Math 3520
Homework 6 Solutions
1. (p84,#2) If a bounded sequence is the sum of a monotone increasing and a monotone decreasing sequence (xn = yn + zn where {yn } is monotone increasing and {zn } is monotone
decreasing) does it follow that the sequence converges? What if {yn } and {zn } are bounded?
Solution: For a counter-example to the unbounded case, consider {yn } = {n2 } and
{zn } = {−n}. In this case, {xn } = {n2 − n} which does not converge. Similarly,
if one is unbounded and the other is bounded. Suppose that both {yn } and {zn } are
bounded sequences. Bounded monotone sequences are convergent. Let y = lim {yn }
n→∞
and z = lim {zn }. Then {xn } = {yn + zn } converges to y + z. Indeed, given N ∈ N
n→∞
pick m ∈ N large enough so that | y − yj | < 1/2N and | z − zj | < 1/2N for all j ≥ m.
Then, given N ∈ N there exists an m ∈ N such that
| y + z − (yj + zj ) | ≤ | y − yj | + | z − zj |
1
1
+
≤
2N
2N
1
=
N
for all j ≥ m. Hence {xn } → y + z.
Note that what we really have shown is that if two sequences are convergent, then so is
their sum (this was shown when we defined addition of real numbers). The assumptions
that the sequences were bounded and monotone (in/de)creasing allowed us to conclude
they were convergent.
2. (p84,#5) Prove lim sup{xn + yn } ≤ lim sup{xn } + lim sup{yn } if both lim sups are finite,
and give an example where equality does not hold.
Solution: Note that xj ≤ sup{xj }j≥k and yj ≤ sup{yj }j≥k for all j ≥ k so that
sup{xj }j≥k + sup{yj }j≥k is an upper bound for {xj + yj }j≥k . In particular,
sup{xj + yj }j≥k ≤ sup{xj }j≥k + sup{yj }j≥k
1
Since limits preserve non-strict inequalities
lim sup{xj + yj }j≥k ≤ lim sup{xj }j≥k + lim sup{yj }j≥k
k→∞
k→∞
k→∞
which yields the desired result.
An example where equality does not hold is {xn } = {0, 1, 0, 1, 0, 1, . . .} and {yn } =
{1, 0, 1, 0, 1, 0, . . .} so that {xn + yn } = {1, 1, 1, 1, 1, 1, . . .}. In this case, lim sup{xn } =
lim sup{yn } = lim sup{xn + yn } = 1,
3. (p84,#7) Construct a sequence whose set of limit points is exactly the integers.
Solution: One possible sequence whose set of limit points is Z will contain the subsequence {n, n, n, . . .} for all n. Keep in mind the sequence must reach every single integer,
and when it does it has to repeat it an infinite number of times afterward. Such a sequence
is given by:
{−1, 0, 1, −2, −1, 0, 1, 2, . . . , −n, −n + 1, . . . , −1, 0, 1, . . . , n − 1, n, . . .}
| {z } |
|
{z
}
{z
}
1
2
n
4. Prove the following theorem.
Theorem. The lim inf of a bounded sequence {xj } is a limit point and, in particular, it is the
infimum of the set of limit points of {xj }.
Solution: Let {xj } be a bounded sequence and let yk = inf{xj }j≥k . In particular, by the
definition of inf, for all N ∈ N, there exists an ` > k such that
1
| y k − x` | < .
(1)
N
Here {yk } is a non-decreasing sequence whose limit is the lim inf{xj }. In particular, for
all N ∈ N, there exists an m ∈ N such that
1
| lim inf{xj } − yk | <
(2)
N
for all k ≥ m. Using the triangle inequality, together with (1) and (2), for all N ∈ N, there
exists an ` ∈ N such that
| lim inf{xj } − x` | = | lim inf{xj } − yk + yk − x` |
≤ | lim inf{xj } − yk | + | yk − x` |
1
1
2
+
=
≤
N
N
N
Page 2
Hence, lim inf{xj } is a limit point of {xj }. To show it is the infimum of the set of limit
points of {xj } it suffices to show that lim inf{xj } is the smallest limit point. If x is any
other limit point, then there is a subsequence {x0j } → x. Since yk = inf{xj }j≥k ≤ x0j for
all j ≥ k
lim inf{xj } = lim y ≤ lim {x0k } = x
k→∞
Page 3
k→∞