SOLUTION OF HW2
MINGFENG ZHAO
February 16, 2013
1. [10 Points] Show that there is an uncountable number of Cauchy sequences of rational numbers
equivalent to any given Cauchy sequence of rational numbers.
Proof. Let {xk : k ≥ 1} ⊂ Q be any Cauchy sequence of rational numbers. Consider the set
Y =
(y1 , · · · , yk , · · · ) : 0 ≤ yk ≤
1
, yk ∈ Q .
k
1 \
Since for all k ∈ N, 0,
Q is countable, then Y is uncountable. Notice that for all
k
1
(y1 , · · · , yk , · · · ) ∈ Y , we know that 0 ≤ yk ≤
for all k ∈ N, so lim yk = 0. Consider the
k→∞
k
set
X = {(x1 + y1 , · · · , xk + yk , · · · , ) : (y1 , · · · , yk , · · · ) ∈ X} .
Then Y is uncountable.
Claim I: For any (x1 + y1 , · · · , xk + yk , · · · ) ∈ X, then {xk + yk : k ≥ 1} is a Cauchy sequence
equivalent to {xk : k ≥ 1}.
In fact, since lim yk = 0, then {yk : k ≥ 1} is Cauchy. Since {xk : k ≥ 1}, then {xk + yk : k ≥ 1}
k→∞
is Cauchy. On the other hand, for all k ≥ 1, we have
|(xk + yk ) − xk | = |yk | ≤
1
→ 0.
k
So we know that {xk + yk : k ≥ 1} is a Cauchy sequence equivalent to {xk : k ≥ 1}.
In summary, we know that all elements in X are Cauchy sequences equivalent to {xk : k ≥ 1}.
Therefore, there is an uncountable number of Cauchy sequences of rational numbers equivalent to any
given Cauchy sequence of rational numbers.
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2
MINGFENG ZHAO
2. [10 Points] Suppose x1 , x2 , · · · and y1 , y2 , · · · are two sequences of rational numbers. Define the
shuffled sequence to be x1 , y1 , x2 , y2 , · · · . Prove that the shuffled sequence is a Cauchy sequence if and
only if x1 , x2 , · · · and y1 , y2 , · · · are equivalent Cauchy sequence.
Proof. Let {zk : k ≥ 1} be the shuffled sequence of {xk : k ≥ 1} and {yk : k ≥ 1}, then
zk =




 xi ,
if k = 2i − 1



 yi ,
if k = 2i
(=⇒) Assume that {zk : k ≥ 1} is Cauchy, that is, for any n ∈ N, there exists some K > 1 such
1
. In particular, we know that for all k1 , k2 ≥ K, then
n
1
1
2k1 , 2k2 , 2k1 − 1, 2k2 − 1 ≥ K, which implies that |z2k1 −1 − z2k2 −1 | <
and |z2k1 − z2k2 | < . By
n
n
1
1
the definition of {zk : k ≥ 1}, then |xk1 − xk2 | <
and |yk1 − yk2 | <
for all k1 , k2 ≥ K. Hence
n
n
that for all k1 , k2 ≥ K, we have |zk1 − zk2 | <
{xk : k ≥ 1} and {yk : k ≥ 1} are Cauchy. On the other hand, for all k ≥ K, then 2k − 1, 2k ≥ K,
which implies that |z2k−1 − z2k | <
1
1
. By the definition of {zk : k ≥ 1}, then |xk − yk | <
for all
n
n
k ≥ K. Hence {xk : k ≥ 1} and {yk : k ≥ 1} are equivalent Cauchy sequence.
(⇐=) Assume that {xk : k ≥ 1} and {yk : k ≥ 1} are equivalent Cauchy sequence, then there exists
some K ≥ 1 such that for all k1 , k2 , k ≥ K, we have
|xk1 − xk2 | <
1
,
n
|yk1 − yk2 | <
1
,
n
and |xk − yk | < .
Now for all m1 , m2 ≥ 2K + 1, look at |zm1 − zm2 |, we have the following three cases:
Case I: Both m1 and m2 are even numbers, that is, m1 = 2k1 and m2 = 2k2 for some k1 , k2 ∈ N,
which implies that |zm1 − zm2 | = |yk1 − yk2 |. Since m1 , m2 ≥ 2K + 1, then k1 , k2 ≥ K, which implies
that |zm1 − zm2 | = |yk1 − yk2 | <
1
.
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SOLUTION OF HW2
3
Case II: Both m1 and m2 are odd numbers, that is, m1 = 2k1 − 1 and m2 = 2k2 − 1 for some
k1 , k2 ∈ N, which implies that |zm1 − zm2 | = |xk1 − xk2 |. Since m1 , m2 ≥ 2K + 1, then k1 , k2 ≥ K,
which implies that |zm1 − zm2 | = |xk1 − xk2 | <
1
.
n
Case III: Only one of m1 and m2 is odd, without loss of generality, assume m1 = 2k1 − 1 and
m2 = 2k2 for some k1 , k2 ∈ N, which implies that |zm1 − zm2 | = |xk1 − yk2 |. Since m1 , m2 ≥ 2K + 1,
then k1 , k2 ≥ K, which implies that |zm1 − zm2 | = |xk1 − yk2 | <
In summary, we know that |zm1 − zm2 | <
1
.
n
1
for all m1 , m2 ≥ 2K + 1. Hence {zk : k ≥ 1} is Cauchy.
n
3. [10 Points] Can a Cauchy sequence of positive rational numbers be equivalent to a Cauchy sequence
of negative rational numbers?
