16th Int. Conf. on IFSs, Sofia, 9–10 Sept. 2012
Notes on Intuitionistic Fuzzy Sets
Vol. 18, 2012, No. 3, 8–15
Absolute value and limit of the function
defined on IF sets
Alžbeta Michalı́ková
Faculty of Natural Sciences, Matej Bel University
Tajovského 40, SK-974 01 Banská Bystrica, Slovakia
e-mail: [email protected]
Abstract: In this paper the set F of all IF sets is embedded to an `-group. Two new functions,
the absolute value and the limit of the function are defined and their properties are studied.
Keywords: Absolute value of IF set, Limit of function.
AMS Classification: XXXXX.
1
Introduction
By an IF -set we consider a pair A = (µA , νA ) of functions µA , νA : Ω → [0, 1] such that
µA + νA ≤ 1.
The function µA is called a membership function of A, νA a nonmembership function of A. If
(Ω, S) is a measurable space and µA , νA are S-measurable, then A is called an IF -event. Denote
by F the family of all IF -events.
In the paper [4] there was proved that we should construct such `-group G that F can be
embedded into G.
Consider the set A = (µA , νA ), where µA , νA : Ω → R. Denote by G the set of all pairs
A = (µA , νA ) and for any A, B ∈ G define following operation
A + B = (µA + µB , νA + νB − 1) =
= (µA + µB , 1 − (1 − νA ) + (1 − νB ))
and relation
A ≤ B ⇐⇒ µA ≤ µB , νA ≥ νB .
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Then the triple G = (G, +, ≤) is the mentioned `-group.
On the set G there are defined some operations. We will need following of them
A − B = (µA − µB , 1 + νA − νB ),
A.B = (µA .µB , νA + νB − νA .νB ).
The element (0, 1) is a neutral element of the operation + and the element −A = (−µA , 2 −
νA ) is an inverse element to the element A. If B > (0, 1), whereby A > B ⇔ µA > µB , νA < νB ,
then
µ A νA − νB
A
=
,
.
B
µ B 1 − νB
Any function on the IF set is defined by the following way
f (A) = (f (µA ), 1 − f (1 − νA ))
for example for any natural number n it holds
n.A = (n.µA , 1 − n.(1 − νA )),
An = ((µA )n , 1 − (1 − νA )n ).
2
Absolute value on IF sets
The motivation for this paper were papers [2], [3], [4] where authors defined and studied some
elementary functions which are important in classical calculus. In our future work we would like
to defined basic terms of calculus on IF sets and also proved that there hold similar properties as
in classical calculus. We need to do it in small steps therefore in this paper we will construct the
theory of absolute value and limits on the `-group G. Then these results we could also use in the
IF set theory.
Definition 2.1 Let A ∈ G. Then |A| = (|µA |, 1 − |1 − νA |) is the absolute value of A.
Lemma 2.2 Let |A| be the absolute value of A. Then |A| = A for each A ≥ (0, 1) and |A| = −A
for each A < (0, 1).
Proof. Let A ≥ (0, 1) i.e. µA ≥ 0 and νA ≤ 1. Then 1 − νA ≥ 0 and |A| = (|µA |, 1 − |1 − νA |) =
(µA , 1 − 1 + νA ) = (µA , νA ) = A.
Let A < (0, 1) i.e. µA < 0 and νA > 1. Then 1 − νA < 0 and |A| = (|µA |, 1 − |1 − νA |) =
(−µA , 1 + 1 − νA ) = (−µA , 2 − νA ) = −A.
Lemma 2.3 For each A, B ∈ G it holds
1. |A.B| = |A|.|B|,
2. |A − B| = |B − A|,
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3. |A + B| ≥ |A| + |B|,
4. |A − B| ≥ |A| − |B|,
5. ||A| − |B|| ≤ |A − B|.
Proof. Let A = (µA , νA ), B = (µB , νB ) then it holds
1. |A|.|B| = (|µA |, 1 − |1 − νA |) . (|µB |, 1 − |1 − νB |) = (|µA .µB |, 1 − |1 − νA | + 1 − |1 −
νB | − (1 − |1 − νA |).(1 − |1 − νB |)) = (|µA .µB |, 1 − |1 − νA − νB + νA .νB |) = |A.B|.
2. |A − B| = (|µA − µB |, 1 − |νB − νA |) = |B − A|.
3. |A + B| = (|µA + µB |, 1 − |2 − νA − νB |) .
|A| + |B| = (|µA | + |µB |, 1 − |1 − νA | − |1 − νB |) .
Since |µA + µB | ≥ |µA | + |µB | and |(1 − νA ) + (1 − νB )| ≥ |1 − νA | + |1 − νB | then also
|A + B| ≥ |A| + |B|.
