1

2
SOLUTIONS FOR HOMEWORK 6
Consequently (Theorem 10.7), lim sup
d
n
= lim inf
d
n
= 6
/
7 (the se-
quence converges, hence, by Theorem 9.1, it is bounded). The set of
subsequential limits of (
d
n
) is
{
6
/
7
}
.
11.9.
(a) We have to show that, for any sequence (
s
n
)
⊂
[
a, b
], conver-
gent to
s
, we have lim
s
n
∈
[
a, b
]. However, by Exercise 8.9 (we proved
this in class),
a
≤
s
≤
b
, which is what we need.
(b) The open interval (0
,
1) is not closed. Consider, for instance, the
sequence
s
n
= 1
/
(
n
+1). Then
s
n
∈
(0
,
1) for any
n
, yet lim
s
n
/
∈
(0
,
1).
By Theorem 11.8, (0
,
1) cannot be the set of subsequential limits for
any sequence.
12.4.
s
k
≤
sup
n>N
s
n
and
t
k
≤
sup
n>N
t
n
for any
k > N
.
Thus,
s
k
+
t
k
≤
sup
n>N
s
n
+ sup
n>N
t
n
for any
k > N
. Taking the supremum
of the left hand side over
k
, we obtain:
sup
k>N
(
t
k
+
s
k
)
≤
sup
n>N
s
n
+ sup
n>N
t
n
.
Therefore,
lim sup(
t
k
+
s
k
) = lim
N
sup
k>N
(
t
k
+
s
k
)
≤
lim
(
sup
n>N
s
n
+ sup
n>N
t
n
)
= lim sup
n>N
s
n
+ lim sup
n>N
t
n
= lim sup
s
n
+ lim sup
t
n
.
12.5.
By Exercise 11.8(a), lim inf
a
n
=
−
lim sup(
−
a
n
) for any se-
quence (
a
n
).
By Exercise 12.4, lim sup(
−
t
n
−
s
n
)
≤
lim sup(
−
t
n
) +
lim sup(
−
s
n
). Therefore,
lim inf(
t
n
+
s
n
) =
−
lim sup(
−
t
n
−
s
n
)
≥ −
(
lim sup(
−
t
n
) + lim sup(
−
s
n
)
)
=
−
lim sup(
−
t
n
)
−
lim sup(
−
s
n
)
= lim inf
s
n
+ lim inf
t
n
.
12.8.
It suffices to show that lim sup
s
n
t
n
≤
st
for any
s >
lim sup
s
n
and
t >
lim sup
t
n
.
Recall that lim sup
s
n
= lim
n
v
n
, where
v
n
= sup
k
≥
n
s
k
.
Therefore,
there exists
N
1
∈
N
s.t.
v
n
< s
whenever
n
≥
N
1
. Therefore,
s
k
< s
for
k
≥
N
1
. Similarly, there exists
N
2
∈
N
s.t.
t
k

#### You've reached the end of your free preview.

Want to read all 4 pages?

- Fall '08
- staff
- Limits, lim, Supremum, lim sup, subsequence, lim sup sn