Proof. Yes, it is possible that a Cauchy sequence of positive rational numbers be equivalent to a
Cauchy sequence of negative rational numbers.
For example, let xk =
1
1
and yk = − for all k ≥ 1. For all n ∈ N, let m = 2n, then for all
k
k
k1 , k2 , k ≥ m = 2n, we know that
|xk1 − xk2 |
≤
|xk1 | + |xk2 | =
1
1
1
1
1
+
=
+
≤
k1
k2
2n 2n
n
|yk1 − yk2 |
≤
|yk1 | + |yk2 | =
1
1
1
1
1
+
≤
+
=
k1
k2
2n 2n
n
|xk − yk |
≤
|xk | + |yk | =
1
1
1
1
1
+ ≤
+
=
k k
2n 2n
n
Then we know that {xk : k ≥ 1} and {yk : k ≥ 1} are equivalent Cauchy sequence.
4. [10 Points] Let x be a real number. Show that there exists a Cauchy sequence of rationals x1 , x2 , · · ·
representing x such that xn ≤ xn+1 for every n.
4
MINGFENG ZHAO
Proof. Let x be a real number, since x − 1 < x < x + 1, then there exists x1 , M ∈ Q such that
x − 1 < x1 < x < M < x + 1. Since x1 < x, then there exists x2 ∈ Q such that x1 <
x + x2
< x3 < x. By induction, there exists
2
Since x2 < x, then there exists x3 ∈ Q such that x2 <
xn+1 ∈ Q such that xn <
x + xn
< xn+1 < x. Hence we know that
2
0 < xn+1 − xn <
Since xn >
x + x1
< x2 < x.
2
x − xn
.
2
x + xn−1
, then
2
x − xn−1
x − xn
<
.
2
22
By induction, we know that x − xn <
x − x1
M −1
<
and
2n
2n
0 < xn+1 − xn <
x − x1
M − x1
<
,
n
2
2n
∀n ≥ 1.
So for all k ∈ N, we have
xn+k − xn =
k−1
X
k−1
X
i=1
i=1
(xn+i − xn+i−1 ) <
M − x1
M − x1
<
,
n+i−1
2
2n−1
∀n ≥ 1.
So we know that {xn : n ≥ 1} is a Cauchy sequence of rational numbers. Since x − xn <
x − x1
<
2n
M −1
, then we know that {xn : n ≥ 1} can represent x with xn ≤ xn+1 for all n ∈ N.
2n
5. [10 Points] Prove that there are an infinite number of rational numbers in between any two distinct
real numbers.
Proof. Let x and y be two distinct real numbers, then x − y 6= 0. Without loss of generality, assume
that x − y > 0. By the Axion of Archimedes, Theorem 2.2.4 on Page 47, then there exists some n ∈ N
such that
x−y ≥
1
.
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SOLUTION OF HW2
5
1
1
>y+
> y, which implies that there exists rational numbers r1 , r2 ∈ Q such
n
2n
r1 − r2
: k ≥ 1 between
that x > r1 > r2 > y. Since there are infinitely many rationals like r2 +
k
That is, x ≥ y +
r2 and r1 , then there are infinitely many rationals between y and x.
6. [10 Points] Prove that a > b > 0 implies
Proof. Since b > 0, then
√
a > 0, then
√
b+
√
√
√
a>
b > 0. Assume that
√
√
b > 0.
a≤
√
b, that is,
√
b−
√
a ≥ 0. Since
√
b > 0 and
√
√ √
√
a > 0. By Theorem 2.2.2 on Page 46, then ( b − a)( b + a) > 0, that is,
b − a > 0, which implies that b > a, we get a contradiction. Therefore, we have
√
a>
√
b > 0.
7. [10 Points] Prove that if lim xk = x and xk ≥ 0 for all k, then lim
k→∞
k→∞
√
xk =
√
x.
Proof. Case I: x = 0. Since lim xk = x = 0, then for all n ∈ N, there exists some m ∈ N such that
k→∞
for all k ≥ m, we have |xk | <
1
n2 ,
which implies that
√
So we know that lim
√
k→∞
x>
xk <
1
1
= ,
n2
n
∀k ≥ m.
xk = 0.
Case II: x > 0. Since x > 0, then
√
r
√
x > 0, which implies that there exists some K ∈ N such that
1
. Since lim xk = x, then for all n ∈ N, there exists some m ∈ N such that for all k ≥ m,
k→∞
K
we have |xk − x| <
1
nK .
Notice that
√ √
√
√
√
| xk − x|| xk + x|
√
|xk − x|
|x − x|
1
1
1
√
√ ≤ k√
=√
<√ ·
| xk − x| =
< .
√
n
| xk + x|
xk + x
x
x nK
Hence we know that lim
k→∞
√
xk =
√
x.
In summary, we can conclude that lim
k→∞
√
xk =
√
x.
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MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu