MAT4500
Exercises
Chapter 1 - Set Theory and
§1 - Fundamental Concepts
§2 - Functions . . . . . . . .
§5 - Cartesian Products . .
§6 - Finite Sets . . . . . . .
Logic
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1
1
12
12
12
Chapter 1 - Set Theory and Logic
§1 - Fundamental Concepts
1
Verify that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
⊂) First we show that A∪(B ∩C) ⊂ (A∪B)∩(A∪C). We begin by assuming
that x ∈ A ∪ (B ∩ C).
First we assume in addition that x ∈ A, in which case x ∈ A ∪ B
and x ∈ A ∪ C or x ∈ (A ∪ B) ∩ (A ∪ C), so when x ∈ A we have
A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C).
Second we assume that x 6∈ A, which means that we must have x ∈
B ∩C. In other words we have x ∈ B and x ∈ C, so obviously x ∈ A∪B
and x ∈ A ∪ C, so for x 6∈ A we have A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C).
Combining both these cases, we can conclude that A ∪ (B ∩ C) ⊂ (A ∪ B) ∩
(A ∪ C). Now we have to show the opposite inclusion.
⊃) Our goal is to show that (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C). We assume
x ∈ (A ∪ B) ∩ (A ∪ C).
If x ∈ A, then it follows that x ∈ A ∪ (B ∩ C), so for x ∈ A we have ⊂.
If x 6∈ A, this means x must be in either B and C, or x ∈ B ∩ C, and
from here it follows that x ∈ A ∪ (B ∩ C), so for x 6∈ A we also have ⊂.
From this it follows that we have (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C), which
concludes the proof of equality.
1
Verify that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
⊂) Assume x ∈ A ∩ (B ∪ C). This means x is in A and either B or C, which
means one of the following is true: x ∈ A ∩ B or x ∈ A ∩ C. This amounts
to x ∈ (A ∩ B) ∪ (A ∩ C), which gives us the required inclusion ⊂.
⊃) Assume x ∈ (A ∩ B) ∪ (A ∩ C). This means x is included in at least one
of the sets: either x ∈ A ∩ B or x ∈ A ∩ C or both, so we can say that x ∈ A
and either x ∈ B or x ∈ C is true. This is just x ∈ A ∩ (B ∪ C) which shows
the other inclusion and concludes the proof.
Verify DeMorgan’s Law: (A ∩ B)c = Ac ∪ B c
We assume x ∈ (A ∩ B)c .
x ∈ (A ∩ B)c ⇐⇒
⇐⇒
⇐⇒
⇐⇒
x 6∈ A ∩ B
x 6∈ A or x 6∈ B
x ∈ Ac or x ∈ B c
x ∈ Ac ∪ B c
Verify DeMorgan’s Law: (A ∪ B)c = Ac ∩ B c
We assume x ∈ (A ∪ B)c .
x ∈ (A ∪ B)c ⇐⇒
⇐⇒
⇐⇒
⇐⇒
x 6∈ A ∪ B
x 6∈ A and x 6∈ B
x ∈ Ac and x ∈ B c
x ∈ Ac ∩ B c
2
2
a)
We are going to determine if the following statements are true, false or if one of
the implications or inclusions are true.
A ⊂ B and A ⊂ C ⇔ A ⊂ (B ∪ C)
Since B ⊂ B ∪C and C ⊂ B ∪C, we get A ⊂ B ⊂ B ∪C and A ⊂ C ⊂ B ∪C,
so the implication ⇒ is true.
The reverse implication is not true in general. If we assume A is nonempty
and set B = A and C = ∅, we get A ⊂ (B ∪ C) = (A ∪ ∅) = A, which is true,
but A 6⊂ C = ∅.
b)
A ⊂ B or A ⊂ C ⇔ A ⊂ (B ∪ C)
As in a) we get A ⊂ B ⊂ B ∪ C and the same for C, so the first implication
is true.
The other implication is not true as demonstrated by the following image.
B
C
A
Figur 1: Counterexample
As we can see, A ⊂ B ∪ C, but A 6⊂ B and A 6⊂ C.
c)
A ⊂ B and A ⊂ C ⇔ A ⊂ (B ∩ C)
⇒) For A ⊂ B and A ⊂ C, A is a subset of both B and C, so A ⊂ B ∩ C.
⇐) Since A ⊂ B ∩ C, A is a subset of both B and C, so A ⊂ B and A ⊂ C.
3
d)
A ⊂ B or A ⊂ C ⇔ A ⊂ (B ∩ C)
Since A ⊂ B or A ⊂ C means A ⊂ B ∪ C, we are left with considering
A ⊂ B ∪ C ⇔ A ⊂ B ∩ C.