4. |A| − |B| = (|µA | − |µB |, 1 − |1 − νA | + |1 − νB |) .
And it holds |µA − µB | ≥ |µA | − |µB | and also |(1 − νA ) − (1 − νB )| ≥ |1 − νA | − |1 − νB |
therefore |A − B| ≥ |A| − |B|.
5. ||A| − |B|| = (||µA | − |µB ||, 1 − |1 − (1 − |1 − νA | + |1 − νB |)|) .
It holds ||µA | − |µB || ≤ |µA − µB | and ||1 − νA | − |1 − νB || ≤ |(1 − νA ) − (1 − νB )|.
Therefore ||A| − |B|| ≤ |A − B|.
Lemma 2.4 Let A, B ∈ G and let δ̃ = (δ, 1 − δ). Then
|A − B| < δ ⇐⇒ A − δ < B < A + δ.
Proof. The inequality |A − B| < δ holds if and only if (|µA − µB |, 1 − |νA − νB |) < (δ, 1 − δ)
and it holds when |µA − µB | < δ and |νA − νB | < δ and that is µA − δ < µB < µA + δ and
νA − δ < νB < νA + δ.
On the other hand A−δ < B if and only if (µA −δ, νA +δ) < (µB , νB ) and that is µA −δ < µB
and νA + δ > νB . Similarly B < A + δ if and only if (µB , νB ) < (µA + δ, νA − δ) and that is
µB < µA +δ and νB > νA −δ. Therefore it holds µA −δ < µB < µA +δ and νA −δ < νB < νA +δ.
3
Limit of the function
To define the limit of the function on `-group G for the first we will give the definition of the
neighborhood of some point and then we will copy the definition of the limit on real-valued
function.
Definition 3.1 Denote ε̃ = (ε, 1 − ε). Let A0 , A, ε̃ be from `-group G. Point A is in a neighborhood of a point A0 if it holds
|A − A0 | < ε̃.
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Definition 3.2 Denote ε̃ = (ε, 1−ε) and δ̃ = (δ, 1−δ). Let f be a function defined on the `-group
G and let A0 , A, L, ε̃, δ̃ be from G. For a function f of a variable A defined on a neighborhood
of a point A0 except possibly the point A0 itself, for each ε̃ > (0, 1) there exists δ̃ > (0, 1) such
that |f (A) − L| < ε̃ holds whenever (0, 1) < |A − A0 | < δ̃. Then we say that the function f (A)
tends to the limit L as A approaches A0 and we write lim f (A) = L.
A→A0
Lemma 3.3 Let lim f (A) = L. Then to the point A0 there exist such δ̃ > (0, 1) that the function
A→A0
f is bounded on the set (0, 1) < |A − A0 | < δ̃.
Proof. Since lim f (A) = L then for ε̃ = (1, 0) there exist such δ̃ > (0, 1) that for each A,
A→A0
(0, 1) < |A − A0 | < δ̃ it holds |f (A) − L| < (1, 0) or else (Lemma 2.4)
L − (1, 0) < f (A) < L + (1, 0).
Therefore f is the bounded function on (0, 1) < |A − A0 | < δ̃.
Theorem 3.4 Let lim f (A) = L and lim g(A) = K. Then there are satisfied the following
A→A0
A→A0
properties
1. lim f (|A|) = |L|.
A→A0
2. lim (f (A) + g(A)) = L + K.
A→A0
3. lim (f (A).g(A)) = L.K
A→A0
f (A)
A→A0 g(A)
4. If K > (0, 1) then lim
=
L
.
K
Proof.
1. Since lim f (A) = L then for each ε̃ > (0, 1) there exists such δ̃ > (0, 1) that for each A ∈
A→A0
G, (0, 1) < |A − A0 | < δ̃ holds |f (A) − L| < ε̃. However ||f (A)| − |L|| ≤ |f (A) − L| < ε̃
and therefore lim f (|A|) = |L|.
A→A0
2. Since lim f (A) = L then for each ε̃ > (0, 1) there exists such δ˜1 > (0, 1) that for each
A→A0
A ∈ G, (0, 1) < |A−A0 | < δ˜1 it holds |f (A)−L| < 2ε̃ . And since lim g(A) = K then for
A→A0
each ε̃ > (0, 1) there exists such δ˜2 > (0, 1) that for each A ∈ G, (0, 1) < |A − A0 | < δ˜2
it holds |g(A) − L| < 2ε̃ . Put δ̃ = min{δ˜1 , δ˜2 } = (δ˜1 ∧ δ˜2 , (1 − δ˜1 ) ∨ (1 − δ˜2 )) and
consider such A that it holds (0, 1) < |A − A0 | < δ̃. Then from Lemma 2.3 it holds
|f (A) + g(A) − (L + K)| ≤ |f (A) − L| + |g(A) − K| < 2ε̃ + 2ε̃ = ε̃.
