Download Solutions to Hw 2- MTH 4350- W13

Solutions to Homework Assignment 2
Real Analysis I
February 12, 2013
Notes: (a) Be aware that there maybe some typos in the solutions. If you
find any, please let me know.
(b) As is usual in proofs, most problems can be proved in many different
ways. These solutions contain only one of these multiple ways, not the only
way.
Section 1.4
Problem 1.4.7 Assume B is a countable set. Thus, there exists f : N → B,
which is 1 − 1 and onto. Let A ⊂ B be an infinite subset of B. Show that A is
countable.
Proof. We will define a 1 − 1 and onto function g from N to A by induction on
k:
• Base Case: When k = 1, let n1 = min{n|f (n) ∈ A} and set g(1) = f (n1 ).
• Induction Step: Assume that g(k) has been defined and we need to
define g(k + 1). Let nk+1 = min{n|f (n) ∈ A and n 6= n1 , . . . , nk } and set
g(k + 1) = f (nk+1 ).
Observe that n1 < n2 < . . . and since A is an infinite set, the above definition
defines g on all natural numbers.
Claim 1. g is a 1 − 1 function.
Proof of Claim. Towards contradiction assume that g is not 1 − 1. I.e. there
are k1 , k2 ∈ N such that k1 < k2 and g(k1 ) = g(k2 ). Equivalently
f (nk1 ) = f (nk2 ). Since f is 1 − 1, it must be that nk1 = nk2 . But,
nk2 = min{n|f (n) ∈ A and n 6= n1 , . . . , nk2 −1 }. Since k1 < k2 , nk2 must be
different than nk1 . Contradiction.
Claim 2. g is an onto function.
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Proof of Claim. Let a ∈ A. Since f is onto, there exists some n ∈ N such that
f (m) = a. Let k be the smallest numbers such that nk > m. I.e. nk > m and
n1 < n2 < . . . < nk−1 ≤ m. If nk−1 = m, then g(k − 1) = f (nk−1 ) = f (m) = a
and a is in the range of g. If nk−1 < m, then
m ∈ {n|f (n) ∈ A and n 6= n1 , . . . , nk−1 }.
So, m ≥ min{n|f (n) ∈ A and n 6= n1 , . . . , nk−1 } = nk , which is a contradiction
since m < nk . So, this case can not happen and we conclude that a is in the
range of g. Equivalently, g is onto.
Since g is both 1 − 1 and onto, A is countable.
Problem 1.4.8 (a) If A1 , A2 are countable, then A1 ∪ A2 is also countable.
Generalize for unions of finitely many A1 , . . . , An .
(b) Explain why induction can not be used for part (b)
S of Theorem 1.4.13.
(c) Prove that if An is countable for all n ∈ N, then n An is also countable.
Proof. (a) If A1 and A2 are not disjoint, i.e. A1 ∩ A2 6= ∅, then let
B2 = A2 \ A1 . B2 is a subset of A2 , B2 ⊂ A2 , and B2 is disjoint from A1 ,
B2 ∩ A1 = ∅. B2 ⊂ A2 implies that A1 ∪ B2 ⊂ A1 ∪ A2 . On the other hand, if
x ∈ A1 ∪ A2 , then x ∈ A1 ∪ B2 . This is obvious if x ∈ A1 . If x ∈
/ A1 , then x
must be in A2 . So, x ∈ A2 \ A1 , which is equivalent to x ∈ B2 . In either case,
x belongs to A1 ∪ B2 , which implies that A1 ∪ A2 ⊂ A1 ∪ B2 . Altogether,
A1 ∪ A2 = A1 ∪ B 2 .
By problem 1.4.7, B2 is either finite or countable.
Case I (a bit easier): B2 is finite. Then there exists a 1-1 and onto function
f : N → A1 and some natural number N (the size of B2 ) and a 1-1 and onto
function g : {1, 2, . . . , N } → B. Consider the function h : N → A1 ∪ B2 defined
by:
(
g(n),
if n ∈ {1, 2, . . . , N }
h(n) =
f (n − N ), if n > N
It is immediate that h is defined for all numbers n ∈ N.