⇐) If A ⊂ B ∩ C, then A is a subset of both sets, and is naturally included
in B ∪ C.
⇒) This is not true in general, as can be seen by the situation in figure 1.
Here A ⊂ B ∪ C, but A 6⊂ ∅ = B ∩ C.
e)
A \ (A \ B) = A
An alternative way of writing the set minus is in terms of intersections and
complements. A \ B = A ∩ B c . Rewriting we get A \ (A \ B) = A \ (A ∩ B c ) =
A∩(A∩B c )c . Applying DeMorgan’s Law, we get A∩(Ac ∪B cc ) = A∩(Ac ∪B).
Distributing the intersection: (A ∩ Ac ) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B.
After rewriting, we see that we are considering A ∩ B = A, where ⊂ is true
but ⊃ generally isn’t (only if A ⊂ B).
f)
A \ (B \ A) = A \ B
Rewriting we are comparing A ∩ (B ∩ Ac )c = A ∩ B c . We can rewrite the left
hand expression by first applying DeMorgan’s Law: A ∩ (B c ∪ A), and then
by using the distributive property we get (A ∩ B c ) ∪ (A ∩ A) = (A ∩ B c ) ∪ A,
so we are comparing (A ∩ B c ) ∪ A with A ∩ B c , and we see that we have ⊃
but not generally ⊂.
g)
A ∩ (B \ C) = (A ∩ B) \ (A ∩ C)
The left hand expression can be written as A ∩ B ∩ C c . For the right hand
expression we get (A ∩ B) ∩ (A ∩ C)c . Apply DeMorgan’s Law and get
(A ∩ B) ∩ (Ac ∪ C c ). Using the distributive property we get (A ∩ B ∩ Ac ) ∪
(A ∩ B ∩ C c ) = ∅ ∪ (A ∩ B ∩ C c ) = A ∩ B ∩ C c . In other words, both these
sets are the same, so we have equality.
4
h)
A ∪ (B \ C) = (A ∪ B) \ (A ∪ C)
The left hand side has the alternative representation A ∪ (B ∩ C c ). For
the right hand side we get (A ∪ B) ∩ (A ∪ C)c , apply DeMorgan’s Law,
(A∪B)∩(Ac ∩C c ), use the distributive property (A∩Ac ∩C c )∪(B∩Ac ∩C c ) =
(B ∩ Ac ∩ C c ) = (B ∩ C c ) \ A. When comparing, we begin with the set B ∩ C c
and on the left side we add A, whereas on the right side we remove anything
in A, so in general we have ⊃ and ⊂ is generally wrong.
i)
(A ∩ B) ∪ (A \ B) = A.
We consider the left hand side, and rewrite the set minus: (A∩B)∪(A\B) as
(A∩B)∪(A∩B c ). We use the distributive law backwards and get A∩(B∪B c ).
Here B ∪ B c becomes the universe U we are working in, i.e everything, so we
have A ∩ U = A. Hence the left hand side is equal to A and we have a valid
equality.
j)
A ⊂ C and B ⊂ D ⇒ (A × B) ⊂ (C × D)
We assume A ⊂ C and B ⊂ D, and choose an arbitrary point (a, b) ∈ A × B,
which means a ∈ A and b ∈ B. From our assumption a ∈ A ⊂ C implies
a ∈ C and likewise b ∈ D. So we have a ∈ C and b ∈ D which means
(a, b) ∈ C × D which proves the implication.
k)
(A × B) ⊂ (C × D) ⇒ A ⊂ C and B ⊂ D
We can set A = B = ∅ and this is always true.
l)
Same as k), but assuming A and B are nonempty.
We assume (A × B) ⊂ (C × D) which means for any (a, b) ∈ (A × B) we
have (a, b) ∈ (C × D), or a ∈ A and b ∈ B implies a ∈ C and b ∈ D, or
a ∈ A ⇒ a ∈ C and b ∈ B ⇒ b ∈ D which are the defining properties of
A ⊂ C and B ⊂ D. The implication is true.
5
m)
(A × B) ∪ (C × D) = (A ∪ C) × (B ∪ D)
Not generally true as indicated by the following image (but we see that ⊂
holds).