3. Since lim g(A) = K, then from Lemma 3.3 follows that there exist such δ˜1 > (0, 1) and
A→A0
such M ∈ G, M > (0, 1) that for each A ∈ G, (0, 1) < |A−A0 | < δ˜1 it holds |g(A)| ≤ M.
Put N = max{M, |L|}, then |f (A).g(A) − L.K| = |f (A).g(A) − L.g(A) + L.g(A) −
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L.K| = |(f (A) − L).g(A) + L.(g(A) − K)| ≤ |(f (A) − L)|.|g(A)| + |L|.|(g(A) − K)| ≤
ε̃
N.(|(f (A) − L)| + |(g(A) − K)|). Put ε̃0 = 2N
and find such δ˜2 > (0, 1), δ˜3 > (0, 1)
such for each A ∈ G, (0, 1) < |A − A0 | < δ˜2 it holds |f (A) − L| < ε̃0 and for A ∈ G,
(0, 1) < |A − A0 | < δ˜3 it holds |g(A) − K| < ε̃0 . Then for δ̃ = min{δ˜1 , δ˜2 , δ˜3 } it holds
N.(|(f (A) − L)| + |(g(A) − K)|) < N.(ε̃0 + ε̃0 ) = ε̃.
L
= L. (0,1)
therefore it is suffice to prove that for K >
4. For each L, K ∈ G it holds K
K
(1,0)
(1,0)
(0, 1) it holds lim g(A) = K . From the point 1. it holds lim g(|A|) = |K| > | K2 | >
A→A0
A→A0
˜
(0, 1). Therefore there exists
such δ˜1 that
for each
A ∈ G, (0, 1) < |A − A0 | < δ1 it
|K−g(A)|
− (1,0)
< |g(A)−K|
= K−g(A)
= |g(A)|.|K|
= |g(A)−K|
holds |g(A)| > K2 and (1,0)
K 2 . Put
g(A)
K g(A).K | K2 |.|K|
2 2
ε̃0 = K2 .ε̃ . Since ε̃ > (0, 1) then also ε̃0 > (0, 1) and there exist such δ˜2 that for each
A ∈ G, (0, 1) < |A − A0 | < δ˜2 it holds |g(A) − K| < ε̃0 . Then for δ̃ = min{δ˜1 , δ˜2 } it
holds
lim
A→A0
|g(A)−K|
K2 2 ε̃0
<
K2
2
(1,0)
A→A0 g(A)
= ε̃ and lim
=
(1,0)
A→A0 g(A)
f (A). (1,0)
= lim f (A). lim
g(A)
A→A0
(1,0)
.
K
f (A)
A→A0 g(A)
From these results we get lim
= L. (1,0)
=
K
=
L
.
K
Consequence 3.5 Let A, A0 , C ∈ G and let for each A there hold f (A) = C. Then lim C = C.
A→A0
Consequence 3.6 Let lim f (A) = L and lim g(A) = K. Then lim (f (A)−g(A)) = L−K.
A→A0
A→A0
A→A0
Consequence 3.7 Let lim f (A) = L, n ∈ N . Then lim f (A)n = Ln .
A→A0
A→A0
Theorem 3.8 Let f (A) ≤ g(A) ≤ h(A) and let lim f (A) = L and also lim h(A) = L. Then
A→A0
A→A0
lim g(A) = L.
A→A0
Proof. Since lim f (A) = L and lim h(A) = L then for each ε̃ > (0, 1) there exists such
A→A0
A→A0
δ˜1 , δ˜2 > (0, 1) that for each A ∈ G, holds (0, 1) < |A − A0 | < δ˜1 =⇒ |f (A) − L| < ε̃ and
(0, 1) < |A−A0 | < δ˜2 =⇒ |h(A)−L| < ε̃. Then from Lemma 2.4 follows L− ε̃ < f (A) < L+ ε̃
and also L − ε̃ < h(A) < L + ε̃. Put δ̃ = min{δ˜1 , δ˜2 } then it holds (0, 1) < |A − A0 | < δ̃ =⇒
L − ε̃ < f (A) ≤ g(A) ≤ h(A) < L + ε̃. And therefore lim g(A) = L.
A→A0
Theorem 3.9 Let lim f (A) = (0, 1) and let g(A) be a bounded function. Then
A→A0
lim (f (A).g(A)) = (0, 1).
A→A0
Proof. Since lim f (A) = (0, 1) then for each ε˜1 > (0, 1) there exists such δ˜1 > (0, 1) that for
A→A0
each A ∈ G, such that (0, 1) < |A − A0 | < δ˜1 there holds |f (A)| < ε˜1 . From the assumption that
the function g is bounded it follows that for each δ˜2 > (0, 1) there exists such K > (0, 1) that for
A ∈ G, (0, 1) < |A − A0 | < δ˜2 there holds |g(A)| < K. Put ε̃ = K.ε˜1 and δ̃ = min{δ˜1 , δ˜2 }.
