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3. A. Karthika and I. Arockiarani / International Journal of Engineering Research and
Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 3, Issue 4, Jul-Aug 2013, pp.2603-2607
2605 | P a g e
Let (X, π) be a IFTS and π b is π΅π π π΄ e a
complement function that satisfies the monotonic and
involutive condition. Then for any IFS A of X,
is π΅π π π΄ = is π΅π π ππ΄ .
Proof
By the definition 3.1, is π΅π π π΄ =is ππ π π΄ β§
is ππ π( ππ΄). Since π satisfies the involutive condition
π ππ΄ = π΄, that implies is π΅π π π΄ =is ππ π(ππ΄) β§
is ππ π π( ππ΄). That is is π΅π π π΄ = is π΅π π ππ΄ .
Proposition 3.3
Let (X, π) be a IFTS and π be a complement
function that satisfies the monotonic and involutive
conditions. If A is intuitionistic fuzzy π-semi closed,
then is π΅π π π΄ β€ π΄.
Proof
Let A be intuitionistic fuzzy π-semi closed,
and π satisfies the involutive and monotonic
conditions we have is π΅π π π΄ β€ is ππ π π΄ = π΄.
Proposition 3.4
Let (X, π) be a IFTS and π be a complement
function that satisfies the monotonic and involutive
conditions. If A is intuitionistic fuzzy π-semi open,
then is π΅π π π΄ β€ π π΄.
Proof
Let A be intuitionistic fuzzy π-semi open.
Since π satisfies the involutive condition, we get,
is π΅π π π΄ β€ ππ΄.
Proposition 3.5
Let (X, π) be a IFTS. Let π be a complement
function satisfies the monotonic and involutive
conditions. Then for any IFS A of X, we have π
is π΅π π π΄ = is πππ‘ π π΄ β¨ is πππ‘ π ππ΄ .
Proof
Using Definition 3.1, is π΅π π π΄ = is ππ π π΄ β§
is ππ π( ππ΄) . Taking complement on both sides, we
get π [is π΅π π π΄ ] = π [is ππ π π΄ β§ is ππ π( ππ΄)] . Since π
the monotonic and involutive conditions, π [
is π΅π π π΄ ] = π[is ππ π π΄] β¨ π[ is ππ π ππ΄ ]. Using
proposition 5.6 in [6], π is π΅π π π΄ = is πππ‘ π π΄ β¨
is πππ‘ π ππ΄ .
Proposition 3.6
Let (X, π) be a IFTS. Let π be a
complement function that satisfies the monotonic and
involutive conditions. Then for any IFS A of X, we
have is π΅π π π΄ =is ππ π π΄ β§ is πππ‘ π π΄ .
Proof
By the Definition 3.1, we have
is π΅π π π΄ =is ππ π π΄ β§ is ππ π ππ΄ . Since π satisfies
the monotonic and involutive conditions, we have
is π΅π π π΄ =is ππ π π΄ β§ is πππ‘ π π΄ .
Proposition 3.7
Let (X, π) be a IFTS. Let π be a complement
function that satisfies the monotonic and involutive
conditions. Then for any IFS A of X, is π΅π π[
Ξ² πππ‘ π π΄ ] β€ is π΅π π π΄ .
Proof
Since π satisfies the monotonic and
involutive conditions, by using proposition 3.6, we
have is π΅π π[ is πππ‘ π π΄ ]= is ππ π π΄ [is πππ‘ π π΄ ] β§
is πππ‘ π[ is πππ‘ π π΄ ]. Since is πππ‘ π π΄ is intuitionistic
fuzzy π-semi- open, is π΅π π[ is πππ‘ π π΄ ]=
is ππ π π΄ [is πππ‘ π π΄ ] β§ π [ is πππ‘ π π΄ ]. Since
is πππ‘ π π΄ β€ π΄, by using proposition 5.6(ii) in [6],
and proposition 5.5 in [6],
is π΅π π[ is πππ‘ π π΄ ] β€ is ππ π π΄ β§ is ππ π[ ππ΄]. By using
the definition 3.1, we have is π΅π π[
is πππ‘ π π΄ ] β€ is π΅π π π΄ .
