EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 2, 2023, 1318-1325
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Bounds on Intersection Number in the Join and Corona
of Graphs

Jesrael B. Palco1,∗, Rolando N. Paluga2

1 Department of Physical Sciences and Mathematics, College of Marine and Allied Sciences,
Mindanao State University at Naawan, 9023, Philippines
2 Department of Mathematics, College of Mathematics and Natural Sciences,
Caraga State University , 8600, Philippines

Abstract. In this paper, we provide an upper bound for the intersection number in the join and
corona of graphs. Moreover, we give formulas for the intersection number of Kn ◦G, Pn ◦G, Cn ◦G
and Crn.

2020 Mathematics Subject Classifications: 05C69

Key Words and Phrases: Intersection number, extreme intersection graph, join and corona

1. Introduction

Let S be a set and F = {S1, S2, · · · , Sp}, for some integer p, a nonempty family
of distinct nonempty subsets of S whose union is S. The intersection graph of F is
denoted by Ω(F ) and defined by V (Ω(F )) = F , with Si and Sj adjacent whenever i ̸= j
and Si ∩ Sj ̸= ∅. A graph G is an intersection graph on S if there exists a family F of
subsets of S for which G ∼= Ω(F ). The intersection number ω(G) of a given graph G is
the minimum number of elements in a set S such that G is an intersection graph on S.
The intersection number has been studied by [1]. They obtained the best possible upper
bound for the intersection number of a graph with a given number of points. In [2], Frank
Harary provided an upper bound for the intersection number of a graph G. He showed
that ω(G) ≤ |E(G)|. In [3], the authors provided a lower bound for the intersection
number of a graph G. They showed that log2(|V (G)|+1) ≤ ω(G). Moreover, the authors
provided formulas for the intersection numbers of Pn, Cn, Wn, Fn, Kn, and G+K1 for any
connected graph G. They also defined the concept of an extreme intersection graph. A
graph G is an extreme intersection graph if for any family F of subsets of
S = {1, 2, 3, ..., ω(G)} such that Ω(F ) ∼= G, then S ∈ F .

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4744

Email addresses: jesrael.palco@msunaawan.edu.ph (Jesrael B. Palco), rnpaluga@carsu.edu.ph
(Rolando N. Paluga)

https://www.ejpam.com 1318 © 2023 EJPAM All rights reserved.



J. B. Palco, R. N. Paluga / Eur. J. Pure Appl. Math, 16 (2) (2023), 1318-1325 1319

2. Results

The join of two graphs G and H, denoted by G + H, is the graph with
V (G+H) = V (G)∪V (H) and E(G+H) = E(G)∪E(H)∪{aibj : ai ∈ V (G) and bj ∈ V (H)}.

Theorem 1. Suppose G is not an extreme intersection graph. Then for any graph H,
ω(G+H) ≤ ω(G)ω(H).

Proof. Let G be not an extreme intersection graph. Then there exists a family F1 of
nonempty subsets of a set S1 such that S1 /∈ F1 and Ω(F1) ∼= G. That is, there is an
isomorphism ϕ1 : V (G) → F1 such that ϕ1(x) ̸= S1, for all x ∈ V (G). Let H be any graph
and suppose ω(H) = m. Let S2 = {1, 2, ..,m} and F2 be a nonempty subset of a set S2

for which Ω(F2) ∼= H. That is, there is an isomorphism ϕ2 : V (H) → F2. Let S = S1×S2,
and F = (∪{A × S2 : A ∈ F1}) ∪ (∪{S1 × B : B ∈ F2}). Let ϕ : V (G + H) → F be a
mapping defined by

ϕ(x) =

{
ϕ1(x)× S2, if x ∈ V (G)

S1 × ϕ2(x), if x ∈ V (H).

Let x1, x2 ∈ V (G+H) such that ϕ(x1) = ϕ(x2). The case x1 ∈ V (G) and x2 ∈ V (H)
is not possible. Since ϕ(x1) = ϕ1(x1)×S2 and ϕ(x2) = S1×ϕ2(x2). Consider the following
cases:
Case 1. Suppose x1, x2 ∈ V (G). Then ϕ(x1) = ϕ1(x1) × S2 and ϕ(x2) = ϕ1(x2) × S2.
Note that ϕ(x1) = ϕ(x2), so we have ϕ1(x1) = ϕ1(x2). Since ϕ1 is one to one, x1 = x2.
Case 2. Suppose x1, x2 ∈ V (H). Then ϕ(x2) = S1 × ϕ2(x1) and ϕ(x2) = S1 × ϕ2(x2).
Note that ϕ(x1) = ϕ(x2), so we have ϕ2(x1) = ϕ2(x2). Since ϕ2 is one to one, x1 = x2.
Therefore, ϕ is one to one.

