Solutions Manual Understanding Analysis by Stephen Abbott

Solutions Manual
Understanding Analysis
By
Stephen Abbott



( All Chapters Included - 100% Verified Solutions )




1

,Chapter 1

The Real Numbers
√
1.1 Discussion: The Irrationality of 2
1.2 Some Preliminaries
Exercise 1.2.1. (a) Assume, for contradiction, that there exist integers p and
q satisfying
µ ¶2
p
(1) = 3.
q
Let us also assume that p and q have no common factor. Now, equation (1)
implies

(2) p2 = 3q 2 .

From this, we can see that p2 is a multiple of 3 and hence p must also be
a multiple of 3. This allows us to write p = 3r, where r is an integer. After
substituting 3r for p in equation (2), we get (3r)2 = 3q 2 , which can be simplified
to 3r2 = q 2 . This implies q 2 is a multiple of 3 and hence q is also a multiple of
3. Thus we have shown p and q have a common factor, namely 3, when they
were originally assumed to have no common
√ factor.
A similar argument will work for 6 as well because we get p2 = 6q 2 which
implies p is a multiple of 2 and 3. After making √ the necessary substitutions, we
can conclude q is a multiple of 6, and therefore 6 must be irrational.
(b) In this case, the fact that p2 is a multiple of 4 does not imply p is also a
multiple of 4. Thus, our proof breaks down at this point.
Exercise 1.2.2. (a) False, as seen in Example 1.2.2.
(b) True. This will follow from upcoming results about compactness in
Chapter 3.
(c) False. Consider sets A = {1, 2, 3}, B = {3, 6, 7} and C = {5}. Note that
A ∩ (B ∪ C) = {3} is not equal to (A ∩ B) ∪ C = {3, 5}.

1




2

,2 Chapter 1. The Real Numbers

(d) True.
(e) True.

Exercise 1.2.3. (a) If x ∈ (A ∩ B)c then x ∈ / (A ∩ B). But this implies x ∈/A
or x ∈/ B. From this we know x ∈ Ac or x ∈ B c . Thus, x ∈ Ac ∪ B c by the
definition of union.
(b) To show Ac ∪ B c ⊆ (A ∩ B)c , let x ∈ Ac ∪ B c and show x ∈ (A ∩ B)c .
So, if x ∈ Ac ∪ B c then x ∈ Ac or x ∈ B c . From this, we know that x ∈ / A or
x∈/ B, which implies x ∈ / (A ∩ B). This means x ∈ (A ∩ B)c , which is precisely
what we wanted to show.
(c) In order to prove (A ∪ B)c = Ac ∩ B c we have to show,

(1) (A ∪ B)c ⊆ Ac ∩ B c and,


(2) Ac ∩ B c ⊆ (A ∪ B)c .

To demonstrate part (1) take x ∈ (A ∪ B)c and show that x ∈ (Ac ∩ B c ). So,
if x ∈ (A ∪ B)c then x ∈ / (A ∪ B). From this, we know that x ∈A / and x ∈ / B
which implies x ∈ Ac and x ∈ B c . This means x ∈ (Ac ∩ B c ).
Similarly, part (2) can be shown by taking x ∈ (Ac ∩ B c ) and showing that
x ∈ (A ∪ B)c . So, if x ∈ (Ac ∩ B c ) then x ∈ Ac and x ∈ B c . From this, we know
that x ∈/ A and x ∈ / B which implies x ∈ / (A ∪ B). This means x ∈ (A ∪ B)c .
Since we have shown inclusion both ways, we conclude that (A ∪ B)c = Ac ∩ B c .

Exercise 1.2.4. (a)When a and b have the same sign, consider the following
two cases:
(i) If a ≥ 0 and b ≥ 0 then we have a + b > 0 which implies |a + b| = a + b.
Furthermore, because |a| = a and |b| = b, we have |a| + |b| = a + b. This implies,
|a + b| = |a| + |b|, which satisfies the triangle inequality.
(ii) If a ≤ 0 and b ≤ 0 then we have a + b ≤ 0 which implies |a + b| = −a − b.
Furthermore, since we know |a| = −a and |b| = −b we have |a| + |b| = −a − b.
This implies, |a + b| = |a| + |b|, which satisfies the triangle inequality.
(b)If a ≥ 0, b < 0, and a + b ≥ 0 then we have |a + b| = a + b = a − (−b) =
|a| − |b| < |a| + |b|. This implies |a + b| ≤ |a| + |b| as desired.