Proposition 3.8
Let (X, π) be a IFTS. Let π be a complement
function that satisfies the monotonic and involutive
conditions. Then for any IFS A of X, is π΅π π[
is ππ π π΄ ] β€ is π΅π π π΄ .
Proof
Since π be a complement function that
satisfies the monotonic and involutive conditions, by
using proposition 3.11, is π΅π π[ is ππ π π΄ ]=
is ππ π π΄ [is ππ π π΄ ] β§ is πππ‘ π[ is ππ π π΄ ]. By using
proposition 5.6(iii) in [6], we have is ππ π[ is ππ π π΄ ] =
is ππ π π΄ , that implies
is π΅π π[ is ππ π π΄ ] = is ππ π π΄ β§ is πππ‘ π[ is ππ π π΄ ]. Since
A β€[ is ππ π π΄ ] , that implies is πππ‘ π π΄ β€
is πππ‘ π[ is ππ π π΄ ]. Therefore is π΅π π[ is ππ π π΄ ] β€
is ππ π π΄ β§ is πππ‘ π(π΄). By using proposition 5.5(ii) in
[6], and using the definition 3.1, we get is π΅π π[
is ππ π π΄ ] β€ is π΅π π π΄ .
Theorem 3.9
Let (X, π) be a IFTS. Let π be a complement
function that satisfies the monotonic and involutive
conditions. Then for any IFS A and B of X,
is π΅π π[π΄ β¨ π΅] β€ is π΅π π π΄ β¨ is π΅π π π΅ .
Proof
From the definition 3.1, is π΅π π[π΄ β¨ π΅]= is ππ π[π΄ β¨
π΅] β§ is ππ π π[π΄ β¨ π΅] . Since π satisfies the
monotonic and involutive conditions, by the
proposition 5.7(i) in [6], that implies is π΅π π[π΄ β¨ π΅]=
{is ππ π[π΄] β¨ [ is ππ π π΅ ]}β§ is ππ π π[π΄ β¨ π΅] . By
using , proposition 5.7 (ii) in [6], is π΅π π[π΄ β¨ π΅] β€
{is ππ π[π΄] β¨ [ is ππ π π΅ ]}β§ { is ππ π ππ΄ β§
is ππ π ππ΅ }. That is is π΅π π[π΄ β¨ π΅] β€ {is ππ π[π΄] β§ [
is ππ π ππ΄ ]}β¨ { is ππ π π΄ β§ is ππ π ππ΅ }. Again by
4. A. Karthika and I. Arockiarani / International Journal of Engineering Research and
Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 3, Issue 4, Jul-Aug 2013, pp.2603-2607
2606 | P a g e
using definition 3.1, is π΅π π[π΄ β¨ π΅] β€ is π΅π π π΄ β¨
is π΅π π π΅ .
Theorem 3.10
Let (X, π) be a IFTS. Let π be a complement
function that satisfies the monotonic and involutive
conditions. Then for any IFS A and B of X,
is π΅π π[π΄ β§ π΅] β€ {is π΅π π[π΄] β§ [
is ππ π π΅ ]}β¨ { is π΅π π π΅ β§ is ππ π π΄ }.
Proof
By 3.1 is π΅π π[π΄ β§ π΅]= is ππ π[π΄ β§ π΅] β§
is ππ π π[π΄ β§ π΅] . Since π satisfies the monotonic
and involutive conditions, by using propositions
5.7(i), in [6] and by using lemma 2.11, we get
is π΅π π[π΄ β§ π΅] β€ {is ππ π[π΄] β§ [ is ππ π π΅ ]}β§ {
is ππ π ππ΄ β¨ is ππ π ππ΅ } is equal to is π΅π π[π΄ β§
π΅] β€ {{is ππ π[π΄] β§ [ is ππ π ππ΄ ]}β§ [is ππ π π΄] } β¨ {[
is ππ π π΅ β§ is ππ π ππ΅ ] β§ [is ππ π π΅]}. Again by
definition 3.1, we get
is π΅π π[π΄ β§ π΅] β€ {is π΅π π[π΄] β§ [ is ππ π π΅ ]}β¨ {
is π΅π π π΅ β§ is ππ π π΄ }.
