EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 3, 2021, 695-705
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On a Topological Space Generated by Monophonic
Eccentric Neighborhoods of a Graph

Anabel E. Gamorez1,∗, Sergio R. Canoy Jr.2

1 Department of Mathematics and Statistics, College of Science and Mathematics,
Western Mindanao State University, 7000, Zamboanga City, Philippines
2 Department of Mathematics and Statistics, College of Science and Mathematics,
Center of Graph Theory, Algebra, and Analysis-Premier Research Institute of Science and
Mathematics, Mindanao State University-Iligan Institute of Technology, 9200, Iligan City,
Philippines

Abstract. In this paper, we present a way of constructing a topology on a vertex set of a graph
using monophonic eccentric neighborhoods of the graph G. In this type of construction, we char-
acterize those graphs that induce the indiscrete topology, the discrete topology, and a particular
point topology.

2020 Mathematics Subject Classifications: 05C12, 54B05

Key Words and Phrases: Topology, graph, monophonic distance, monophonic eccentric neigh-
borhood

1. Introduction

A metric or distance function in a non-empty set is known to generate a topology on
the set via the family of open balls the metric induces. Indeed, it is well known that every
metric space is a topological space. Topologizing a non-empty set can well be done by
using a family of subsets of the set (as done in a metric space) that will serve as a base of
some topology on the given set. Recently, topologizing the vertex set of a given graph was
done to obtain topological spaces from a given graph. Gervacio and Diesto [2] used the
standard neighborhoods of a graph to construct a topology on its vertex set. Admittedly,
due to its limited circulation, the work is not so popular. This construction, however, was
further studied in [3], [6] and [1].

Nianga and Canoy in [8] presented another way of generating a topology on a graph
using the hop or 2-step neighborhoods of a graph. They further investigated in [9], the
topologies induced by the complement of a graph, the join, corona, composition and the

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.3990

Email addresses: anabel.gamorez@wmsu.edu.ph (A. Gamorez),
sergio.canoy@g.msuiit.edu.ph (S. Canoy Jr.)

http://www.ejpam.com 695 © 2021 EJPAM All rights reserved.



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 696

Cartesian product of graphs. The same construction was also studied by Canoy and
Gimeno [4].

In this paper we construct a topology on a vertex set of a graph using its mono-
phonic eccentric neigbhorhoods and investigate some of the topological structures and
properties of the space generated. Under this construction we, among others, characterize
those graphs that induced the indiscrete topology, the discrete topology and a particular
point topology. For any two vertices u and v in a graph G, the distance dG(u, v) is the
length of a shortest path joining u and v. The open neighborhood of a point u is the set
NG(u) consisting of all points v which are adjacent to u. The closed neighborhood of u is

NG[u] = NG(u)∪{u}. For any A ⊆ V (G), NG(A) =
⋃
v∈A

NG(v) is called the open neighbor-

hood of A and NG[A] = NG(A)∪A is called the closed neighborhood of A. The complement
of NG[A] is denoted by FG[A] that is, FG[A] = V (G)\NG[A]. If A = {v}, then we write
FG[A] = FG[v]. For each v ∈ V (G), N2

G(v) = {u ∈ V (G) : dG(u, v) = 2} is called the open
hop neighborhood of v and N2

