SOLUTION OF HW3
MINGFENG ZHAO
March 01, 2013
1. [10 Points] Compute the sup, inf, lim sup, lim inf, and all the limit points of xn = (−1)n +
2 sin
1
+
n
nπ
for all n ∈ N.
2
Proof. Notice that




1, if n ≡ 1 (mod 4)





nπ
=
sin
0, if 2 | n

2






 −1, if n ≡ 3 (mod 4).
So we get




1 + n1 , if n ≡ 1 (mod 4)





xn =
1 + n1 , if 2 | n







 −3 + 1 , if n ≡ 3 (mod 4).
n
Hence we have
sup xn
=
2
n≥1
inf xn
n≥1
lim sup xn
= −3
=
1
n→∞
lim inf xn
n→−∞
= −3
All limit points of {xn } are 1 and −3.
1
2
MINGFENG ZHAO
2. [10 Points] If E is a set and y a point that is the limit of two sequences, {xn } and {yn } such that
xn is in E and yn is an upper bound for E, prove that y = sup E. Is the converse true?
Proof. Since yn is an upper bound for E for all n ≥ 1, then
x ≤ yn ,
∀x ∈ E, ∀n ≥ 1.
Since yn → y as n → ∞, then
x ≤ y,
(1)
∀x ∈ E.
That is, y is an upper bound of E. On the other hand, since xn → y as n → ∞, then for any > 0,
there exists some N ∈ N such that for all n ≥ N , we have
|y − xn | < .
That is, we get y < xn + for all n ≥ N . Therefore, we know that
y = sup E.
Claim I: The converse is not true.
Let E = R, then ∞ = sup E, but we can not find any yn ∈ R such that yn → ∞, and yn is upper
bound for E.
3. [10 Points] Prove lim sup (xn + yn ) ≤ lim sup xn + lim sup yn if both limsups are finite, and give
an example where equality does not hold.
Proof. Let A = lim sup xn and B = lim sup yn , by the assumption, we know that A, B ∈ R. For any
n→∞
n→∞
> 0, by the definition of A and B, there exists some N ∈ N such that for all n ≥ N , we have
xn < A + ,
and yn < B + .
SOLUTION OF HW3
3
Hence we get
xn + yn < A + B + 2.
Which implies that
lim sup (xn + yn ) ≤ A + B + 2.
n→∞
By taking → 0, we get
lim sup (xn + yn ) ≤ A + B = lim sup xn + lim sup yn .
n→∞
Claim I: In general, the equality does not hold.
For example, let xn = (−1)n and yn = (−1)n+1 for all n ≥ 1, then xn + yn = (−1)n + (−1)n+1 = 0
for all n ≥ 1, and
lim sup xn = lim sup yn = 1.
n→∞
n→∞
But we have
lim sup (xn + yn ) = 0 < 2 = lim sup xn + lim sup yn .
n→∞
4. [10 Points] Can there exists a sequence whose set of limit points is exactly 1, 12 , 31 , · · · ?
Proof. Claim I: There is no sequence whose set of limit points is exactly 1, 12 , 13 , · · · .
If the Claim I is not true, say {xn } is a sequence whose set of limit points is exactly 1, 21 , 13 , · · · .
For any > 0, for any k ∈ N, since k is a limit point of {xn }, then there exists some nk > k such that
xn − 1 < .
k k
So we get
|xnk | <
1
+ ,
k
∀k ≥ 1.
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MINGFENG ZHAO
Now take K ∈ N such that
1
K
< , then for all k ≥ K, we have
|xnk | < 2.
That is, xnk → 0, as k → ∞. Hence 0 is a limit point of {xn }, contradiction.
5. [10 Points] Consider a sequence obtained by diagonalizing a rectangular array.
a11
% a12
%
a13
%
···
a21
% a22
%
a23
%
···
% a31
% a32
%
a33
%
···
%
%
%
..
.
Prove that any limit-point of any row or column of the array is a limit point of the sequence. Do you
necessarily get all limit points this way?
Proof. Claim I: any limit-point of any row of the array is a limit point of the sequence.
Let x be a limit-point of n-th row, then there exists some subsequence {an,nk : k ≥ 1} such that
an,nk → x, as k → ∞. By the definition of limit points of sequence {ai,j : i, j ∈ N}, we know that x
is a limit point of {ai,j : i, j ∈ N}.
Claim II: any limit-point of any column of the array is a limit point of the sequence.
Let x be a limit-point of n-th column, then there exists some subsequence {ank ,n : k ≥ 1} such
that ank ,n → x, as k → ∞. By the definition of limit points of sequence {ai,j : i, j ∈ N}, we know
that x is a limit point of {ai,j : i, j ∈ N}.
Claim III: In general, we can not get all limit points this way.
For example, let ai,j = δij , that is, ai,j = 1 if i = j, and ai,j = 0 if i 6= j. It is easy to see
that for each n-th row, we know that {an,k = δnk : k ≥ 1}, we have an,k → 0, as k → ∞, that
is, 0 is the only limit point of the n-th row {an,k : k ≥ 1}. For each n-th column, we know that
SOLUTION OF HW3
5
{ak,n = δkn : k ≥ 1}, we have ak,n → 0, as k → ∞, that is, 0 is the only limit point of the n-th column
{ak,n : k ≥ 1}. On the other hand, we know that since an,n = δnn = 1, that is, 1 is a limit point of
sequence {ai,j : i, j ≥ 1}. Therefore, we know that we can not get all limit points this way.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu