622 Appendix B • Sets and Functions(e) VDi, Dz<:;; B, f-i(Di n Dz)= f-i(Di) n f-i(Dz).
(f) VDi, Dz<:;; B , ri(Di - Dz)= 1-i(Di) - 1-i(Dz).
Note that f-i satisfies certain universal properties that f does not.
Proof of (a): Suppose f: A---+ Bis a function, and Ci, Cz <:;;A.Part 1: Let y E f(Ci U Cz). By definition, this means :Jx E Ci U Cz 3
f(x) = y. But then, :Jx E Ci 3 f(x) = y, or :Jx E Cz 3 f(x) = y. That is,
y E f(Ci) or y E f(Cz). Thus, y E f(Ci) U f(Cz). Therefore, J(Ci U Cz) <:;;
f(Ci) u J(Cz).
P art 2: Let y E f(Ci) U l(Cz). Then y E f(Ci) or y E f(Cz). That is,
3x E Ci 3 f(x) = y, or :Jx E Cz 3 f(x) = y. In both of these two cases,
we can say :Jx E Ci U Cz 3 l(x) = y. Thus, y E J(Ci U Cz). Therefore,
f(Ci) u f(Cz) <:;; f(Ci u Cz).
By Parts 1 and 2 and Theorem B .1.7 (a), f(Ci U Cz) = l(Ci) U J(Cz).
Proof of ( e): The above proof is in two parts. We can often combine the
two parts into a single-part proof using a string of equivalent statements, as we
do here. Suppose f: A---+ Bis a function, and Di, Dz <:;; B. Then
x E f-i(Di n Dz){:} f(x) E Din Dz
{:} f(x) E Di and f(x) E Dz
{:} x E f-i(Di) and x E 1-i(Dz)
{:} x E f-i(Di) n 1-i(Dz).Therefore, 1-i(Di n Dz)= f-i(Di) n 1-i(Dz). •
Example B.2.12 Consider the function l : IR---+ IR defined by f(x) = xz + 3.
Then (see Figure B .5).
(a) f ((-1, 0) U (0, 2)) = f((-1, 2) - {O}) = [3, 7) - {3} = (3, 7), and
f(-1, 0) U J(O, 2) = (3, 4) u (3, 7) = (3, 7).(b) f ((-1, 0) n (0, 2)) = 1(0) = 0, while
f(-1, O) n f(O, 2) = (3, 4) n (3, 7) = (3, 4).
Note: in this case, l(Ci n Cz) 'I-f(Ci) n l(Cz).(c) f ((-1, 1) - (0, 1))) = f(-1,0J = [3,4), while
f(-l, 1) - J(O, 1) = [3, 4) - (3, 4) = {3}.
Note: in this case, f(Ci - Cz) 'I-f(Ci) - f(Cz).(d) f-i ((-oo,3] U (2,4)) = f-i(-oo,4) = (-1, 1), and
f-i(-oo, 3] u f-i(2, 4) = {O} U (-1, 1) = (-1, 1).