G[v] = {v} ∪N2
G(v) is called the closed hop neighborhood of

v. For any A ⊆ V (G), N2
G(A) =

⋃
a∈A

N2
G(a) = {v ∈ V (G) : N2

G(v) ∩ A 6= ∅} is called the

open hop neighborhood of A and N2
G[A] = A ∪ N2

G(A) is the closed hop neighborhood of
A. Denote by F 2

G[A] the complement of N2
G[A], that is, F 2

G[A] = V (G)\N2
G[A]. Recently,

Titus [10] introduced some concepts related to monophonic paths in a graph. A chord of a
path P in a graph G is an edge joining two non-adjacent vertices of P . A P in a graph G
is called a monophonic path if it is chordless. For any two vertices u and v in a connected
graph G, the monophonic distance dmG (u, v) from u to v is defined as the length of a longest
u-v monophonic path in G. The monophonic eccentricity emG (v) of a vertex v in G is the
maximum monophonic distance from v to a vertex of G. The monophonic radius radm(G)
of graph G is radm(G) = min{emG (v) : v ∈ V (G)}. A vertex w in G is a monophonic
eccentric vertex of a vertex v in G if emG (v) = dmG (w, v). In this case, we say that w is a
monophonic eccentric neighbor of v. The set of all monophonic eccentric vertices (neigh-
bors) of v is denoted by N em

G (v). That is, N em
G (v) = {w ∈ V (G) : dmG (w, v) = emG (v)}. The

monophonic eccentric open neighborhood of A ⊆ V (G) given by N em
G (A) =

⋃
a∈A

N em
G (a).

The monophonic eccentric closed neighborhood of A is N em
G [A] = A ∪N em

G (A). The com-
plement of N em

G [A] is F em
G [A] = V (G)\N em

G [A]. If A = {v}, we write F em
G [A] = F em

G [v].
For other basic concepts not defined here, we refer the readers to [5] and [7].

2. Results

The first few results show how a topological space from a given graph G is being
constructed using the monophonic eccentric neighborhoods of the graph.

Lemma 1. Let G be any graph and let A,B ⊆ V (G). Then

N em
G (A ∪B) = N em

G (A) ∪N em
G (B).



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 697

Proof. Clearly, N em
G (A) ⊆ N em

G (A∪B) and N em
G (B) ⊆ N em

G (A∪B). Hence, N em
G (A)∪

N em
G (B) ⊆ N em

G (A ∪ B). Next, let w ∈ N em
G (A ∪ B). Then there exists v ∈ A ∪ B such

that dmG (w, v) = emG (v). Thus, w ∈ N em
G (A) or w ∈ N em

G (B) showing that N em
G (A ∪ B) ⊆

N em
G (A) ∪N em

G (B). Therefore, equality holds.

Lemma 2. Let G be any graph. If A,B ⊆ V (G) and A ⊆ B, then F em
G [B] ⊆ F em

G [A].

Proof. Let v ∈ F em
G [B]. Then v /∈ B and v is not a monophonic eccentric vertex of

any vertex in B, that is dmG (v, b) 6= emG (b) for all b ∈ B. Since A ⊆ B, v /∈ A and v is
not a monophonic eccentric vertex of A, that is, dmG (v, a) 6= emG (a) for all a ∈ A. Thus,
v ∈ F em

G [A]. Therefore, F em
G [B] ⊆ F em

G [A].

Lemma 3. Let G be any graph. If A,B ⊆ V (G) then

F em
G [A ∪B] = F em

G [A] ∩ F em
G [B].

Proof. Since A ⊆ A∪B and B ⊆ A∪B,F em
G [A∪B] ⊆ F em

G [A] and F em
G [A∪B] ⊆ F em

G [B]
by Lemma 2. Thus,

F em
G [A ∪B] ⊆ F em

G [A] ∩ F em
G [B].

Now, let v ∈ F em
G [A] ∩ F em

G [B]. Then v ∈ F em
G [A] and v ∈ F em

G [B]. It follows that v /∈ A,
v /∈ B, v /∈ N em

G (A) and v /∈ N em
G (B). Hence, by Lemma 1, v /∈ A∪B and v /∈ N em

G (A∪B).
Therefore, v ∈ F em

G [A ∪B] and F em
G [A] ∩ F em

G [B] ⊆ F em
G [A ∪B]. Accordingly,

F em
G [A ∪B] = F em

G [A] ∩ F em
G [B].

Note that Lemma 3 can also be proved using Lemma 1. By induction on the number
of sets involved, the next is immediate.

