1=2c(X)≤ω(L(Γ(C(X))))≤2d(X)=ℵm+1. Also ω(L(Γ(C(ℝ))))=ℵ1, since c(ℝ)=d(ℝ)=ℵ0.



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