(4) Ais (Λ, α)-open and p(Λ, α)-closed.
(5) Ais α(Λ, α)-open and p(Λ, α)-closed.
(6) Ais α(Λ, α)-open and (Λ, α)-closed.
(7) Ais p(Λ, α)-open and (Λ, α)-closed.
(8) Ais β(Λ, α)-open and α(Λ, α)-closed.
Proof. (1) ⇒(2): Let Abe a (Λ, α)-clopen set.
Then, we have A=A(Λ,α)=A(Λ,α)and hence
A= [A(Λ,α)](Λ,α)= [A(Λ,α)](Λ,α). This shows
that Ais r(Λ, α)-open. Thus, Ais r(Λ, α)-open and
r(Λ, α)-closed.
(2) ⇒(3): Let Abe r(Λ, α)-open and
r(Λ, α)-closed. Then, A= [A(Λ,α)](Λ,α)=
[A(Λ,α)](Λ,α). Thus, A(Λ,α)= [[A(Λ,α)](Λ,α)](Λ,α)=
[A(Λ,α)](Λ,α)=Aand hence
[[A(Λ,α)](Λ,α)](Λ,α)= [[A(Λ,α)](Λ,α)](Λ,α)
= [A(Λ,α)](Λ,α)=A.
Consequently, we obtain Ais (Λ, α)-open and
α(Λ, α)-closed.
(3) ⇒(4): Suppose that Ais (Λ, α)-
open and α(Λ, α)-closed. Then, we have A=
A(Λ,α)and [[A(Λ,α)](Λ,α)](Λ,α)⊆A, by Lemma
11, [A(Λ,α)](Λ,α)= [[[A(Λ,α)](Λ,α)](Λ,α)](Λ,α)=
[[A(Λ,α)](Λ,α)](Λ,α)⊆A. Thus, Ais p(Λ, α)-closed.
This shows that Ais (Λ, α)-open and p(Λ, α)-closed.
(4) ⇒(5): Let Abe (Λ, α)-open and p(Λ, α)-
closed. Then, A=A(Λ,α)and [A(Λ,α)](Λ,α)⊆A.
Thus, A=A(Λ,α)⊆[[A(Λ,α)](Λ,α)](Λ,α)⊆A(Λ,α)
and hence [[A(Λ,α)](Λ,α)](Λ,α)=A(Λ,α)=A. There-
fore, Ais α(Λ, α)-open. Thus, Ais α(Λ, α)-open and
p(Λ, α)-closed.
(5) ⇒(6): Let Abe α(Λ, α)-open
and p(Λ, α)-closed. Then, we have A⊆
[[A(Λ,α)](Λ,α)](Λ,α)and [[A(Λ,α)](Λ,α)](Λ,α)⊆A.
Thus, A= [[A(Λ,α)](Λ,α)](Λ,α)and hence
A(Λ,α)= [[[A(Λ,α)](Λ,α)](Λ,α)](Λ,α). By Lemma
11, we have A(Λ,α)= [A(Λ,α)](Λ,α). Since
[A(Λ,α)](Λ,α)⊆A, we have A(Λ,α)⊆Aand
hence A(Λ,α)=A. Therefore, Ais (Λ, α)-closed and
α(Λ, α)-open.
(6) ⇒(7): Let Abe α(Λ, α)-open and (Λ, α)-
closed. Then, A⊆[[A(Λ,α)](Λ,α)](Λ,α)and A=
A(Λ,α), by Lemma 11, A⊆[[A(Λ,α)](Λ,α)](Λ,α)⊆
[[[A(Λ,α)](Λ,α)](Λ,α)](Λ,α)= [A(Λ,α)](Λ,α). This
shows that Ais p(Λ, α)-open. Thus, Ais p(Λ, α)-
open and (Λ, α)-closed.
(7) ⇒(8): Let Abe p(Λ, α)-open and (Λ, α)-
closed. Then, we have A⊆[A(Λ,α)](Λ,α)and A=
A(Λ,α). Thus, [[A(Λ,α)](Λ,α)](Λ,α)⊆A(Λ,α)=A.
Therefore, Ais p(Λ, α)-open and α(Λ, α)-closed.
(8) ⇒(1): Let Abe p(Λ, α)-open and
α(Λ, α)-closed. Then, A⊆[A(Λ,α)](Λ,α)and
[[A(Λ,α)](Λ,α)](Λ,α)⊆A. Therefore, A(Λ,α)⊆
[[A(Λ,α)](Λ,α)](Λ,α)⊆Aand hence A(Λ,α)⊆
A. This shows that A=A(Λ,α). Thus, A
is (Λ, α)-closed. Since [[A(Λ,α)](Λ,α)](Λ,α)⊆A,
[[[A(Λ,α)](Λ,α)](Λ,α)](Λ,α)⊆A(Λ,α), by Lemma 11,
we have A⊆[A(Λ,α)](Λ,α)⊆A(Λ,α)and hence
A⊆A(Λ,α). This implies that A=A(Λ,α). There-
fore, Ais (Λ, α)-open. Consequently, we obtain Ais
(Λ, α)-clopen.
Definition 16. A subset Aof a topological space
(X, τ)is called α(Λ, α)-⋆-open (resp. β(Λ, α)-⋆-
open) if A= [[A(Λ,α)](Λ,α)](Λ,α)(resp. A=
[[A(Λ,α)](Λ,α)](Λ,α)).
Proposition 17. A subset Aof a topological space
(X, τ)is r(Λ, α)-open if and only if Ais α(Λ, α)-⋆-
open.
Proof. Suppose that Ais a r(Λ, α)-open set. Then,
A= [A(Λ,α)](Λ,α). Thus, Ais (Λ, α)-open and hence
A= [[A(Λ,α)](Λ,α)](Λ,α). Therefore, Ais α(Λ, α)-⋆-
open.
Conversely, suppose that Ais a α(Λ, α)-⋆-open
set. Then, A= [[A(Λ,α)](Λ,α)](Λ,α). By Lemma 11,
[A(Λ,α)](Λ,α)= [[[[A(Λ,α)](Λ,α)](Λ,α)](Λ,α)](Λ,α)
= [[A(Λ,α)](Λ,α)](Λ,α)=A.
This shows that Ais r(Λ, α)-open.
Proposition 18. A subset Aof a topological space
(X, τ)is r(Λ, α)-closed if and only if Ais β(Λ, α)-⋆-
open.
Proof. Suppose that Ais a r(Λ, α)-closed set.
Then, we have A= [A(Λ,α)](Λ,α)and hence A
is (Λ, α)-closed. Thus, A= [A(Λ,α)](Λ,α)=
[[A(Λ,α)](Λ,α)](Λ,α). Therefore, Ais β(Λ, α)-⋆-open.
Conversely, suppose that Ais a β(Λ, α)-⋆-open
set. Then, A= [[A(Λ,α)](Λ,α)](Λ,α)and by Lemma
11, [A(Λ,α)](Λ,α)= [[[[A(Λ,α)](Λ,α)](Λ,α)](Λ,α)](Λ,α)=
[[A(Λ,α)](Λ,α)](Λ,α)=A. Thus, Ais r(Λ, α)-closed.
Proposition 19. For a subset Aof a topological space
(X, τ), the following properties are equivalent:
4
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2023.22.2
Jeeranunt Khampakdee, Chawalit Boonpok