Theorem 1. Let G be any graph. If A1, A2, ..., An are subsets of V (G), then

F em
G

[
n⋃

i=1

Ai

]
=

n⋂
i=1

F em
G [Ai].

Theorem 2. Let G be any graph. The family BemG = {F em
G [A] : A ⊆ V (G)} is a base for

some topology on V (G).

Proof. Note that N em
G [∅] = ∅ and so F em

G [∅] = V (G) ∈ BemG . Now let A,B ⊆ V (G).
By Lemma 3, F em

G [A] ∩ F em
G [B] = F em

G [A ∪ B] ∈ Bem
G . Therefore, BemG is a base for some

topology on V (G).

Henceforth, we denote by τ emG the topology generated by BemG . Also we denote by IG
and DG the indiscrete and the discrete topologies on V (G), respectively.

Theorem 3. Let G be any graph. The family SemG = {F em
G [v] : v ∈ V (G)} forms a subbase

for τ emG .



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 698

Proof. Let SemG = {F em
G [v] : v ∈ V (G)} and let A = {a1, a2, ..., an}. By Lemma

3, F em
G [a1] ∩ F em

G [a2] ∩ ... ∩ F em
G [an] = F em

G [A]. Thus, every element of BemG is a finite
intersection of members of SemG . Therefore, BemG is a subbase of τ emG .

Theorem 4. Let G be any graph of order n ≥ 1. Then τ emG is the indiscrete topology if
and only if G = Kn.

Proof. Suppose that τ emG is the indiscrete topology. Suppose further that G 6= Kn.
Then there exist x, y ∈ V (G) such that dmG (x, y) = emG (x) ≥ 2. Let P = [x1, x2, ..., xk],
where x1 = x and xk = y, be an x-y monophonic path. Then k ≥ 3 and x2 /∈ N em

G [x].
Hence, x2 ∈ F em

G [x] 6= ∅. Since x, y /∈ F em
G [x], it follows that F em

G [x] 6= V (G). Therefore,
τ emG is not the indiscrete topology, a contradiction. Thus, G = Kn. Let G = Kn and let
A be a non empty subset of V (G). Then N em

G [A] = V (G). Hence, F em
G [A] = ∅. Therefore,

τ emG is the indiscrete topology on V (G).

Theorem 5. Let G be any graph. Then τ emG is the discrete topology on V (G) if and only
if for each a ∈ V (G) and for each v ∈ V (G) with a ∈ N em

G (v), there exists w ∈ V (G)\{a}
such that v ∈ N em

G (w) but a /∈ N em
G (w).

Proof. Suppose that τ emG is the discrete topology DG on V (G). Let a ∈ V (G) and let
v ∈ V (G) with a ∈ N em

G (v). Since τ emG is the discrete topology, {a} ∈ BemG , that is, there
exists A ⊆ V (G) such that F em

G [A] = {a}. Since a ∈ N em
G (v), v /∈ A. Also, v /∈ F em

G [A]
implies that there exists w ∈ A such that dmG (w, v) = emG (w), that is, v ∈ N em

G (w).
Moreover, because a ∈ F em

G [A], a /∈ N em
G (w). Thus, G satisfies the desired property. For

the converse, suppose that the given condition is satisfied by G. If G = K1, then clearly,
τ emG = DG. Suppose G 6= K1. Let a ∈ V (G) and let Aa = {v ∈ V (G) : a ∈ N em

G (v)}. Set
A = V (G)\(Aa ∪ {a}). Then, by assumption, A 6= ∅. Since a /∈ A and a /∈ N em

G (w) for all
w ∈ A, it follows that a ∈ F em

G [A]. Suppose there exists q ∈ F em
G [A]\{a}. Then q /∈ A∪{a}.

Hence, q ∈ Aa, that is, a ∈ N em
G (q). By assumption, there exists w /∈ Aa ∪ {a} such that

q ∈ N em
G (w), that is, dmG (q, w) = emG (w). This contradicts the fact that q ∈ F em

G [A].
Therefore, F em

G [A] = {a}. Since a was arbitrarily chosen, it follows that {a} ∈ τ emG for all
a ∈ V (G). Thus, τ emG is the discrete topology.