Let u ∈ F . If u = S1×B, B ∈ F2. Since ϕ2 is onto, there exists x ∈ V (H) ⊆ V (G+H)
such that ϕ2(x) = B. Thus, ϕ(x) = S1 × ϕ2(x) = S1 ×B = u. Therefore, ϕ is onto.

If u = A× S2, A ∈ F1. Since ϕ1 is onto, there exists x ∈ V (G) ⊆ V (G+H) such that
ϕ1(x) = A. Thus, ϕ(x) = ϕ1(x)× S2 = A× S2 = u. Therefore, ϕ is onto.

Let x1 and x2 be adjacent in G+H. Consider the following cases:
Case 1. Suppose x1 and x2 are adjacent in G. Then ϕ(x1) = ϕ1(x1) × S2 and
ϕ(x2) = ϕ1(x2)× S2. Now,

ϕ(x1) ∩ ϕ(x2) = (ϕ1(x1)× S2) ∩ (ϕ1(x2)× S2)

= (ϕ1(x1) ∩ ϕ1(x2))× S2

̸= ∅, since ϕ1 preserves adjacency.

Therefore, ϕ(x1) and ϕ(x2) are adjacent in Ω(F ).
Case 2. Suppose x1 and x2 are adjacent in H. Then ϕ(x1) = S1 × ϕ2(x1) and
ϕ(x2) = S1 × ϕ2(x2). Now,

ϕ(x1) ∩ ϕ(x2) = (S1 × ϕ2(x1)) ∩ (S1 × ϕ2(x2))

= S1 × (ϕ2(x1) ∩ ϕ2(x2))



J. B. Palco, R. N. Paluga / Eur. J. Pure Appl. Math, 16 (2) (2023), 1318-1325 1320

̸= ∅, since ϕ2 preserves adjacency.

Therefore, ϕ(x1) and ϕ(x2) are adjacent Ω(F ).
Case 3. Suppose x1 ∈ V (G) and x2 ∈ V (H). Then ϕ(x1) = ϕ1(x1) × S2 and
ϕ(x2) = S1 × ϕ2(x2). Now,

ϕ(x1) ∩ ϕ(x2) = (ϕ1(x1)× S2) ∩ (S1 × ϕ2(x2))

= (ϕ1(x1) ∩ S1)× (S2 ∩ ϕ2(x2))

= ϕ1(x1)× ϕ2(x2), since ϕ1(x1) ⊆ S1 and ϕ2(x2) ⊆ S2

̸= ∅.

Therefore, ϕ(x1) and ϕ(x2) are adjacent Ω(F ).
Let u, v ∈ F . If u = A × S2 and v = S1 × B for some A ∈ F1 and B ∈ F2, then

u = ϕ1(x) × S2 and v = S1 × ϕ2(y) for some x ∈ V (G) and y ∈ V (H). Consequently,
ϕ−1(u) = x ∈ V (G) and ϕ−1(v) = y ∈ V (H). It follows that x and y are adjacent in
G+H.

If u = A1 × S2 and v = A2 × S2, for some A1, A2 ∈ F1 then u = ϕ1(x1)× S2 = ϕ(x1)
and v = ϕ1(x2)×S2 = ϕ(x2), for some x1, x2 ∈ V (G). Consequently, ϕ−1(u) = x1 ∈ V (G)
and ϕ−1(v) = x2 ∈ V (G). Thus, x1 and x2 are adjacent in G.

If u = S1 × B1 and v = S1 × B2, for some B1, B2 ∈ F2 then u = S1 × ϕ1(y1) = ϕ(y1)
and v = S1 × ϕ1(y2) = ϕ(y2) for some y1, y2 ∈ V (H). Consequently, ϕ−1(u) = y1 ∈ V (H)
and ϕ−1(v) = y2 ∈ V (H). Thus, y1 and y2 are adjacent in H. Therefore, ϕ preserves
adjacency.

Hence, Ω(F ) ∼= G+H
Accordingly, ω(G+H) ≤ |S|, since S = S1 × S2. Then |S| = |S1||S2| = ω(G)ω(H).
Hence, ω(G+H) ≤ ω(G)ω(H).

