Math 317 HW #1 Solutions
1. Exercise 1.2.3. Let A and B be subsets of R.
(a) If x ∈ (A ∩ B)c , explain why x ∈ Ac ∪ B c . This shows that (A ∩ B)c ⊆ Ac ∪ B c .
Proof. Suppose x ∈ (A ∩ B)c . Then x is not an element of A ∩ B, meaning that x can’t
be in both A and B. Therefore, either x ∈
/ A (meaning that x ∈ Ac ) or x ∈
/ B (meaning
c
that x ∈ B ), or both. This means that
x ∈ Ac ∪ B c .
Since the choice of x ∈ (A∩B)c was arbitrary, we see that the same holds for all elements
of (A ∩ B)c , and so (A ∩ B)c ⊆ Ac ∪ B c .
(b) Prove the reverse inclusion (A ∩ B)c ⊇ Ac ∪ B c , and conclude that (A ∩ B)c = Ac ∪ B c .
Proof. Suppose x ∈ Ac ∪ B c ; our goal is to show that x ∈ (A ∩ B)c . Since x ∈ Ac ∪ B c ,
it must be the case that x ∈ Ac or x ∈ B c (or both). This certainly implies that x is not
in both A and B, so x cannot be in A ∩ B. Which, of course, means that
x ∈ (A ∩ B)c .
Since our choice of x ∈ Ac ∪B c was arbitrary, the same holds for every element of Ac ∪B c ,
and so we can conclude that Ac ∪ B c ⊆ (A ∩ B)c .
On the other hand, we saw in (a) that (A ∩ B)c ⊆ Ac ∪ B c . The only way both inclusions
can be true is if (A ∩ B)c = Ac ∪ B c .
(c) Show (A ∪ B)c = Ac ∩ B c by demonstrating inclusion both ways.
Proof. First, we’ll show that (A∪B)c ⊆ Ac ∩B c . To that end, suppose that x ∈ (A∪B)c .
This means that x is not in A∪B, meaning that x is in neither A nor B. Said differently,
x ∈ Ac and x ∈ B c , which means that
x ∈ Ac ∩ B c .
Since our choice of x ∈ (A ∪ B)c was arbitrary, we conclude that (A ∪ B)c ⊆ Ac ∩ B c .
To show the reverse inclusion, suppose that x ∈ Ac ∩ B c . This means that x ∈ Ac
and x ∈ B c or, equivalently, x ∈
/ A and x ∈
/ B. Hence, x is not in A ∪ B, meaning
c
that x ∈ (A ∪ B) . Again, the choice of x ∈ Ac ∩ B c was arbitrary, so we see that
Ac ∩ B c ⊆ (A ∪ B)c .
Since we’ve shown inclusions both ways, we can conclude that (A ∪ B)c = Ac ∩ B c .
2. Exercise 1.2.7. Given a function f : D → R and a subset B ⊆ R, let f −1 (B) be the set of all
points from the domain D that get mapped into B; that is, f −1 (B) = {x ∈ D : f (x) ∈ B}.
This set is called the preimage of B.
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(a) Let f (x) = x2 . If A is the closed interval [0, 4] and B is the closed interval [−1, 1],
find f −1 (A) and f −1 (B). Does f −1 (A ∩ B) = f −1 (A) ∩ f −1 (B) in this case? Does
f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B)?
Answer. If x ∈ [−2, 2], then it’s certainly true that f (x) ∈ A. Moreover, these are the
only possible values of x such that f (x) ∈ A, so we can see that
f −1 (A) = [−2, 2].
Turning to B, notice that, if x ∈ [−1, 1], then f (x) ∈ B and, in fact, these are the only
real numbers that map to B under f (of course, there is no real number x such that
f (x) < 0, so half of B is completely missed by f ). Therefore,
f −1 (B) = [−1, 1]
(which is, coincidentally, B again, so f −1 (B) = B).
Now, A ∩ B = [0, 4] ∩ [−1, 1] = [0, 1] and f −1 ([0, 1]) = [−1, 1], so we see that
f −1 (A ∩ B) = [−1, 1] = [−2, 2] ∩ [−1, 1] = f −1 (A) ∩ f −1 (B).
On the other hand, A ∪ B = [−1, 4] and f −1 ([−1, 4]) = [−2, 2], so we also have that
f −1 (A ∪ B) = [−2, 2] = [−2, 2] ∪ [−1, 1] = f −1 (A) ∪ f −1 (B).
(b) The good behavior of preimages demonstrated in (a) is completely general. Show that
for an arbitrary function g : R → R, it is always true that g −1 (A ∩ B) = g −1 (A) ∩ g −1 (B)
and g −1 (A ∪ B) = g −1 (A) ∪ g −1 (B).
Proof. Suppose g : R → R is a function and that A, B ⊆ R. There are two things to
show, so we proceed in two parts:
Part I: First, we want to show that g −1 (A ∩ B) = g −1 (A) ∩ g −1 (B). As usual, we’ll
show inclusion both ways. To show that g −1 (A ∩ B) ⊆ g −1 (A) ∩ g −1 (B), suppose
x ∈ g −1 (A ∩ B). By definition, this means that
g(x) ∈ A ∩ B.
In particular, this means that g(x) ∈ A and g(x) ∈ B or, equivalently, x ∈ g −1 (A)
and x ∈ g −1 (B). This means that x ∈ g −1 (A) ∩ g −1 (B). Since our choice of x was
arbitrary, we see that g −1 (A ∩ B) ⊆ g −1 (A) ∩ g −1 (B).