Corollary 1. Let G1, G2, ..., Gn be graphs such that τ emGi
= DGi for each i ∈ {1, 2, ..., n}.

If G =

n⋃
i=1

Gi, then τ
em
G = DG.

Proof. Let G =
n⋃

i=1

Gi and let a, v ∈ V (G) such that a ∈ N em
G (v). Then there exists a

unique i ∈ {1, 2, ..., n} such that a, v ∈ V (Gi). Hence, a ∈ N em
Gi

(v). Since τ emGi
= DGi , there

exists w ∈ V (Gi)\{a} such that v ∈ N em
Gi

(w) and a /∈ N em
Gi

(w) by Theorem 5. Thus, there
exists w ∈ V (G)\{a} such that v ∈ N em

G (w) and a /∈ N em
G (w). Therefore, τ emG = DG.



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 699

Corollary 2. If G = Kn, then τ
em
G = DG.

Proof. Let a ∈ V (G). Then Aa = {v ∈ V (G) : a ∈ N em
G (v)} = ∅. Let

A = V (G)\[Aa ∪ {a}] = V (G)\{a}.

Then, F em
G [A] = {a} ∈ τ emG . Thus, τ emG = DG.

Lemma 4. Let G = Cn = [v1, v2, ...vn, v1] be a cycle with n ≥ 3. Then emG (v) = n− 2 for
all v ∈ V (G).

Proof. Suppose w ∈ V (Cn). Without loss of generality, let w = v1. Since

emCn
(w) = max{dmCn

(w, v) : v ∈ V (Cn)},

it follows that emCn
(w) = n− 2.

Example 1. Let C5 = [v1, v2, v3, v4, v5, v1]. Then by Lemma 4, we have

N em
C5

[v1] = {v1, v3, v4} F em
C5

[v1] = {v2, v5}
N em

C5
[v2] = {v2, v4, v5} F em

C5
[v2] = {v1, v3}

N em
C5

[v3] = {v1, v3, v5} F em
C5

[v3] = {v2, v4}
N em

C5
[v4] = {v1, v2, v4} F em

C5
[v4] = {v3, v5}

N em
C5

[v5] = {v2, v3, v5} F em
C5

[v5] = {v1, v4}.

Note that

F em
C5

[v2] ∩ F em
C5

[v5] = {v1} F em
C5

[v3] ∩ F em
C5

[v5] = {v4}
F em
C5

[v1] ∩ F em
C5

[v3] = {v2} F em
C5

[v1] ∩ F em
C5

[v4] = {v5}.
F em
C5

[v1] ∩ F em
C5

[v2] = {v3}

Since {v1}, {v2}, {v3}, {v4}, {v5}, {v6} ∈ BemC5
, it follows that τ emC5

= DC5.

Example 2. Consider now C6 = [v1, v2, v3, v4, v5, v6, v1]. Then by Lemma 4, we have

N em
C6

[v1] = {v1, v3, v5} F em
C6

[v1] = {v2, v4, v6}
N em

C6
[v2] = {v2, v4, v6} F em

C6
[v2] = {v1, v3, v5}

N em
C6

[v3] = {v1, v3, v5} F em
C6

[v3] = {v2, v4, v6}
N em

C6
[v4] = {v2, v4, v6} F em

C6
[v4] = {v1, v3, v5}

N em
C6

[v5] = {v1, v3, v5} F em
C6

[v5] = {v2, v4, v6}
N em

C6
[v6] = {v2, v4, v6} F em

C6
[v6] = {v1, v3, v5}.

Note that {v1}, {v2}, {v3}, {v4}, {v5}, {v6} /∈ BemC6
. Hence, τ emC6

6= DC6.



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 700

Theorem 6. τ emCn
6= DCn for n = 3, 4, 6 and τ emCn

= DCn for n ∈ {5, 7, 8, ...}.