Let G be a connected graph. A subset S of V (G) is a clique if ⟨S⟩ is a complete
graph. A clique M is maximal if a ∈ V (G) −M , then M ∪ {a} is no longer a clique in
G. The clique graph of G, denoted by ζ(G), is the intersection graph of the set of all
maximal cliques of G. The clique order of G, denoted by co(G), is |V (ζ(G))|. That is,
co(G) is the number of maximal cliques in G.

Theorem 2. Let Kn, Pn and Cn be a complete graph, path and cycle, respectively. Then

(i) co(Kn) = 1 , n ≥ 1

(ii) co(Pn) = n− 1 , n ≥ 2

(iii) co(Cn) =

{
1, if n = 3

n, if n ≥ 4

The corona G ◦H of two graphs G and H, is the graph obtained by making n copies
(n is the ordered of G) of H and joining every vertex of the ith copy of H with the vertex
vi of G. For each a ∈ V (G), we denote by Ha the copy of H corresponding to the vertex
a.



J. B. Palco, R. N. Paluga / Eur. J. Pure Appl. Math, 16 (2) (2023), 1318-1325 1321

Theorem 3. Let G be a connected graph and H be any graph. Then

ω(G ◦H) ≤ co(G) + |V (G)| · ω(H).

Proof. Let V (G) = {a1, a2, a3, ..., an} and V (ζ(G)) = {B1, B2, ..., Bco(G)}. For each
i = 1, 2, ..., n, let Fi be a collection of nonempty subsets of Si = {(i, j) : 1 ≤ j ≤ ω(H)}
such that Ω(Fi) ∼= Hai. For each i = 1, 2, ..., n, let ϕi : V (Hai) → Fi be an isomorphism.
Let So = {(0, j) : 1 ≤ j ≤ co(G)} and S =

⋃n
i=0 Si. For each i = 1, 2, ..., n, let Ti = {(0, j)

: ai ∈ Bj , for some j}. Let F = (
⋃n

i=1 Fi)
⋃
{Si

⋃
Ti : 1 ≤ i ≤ n}.

Define a mapping ϕ : V (G ◦H) → F as follows

ϕ(x) =

{
ϕi(x), if x ∈ V (Hai), for some i

Si ∪ Ti, for some i.

Let x1, x2 ∈ V (G ◦H) such that ϕ(x1) = ϕ(x2). Suppose x1 ∈ V (G) and x2 ∈ V (Hai)
for some i. Then x1 ∈ Bj for some j. Thus, (0, j) ∈ ϕ(x1). Now, ϕ(x2) = ϕi(x2) ⊆ Si,
so (0, j) /∈ Sj . This is a contradiction. Therefore, the case x1 ∈ V (G) and x2 ∈ V (Hai) is
not possible. Consider the following cases:
Case 1. Suppose x1, x2 ∈ V (G). Then x1 = ai and x2 = aj . Thus, ϕ(x1) = Si ∪ Ti and
ϕ(x2) = Sj ∪ Tj . Note that (i, 1) ∈ Si ⊆ ϕ(x1) = ϕ(x2). It follows that
(i, 1) ∈ Sj = {(j, 1), (j, 2), ..., (j, ω(H))}. Consequently, i = j. In effect x1 = x2.
Case 2. Suppose x1 ∈ V (Hai) and x2 ∈ V (Haj ). Suppose i ̸= j. Then
ϕ(x1) ∩ ϕ(x2) = ϕi(x1) ∩ ϕj(x2) ⊆ Si ∩ Sj ̸= ∅. This is a contradiction. Hence, i = j.
Consequently, ϕi(x1) = ϕ(x1) = ϕ(x2) = ϕj(x2) = ϕi(x2). Since ϕi is one to one, x1 = x2.
Therefore, ϕ is one to one.

Suppose B ∈ Fi for some i. Since ϕi : V (Hai) → Fi is onto, there exists x ∈ V (Hai)
such that ϕi(x) = B. Consequently, ϕ(x) = ϕi(x) = B. Suppose B = Si ∪ Ti, for some i.
Take x = ai. Then ϕ(x) = ϕ(ai) = B. Hence, ϕ is onto.

