Math 131A Analysis Summer Session A
Homework 4 Solutions
1. Suppose that (
s
n
) is a bounded sequence such that all convergent
subsequences have limit
s
∈
R
. Prove (
s
n
) converges to
s
.
Proof.
Since (
s
n
) is bounded, any subsequence of (
s
n
) is bounded. By Corollary
11.4, (
s
n
) contains a bounded monotonic subsequence that has limit lim sup
s
n
and one that has limit lim inf
s
n
. Since bounded monotonic sequences converge
and by hypothesis every convergent subsequence has limit
s
,
lim sup
s
n
=
s
= lim inf
s
n
. Thus
lim
n
→∞
s
n
=
s
.
2. (a) Let
S
be a nonempty subset of
R
and let

S
=
{
s
:
s
∈
S
}
.
Prove inf
S
=

sup(

S
). (Handle the cases inf
S
=
∞
and
inf
S >
∞
separately.)
(b) Let (
s
n
) be any sequence. Prove lim inf
s
n
=

lim sup(

s
n
).
Proof.
(a) Suppose inf
S
∈
R
. Then sup

S
∈
R
since inf
S
≤
s
for all
s
∈
R
implies

s
≤ 
inf
S
for all
s
∈
R
. Therefore, sup

S
≤ 
inf
S <
∞
. To show

sup

S
is a lower bound for
S
, note

s
≤
sup

S
for all
s
∈
S
by definition.
Thus

sup

S
≤
s
for all
s
∈
S
. Let
a
be any lower bound for
S
. Thus for all
s
∈
S
,
a
≤
s
or equivalently,

s
≤ 
a
. So

a
is an upper bound for the set

S
. Hence, sup

S
≤ 
a
by definition of sup and we get
a
≤ 
sup

S
.
Suppose inf
S
=
∞
.
Then
S
is not bounded below which means

S
is not
bounded above. For if there exists
M >
0 such that

s
≤
M
for all
s
∈
S
, then

M
≤
s
for all
s
∈
S
which contradicts
S
not having a lower bound.
Thus
sup

S
=
∞
which is equivalent to

sup

S
=
∞
. Recall sup

S > M
for
any
M >
0 if and only if

sup

S < C
for any
C <
0.
(b) Letting
S
N
=
{
s
n
:
n > N
}
, we see

S
N
=
{
s
n
:
n > N
}
.
By (a)
inf
S
N
=

sup

S
N
. Thus
lim inf
s
n
=
lim
N
→∞
inf
S
N
=
lim
N
→∞

sup

S
N
=

lim
N
→∞
sup

S
N
=

lim sup

s
n
1
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3. Let (
a
n
) and (
b
n
) be bounded sequences and
c
∈
R
such that
c
≥
0.
Prove
(a) lim sup(
a
n
+
b
n
)
≤
lim sup
a
n
+ lim sup
b
n
(b) lim inf(
a
n
+
b
n
)
≥
lim inf
a
n
+ lim inf
b
n
(c) lim sup(
ca
n
) =
c
lim sup
a
n
(d) lim inf(
ca
n
) =
c
lim inf
a
n
Proof.
(a) Claim for every
N
∈
N
,
sup
{
a
n
+
b
n
:
n > N
} ≤
sup
{
a
n
:
n > N
}
+ sup
{
b
n
:
n > N
}
Since (
a
n
) and (
b
n
) are bounded, sup
{
a
n
+
b
n
:
n > N
}
, sup
{
a
n
:
n > N
}
and sup
{
b
n
:
n > N
}
are real for every
N
and are bounded monotonic sequences,
hence they converge: If for some positive constants
M
1
and
M
2
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 Summer '08
 hitrik
 Math, 1 m, subsequence, 0 205 1 104 103k, 1 103k

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