8.2 MORE ON FUNCTIONS AND MAPPINGS 275(6) Prove (b) of Theorem 6; that is, iff: A -, B, then f (MI u M,) = f (MI) u f (M,)
for any subsets M, and M, of A.
(c) Prove the "only if" part of (d) of Theorem 6; that is, iff: A -+ B has the prop-
erty that f(M1 n M,) = f(M,) n f(M,) for any subsets MI and M, of A, then
f is injective.- Let f: A -+ B, let k be a positive integer. Let MI, M,,... , Mk and N,, N,,... ,
Nk be k subsets of A and B, respectively. Use induction (recall Example 6, Article
5.4) to prove:
11. Let f: A -+ B. Let (M,J~ = 1,2, 3,.. .) and (Nklk = 1, 2, 3,.. .} beinfinitecol-
lections, indexed by N, of subsets of A and B, respectively (recall Article 4.2). Prove:
12. Let f: A -+ B, X c A, and Y G B. Prove that:
*(a) f (f - '(Y)) = Y if and only if Y c rng f. Is this result consistent with the result
in (d) of Theorem 5?
(b) f - '( f(X)) = X if and only if the restriction off to the subset f - '( f(X)) of
A is one to one. Is this result consistent with the result in (c) of Theorem 5?
- Let f: A -+ B, let M G A, and N c B. Prove:
(a) f-'(B- N)= A - f-'(N)
(6) f(M) c N if and only if M E f -'(N)
(c) Iff is a bijection, then f (M) = N if and only if M = f - '(N) - Let f: A -+ B, let MI, M, E A, and N,, N, c B. Prove:
(a) f-1(N1-N2)=f-'(N1)-f-1(N2)
(b) f(M,)-f(M,)cf(M, -M,)
*(c) Iff is one to one, then f(M ,) - f (M,) = f(M1 - M,). - (a) Prove that iff and g are functions, then dom (g 0 f) = dom f n f - '(dorn g)
[recall Exercise 8(a), Article 8.11.
(b) Prove that iff: R -+ R, if a, L E K, and if lim,,, f(x) = L, then to every E > 0,
there corresponds 6 > 0 such that (a - 6, a + 6) - (a) c f -'((L - E, L + E)).
(c) Supposel: R -+ R and that f is continuous at a E R. Prove that the inverse
image of every E neighborhood (f (a) - E, f (a) + E) of f (a) contains as a subset
some 6 neighborhood (a - 6, a + 6) of a. - Image and inverse image of sets under a mapping can be used to define the
so-called induced set functions. Suppose f: A -+ B. We define the function from
A) to 9(B), Sf: P(A) - P(B), by the rule SAX) = f(X), where X is any subset
of A; that is, X is any element of P(A). We define the mapping Sf - ,: 9(B) + *A)
by the rule Sf- ,(Y) = f -'(Y), for any subset Y of B.
(a) Prove that Sf -, is a function, even if f is not one to one (i.e., even if the
relation f - ' is not a function).