If x ∈ A1 , then by virtue of f being onto, there exists some n ∈ N such that
f (n) = x. Then h(n + N ) = f (n) = x, i.e. x is in the range of h. If x ∈ B2 ,
then by virtue of g being onto, there exists some n ∈ {1, 2, . . . , N } such that
g(n) = x. Then h(n) = g(n) = x, i.e. x is in the range of h. Overall,
range(h) = A1 ∪ B2 and h is an onto function.
Moreover, h is also a 1-1 function. Let n1 , n2 ∈ N be such that h(n1 ) = h(n2 ).
Since h(n1 ), h(n2 ) are equal, they are both in A1 , or both in B2 (A1 is disjoint
from B2 ). If they are both in B2 , then g(n1 ) = h(n1 ) = h(n2 ) = g(n2 ) and
since g is a 1-1 function, n1 = n2 . If both h(n1 ), h(n2 ) are in A1 , then
f (n1 − N ) = h(n1 ) = h(n2 ) = f (n2 − N ) and since f is a 1-1 function,
n1 − N = n2 − N and n1 = n2 . So, in either case, n1 = n2 which proves that h
is a 1-1 function.
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Case II (a bit harder): B2 is countable. Then there exist 1-1 and onto
functions f : N → A1 and g : N → B2 . Define h : N → A1 ∪ B2 by:
(
g( n2 ),
if n is even
h(n) =
n+1
f ( 2 ), if n is odd
Again, h is defined for all natural numbers n ∈ N and working as before we
can prove that h is 1-1 and onto.
Now, we can generalize the result to the unions of arbitrary A1 , A2 , . . . , An by
induction on n:
• n=2. We just proved that.
Sn
• n > 2. Assume that for A1 , . . . , An countable, i=1 Ai is countable. We
Sn+1
need to prove that for A1 , . . . , AnS, An+1 countable, i=1 Ai is countable.
n
Using the induction hypothesis, i=1 Ai is countable. Then
!
n+1
n
[
[
Ai =
Ai ∪ An+1 ,
i=1
i=1
and the union of two countable sets is countable.
(b) Induction is good for proving statements
that hold “for all n ∈ N”. Part
S
(ii) of Theorem 1.4.13, states that n An is countable. Therefore, induction is
not good for proving part (ii) of Theorem 1.4.13.
(c) The proof here uses a diagonal argument similar to the argument in the
proof that Q is countable. We describe the idea:
First we can assume that all An ’s are disjoint. If this is not the case, then we
replace A2 by B2S= A2 \ A1 , A3Sby B3 = A3 \ (A1 ∪ A2 ), etc. We can prove as
in part (a) that n An = A1 ∪ n Bn and that A1 , B2 , B3 , . . . are all disjoint
sets. Thus, without loss of generality, we will assume that all An ’s are disjoint.
Now, for each n ∈ N, An is countable. I.e. there exists a 1-1 and onto function
fn : N → An . Using these functions fn , we write:
A1 = {f1 (1), f1 (2), . . .}
A2 = {f2 (1), f2 (2), . . .}
..
..
..
.
.
.
An = {fn (1), fn (2), . . .}
S
The idea is to define a function h : N → n An that is 1-1 and onto, tracing all
the values fn (m), n, m ∈ N through the diagonals. More specifically, we follow
the following pattern:
1
3
6
10
15 . . .
2
5
9
14 . . .
4
8
13 . . .
7
12 . . .
11 . . .
..
.
3
This means that h(1) = f1 (1), h(2) = f2 (1), h(3) = f1 (2), h(4) = f3 (1),
h(5) = f2 (2), h(6) = f1 (3) etc. It follows that h is onto (we trace all the
values) and that h is 1-1 (no values are repeated). We could actually write a
formula for h and then use it to prove that it is 1-1 and onto, but it becomes
unnecessarily complicated.