Let x1 and x2 be adjacent in G ◦H. Consider the following cases:
Case 1. Suppose x1 and x2 are adjacent in G. Then x1 = ai and x2 = aj , for some i and
j. In effect, ϕ(x1) = Si ∪ Ti and ϕ(x2) = Sj ∪ Tj . Since ai and aj are adjacent in G, there
exists k such that ai, aj ∈ Bk. This implies that (0, k) ∈ Ti and (0, k) ∈ Tj . It follows
ϕ(x1) ∩ ϕ(x2) ̸= ∅. Therefore, ϕ(x1) and ϕ(x2) are adjacent in Ω(F ).
Case 2. Suppose x1 and x2 are adjacent in Hai for some i. Then x1, x2 ∈ V (Hai). It
follows ϕ(x1) = ϕi(x1) and ϕ(x2) = ϕi(x2). Since ϕi preserves adjacency, ϕ(x1) ∩ ϕ(x2) =
ϕi(x1) ∩ ϕi(x2) ̸= ∅. Thus, ϕ(x1) and ϕ(x2) are adjacent in Ω(F ).
Case 3. Suppose x1 = ai and x2 ∈ V (Hai). Then ϕ(x1) = ϕ(ai) = Si ∪ Ti and
ϕ(x2) = ϕi(x2). Since ϕi(x2) ⊆ Si, ϕ(x1) ∩ ϕ(x2) ̸= ∅. Thus, ϕ(x1) and ϕ(x2) are
adjacent in Ω(F ).

Suppose A and B are adjacent in Ω(F ). That is, A ∩ B ̸= ∅. The case A ∈ Fi and
B ∈ Fj , where i, j ̸= 0 and i ̸= j, is not possible, since Si ∩ Sj = ∅ in this case. Consider
the following cases:
Case 1. Suppose A,B ∈ Fi, for some i. Since ϕi is onto, there exists
x1, x2 ∈ V (Hai) such that ϕi(x1) = A and ϕi(x2) = B. Since ϕi preserves adjacency, x1



J. B. Palco, R. N. Paluga / Eur. J. Pure Appl. Math, 16 (2) (2023), 1318-1325 1322

and x2 are adjacent in Hai . It follows that x1 and x2 are adjacent in G ◦H.
Case 2. Suppose A = Si ∪ Ti and B = Sj ∪ Tj for some i, j = 1, 2, 3, ..., n, i ̸= j. Since
A ∩ B ̸= ∅, (Si ∩ Sj) ∪ (Si ∩ Tj) ∪ (Ti ∩ Sj) ∪ (Ti ∩ Tj) ̸= ∅. Note that Si ∩ Sj = ∅,
Si∩Tj = ∅, Ti∩Sj = ∅. Consequently, (Ti∩Tj) ̸= ∅. Moreover, ϕ(ai) = A and ϕ(aj) = B.
Let t ∈ Ti ∩ Tj . Then t ∈ Ti and t ∈ Tj . This implies that t = (0, r) where ai ∈ Br and
t = (0, s) where aj ∈ Bs. Obviously, r = s and ai, aj ∈ Br. It follows that ai and aj are
adjacent in G. Accordingly, ai and aj are adjacent in G ◦H.
Case 3. Suppose A ∈ Fi and B = Sj ∪ Tj for some i and j. Suppose i ̸= j. Then ϕ(a) =
ϕi(a) = A for some a ∈ V (Hai) and ϕ(aj) = B. Since A∩B ̸= ∅, (A∩ Sj)∪ (A∩ Tj) ̸= ∅.
Since A ⊆ Si, A ∩ Tj ⊆ Si ∩ Tj = ∅ and A ∩ Sj ⊆ Si ∩ Sj = ∅. This is a contradiction.
Thus, i = j. Consequently, a ∈ V (Haj ). It follows that a and aj are adjacent in G ◦H.
Hence ϕ preserves adjacency.

Therefore, Ω(F ) ∼= G ◦H.
Accordingly,

ω(G ◦H) ≤ |S|

=
n∑

i=0

|Si|

= |So|+
n∑

i=1

|Si|

= co(G) +
n∑

i=1

ω(H)

= co(G) + n · ω(H)

= co(G) + |V (G)| · ω(H).

Therefore, ω(G ◦H) ≤ co(G) + |V (G)| · ω(H).

Corollary 1. Let G be a connected graph and n ≥ 2. Then ω(Kn ◦G) = 1 + n · ω(G).

Proof. By Theorem 3, ω(Kn◦G) ≤ co(Kn)+|V (Kn)|·ω(G). By Theorem 2, co(Kn) = 1.
Thus,

ω(Kn ◦G) ≤ co(Kn) + |V (Kn)| · ω(G)

= 1 + n · ω(G).

