36
Mohanapriya Rajagopal and Durgadevi Shanmugasudaram
Hence, the neutrosophic block topology is
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{(v3, η(v3), κ(v3), λ(v3))},
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{(v1, η(v1), κ(v1), λ(v1)), (v2, η(v2), κ(v2), λ(v2)), (v3, η(v3), κ(v3), λ(v3))},
{(v3, η(v3), κ(v3), λ(v3)), (v4, η(v4), κ(v4), λ(v4)), (v5, η(v5), κ(v5), λ(v5))},
{(v1, η(v1), κ(v1), λ(v1)), (v2, η(v2), κ(v2), λ(v2)), (v3, η(v3), κ(v3), λ(v3)), (v4, η(v4), κ(v4), λ(v4))},
{(v1, η(v1), κ(v1), λ(v1)), (v2, η(v2), κ(v2), λ(v2)), (v3, η(v3), κ(v3), λ(v3)), (v5, η(v5), κ(v5), λ(v5))},
V (HG)
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τB(HN ) =
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2.1. Continuity in Neutrosophic Block Topological Space.
Definition 2.3. A function f : (V (NG), τB(NG)) → (V (NH), τB(NH)) is said to be
continuous if and only if for every U ∈ τB(NH), we have f−1(U) ∈ τB(NG), where
f
−1(U) = { (v, η(v), κ(v), λ(v)) ∈ V (NG) : f(v, η(v), κ(v), λ(v)) ∈ U }.
Theorem 2.4. Let NG and NH be neutrosophic graphs. A function f : (V (NG), τB(NG)) →
(V (NH), τB(NH)) is continuous if and only if for every neutrosophic block B of NH, the
set f−1(V (B)) belongs to τB(NG).
Proof. Assume f is continuous. Let B be any neutrosophic block of NH. Since
V (B) ∈ ΣB(NH), it is τB(NH)-open. By continuity, f−1(V (B)) ∈ τB(NG).
Conversely, suppose that for every neutrosophic block B of NH, f−1(V (B)) ∈ τB(NG).
Let U be an arbitrary τB(NH)-open set. By definition of neutrosophic block topology,
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U =
V (Bjα) ,
α
j
where each Bα is a neutrosophic block of NH. Taking preimages,
j
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f
−1(U) =
f−1
V (Bjα)
=
f
−1(V (Bjα)) .
α
j
α
j
By hypothesis, each f−1(V (Bα)) is τB(NG)-open. Since τB(NG) is closed under finite
j
intersections and arbitrary unions, we have f−1(U) ∈ τB(NG). Hence, f is continuous.
Theorem 2.5. Let NG and NH be neutrosophic graphs. If (v, η(v), κ(v), λ(v)) is a neu-
trosophic cut-vertex of NG and f : (V (NG), τB(NG)) → (V (NH), τB(NH)) is continuous,
then f((v, η(v), κ(v), λ(v))) is either a neutrosophic cut-vertex of NH or an isolated vertex
of NH.
Proof. Since (v, η(v), κ(v), λ(v)) is a cut-vertex of the neutrosophic graph NG, and
since every cut-vertex induces a singleton which is τB(NG)-open (see [7], Theorem 3),
it follows that {(v, η(v), κ(v), λ(v))} ∈ τB(NG).
Assume f
is continuous.
If
f((v, η(v), κ(v), λ(v)))
is
neither
a cut-vertex nor an isolated vertex in NH, then
{ f((v, η(v), κ(v), λ(v))) } ∈/ τB(NH).
However,
by continuity,
the preimage of
{ f((v, η(v), κ(v), λ(v))) } under f must be τB(NG)-open, which contains (v, η(v), κ(v), λ(v)).