Then for each A ∈ G, (0, 1) < |A − A0 | < δ̃ it holds |f (A).g(A)| = |f (A)|.|g(A)| < K.ε˜1 = ε̃.
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Theorem 3.10 Let A = (µA , νA ), A0 = (µA0 , νA0 ), L = (µL , νL ). Then
lim f (A) = L
A→A0
if and only if
lim f (µA ) = µL
µA →µA0
and at the same time
lim f (1 − νA ) = 1 − νL .
νA →νA0
Proof. Let lim f (A) = L. Then it holds (0, 1) < |A − A0 | < δ̃ if and only if 0 < |µA − µA0 | < δ
A→A0
and 0 < |νA − νA0 | < δ.
Similarly |f (A) − L| < ε̃ if and only if |f (µA ) − µL | < ε and |f (1 − νA ) − (1 − νL )| < ε.
Therefore lim f (µA ) = µL and lim f (1 − νA ) = 1 − νL .
µA →µA0
νA →νA0
Example 3.11 In the real valued analysis it holds lim sinx x = 1. In the paper [4] there was given
x→0
the definition of the function sinA. Therefore we could show that for each A ∈ G it holds
sin A
= (1, 0).
A→(0,1) A
sin µA 1−νA −sin(1−νA )
, then lim sinµAµA = 1 and lim
=
,
µA
1−νA
lim
Proof. Since
sin A
A
1 − 0 hence lim
νA −1→0
µA →0
− sin(1−νA )
1−νA
=
lim
1−νA →0
sin(1−νA )
1−νA
=
= 0 what is correct for any real number νA .
ah −1
h→0 h
Example 3.12 Let a, h ∈ R and a > 0 then lim
1−νA −sin(1−νA )
1−νA
νA →1
= ln a. Now we will show that for each
A, H ∈ G, A > (0, 1) it holds
AH − (1, 0)
lim
= ln A
H→(0,1)
H
while
ln A = (ln µA , 1 − ln(1 − νA ))
and
AH = (µA )µH , 1 − (1 − νA )(1−νH ) .
Proof. If A > (0, 1) then µA > 0 and νA < 1 i.e.
1 − νA > 0 and therefore ln A is defined.
AH −(1,0)
µA µH −1 2−(1−νA )(1−νH ) −νH
Moreover
=
,
. Hence
H
µH
1−νH
µA µH − 1
= ln µA
µH →0
µH
lim
and
(1 − νA )(1−νH ) − 1
lim 1 −
= 1 − ln(1 − νA )
νH →1
1 − νH
or after the modification
(1 − νA )(1−νH ) − 1
= ln(1 − νA ).
1−νH →0
1 − νH
lim
13
Definition 3.13 The function f is continuous at the point A0 if it holds lim f (A) = f (A0 ).
A→A0
Function f is continuous if it is continuous at each point in its domain.
Theorem 3.14 The function f is continuous at the point A0 if and only if
lim f (µA ) = f (µA0 )
µA →µA0
and at the same time
lim f (1 − νA ) = f (1 − νA0 ).
νA →νA0
Proof. Let f (A) = (f (µA ), 1 − f (1 − νA )), f (A0 ) = (f (µA0 ), 1 − f (1 − νA0 )). Then
lim f (A) = f (A0 )
A→A0
if and only if
lim f (µA ) = f (µA0 )
µA →µA0
and at the same time
lim f (1 − νA ) = 1 − (1 − f (1 − νA0 )) = f (1 − νA0 ).
νA →νA0
Example 3.15 The function y = sin A where A ∈ G is continuous.
Proof. Since sin A = (sin µA , 1 − sin(1 − νA )) then
lim sin A =
lim sin µA , lim (1 − sin(1 − νA )) =
µA →µA0
A→A0
νA →νA0
= (sin µA0 , 1 − sin(1 − νA0 )) = sin A0
for any A0 ∈ G.
Example 3.16 The function cotangent is not continuous at the point A0 = (0, 1).
Proof. cot A = (cot µA , 1 − cot(1 − νA )). Therefore
lim cot A = lim cot µA , lim (1 − cot(1 − νA ))
A→(0,1)
µA →0
νA →1
but this expression has in the both coordinates the number cot 0 which is not defined. Therefore
the function cotangent is not continuous at the point A0 = (0, 1).
Acknowledgements
This paper was supported by Grant VEGA 1/0621/11.
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References
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[3] Hollá, I. On exponencial and logarothmics functions on IF sets. Notes on Intuitionistic
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