Problem 1.4.10 Show that the set of finite sequences of natural numbers is
countable.
Proof. 1 Let S be the set of finite sequences of natural numbers. E.g.
(0, 2, 143, 2) ∈ S. The idea is to associate to each such sequence a natural
number that serves as “code” for the sequence. Given the code we must be
able to tell how long the sequence is and what the numbers in the sequence are.
We need some number theory: A natural number p is prime if its only divisors
are 1 and p itself. E.g. 2, 3, 5, 7 are primes, but 4, 6, 8 are not primes. For
every natural number n, we can write n uniquely as the product of prime
numbers p1 , p2 , . . . , pk :
n = pe11 · pe22 · . . . pekk ,
where the exponents e1 , e2 , . . . , ek are natural numbers.
Let’s see how we can use this theorem to code finite sequences. Map the
sequence (0, 2, 143, 2) to the number
20+1 · 32+1 · 5143+1 · 72+1 = 21 · 33 · 5144 · 73 .
We use here 2, 3, 5, 7 because they are the first four prime numbers. Reservely,
given the number 21 · 33 · 5144 · 73 we can determine that the length of the
corresponding sequence to this number is 4, since the first four prime numbers
are being used, and we can determine the numbers in the sequence by
subtracting 1 from each exponent. I.e. (0, 2, 143, 2).
More generally now, map every finite sequence s = (n1 , n2 , . . . , nk ) ∈ S to
f (s) = pn1 1 +1 · pn2 2 +1 · . . . pnk k +1 ∈ N,
where p1 , p2 , . . . , pk are the first k prime numbers.
Using the fact that every natural numbers can be written uniquely as the
product of prime numbers, if f (s1 ) = f (s2 ), then s1 must be equal to s2 .
Thus, f is a 1-1 function, but it is not onto N. Let B = range(f ). Then B is
an infinite subset of the natural numbers. By problem 1.4.7, B is countable.
Therefore, there exists some 1-1 and onto function g : B → N. Then the
composition g ◦ f : S → N is a 1-1 and onto function2 from S to N, which
proves that S is countable.
1 This
problem is the hardest in this assignment.
use here without proof the fact the composition of two 1-1 and onto functions is 1-1
and onto.
2 We
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Section 1.5
Problem 1.5.9
Proof. (a) A function from {0, 1} to N is just a pair of natural numbers. So,
the set of all functions from {0, 1} to N is the same as the set of all pairs of
natural numbers. Associate to each such pair (n, m) a rational number, e.g.
2n
3m . The rational numbers are countable. So, the same is true for the set of all
functions from {0, 1} to N.
(b) The situation is different for all functions from N to {0, 1}. Every such
function defines an infinite sequence of 0’s and 1’s. By Exercise 1.5.4 there are
uncountably many such sequences.
(c) P (N) does contain an uncountable antichain.
Section 2.2
Problem 2.2.1
Proof. (a) For every > 0, we need to find some N ∈ N such that for all
n ≥ N,
1
6n2 + 1 − 0 < .
q
1
Let N be bigger than 6
(we found this expression by solving for n the
inequality
1
6n2 +1
< ). Then,
r
N
>
N2
>
>
>
>
1
⇒
6
1
⇒
6
1
⇒
6N 2
1
⇒
6N 2 + 1
1
, for all n ≥ N .
2
6n + 1
(b) Let > 0 and we need some N ∈ N such that for all n ≥ N ,
3n + 1 3 −
2n + 5 2 < .
5
Equivalently,
3n + 1 3 2n + 5 − 2 2(3n + 1) − 3(2n + 5) (2n + 5) · 2
13
(2n + 5) · 2
13
4 .
Let N be a natural number bigger than
N
>
>
>
>
>
<
⇔
<
⇔
<
Then,
13
⇒
4
13
⇒
4N
13
⇒
4N + 10
13
, for all n ≥ N . ⇔
(2n + 5) · 2
3n + 1 3 2n + 5 − 2 , for all n ≥ N .