To show the reverse inclusion, suppose x ∈ g −1 (A) ∩ g −1 (B). Then x ∈ g −1 (A) and
x ∈ g −1 (B) and so, by definition, this means that g(x) ∈ A and g(x) ∈ B. But
that’s the same as saying that g(x) ∈ A ∩ B, which implies that x ∈ g −1 (A ∩ B).
Since our choice of x was arbitrary, we conclude that g −1 (A)∩g −1 (B) ⊆ g −1 (A∩B).
Having shown inclusion both ways, we can conclude that g −1 (A ∩ B) = g −1 (A) ∩
g −1 (B).
Part II: Now, we want to show that g −1 (A ∪ B) = g −1 (A) ∪ g −1 (B). Following the
usual pattern, we want to show first that g −1 (A ∪ B) ⊆ g −1 (A) ∪ g −1 (B), so suppose
that x ∈ g −1 (A ∪ B). By definition, this means that
g(x) ∈ A ∪ B,
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so g(x) ∈ A or g(x) ∈ B. Stated equivalently, x ∈ g −1 (A) or x ∈ g −1 (B), which
implies that x ∈ g −1 (A) ∪ g −1 (B). Since our choice of x was arbitrary, we see that
g −1 (A ∪ B) ⊆ g −1 (A) ∪ g −1 (B).
To show the other inclusion, suppose x ∈ g −1 (A) ∪ g −1 (B). This means that x ∈
g −1 (A) or x ∈ g −1 (B), which is to say that g(x) ∈ A or g(x) ∈ B. Either way,
g(x) ∈ A ∪ B,
and so x ∈ g −1 (A ∪ B). Since the choice of x was arbitrary, we see that g −1 (A) ∪
g −1 (B) ⊆ g −1 (A ∪ B).
Having shown both inclusions, we can conclude that g −1 (A ∪ B) = g −1 (A) ∪ g −1 (B),
as desired.
3. Exercise 1.2.8. Form the logical negation of each claim. One way to do this is to simply add
“It is not the case that...” in front of each assertion, but for each statement, try to embed
the word “not” as deeply into the resulting sentence as possible (or avoid using it altogether).
(a) For all real numbers satisfying a < b, there exists an n ∈ N such that a + 1/n < b.
Negation: There are two real numbers satisfying a < b such that, for all n ∈ N, a+1/n ≥
b.
(b) Between every two distinct real numbers, there is a rational number.
Negation: There are two distinct real numbers with no rational numbers between them.
√
(c) For all natural numbers n ∈ N, n is either a natural number or an irrational number.
√
Negation: There is some n ∈ N such that n is a rational number which isn’t a natural
number.
(d) Given any real number x ∈ R, there exists n ∈ N satisfying n > x.
Negation: There is some x ∈ R such that x ≥ n for all n ∈ N.
4. Exercise 1.2.10. Let y1 = 1, and for each n ∈ N define yn+1 = (3yn + 4)/4.
(a) Use induction to prove that the sequence satisfies yn < 4 for all n ∈ N.
Proof. Our proof is by induction, so we need to show the base case and the inductive
step.
Base Case: Since y1 = 1, it’s clearly true that y1 < 4.
Inductive Step: Suppose yk < 4. Our goal is to show that yk+1 < 4. To that end,
note that
3
3yk + 4
yk+1 =
= yk + 1.
4
4
Since, by the inductive hypothesis, yk < 4, we know that
3
3
yk+1 = yk + 1 < · 4 + 1 = 3 + 1 = 4,
4
4
so indeed yk+1 < 4.
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Therefore, since both the base case and the inductive step are true, we can conclude, by
induction, that yn < 4 for all n ∈ N.
(b) Use another induction argument to show the sequence (y1 , y2 , y3 , . . .) is increasing.
Proof. We want to show that yn < yn+1 for all n ∈ N. Again, we’re going to use
induction, so we need to show the base case and the inductive step.
Base Case: The base case is the statement that y1 < y2 . Since y1 = 1 and
y2 =
3y1 + 4
7
= ,
4
4
the statement y1 < y2 is clearly true.
Inductive Step: Suppose that yk < yk+1 . Then our goal is to show that yk+1 < yk+2
follows. Since yk < yk+1 , we have that
yk+1 =
3yk+1 + 4
3yk + 4
<
= yk+2 ,
4
4
where the equalities each follow from the definition of the yi ’s.
Therefore, we conclude by induction that yn < yn+1 for all n ∈ N.
5. Exercise 1.2.11. If a set A contains n elements, prove that the number of different subsets of
A is equal to 2n .
Proof. Let A = {a1 , a2 , . . . , an } (in other words, we’re just assigning the names ai to the
elements of A). Each subset B ⊆ A is uniquely specified by the answers to the following list
of n questions:
Is a1 an element of B?
Is a2 an element of B?
..
.
Is an an element of B?
On the other hand, since listing its elements specifies a set, any arbitrary set of answers to
the above n questions determines a subset of A. Therefore, the number of subsets of A is
equal to the number of possible ways that the above list of questions can be answered. Since
each of the n questions has 2 possible answers, this gives a total of
2| · 2 ·{z. . . · 2} = 2n
n times
possible ways of answering and, thus, 2n subsets of A.
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