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Math 131A  Section 2
Spring 2010
Homework 4
All problems graded (out of 3 points for this short assignment).
10.1(b). bounded; (d) bounded; (f) bounded, nonincreasing
10.2. Let (
s
n
) be a bounded nonincerasing sequence. Let
S
=
{
s
n

n
∈
N
}
and
u
= inf
S
, which is a real number since
S
is bounded. Let
± >
0. Then,
u
+
±
is not a lower bound of
S
, so there exists
N
such that
s
N
< u
+
±
. Then, since
(
s
n
) is nonincreasing, we have that
u
≤
s
n
≤
s
N
< u
+
±
for all
n > N
, and so

s
n

u

< ±
for
n > N
. Therefore,
lim
n
→∞
s
n
=
u
= inf
S
, which in particular shows (
s
n
) converges.
NOTES: This is almost exactly like the proof of Theorem 10.2 given in the textbook. Some care needed to be made in
making the needed changes (make sure the inequalities all go in the correct direction!).
10.5. Let (
s
n
) be an unbounded nonincerasing sequence. Then, since
{
s
n

n
∈
N
}
is bounded above by
s
1
, we must
have that it is unbounded below. Let
M <
0. Then, there exists
N
such that
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This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.
 Spring '10
 Kim

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