(c) Let > 0 and we need some N ∈ N such that for all n ≥ N ,
1
√
n + 3 − 0 < .
Equivalently,
1
√
n + 3 − 0 <
1
1
2
<
<
Let N be a natural number bigger than
N
>
N +3
>
n+3
1
√
−
0
n+3
>
<
1
2 .
⇔
√
n+3⇔
n+3
Then,
1
⇒
2
1
⇒
2
1
, for all n ≥ N ⇔
2
, for all n ≥ N
6
Problem 2.2.2
Proof. Consider the sequence (0, 1, 0, 1, 0, 1, . . .). It is divergent. We will prove
that it is vercongent to 0. Let = 2 > 0. Then for all n ∈ N,
|xn − 0| ≤ 1 < = 2.
The strange definition of vercongence describes sequences that are
bounded.
Problem 2.2.4
Proof. Let = .5. We will prove that for all N there exists some n ≥ N such
that |xn − 0| > (that’s the negation of the definition of the limit).
Since the sequence contains an infinitely many number of 0’s and 1’s, given N
find some n ≥ N such that xn = 1. Then |xn − 0| = |1 − 0| > .
For > 1, we can find some N . For ≤ 1, we can not find any N .
Problem 2.2.6
Proof. (a) In the definition of the limit, if we found a value N that works for
> 0, then any larger value also works for the same .
(b) This same N also works for any larger value of .
Problem 2.2.8
Proof. (a) The sequence (−1)n is frequently in the set {1}.
(b) Eventually implies frequently. I.e. eventually is stronger.
(c) A sequence (an ) converges to a iff for every > 0 the sequence (an ) is
eventually in the neighborhood V (a). Eventually is the term we need.
(d) The sequence (1, 2, 1, 2, 1, 2, . . .) takes the value 2 infinitely many times but
is not eventually in the interval (1.9, 2.1). It is frequently in this interval.
Section 2.3
Problem 2.3.2 Let xn ≥ 0, for all n ∈ N. (a) Show that if (xn ) → 0, then
√
( xn ) → 0.
√
√
(b) Show that if (xn ) → x, then ( xn ) → x.
Proof. (a) We have that
∀ > 0 ∃N1 ∈ N ∀n ≥ N1 |xn − 0| < .
(1)
Equivalently, xn < .
√
√
We need that ∀ > 0 ∃N2 ∈ N ∀n ≥ N2 , | xn − 0| < . Equivalently, xn < .
Let be a positive number and use property (1) for 2 to find some N1 such
√
that for all n ≥ N1 , xn < 2 . Then xn < and we can
√ take N2 = N1 .
(b) If (xn ) → x, then (xn − x) → 0 and by part (a), xn − x → 0. In symbols,
√
∀ > 0 ∃N ∈ N ∀n ≥ N | xn − x| < .
(2)
7
Using some algebra we can prove that for xn , x ≥ 0,
√
√
√
0 ≤ | xn − x| ≤ xn − x.
(3)
The proof is similar to the proof of the triangular inequality.
Combining properties (2) and (3), we conclude that there exists some N such
that for all n ≥ N ,
√
√
√
| xn − x| ≤ xn − x < .
√
√
This proves that ( xn ) → x.
Problem 2.3.3 Prove the Squeeze Theorem.
Proof. We know that
xn ≤ yn ≤ zn for all n ∈ N
and
lim xn = lim zn = l.
n
n
From this last equation we can also derive that limn (zn − xn ) = 0. In symbols,
∀ > 0 ∃N1 ∈ N ∀n ≥ N1 , |xn − l| < (1)
∀ > 0 ∃N2 ∈ N ∀n ≥ N2 , |zn − l| < (2)
∀ > 0 ∃N3 ∈ N ∀n ≥ N3 , |zn − xn | < .
