370 ANSWERS AND SOLUTIONS TO SELECTED EXERCISES
3m2 + 3m + 1 < 3m2 + 3mZ + 3m2 = 9m2 < m(m2) = m3. Hence
m3 + 3m2 + 3m + 1 < m3 + m3 c 2" + 2" = 2(2") = 2"+', as required.- (a) Proof "If m E 0, then m + 1 E 0'' is true since "m E 0'' is false for any
mEN.
Article 6.1
(b) (i) [ - (p o q)] t, Cp t, - q]. You should verify, by using the truth
table, that this statement form is a tautology.
(a) Proof Let X, Y, and Z be sets. We establish equality by proving mutual
inclusion. Let t E (X u Y) x Z. Then t = (p, q) where p E X u Y and q E Z.
Now p E X u Y means either p E X or p E Y. In the case p E X, we have
t=(p,q)~X x Zc(X x Z)u(Y x 2). Ifp~ Y, thent=(p,q)~ Y x ZE
(X x Z) u (Y x 2). These two cases are exhaustive and, in either case, we
have t E (X x Z) u (Y x Z), as desired. Conversely, suppose
t E (X x Z) u (Y x Z), so that either t E (X x 2) (in which case t = (x, 2,)
where x E X and z, E Z) or t E (Y x Z) [so that t = (y, z,) where y E Y and
2, E Z]. In the first case t = (x, z,) where x E X E X u Y and 2, E Z. In
the second case t = (y, z,) where y E Y E X u Y and z, E Z. In either case
t E (X u Y) x Z, as desired.
(b) Proof Assume S is a convex subset of R. To prove S is an interval,
suppose x, y, and z are real numbers with x < y c z, x E S, and z E S. We
must prove y E S. Now, by (a), there must exist a real number t, with
0 c t < 1, such that z = tx + (1 - t)y. Since S is convex, then for any x, z E S
and t E (0, l), we may conclude tx + (1 - t)y E S. Hence z = tx + (1 - t)y E S,
as desired. 0
(b) A real number a is not a point of accumulation of S if and only if 36 > 0
such that N1(a; 6) n S = 0; in other words, there exists a 6 neighborhood
of the point a containing no points of S other than, possibly, a itself. To
prove that 2 is not a point of accumulation of S = (0, 1) u (21, let 6 = 3
(or any other specific positive real number less than 1). Clearly
N(2,i) = ($, 5) contains no points of S, other than 2 itself, as required.
(b) Proof Let e > 0 be given. We must produce 6 > 0 such that, whenever
0 c 1x1 c 6, then (g(x)( < E. Let 6 = E. Suppose now that x is a real number
satisfying 0 < 1x1 < 6. If x is irrational, then Ig(x)l = 0 < e. If x is rational,
then lg(x)I = 1x1 < 6 < E, SO that lg(x)l c e. In either case we have the
desired conclusion. 0Article 6.2
- (e) Proof Let A, B, and C be sets. Let t E (A x C) - (B x C). To prove
t E (A - B) x C, we must prove that there exist x E A - B and y E C such
that t = (x, y). Now t E (A x C) - (B x C) means t E A x C so 3x E A, y E C
such that t = (x, y). We will be finished if we can prove x 4 B, so that