Suppose ω(Kn◦G) < 1+n·ω(G). Let V (Kn) = {a1, a2, ..., an} and for each i, 1 ≤ i ≤ n,
let Gi be the ith copy of G corresponding to the vertex ai. Let F be a collection of subsets
of S = {1, 2, 3, ..., ω(Kn ◦ G)} such that Ω(F ) ∼= Kn ◦ G. Let ϕ : V (Kn ◦ G) → F be an
isomorphism. For each i, 1 ≤ i ≤ n, {ϕ(x) : x ∈ V (Gi)} is a set representation for Gi.
Thus, | ∪x∈V (Gi) ϕ(x)| ≥ ω(Gi) = ω(G). Note that for each i, j, i ̸= j, and each a ∈ Gi



J. B. Palco, R. N. Paluga / Eur. J. Pure Appl. Math, 16 (2) (2023), 1318-1325 1323

and b ∈ Gj , ab /∈ E(Kn ◦ G). Consequently, Ei = ∪x∈V (Gi)ϕ(x) and Ej = ∪x∈V (Gj)ϕ(x)
are disjoint whenever i ̸= j. Now,

| ∪n
i=1 Ei| =

n∑
i=1

|Ei|

≥
n∑

i=1

ω(G)

= n · ω(G).

It follows that the elements of S− (∪n
i=1Ei) are used for the set representation of G. Note

that

|S − (∪n
i=1Ei)| = |S| − |(∪n

i=1Ei)|
≤ ω(Kn ◦G)− n · ω(G), since we suppose ω(Kn ◦G) < 1 + n · ω(G).

< 1.

That is, |S − (∪n
i=1Ei)| = 0. This implies, S = ∪n

i=1Ei. Since a1 and a2 are adjacent,
ϕ(a1)∩ϕ(a2) ̸= ∅. Let t ∈ ϕ(a1)∩ϕ(a2). Then t ∈ ϕ(a1) and t ∈ ϕ(a2) . Since S = ∪n

i=1Ei,
t ∈ Er for some r. Thus t ∈ ϕ(x) for x ∈ V (Gr). Therefore, ⟨{x, a1, a2}⟩ is complete. This
is a contradiction.

Therefore, ω(Kn ◦G) = 1 + n · ω(G).

Corollary 2. Let G be a connected graph and n ≥ 2. Then ω(Pn◦G) = (n−1)+n ·ω(G).

Proof. By Theorem 3, ω(Pn ◦ G) ≤ co(Pn) + |V (Pn)| · ω(G). By Theorem 2,
co(Pn) = n− 1. Thus,

ω(Pn ◦G) ≤ co(Pn) + |V (Pn)| · ω(G)

= (n− 1) + n · ω(G).

Suppose ω(Pn ◦G) < (n− 1) + n · ω(G). Let V (Pn) = {a1, a2, ..., an} , E(Pn) = {aiai+1 :
1 ≤ i ≤ n − 1} and for each i, 1 ≤ i ≤ n, let Gi be the ith copy of G corresponding to
the vertex ai. Let F be a collection of subsets of S = {1, 2, 3, ..., ω(Pn ◦ G)} such that
Ω(F ) ∼= Pn ◦ G. Let ϕ : V (Pn ◦ G) → F be an isomorphism. For each i, 1 ≤ i ≤ n,
{ϕ(x) : x ∈ V (Gi)} is a set representation for Gi. Thus, | ∪x∈V (Gi) ϕ(x)| ≥ ω(Gi) = ω(G).
Note that for each i, j, i ̸= j, and each a ∈ Gi and b ∈ Gj , ab /∈ E(Pn ◦G). Consequently,
Ei = ∪x∈V (Gi)ϕ(x) and Ej = ∪x∈V (Gj)ϕ(x) are disjoint whenever i ̸= j. Now,

| ∪n
i=1 Ei| =

n∑
i=1

|Ei|

≥
n∑

i=1

ω(G)



J. B. Palco, R. N. Paluga / Eur. J. Pure Appl. Math, 16 (2) (2023), 1318-1325 1324

= n · ω(G).

It follows that the elements of S− (∪n
i=1Ei) are used for the set representation of G. Note

that

|S − (∪n
i=1Ei)| = |S| − |(∪n

i=1Ei)|
≤ ω(Pn ◦G)− n · ω(G), since we suppose ω(Pn ◦G) < (n− 1) + n · ω(G).

< n− 1.

