ANSWERS AND SOLUTIONS TO SELECTED EXERCISES^3718.9.IS.
x E A - B. To do this, suppose x were an element of B. Then we would have
t = (x, y) E B x C, contradicting the fact that t E (A x C) - (B x C). Hence
x 4 B, so x E A - B and t E (A - B) x C, as desired. 0
(g) Proof Let A and B be subsets of U such that A' u B = U. To prove
A ,c B, let x E A. Hence x 4 A'. If x 4 B, then x 4 A' u B. This contradicts
the hypothesis A' u B = U (which says that every element of U is either in B
or in A'). 0
Proof Suppose f'(x,) # 0. Then since f'(x,) exists, we may assert that either
f'(xo) > 0 or f' (x,) < 0. Iff' (x,) > 0, then by Exercise lqc), Article 6.1,
there exists a neighborhood N(xo; 6) such that if x E N(x,; 6) and x < x,, then
f (x) < f(xo) and if x E N(xo; 6) and x > x,, then f (x) > f (x,). This contradicts
the hypothesis that f has a relative maximum at x,. A similar argument
can be formulated to reach a contradiction in the case f'(x,) < 0.
(c) Proof If B were a subset of C, we would have A G B and B c C, so that
A E C, contradicting part of the hypothesis. Cl
(d) Proof If not, then both ,h + x and a - x are rational, so that their
sum 2& is rational. But by (c), since 2 E Q and q! Q, 2& q! Q, so that
we have a contradiction. 0
(c) Proof If 0 is not open, then 3x E 0 and 6 > 0 such that N(x; 6) is not
a subset of 0. But the statement "3x E @" is false, so we have a
contradiction.
(b) Proof Assume A, B, and C are sets with A c B and 3 c C. If A $ C,
then there exists an element x E A such that x $ C. Now this x is either in B
or not in B. If x E B, then the fact that x E B, but x 4 C contradicts B E C.
If x 4 B, then the fact that x e A, but x 4 B contradicts the fact that A E B.
In either case we arrive at a contradiction, so that our assumption A $ C must
be incorrect, as desired. (Compare this proof with the proof in Example 7,
Article 4.1.)Article 6.3
I. (c) proof J(4x+ 36) = x + 8 -4x + 36 =(x+ 812 = x2 + 16x+ 64 -
x2 + 12x + 28 = 0 * x = -6 f 2,h. Hence -6 + 2,h and -6 - 2a are
the only possible solutions. Substitution shows that only - 6 + 2& is an
actual solution. (Note: In substituting, we must compare 2 + 23 with
2J- (equal, since both are positive and both have square 12 + 8 &)
and 2 - 2 & with 2 Js [negatives of each other since both have
square 12 - 8&, but 2- 2,h<O, whereas ~J~>oI.- (b) Proof First, note that X = 0 satisfies the statement A n X = X for all
sets A in U. For uniqueness, suppose X is any set satisfying "A n X = X for
all subsets A of U." In particular (using specialization), letting A = 0, we
have 0 n X = X. But again, it is known that 0 = 0 n X so that we have
0 = 0 n X = X, so that X = 0, our desired conclusion. 0 - (c) Proof First, it is clear that if x, + x, then x is a cluster point of {x,).
In detail, let E > 0 and N E N be given; we must produce m > N such that
Ix, - XI < E. Now since x, -* x, then corresponding to this e, there exists a