Download NESTED INTERVALS

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.