Proof. Since C3
∼= K3, τ

em
C3

= IC3 6= DC3 by Theorem 4. Let C4 = [v1, v2, v3, v4, v1] and
let a = v1. Set Aa = {v ∈ V (C4) : a ∈ N em

C4
(v)}. Then by Lemma 4, Aa = {v3}. Note that

v3 /∈ N em
G (v2)∩N em

G (v4). Hence, we could not find w 6= a such that v3 ∈ N em
C4

(w). Therefore,
C4 does not induce the discrete topology. Suppose C6 = [v1, v2, v3, v4, v5, v6, v1] and let
a = v1. Set Aa = {v ∈ V (C6) : a ∈ N em

C6
(v)}. Again, by Lemma 4, Aa = {v3, v5}. Note

that the only vertex w 6= a with v3 ∈ N em
C6

(w) is v5. However, a = v1 ∈ N em
C6

(v5). Thus, by
Theorem 5, C6 does not induce the discrete topology. Next let n = 5 and let a ∈ V (C5). We
may assume that a = v1. Let Aa = {v ∈ V (Cn) : a ∈ N em

Cn
(v)}. Then Aa = {v3, v4}. Since

v3 ∈ N em
Cn

(v5), v4 ∈ N em
Cn

(v2) where v2, v5 /∈ Aa, it follows from Theorem 5 that τ emC5
= DC5 .

Suppose n ≥ 7. Let a = v1. Then, Aa = {v ∈ V (Cn) : a ∈ N em
Cn

(v)}. Thus, Aa = {v3, vn−1}.
Note that v3 ∈ N em

Cn
(v5) and vn−1 ∈ N em

Cn
(vn−3) but v1 /∈ N em

Cn
(v5) ∩N em

Cn
(vn−3). Thus, by

Theorem 5, τ emCn
= DCn .

Theorem 7. Let G = Cn be a cycle with ≥ 4. Then

F em
G [vi] =


V (G)\{vi, vi+2, vi+n−2}, if i = 1, 2
V (G)\{vi−2, vi, vi+2}, if 3 ≤ i ≤ n− 2
V (G)\{vi−n+2, vi−2, vi}, if i = n, n− 1

where vi+2 = vi+n−2 and vi−2 = vi−n+2 if n = 4.

Proof. Let i = 1. By Lemma 4, emG (v) = 2. Thus, N em
C4

[v1] = {v1, v3}. Hence,
F em
C4

[v1] = V (C4)\{vi, vi+2}. Similarly, if i = 2, then F em
C4

[v2] = V (C4)\{vi, vi+2}. If
i = n, then N em

C4
[v4] = {v2, v4}. Thus, F em

C4
[v4] = V (C4)\{vi, vi−2}. Similarly, if i = n− 1,

then F em
C4

[v3] = V (C4)\{vi, vi−2}. Let i ∈ {1, 2}. By Lemma 4, N em
Cn

[v1] = {v1, v3, vn−1}
and N em

Cn
[v2] = {v2, v4, vn}. Thus, F em

Cn
[vi] = V (Cn)\{vi, vi+2, vi+n−2}. Suppose that i ∈

{3, 4, ..., n− 2}. Then, N em
Cn

[vi] = {vi−2, vi, vi+2}. It follows that

F em
Cn

[vi] = V (Cn)\{vi−2, vi, vi+2}.

Next, suppose, i ∈ {n, n − 1}. By Lemma 4, N em
Cn

[vn] = {v2, vn−2, vn} and N em
Cn

[vn−1] =
{v1, vn−3, vn−1}. Therefore,

F em
Cn

[vi] = V (Cn)\{vi−2, vi, vi−n+2}.

This proves the assertion.

Lemma 5. F em
C3

[v] = ∅ for all v ∈ V (C3).

Theorem 8. Let G = Pn = [v1, v2, ...vn] be a path of order n ≥ 3.