(A×B)∪(C×D)
(A∪C)×(B∪D)
A
A
C
C
B
D
B
D
Figur 2: Counterexample
n)
(A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D)
(a, b) ∈ (A × B) ∩ (C × D) ⇔
⇔
⇔
⇔
⇔
o)
(a, b) ∈ A × B and (a, b) ∈ C × D
(a ∈ A, b ∈ B) and (a ∈ C, b ∈ D)
(a ∈ A and a ∈ C) and (b ∈ B and b ∈ D)
a ∈ A ∩ C and b ∈ B ∩ D
(a, b) ∈ (A ∩ C) × (B ∩ D)
A × (B \ C) = (A × B) \ (A × C)
(a, b) ∈ A × (B \ C) ⇔
⇔
⇔
⇔
⇔
⇔
a ∈ A and b ∈ B and b 6∈ C
a ∈ A and a ∈ A and b ∈ B and b 6∈ C
a ∈ A and b ∈ B and a ∈ A and b 6∈ C
(a, b) ∈ A × B and (a, b) 6∈ A × C
(a, b) ∈ (A × B) ∩ (A × C)c
(a, b) ∈ (A × B) \ (A × C)
6
p)
(A \ B) × (C \ D) = (A × C \ B × C) \ A × D
(a, b) ∈ (A × C \ B × C) \ (A × D) ⇐⇒
(a, b) ∈ (A × C) and (a, b) 6∈ (B × C) and (a, b) 6∈ (A × D) ⇐⇒
For (a, b) 6∈ (B × C) the defining condition fails, which is (a ∈ B ∩ b ∈ C)c .
Applying DeMorgan’s Law, this becomes a ∈ B c ∪ b ∈ C c . Collecting what we
know:
a ∈ A ∩ b ∈ C and a ∈ B c ∪ b ∈ C c and a ∈ Ac ∪ b ∈ D c ⇐⇒
We know that a ∈ A and b ∈ C, so from the unions we can discard a ∈ Ac and
b ∈ C c.
a ∈ A and b ∈ C and a ∈ B c and b ∈ D c
a ∈ A and a ∈ B c and b ∈ C and b ∈ D c
a ∈ A ∩ B c and b ∈ C ∩ D c
a ∈ A \ B and b ∈ C \ D
(a, b) ∈ (A \ B) × (C \ D)
(A × B) \ (C × D) = (A \ C) × (B \ D)
Not true in general, but valid for ⊃ as shown in the following image:
(A×B)\(C×D)
(A\C)×(B\D)
A
A
C
{
{
{
{
C
{
{
{
{
q)
⇐⇒
⇐⇒
⇐⇒
⇐⇒
B
B
D
Figur 3: Counterexample
7
D
3
Finding the converse and contrapositive and determining which are true.
a) Statement: If x < 0, then x2 −x > 0 which we abbreviate as “P ⇒ Q”. The converse
is simply “Q⇒P” and the contrapositive is “¬Q⇒ ¬P”.
If x < 0 then x2 − x > 0.
If we define y = −x, then y > 0. Then x2 − x = (−y)2 − (−y) = y 2 + y > 0 which
is obviously true since y > 0.
Converse: If x2 − x > 0, then x < 0.
This is not true as we can verify with the following counterexample. Choosing
x = 3 gives x2 − x = 9 − 3 = 6 > 0.
Contrapositive: If x2 − x ≤ 0, then x ≥ 0.
We assume x2 ≤ x. We assume for contradiction that x < 0. In this case x2 > 0
which leads to x2 > x, which violates our first assumption. Hence we must have
x ≥ 0. As for the original statement, this is true.
b) Statement: If x > 0, then x2 − x > 0.
This is untrue as we can verify by choosing x = 1/2 with x2 = 1/4. In this case
x > 0 but x2 − x = 1/4 − 1/2 = −1/4 < 0.
Converse: If x2 − x > 0, then x > 0.
Untrue. If we choose x = −1/2, then x < 0 but x2 − x = 1/4 − (−1/2) = 3/4.
Contrapositive: If x ≤ 0 then x2 − x ≤ 0.
Untrue. For x = −2, we have x2 − x = 4 − (−2) = 6 > 0.
4
Finding the negation to the statement.
a)
∀a ∈ A → a2 ∈ B. Negated: ∃a ∈ A → a2 6∈ B.
b)
∃a ∈ A → a2 ∈ B. Negated: ∀a ∈ A → a2 6∈ B.
c)
∀a ∈ A → a2 6∈ B. Negated: ∃a ∈ A → a2 ∈ B.
d)
∃a 6∈ A → a2 ∈ B. Negated: ∀a 6∈ A → a2 6∈ B
8
5
a)
b)
c)
d)
6
a)
b)
c)
d)
We will determine if the statements and/or their converses are true. We let A be
a nonempty collection of sets.
S
x ∈ A∈A A =⇒ x ∈ A for at least one A ∈ A. This is true.
S
Converse: x ∈ A for at least one A ∈ A =⇒ x ∈ A∈A A. Also true.