(3)
We need to prove that ∀ > 0 ∃N4 ∈ N ∀n ≥ N4 , |yn − l| < . By the
triangular inequality,
|yn − l| ≤ |yn − xn | + |xn − l|
≤ |zn − xn | + |xn − l|
Using properties (1) and (3) for 2 , we can find N1 , N3 such that for all
n ≥ N1 , N3 , |zn − xn | < 2 and |xn − l| < 2 . Combining together we get that
for all n ≥ N4 = max{N1 , N3 },
|yn − l| ≤ |zn − xn | + |xn − l| <
+ = .
2 2
This proves that (yn ) → l and finishes the proof.
Problem 2.3.5 Prove that (xn ), (yn ) are both convergent at the same limit iff
the “shuffled” sequence zn = (x1 , y1 , x2 , y2 , . . .) is convergent.
Proof. Since (xn ), (yn ) are subsequences of (zn ) the right-to-left direction is
immediate. We need to prove the left-to-right direction.
Assume that limn xn = limn yn = l. In symbols,
∀ > 0 ∃N1 ∈ N ∀n ≥ N1 , |xn − l| < (1)
∀ > 0 ∃N2 ∈ N ∀n ≥ N2 , |yn − l| < (2)
8
We need to conclude that ∀ > 0 ∃N3 ∈ N ∀n ≥ N3 , |zn − l| < . Using (1)
and (2) we can find some N1 , N2 such that for all n ≥ max N1 , N2 , |xn − l| < and |yn − l| < .
Observe that z1 = x1 , z3 = x2 , z5 = x3 , . . . and in general zn = x n+1 , for n:
2
odd. Similarly, z2 = y1 , z4 = y2 , z6 = y3 , . . . and in general zn = y n2 , for n:
even.
n
Let N3 = max{2N1 , 2N2 }. Then for all n ≥ N3 , n+1
2 ≥ 2 ≥ max{N1 , N2 }.
n
Since zn is either equal to x n+1 or y n2 and both indexes n+1
2 , 2 are greater
2
than max{N1 , N2 }, it follows that for all n ≥ N3 , |zn − l| < , which proves
that (zn ) converges to l.
Problem 2.3.6 (a) Prove that if bn → b, then |bn | → |b|.
(b) Can we conclude that if |bn | → |b|, then bn → b?
√
p
2
2
2 →
Proof. (a) Since bn →
b,
b
→
b
.
By
Exercise
2.3.2,
part
(b),
b
b2 .
n
n
√
p
2
2
But bn = |bn | and b = |b| and the result follows.
(b) The conclusion is false. Consider bn = (−1, 1, −1, 1, −1, 1, . . .) and b = 1.
Then |bn | = (1, 1, 1, 1, 1, . . .) which converges to |b|, but bn is a divergent
sequence.
Problem 2.3.8
Proof. (a) Let xn = (−1, 1, −1, 1, −1, 1, . . .) and yn = (1, −1, 1, −1, 1, −1, . . .).
Then both (xn ) and (yn ) are divergent, but (xn + yn ) = (0, 0, 0, 0, 0, 0, . . .) and
(xn + yn ) converges to 0.
(b) This is impossible, by theorem 2.3.3, part (ii).
(c) Let bn = n1 . Then (bn ) → 0, but b1n = n diverges.
(d) By theorem 2.3.2 (bn ) is bounded. Since (an ) is unbounded, (an − bn ) can
not be bounded.
(e) Let (an ) = (0, 0, 0, 0, 0, 0, . . .) and (bn ) = (−1, 1, −1, 1, −1, 1, . . .). Then
(an bn ) = (0, 0, 0, 0, 0, 0, . . .) and both (an ) and (an bn ) are convergent, while
(bn ) is divergent.
Problem 2.3.9
Answer. Theorem 2.3.4 does not remain true is we replace all inequalities with
strict inequalities. E.g. for an = n1 > 0, we can not conclude that
limn an > 0.
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