(a) If n is even, then τ emG has a subbase consisting of all sets of the form

F em
G [vi] =

{
V (G)\{vi, vn} if i ≤ n

2
V (G)\{v1, vi} if i > n

2 .



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 701

(b) If n is odd, then τ emG has a subbase consisting of all sets of the form

F em
G [vi] =


V (G)\{vi, vn} if i < n+1

2
V (G)\{v1, vi, vn} if i = n+1

2
V (G)\{v1, vi} if i > n+1

2 .

Proof. Suppose n is even. Let i ≤ n
2 . Then N em

G [vi] = {vi, vn}. Hence,
F em
G [vi] = V (G)\{vi, vn}. If i > n

2 , then N em
G [vi] = {v1, vi}. Thus, F em

G [vi] = V (G)\{v1, vi}.
Suppose n is odd. Let i < n+1

2 . Then N em
G [vi] = {vi, vn}. Thus, F em

G [vi] = V (G)\{vi, vn}.
Suppose i = n+1

2 . Then N em
G [vi] = {v1, vi, vn}. Hence, F em

G [vi] = V (G)\{v1, vi, vn}. Let
i > n+1

2 . Then N em
G [vi] = {v1, vi}. Therefore, F em

G [vi] = V (G)\{v1, vi}.

Theorem 9. Let G = Pn = [v1, v2, ...vn] be a path of order n ≥ 3. Then {v} ∈ τ emG if and
only if v 6= v1, vn.

Proof. Suppose {v} ∈ τ emG . Suppose further that v = v1. Then there exists ∅ 6= A ⊆
V (G) such that F em

G [A] = {v1}. This means that v1 /∈ A and dmG (v1, a) 6= emG (a) for all
a ∈ A. Since N em

G (vn) = {v1}, vn /∈ A. First, suppose that n is odd. From Theorem 8 (b)
it follows that vi /∈ A for all i ≥ n+1

2 . Hence,

A ⊆ {vj : 1 < j <
n+ 1

2
}.

Thus, by Theorem 8, vn+1
2
∈ F em

G [A], a contradiction. Suppose n is even. From Theorem

8 (a), vi /∈ A for all i > n
2 . Thus,

A ⊆ {vj : 1 < j ≤ n

2
}.

Hence, by Theorem 8, vn
2
+1 ∈ F em

G [A], a contradiction. Therefore, {v1} /∈ τ emG . Similarly,
{vn} /∈ τ emG . For the converse, suppose that v 6= v1, vn and let vj ∈ Pn. Consider the
following cases:
Case 1. 1 < j < dn2 e. Let A = V (G)\{vj , vn}. Then F em

G [A] = {vj}.
Case 2. j = dn2 e. If n is odd and j = n+1

2 , then set B = V (G)\{v1, vj , vn}. Then F em
G [B] =

{vj}. If n is even, set B = V (G)\{vj , vn}.
Case 3. dn2 e < j < n. Let D = V (G)\{v1, vj}. Then F em

G [D] = {vj}. Therefore, {vj} ∈ τ emG
for all j ∈ {2, 3, ..., n− 1}.

Definition 1. The join G+H of graphs G and H is the graph K with V (K) = V (G) ∪
V (H) and E(K) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G) and v ∈ V (H)}.

Theorem 10. Let G be any graph and let K1 = 〈v〉.

(i) If G is connected, then

F em
K1+G[w] =


∅, if [w ∈ V (G) and emG (w) = 1]

or w = v
F em
G [w] ∪ {v}, if w ∈ V (G)andemG (w) ≥ 2.



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 702

(ii) If G is disconnected, then

F em
K1+G[w] =


∅, if w = v

NG(w) ∪ {v} if w ∈ V (G) and
1 ≤ emG (w) ≤ 2

F em
G [w] ∪ {v}, if w ∈ V (G) and emG (w) ≥ 3.

Proof. (i) Let G be a connected graph. Suppose w ∈ V (G) and emG (w) = 1. Then,
N em

K1+G[w] = V (K1+G). It follows that F em
K1+G[w] = ∅. Clearly, if w = v then F em

K1+G[w] =
∅. Next, suppose w ∈ V (G) and emG (w) ≥ 2. Then [w, v, g] is a monophonic path in K1+G
for all g /∈ NG(w). Moreover, since every monophonic path in G is a monophonic path in
K1 +G. It follows that N em

K1+G[w] = N em
G [w]. Thus, F em

K1+G[w] = F em
G [w] ∪ {v}.

(ii) Let G be a disconnected graph. Clearly, if w = v, then N em
K1+G[w] = V (K1 + G)

and F em
K1+G[w] = ∅. Suppose w ∈ V (G) and 1 ≤ emG (w) ≤ 2. Observe that N em

K1+G[w] =
V (G)\NG(w). Hence, F em

K1+G[w] = NG(w)∪{v}. Suppose w ∈ V (G) and emG (w) ≥ 3. Since
every monophonic path in G is a monophonic path in K1 +G. It follows that N em

K1+G[w] =
N em

G [w]. Therefore, F em
G [w] ∪ {v}.

Corollary 3. Let K1 = 〈v0〉 and let G be any graph. Then,

(i) SK1+G = {∅} ∪ {F em
G [w] ∪ {v0} : w ∈ V (G) and emG (w) ≥ 2} if G is connected,

(ii) SK1+G = {∅} ∪ {NG(w) ∪ {v0} : w ∈ V (G) and degG(w) = 0 or 1 ≤ emG (w) ≤
2} ∪ {F em

G [w] ∪ {v0} : w ∈ V (G) and emG (w) ≥ 3} if G is disconnected.

(iii) {v} /∈ τ emK1+G for all v ∈ V (G).

Proof. Set H = K1. By Theorem 10 (i) and Theorem 10 (ii), (i) and (ii) hold. By (i)
and (ii), (iii) holds.

Lemma 6. Let K1 = 〈v0〉 and let G be any graph with radm(G) ≥ 2. Then {v0} ∈ τ emK1+G.

Proof. Suppose G is any graph with radm(G) ≥ 2. Then, emG (z) ≥ 2 for all z ∈ V (G).

Let v ∈ V (G). Since v /∈ (F em
G [v] ∩ {v0}), it follows that v /∈

⋂
z∈V (G)

(F em
G [z] ∩ {v0}). Since

v was arbitrarily chosen, {v0} =
⋂

z∈V (G)

(F em
G [z] ∪ {v0}). By Corollary 3,

(F em
G [z] ∪ {v0}) ∈ SemK1+G ⊆ τ

em
K1+G.

Therefore, {v0} ∈ τ emK1+G.

Definition 2. Let X 6= ∅ and p ∈ X. The particular point p topology on X is the class
τp = {∅} ∪ {A ⊆ X : p ∈ A}.



A. Gamorez, S. Canoy Jr. / Eur. J. Pure Appl. Math, 14 (3) (2021), 695-705 703

Theorem 11. Let K1 = 〈v0〉 and let G be a connected graph with radm(G) ≥ 2. Then
τ emK1+G is the particular point topology τv0 if and only if τ emG is the discrete topology on
V (G).

Proof. Suppose τ emG is the discrete topology on V (G). Note that {v0} ∈ τ emK1+G. Now,
since {v0} ∈ (F em

G [w] ∩ {v0}) for all w ∈ V (G), it follows that {v} /∈ BemK1+G ⊆ τ emK1+G.
Next, since τ emG is a discrete topology, {v} ∈ BemG for all v ∈ V (G). Hence, there exist

vj1 , vj2 , ..., vjk ∈ V (G) such that {v} =

k⋂
s=1

F em
G [vjs ]. Therefore,

{v0, v} =

(
k⋂

s=1

F em
G [vjs ]

)
∪ {v0} =

k⋂
s=1

(
F em
G [vjs ] ∪ {v0}

)
∈ BemK1+G ⊆ τ

em
K1+G.

Accordingly, τ emK1+G = τv0 . For the converse, suppose that τ emK1+G = τv0 . Let v ∈ V (G).
Since τ emK1+G = τv0 , {v0, v} ∈ τ emK1+G. Hence, there exists a basic open set B ∈ BemK1+G

such that v ∈ B ⊆ {v0, v}. Since {v} cannot be a finite intersection of subbasic open
sets of the form F em

G [w] ∪ {v0}, it follows that B 6= {v}. Thus, B = {v0, v}. This means

that there exist vj1 , vj2 , ..., vjk ∈ V (G) such that {v0, v} =
t⋂

k=1

(F em
G [vjk ]∪{v0}). Therefore,

{v} =
t⋂

k=1

F em
G [vjk ], that is, {v} ∈ BemG ⊆ τ emG . This shows that τ emG is the discrete topology

on V (G).

Theorem 12. Let H be a graph with radm(〈V (H)\{v0}〉) ≥ 2 and let v0 ∈ V (H). Then
τ emH = τv0 if and only if H = 〈v0〉+G for some graph G such that radm(G) ≥ 2 and τ emG
is the discrete topology on V (G).

Proof. Suppose τ emH = τv0 . Suppose further that there exists v ∈ V (H)\{v0} such that
v0v /∈ E(H). Then emH(v0) ≥ 2. This implies that NH(v0) 6= ∅ and NH(v0)∩N em

H (v0) = ∅.
Hence, NH(v0) ⊆ F em

H [v0]. This gives a contradiction because v0 /∈ F em
H [v0] and F em

H [v0] ∈
τ emH . Therefore, v0v ∈ E(H) for all v ∈ V (H)\{v0}. Let G = 〈V (H)\{v0}〉. Then H =
〈v0〉+G. By Theorem 11, τ emG is the discrete topology on V (G). For the converse, suppose
H = 〈v0〉+G for some graph G such that radm(G) ≥ 2 and τ emG is the discrete topology
on V (G). Then by Theorem 11, τ emH = τv0 .

Corollary 4. Let G = Wn = 〈v0〉+Cn, where n ∈ {5, 7, 8, ...}. Then τ emWn
is the particular

point topology τv0 .

Proof. Let n ∈ {5, 7, 8, ...}. Then τ emCn
is the discrete topology on V (Cn) by Theorem

6. Thus, by Theorem 11, τ emWn
is the particular point topology τv0 .

Theorem 13. Let G = Fn = 〈v0〉+ Pn (n ≥ 4). Then {v0, v} ∈ τ emFn
for all

v ∈ V (Pn)\{v1, vn}.



REFERENCES 704

Proof. Suppose v ∈ V (Pn)\{v1, vn}. By Theorem 9, {v} ∈ τ emPn
. Thus, {v} ∈ BemPn

.

Hence, there exist vi1 , vi2 , ..., vik ∈ V (Pn) such that {v} =

k⋂
s=1

F em
Pn

[vis ]. Therefore,

{v0, v} =
k⋂

s=1

(F em
Pn

[vis ] ∪ {v0}) ∈ BemPn
⊆ τ emFn

,

proving our assertion.

Theorem 14. If n is a positive integer and K1 = 〈v0〉, then

τ emK1,n
=

{
{∅, V (K1,n)} if n = 1

{∅, {v0}, V (K1,n)} if n ≥ 2.

Proof. By Corollary 3,

SemK1,n
=

{
{∅} if n = 1

{∅} ∪ {v0} otherwise.

Hence,

τ emK1,n
=

{
{∅, V (K1,n)} if n = 1

{∅, {v0}, V (K1,n)} if n ≥ 2.

This proves the assertion.

Acknowledgements

The authors would like to thank Western Mindanao State University (WMSU) and
Mindanao State University-Iligan Institute of Technology (MSU-IIT)for funding this re-
search.

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