Exercise 1.2.5. (a) Observe that |a − b| = |a + (−b)| ≤ |a| + | − b| = |a| + |b|
which implies |a − b| ≤ |a| + |b|.
(b) First note that |a| = |a − b + b| ≤ |a − b| + |b|. Taking |b| to the left
side of the inequality we get |a| − |b| ≤ |a − b|. Reversing the roles of a and b in
the previous argument gives |b| − |a| ≤ |b − a|, and because |a − b| = |b − a| the
result follows.

Exercise 1.2.6. (a) f (A) = [0, 4] and f (B) = [1, 16]. In this case, f (A ∩ B) =
f (A) ∩ f (B) = [1, 4] and f (A ∪ B) = f (A) ∪ f (B) = [0, 16].
(b) Take A = [0, 2] and B = [−2, 0] and note that f (A ∩ B) = {0} but
f (A) ∩ f (B) = [0, 4].




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, 1.2. Some Preliminaries 3

(c) We have to show y ∈ g(A ∩ B) implies y ∈ g(A) ∩ g(B). If y ∈ g(A ∩ B)
then there exists an x ∈ A ∩ B with g(x) = y. But this means x ∈ A and x ∈ B
and hence g(x) ∈ g(A) and g(x) ∈ g(B). Therefore, g(x) = y ∈ g(A) ∩ g(B).
(d) Our claim is g(A ∪ B) = g(A) ∪ g(B). In order to prove it, we have to
show,

(1) g(A ∪ B) ⊆ g(A) ∪ g(B) and,


(2) g(A) ∪ g(B) ⊆ g(A ∪ B).

To demonstrate part (1), we let y ∈ g(A ∪ B) and show y ∈ g(A) ∪ g(B). If
y ∈ g(A ∪ B) then there exists x ∈ A ∪ B with g(x) = y. But this means
x ∈ A or x ∈ B, and hence g(x) ∈ g(A) or g(x) ∈ g(B). Therefore, g(x) = y ∈
g(A) ∪ g(B).
To demonstrate the reverse inclusion, we let y ∈ g(A) ∪ g(B) and show
y ∈ g(A ∪ B). If y ∈ g(A) ∪ g(B) then y ∈ g(A) or y ∈ g(B). This means we
have an x ∈ A or x ∈ B such that g(x) = y. This implies, x ∈ A ∪ B, and
hence g(x) ∈ g(A ∪ B). Since we have shown parts (1) and (2), we can conclude
g(A ∪ B) = g(A) ∪ g(B).
Exercise 1.2.7. (a) f −1 (A) = [−2, 2] and f −1 (B) = [−1, 1]. In this case,
f −1 (A ∩ B) = f −1 (A) ∩ f −1 (B) = [−1, 1] and f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B) =
[−2, 2].
(b) In order to prove g −1 (A ∩ B) = g −1 (A) ∩ g −1 (B), we have to show,

(1) g −1 (A ∩ B) ⊆ g −1 (A) ∩ g −1 (B) and,


(2) g −1 (A) ∩ g −1 (B) ⊆ g −1 (A ∩ B).

To demonstrate part (1), we let x ∈ g −1 (A ∩ B) and show x ∈ g −1 (A) ∩ g −1 (B).
So, if x ∈ g −1 (A ∩ B) then g(x) ∈ (A ∩ B). But this means g(x) ∈ A and
g(x) ∈ B, and hence g(x) ∈ A ∩ B. This implies, x ∈ g −1 (A) ∩ g −1 (B).
To demonstrate the reverse inclusion, we let x ∈ g −1 (A) ∩ g −1 (B) and show
x ∈ g −1 (A ∩ B). So, if x ∈ g −1 (A) ∩ g −1 (B) then x ∈ g −1 (A) and x ∈ g −1 (B).
This implies g(x) ∈ A and g(x) ∈ B, and hence g(x) ∈ A ∩ B. This means,
x ∈ g −1 (A ∩ B).
Similarly, in order to prove g −1 (A ∪ B) = g −1 (A) ∪ g −1 (B), we have to show,

(1) g −1 (A ∪ B) ⊆ g −1 (A) ∪ g −1 (B) and,


(2) g −1 (A) ∪ g −1 (B) ⊆ g −1 (A ∪ B).

To demonstrate part (1), we let x ∈ g −1 (A ∪ B) and show x ∈ g −1 (A) ∪ g −1 (B).
So, if x ∈ g −1 (A ∪ B) then g(x) ∈ (A ∪ B). But this means g(x) ∈ A or
g(x) ∈ B, which implies x ∈ g −1 (A) or x ∈ g −1 (B). From this we know
x ∈ g −1 (A) ∪ g −1 (B).




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