Problem Set Seven: Cluster Points and the Bolzano-Weierstrass Theorem

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Problem Set Seven: Cluster Points and the Bolzano-Weierstrass Theorem
Definition: A real number p is a cluster point of ( a n ) iff  ε  0  m  N  n  m so that a n  p  ε .
Equivalently,  ε  0 { n  N : a n  p  ε } is infinite.
Examples: If a n  a then a is a cluster point of ( a n ) . The converse is false. The sequence alternating
sign sequence a n  (  1) n has two cluster points, p  1 and p   1 .
Example: The next lines indicate one way to make a list including every rational number in [0,1].
n 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
...
rn 
0/1
1/1
0/2
1/2
2/2
0/3
1/3
2/3
3/3
0/4
1/4
2/4
3/4
4/4
0/5
...
Fractions are organized in groups determined by their denominators, with the groups listed from left to
right and the fractions increasing within each group. This list displays the terms of a sequence ( r n ) that
takes on every rational value in [0,1] infinitely often and thus has every rational in [0,1] as a cluster point.
Theorem: If p is a cluster point of a sequence ( a n ) then some subsequence of ( a n ) converges to p.
Bolzano-Weierstrass Theorem: Every bounded sequence has at least one cluster point and at least one
convergent subsequence.
Theorem (The Cauchy Criterion): A Cauchy sequence must converge.
PROBLEMS
Problem 7-1: Let r n be the integer remainder when n 2 is divided by 6, e.g., r15  3 because
15
2
 37 ( 6 )  3 . Without giving formal proofs explain why ( r n ) doesn’t converge, find all cluster points
of ( r n ) , and give a subsequence converging to each cluster point.
Problem 7-2: Prove that a bounded sequence with exactly one cluster point must converge. Can you drop
the requirement that the sequence is bounded?
Problem 7-3: Let s n 
use the inequality
1
k
2

 k2 k
n
1
k 1

1
k
2
. Prove directly that ( s n ) is a Cauchy sequence. You might again
, k > 1.
Problem 7-4: Prove that if ( b n ) is a bounded sequence then s n 
 k 1 b k 2
n
k
is a Cauchy sequence.
The next two problems refer to the sequence ( r n ) of rational numbers described in the second example of
this section.
Problem 7-5: The sequence ( r n ) takes on every rational value in [0,1] infinitely often. Prove that
implies that every number in [0,1] (rational or not) is a cluster point of ( r n ) .
Problem 7-6: Let ( a k ) be any sequence of rational numbers in [0,1]. Prove there is a strictly increasing
sequence ( n(k))
k 1
of indices such a k  r n(k) . In other words every sequence of rationals in [0,1] is a
subsequence of ( r n ) .
Problem 7-7: For a bounded sequence ( a n ) the limit superior of ( a n ) defined by
limsup a n  lub { x : x is a cluster point of (a n ) } .
 m  0 so that n  m  a n  u
.
Prove that if ( a n ) is bounded and limsup a n  u then
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