S
x ∈ A∈A A =⇒ x ∈ A for every A ∈ A. Not true in general, even though it
can be true. Based on the assumption, we only know for sure that x ∈ A for at
least one A ∈ A.
S
Converse: x ∈ A for every A ∈ A =⇒ x ∈ A∈A A. This is true. If x is included
in all the sets A ∈ A, then it is also included in their joint union.
T
x ∈ A∈A A =⇒ x ∈ A for at least one A ∈ A. This is of course true, and the
premise supports the stronger consequence as shown in d).
T
Converse: x ∈ A for at least one A ∈ A =⇒ x ∈ A∈A A. Untrue. x must be
included in all A ∈ A in order to be included in their joint intersection, as shown
in d).
T
x ∈ A∈A A =⇒ x ∈ A for every A ∈ A. True.
T
Converse: x ∈ A for every A ∈ A =⇒ x ∈ A∈A A. True.
Writing the contrapositive statement for each of the cases in exercise 5.
S
Statement: x ∈ A∈A A =⇒ x ∈ A for at least one A ∈ A.
S
Contrapositive: If x 6∈ A for all A ∈ A =⇒ x 6∈ A∈A A
S
Statement: x ∈ A∈A A =⇒ x ∈ A for every A ∈ A
S
Contrapositive: If x 6∈ A for at least one A ∈ A =⇒ x 6∈ A∈A A
T
Statement: x ∈ A∈A A =⇒ x ∈ A for at least one A ∈ A.
T
Contrapositive: If x 6∈ A for all A ∈ A =⇒ x 6∈ A∈A A
T
Statement: x ∈ A∈A A =⇒ x ∈ A for every A ∈ A.
T
Contrapositive: If x 6∈ A for at least one A ∈ A =⇒ x 6∈ A∈A A
9
7
For given sets A, B and C, we will rewrite the following sets using ∪, ∩ and \.
D = x x ∈ A and (x ∈ B or x ∈ C)
= x x ∈ A and (x ∈ B ∪ C)
= x x ∈ A ∩ (B ∪ C)
E = x (x ∈ A and x ∈ B) or x ∈ C
= x x ∈ (A ∩ B) or x ∈ C
= x x ∈ (A ∩ B) ∪ C
F = x x ∈ A and (x ∈ B ⇒ x ∈ C)
= x x ∈ A and (x ∈ B c ∪ C)
= x x ∈ A ∩ (B c ∪ C)
= x x ∈ (A ∩ B c ) ∪ (A ∩ C)
= x x ∈ (A \ B) ∪ (A ∩ C)
8
We have the set A = {a, b}, a set with two elements. We will show that the power
set P(A) has four elements. By the definition of the power set:
P(A) = {x | x ⊂ A}.
We note that ∅ and A = {a, b} itself both satisfy the defining property of the
power set. In addition the single element sets {a} and {b} are also subsets of A,
and we have found four sets.
P(A) = ∅, {a}, {b}, {a, b}
If A only had one element, A = {a}, then we would only have two subsets: ∅ and
A. If A had no elements, then ∅ = A and the power set would only have one single
element. For three elements, A = {a, b, c} we would have ∅ and A as elements
along with the three single elements alone, {a}, {b} and {c} and the different
combinations of these: {a, b}, {a, c} and {b, c}, which total 8 sets.
We see that for n sets, the number of elements in the power set is 2n , and this
is the origin to the word ’power’.
10
9
DeMorgan’s Laws for n sets A collected in the family A.
[ c
\
(i)
A =
Ac
A∈A
A∈A
Proof
x∈
[
A∈A
c
[
A ⇐⇒ x 6∈
A
A∈A
⇐⇒ x 6∈ A for all A ∈ A
⇐⇒ x ∈ Ac for all A ∈ A
\
⇐⇒ x ∈
Ac
A∈A
(ii)
\
A∈A
c
[
A =
Ac
A∈A
Proof
x∈
\
A∈A
c
\
A
⇐⇒ x 6∈
A
A∈A
⇐⇒ x 6∈ A for at least one A ∈ A
⇐⇒ x ∈ Ac for at least one A ∈ A
[
⇐⇒ x ∈
Ac
A∈A
10
a)
Determine if the following can be expressed as a cartesian product of subsets of
R × R.
{(x, y) | x is an integer }. This is an expression for the cartesian product of Z × R.
b) {(x, y) | 0 < y ≤ 1}. Corresponds to R × (0, 1]
c) {(x, y) | y > x}. This is not a cartesian product of subsets of R.
d) {(x, y) | x is not an integer and y is an integer }. Corresponds to R \ Z × Z
e) {(x, y) | x2 + y 2 < 1}. This is not a cartesian product of subsets of R.
11
§2 - Functions
12