p17ru7ui281l5e1qvp1c09a1fpo4.pdf

1 • Fuzzy Set Theory 19
Then fI (X) is a continuous interval-variable and interval-valued function.
This corollary can be easily verified by using the continuity of the real
function f, which guarantees the continuity of all important interval-variable
and interval-valued functions like X n , e X , sin(X), | X | , etc.
Theorem 1.8. Let X = [x, x ], Y = [y, y ], Z = [z, z ], and S = [s, s ] be
intervals in I. Then
(1) d(X+Y,X+Z) = d(Y,Z);
(2) d(X+Y,Z+S) ≤ d(X,Z) + d(Y,S);
(3) d(λX,λY) = |λ| d(X,Y), λ∈ R;
(4) d(XY,XZ) ≤ |X| d(Y,Z).
Proof. For (1), it follows from the definition of d( . , . ) that
d(X+Y,X+Z) = max{ | (x + y) − (x + z) |, | ( x + y ) − ( x + z ) | }
= max{ | y − z |, | y − z | }
= d(Y,Z).
For (2), using the triangular inequality, part (1) above, and the symmetry of
d( . , . ), we have
d(X+Y,Z+S) ≤ d(X+Y,Y+Z) + d(Z+S,Y+Z)
≤ d(X,Z) + d(Y,S).
For (3), for any real number λ∈ R, we have
d(λX,λY) = max{ | λx −λy |, | λ x −λy | }
= |λ| max{ | x − y |, | x − y | }
= |λ| d(X,Y).
For (4), for an interval A = [a, a ], we will use l(A) = a and u(A) = a for
convenience in this proof. Then, what we need to show is
max{ | l(XY) − l(XZ) |, | u(XY) − u(XZ) | } ≤ |X| d(Y,Z).
We only show that
| l(XY) − l(XZ) | ≤ |X| d(Y,Z)
and the inequality | u(XY) − u(XZ) | ≤ |X| d(Y,Z) can be verified in the same
manner. Without loss of generality, assume that
l(XY) ≥ l(XZ);
the case of l(XY) < l(XZ) can be similarly analyzed. Then, since
XZ = { xz | x ∈ X, z ∈ Z },
there exists an x ∈ X such that
l(XZ) = l(xZ).
On the other hand, we have
xY ⊆ XY,
which implies that
l(xY) ≥ l(XY).
Hence, we have
l(xY) − l(xZ) ≥ l(XY) − l(XZ) ≥ 0,
so that
| l(XY) − l(XZ) | = l(XY) − l(XZ)