Proposition 3.11
Let Let (X, π) be a IFTS. Let π be a
complement function that satisfies the monotonic and
involutive conditions. Then for any IFS A of X,we
have
1. is π΅π π[ is π΅π π π΄ ] β€ is π΅π π π΄ .
2. [ is π΅π π π΄ ]{is π΅π π[ is π΅π π π΄ ]} β€ is π΅π π π΄ .
Proof
From the definition 3.1, is π΅π π π΄= is π΅π π π΄ β§
is π΅π π ππ΅. We have is π΅π π is π΅π π π΄ = is ππ π [
is π΅π π π΄] β§ is ππ π[is π΅π π π΄ ] β€ is π΅π π[π΄ β¨ π΅]
is ππ π[is π΅π π π΄]. Since π satisfies the monotonic and
involutive conditions, we have [ is ππ π π΄ ]= A, where
A is A is intuitionistic fuzzy π-semi closed. Hence
the proof.
Proposition 3.12
Let A be a intuitionistic fuzzy π-semi closed
subset of a IFTS X and B be a intuitionistic fuzzy π-
semi closed subset od a IFTS Y, then A x B is a
intuitionistic fuzzy π-seml closed set of the
intuitionistic fuzzy product space X x Y where the
complement function π satisfies the monotonic and
involutive conditions.
Proof
Let A be a intuitionistic fuzzy π-semi closed
subset of a IFTS X. Then π π΄ is intuitionistic fuzzy
semi open in X. Then ππ΄ x 1 is intuitionistic fuzzy
semi open in X x Y. Similarly let B be a intuitionistic
fuzzy π-semi closed subset od a IFTS Y, then 1 x ππ΅
is intuitionistic fuzzy semi open in X x Y. Since the
arbitrary union of intuitionistic fuzzy π-semi open
sets is intuitionistic fuzzy π-semi open , ππ΄ π₯ 1 β¨
1 π₯ ππ΅ is intuitionistic fuzzy π-semi open in X x Y.
Then by lemma 2.19 π (A x B) = π A x 1 β¨ 1 x π B,
hence π(A x B ) is intuitionistic fuzzy π-semi open.
Therefore A x B is intuitionistic fuzzy π-semi closed
of intuitionistic fuzzy product space X x Y.
Theorem 3.13
Let (X, π ) be a IFTS. Let π be a
complement function that satisfies the monotonic and
involutive conditions. Then for any IFS A and B of
X,
(i) is ππ π π΄ x is ππ π π΅ β₯ is ππ π(A x B)
(ii) is ππ‘ π π΄ x is πππ‘ π π΅ β€ is πππ‘ π(A x B)
proof
By the definition 2.17,[ is ππ π(A) x
is ππ π π΅ ] (π₯, π¦)= min { is ππ π π΄(x), is ππ π π΅(y)}β₯
min { A(x), B(y)} = (A x B ) (x,y). This shows
that is ππ π π΄ x is ππ π π΅ β₯ (A x B).
By using proposition 5.6 in [6], is ππ π(A x B) β€
is ππ π ( is ππ π π΄ x is ππ π π΅ )= is ππ π π΄ x
is ππ π π΅ . By using definition 2.17 ,[ is πππ‘ π(A) x
is πππ‘ π π΅ ] (π₯, π¦)= min { is πππ‘ π π΄(x),
is πππ‘ π π΅(y)} β€ min { A(x), B(y)} = (A x B )
(x,y). This shows that
is πππ‘ π π΄ x is πππ‘ π π΅ β€ is πππ‘ π(A x
B).
