646 Appendix C • Answers & Hints for Selected ExercisesEXERCISE SET 2. 7(^1) • (a) m n ' > l g =? 1 i m - .l n I < min{m,n} 1 - max{m,n} 1 < min{m (^1) ,n} < c: ·
( ) l I m _ n 1-1 mn
(^2) +m-nm (^2) -n I - l(n-m)(mn-1)1
c n > m > g =} m2+1 n2+1 - (m2+1)(n2+1) - (m2+l)(n2+1)
< n(mn) rn:2n2 = -1. m < c:.
- (b) I m:,;-^1 - n
2
n+l I = I (m - n) + ( ~ - ~)I· Note that if n 2:: 2 and m = n+k
for some k > 1 then I m
2
+^1 - n
2- ' m n +^11 = k + -n+k^1 - - .l n = k _ n(n+k) k > k _ .l n > -
1 - ~ = ~-Thus, 1'no EN 3 m, n 2:: no=? lxn - Xml < ~-
- Form< n, lxm - Xnl::::; lxm - Xm+il + lxm+l - Xm+2I + · · · + lxn-1 - Xnl
<cm+ cm+i + ... +en< cm(l + c + +cn-m) <cm (l~C) (see
Exercise 2.3.6). By Thm. 2.3.7, cm-> 0. - Suppose a Cauchy sequence {xn} has c as a cluster point. Then it has a
subsequence {xnk} converging to c. Follow the proof of Thm. 2.7.4 from the
second paragraph, using c instead of L. - (a) Modify t he proofs of Thm. 2.2.13. For example, for Cauchy sequences
{xn}, {Yn}, l(xm + Ym) - (xn - Yn)I ::::; lxm - Xnl + IYm - Ynl < ~ + ~-To see
that {h -"'n.} is not necessarily Cauchy, take Xn = .l n and Yn = ~. n
(b) A sufficient condition: {yn} f+ 0.(^9) • ( ) a I Xn+2 - Xn+l 1 -1 - Xn+1+Xn 2 - Xn+l I --lxn+1+xn 2 -2Xn+1I _ - 2 11 Xn+l - Xn I.
(b) Xn+l = ~Xn + ~Xn-1 =? Xn+l + ~Xn = Xn + ~Xn-1 = · · · = b +~a.
(c) Let Xn-> L. Then L+ ~L = b+ ~a, so L = a~^2 b.
- By hypothesis, :le 3 0 < c < 1 and \:/x, y EI, lf(x)-f(y)I < clx-yj. Pick
any X1 EI. \:/n EN, define Xn+i = f(xn)· Then lxn+2 - Xn+1I = lf(xn+1) -
f(xn)I ::::; clxn+l - Xnl· By Ex. 8, {xn} converges; say Xn -> L. By Exercise
2.3.18, LE [a,b]. Let c: > 0. Then 3no EN 3 n 2:: no=? lxn - LI<~< {c =?
lf(L)-LI ::::; lf(L)-f(xn)I + IJ(xn)-LI ::::; clL- xnl + lxn+l -LI < c{c + ~ = c:.
By the forcing principle, J(L) = L.
If f(L') = L' for some L -=f. L' E A , then IL - L'I = lf(L) - J(L')I ::::;
clL - L'I < IL - L'I, a contradiction. - For a 2:: 1 define X1 =a and Xn+l =a+ L. First prove by induction that
a+ 21 a ::::; Xn ::::; 2a, and use this to prove that }n ::::; 2 a;".j.. 1 < ~ ::::; 1. Show that
Jxn+l - Xnl = lx;n~:':._~^11 ::::; ( 2 a;".j.. 1 f lxn - Xn-1 I· Apply Exercise 8 to conclude
that {xn} converges, say Xn-> L. Show L =a+ -f;, and hence L = a+v;p+4.