* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download NESTED INTERVALS
Line (geometry) wikipedia , lookup
List of important publications in mathematics wikipedia , lookup
Law of large numbers wikipedia , lookup
Infinitesimal wikipedia , lookup
Elementary mathematics wikipedia , lookup
Fermat's Last Theorem wikipedia , lookup
Interval arithmetic wikipedia , lookup
Nyquist–Shannon sampling theorem wikipedia , lookup
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Four color theorem wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Fundamental theorem of calculus wikipedia , lookup
Non-standard calculus wikipedia , lookup
NESTED INTERVALS 22. Prove that to every set of nested intervals ½an; bn, n ¼ 1; 2; 3; . . . ; there corresponds one and only one real number. By definition of nested intervals, an1Aan; bn1@bn; n ¼ 1; 2; 3; . . . and lim ًanbn ¼ ق0. n!1 Then a1@ an@ bn@ b1, and the sequences fang and fbng are bounded and respectively monotonic increasing and decreasing sequences and so converge to a and b. To show that a ¼ b and thus prove the required result, we note that a ¼ً b bnً قbnanً قanaق ً1ق aj @ jb bnj jbnanj janaj ً2ق b jb Now given any > 0, we can find N such that for all n > N jb so that from (2), jb aj < . bnj < =3; Since jbnanj < =3; ً3ق janaj < =3 is any positive number, we must have b a ¼ 0 or a ¼ b. 2.23. Prove the Bolzano–Weierstrass theorem (see Page 6). Suppose the given bounded infinite set is contained in the finite interval ½a; b . Divide this interval into two equal intervals. Dividing ½a1; many points. Then at least one of these, denoted by ½a1; b1, contains infinitely many points. b1into two equal intervals, we obtain another interval, say, ½a2; b2, containing infinitely Continuing this process, we obtain a set of intervals ½ an ; bn , n ¼ 1; 2; 3; . . . ; contained in the preceding one and such that b1a1¼ً b a=ق2; b2a2¼ً b1a1=ق2 ¼ً b from which we see that lim ًbnan ¼ ق0. a=ق22; . . . ; bnan¼ً b a=ق2n each interval n!1 This set of nested intervals, by Problem 2.22, corresponds to a real number which represents a limit point and so proves the theorem. CAUCHY’S CONVERGENCE CRITERION 2.24. Prove Cauchy’s convergence criterion as stated on Page 25. Necessity. Then given any > 0, we can find N such that Suppose the sequence fung converges to l. juplj < =2 for all p > N and juqlj < =2 for all q > N Then for both p > N and q > N, we have jupuqj ¼ jًuplً قl uqقj @ juplj jl uqj < =2 =2 ¼ Sufficiency. Suppose jupuqj < for all p; q > N and any > 0. Then all the numbers u N; uN1;. . . lie in a finite interval, i.e., the set is bounded and infinite. Hence, by the Bolzano–Weierstrass theorem there is at least one limit point, say a. If a is the only limit point, we have the desired proof and lim u n¼ a. n!1 Suppose there are two distinct limit points, say a and b, and suppose b > a (see Fig. 2-1). By definition of limit points, we have jupaj < ًb a=ق3 for infinnitely many values of p juqbj < ًb a=ق3 Then since b jb a ¼ً b aj ¼ b for infinitely many values of ً1ق q ً2ق uqj jupuqj jupaj b_a 3 a uqً قuqupً قupaق, we have a @ jb b_a 3 ً3ق b Fig. 2-1 Using (1) and (2) in (3), we see that jupuqj > ًb a=ق3 for infinitely many values of p and q, thus contradicting the hypothesis that jupuqj < for p; q > N and any > 0. Hence, there is only one limit point and the theorem is proved.