Theorem 3.14
Let X and Y be π-product related
IFTSs. Then for a IFS A of X and a IFS B of Y,
(i) is ππ π(A x B) = is ππ π π΄ x is ππ π π΅
(ii) is πππ‘ π(A x B) = is πππ‘ π π΄ x
is πππ‘ π π΅ .
Proof
By the theorem 3.13, it is sufficient to
show that is ππ π π΄ x is ππ π π΅ β€ is ππ π(A x B).
By using the definition in [6], we have is ππ π(A x
B) = inf { π( π΄ πΌ x π΅π½ ) βΆ π( A x B) β₯ A x B
where π΄ πΌ and π΅π½ are intuitionistic fuzzy π-semi
open }. By using lemma 2.17, we have,
is ππ π(A x B) = inf { π( π΄ πΌ) x 1 β¨ 1 x π(π΅π½ ) βΆ
π( π΄ πΌ) x 1 β¨ 1 x π(π΅π½ ) β₯ A x B}
=inf { π( π΄ πΌ) x 1 β¨ 1 x π(π΅π½ ) :
π( π΄ πΌ) β₯ A or π π΅π½ β₯B}
= min ( inf { π( π΄ πΌ) x 1 β¨ 1 x
π(π΅π½ ) : π( π΄ πΌ) β₯ A }, inf { π( π΄ πΌ) x 1 β¨ 1 x
π(π΅π½ ) : π π΅π½ β₯B}).
Now inf{ π( π΄ πΌ) x 1 β¨ 1 x π(π΅π½ ) : π( π΄ πΌ) β₯
A} β₯ inf { π( π΄ πΌ) x 1 : π( π΄ πΌ) β₯ A}
= inf { π( π΄ πΌ) : π( π΄ πΌ) β₯ A} x 1
= [ is ππ π π΄ ] x 1.
5. A. Karthika and I. Arockiarani / International Journal of Engineering Research and
Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 3, Issue 4, Jul-Aug 2013, pp.2603-2607
2607 | P a g e
Also inf { π( π΄ πΌ) x 1 β¨ 1 x (π΅π½ ) : π π΅π½ β₯B}
β₯ inf { 1 x π π΅π½ : π π΅π½ β₯B}
= 1 x inf { π π΅π½ : π π΅π½ β₯B}
= 1 x [ is ππ π π΅ ].
Thus we conclude the result as ,
is ππ π(A x B) β₯ min ([ is ππ π π΄] x 1 , 1 x [ is ππ π π΅]
= [ is ππ π π΄]x [ is ππ π π΅].
Theorem 3.15
Let ππ, i= 1, 2, 3, β¦.. n be a family of
π-product related IFTSs. If π΄π is a IFS of ππ, and
the complement function π satisfies the
monotonic and involutive conditions, then
is π΅π π[ π΄π
π
π=1 ]={ is π΅π π(π΄1) x is ππ π π΄2 x
β¦β¦β¦ x is ππ π π΄ π } β¨ { is ππ π(π΄1) x
is π΅π π π΄2 x β¦β¦β¦ x is ππ π π΄ π } β¨ β¦β¦β¦. β¨
{ is ππ π(π΄1) x is ππ π π΄2 x β¦β¦β¦ x
Ξ²is π΅π π π΄ π }.
Proof
Using lemma 2.14, lemma 2.19, we get
the result.
Theorem 3.16
Let π βΆ π β π be a intutionistic fuzzy
continuous function. Suppose the complement
function π satisfies the monotonic and
involutive conditions, then
is π΅π π(πβ1
π΅ β€ πβ1
(is π΅π π[B]), for any IFS B
in Y.
Proof
Let f be a IF continuous function and B
be a IFS in Y. By using definition 3.1, we have
is π΅π π(πβ1
(π΅)) = is ππ π πβ1
(π΅ ) β§ [
is ππ π π (πβ1
(π΅ )). Then using the lemmaβs
2.16, 2.18, finally we get is π΅π π(πβ1
π΅ β€
πβ1
(is π΅π π[B]). Hence the result.
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