Since ai and ai+1 are adjacent, ϕ(ai) ∩ ϕ(ai+1) ̸= ∅, for every i, 1 ≤ i ≤ n − 1. Let
Ai = ϕ(ai) ∩ ϕ(ai+1), 1 ≤ i ≤ n − 1. Since |S − (∪n

i=1Ei)| < n − 1, there exist i, j with
i < j, such that Ai ∩ Aj ̸= ∅. Let t ∈ Ai ∩ Aj . Then t ∈ Ai and t ∈ Aj . It follows that
t ∈ ϕ(ai) and t ∈ ϕ(aj+1). Note that j ≥ i + 1, it follows ai and aj+1 are adjacent. This
is a contradiction.

Hence, ω(Pn ◦G) = (n− 1) + n · ω(G).

Corollary 3. Let G be a connected graph. Then

ω(Cn ◦G) =

{
1 + 3ω(G), if n = 3

n+ n · ω(G), if n ≥ 4.

Proof. By Theorem 3, ω(Cn ◦G) ≤ co(Cn) + |V (Cn)| · ω(G). By Theorem 2,

co(Cn) =

{
1, if n = 3

n, if n ≥ 4

The case n = 3, follows from Corollary 1 and for n ≥ 4,

ω(Cn ◦G) ≤ co(Cn) + |V (Cn)| · ω(G)

= n+ n · ω(G).

Suppose ω(Cn ◦G) < n+ n · ω(G). Let V (Cn) = {a1, a2, ..., an},
E(Cn) = {aiai+1 : 1 ≤ i ≤ n − 1} ∪ {a1an} and for each i, 1 ≤ i ≤ n, let Gi be
the ith copy of G corresponding to the vertex ai. Let F be a collection of subsets of
S = {1, 2, 3, ..., ω(Cn ◦ G)} such that Ω(F ) ∼= Cn ◦ G. Let ϕ : V (Cn ◦ G) → F be an
isomorphism. For each i, 1 ≤ i ≤ n, {ϕ(x) : x ∈ V (Gi)} is a set representation for Gi.
Thus, | ∪x∈V (Gi) ϕ(x)| ≥ ω(Gi) = ω(G). Note that for each i, j, i ̸= j, and each a ∈ Gi

and b ∈ Gj , ab /∈ E(Cn ◦ G). Consequently, Ei = ∪x∈V (Gi)ϕ(x) and Ej = ∪x∈V (Gj)ϕ(x)
are disjoint whenever i ̸= j. Now,

| ∪n
i=1 Ei| =

n∑
i=1

|Ei|



REFERENCES 1325

≥
n∑

i=1

ω(G)

= n · ω(G).

It follows that the elements of S− (∪n
i=1Ei) are used for the set representation of G. Note

that

|S − (∪n
i=1Ei)| = |S| − |(∪n

i=1Ei)|
≤ ω(Cn ◦G)− n · ω(G), since we suppose ω(Cn ◦G) < n+ n · ω(G).

< n.

Since ai and ai+1 are adjacent, ϕ(ai) ∩ ϕ(ai+1) ̸= ∅, for every i, 1 ≤ i ≤ n. Let
Ai = ϕ(ai)∩ ϕ(ai+1), 1 ≤ i ≤ n. Since |S − (∪n

i=1Ei)| < n, there exist i, j with i < j, such
that Ai ∩ Aj ̸= ∅. Let t ∈ Ai ∩ Aj . Then t ∈ Ai and t ∈ Aj . It follows that t ∈ ϕ(ai)
and t ∈ ϕ(aj+1). Note that j ≥ i + 1, it follows ai and aj+1 are adjacent. This is a
contradiction.

Hence, ω(Cn ◦G) = n+ n · ω(G).

Corollary 4. Let n ≥ 3. Then

ω(Crn) =

{
4, if n = 3

2n, if n ≥ 4.

Proof. The proof follows from Corollary 3.

Acknowledgements

The author would like to thank the peer reviewers of the paper and this research is
funded by the Mindanao State University at Naawan.

References

[1] Paul Erdos, A Goodman, and Louis Posa. The representation of a graphing by set
intersections. Canadian Journal of Mathematics, 18:106–112, 1966.

[2] Frank Harary. Graph Theory. Addison-Wesly Publishing Company, Massachusetts,
1972.

[3] Palco J and Paluga R. Intersection number of some graphs. The Mindanawan Journal
of Mathematics, 3:63–75, 2012.