set theory - test bank and solution manual cafe · 1 j set theory 1.1 { sets and functions ......

1 | Set Theory

1.1 – Sets and Functions

Exercise: 1 Section 1.1Question: Let U = {n ∈ N |n ≤ 10} and consider the subsets A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 4, 5}, andC = {1, 2, 5, 7, 8}. Calculate the following operations.

a) A ∩Bb) (B ∪ C)−Ac) (A ∩B) ∩ (A ∩ C) ∩ (B ∩ C)

d) ((A−B)− C) ∩ (A− (B − C))

Solution: We apply the definitions of set operations:

a) A ∩B = {1, 3, 5}b) (B ∪ C)−A = {1, 2, 3, 4, 5, 7, 8} − {1, 3, 5, 7, 9} = {2, 4, 8}c) A ∩B∩A ∩ C∩B ∩ C = {1, 3, 5}∩{1, 5, 7}∩{1, 2, 5} = {2, 4, 6, 7, 8, 9, 10}∩{2, 3, 4, 6, 8, 9, 10}∩{4, 6, 8, 9, 10}d) ((A−B)− C) ∩ (A− (B − C)) = ({7, 9} − C) ∩ (A− {3, 4}) = {9} ∩ {1, 5, 7, 9} = {9}

Exercise: 2 Section 1.1Question: Let U = {a, b, c, d, e, f, g} and consider the subsets A = {a, b, d}, B = {b, c, e}, and C = {c, d, f}.Calculate the following operations.

a) C ∩ (A ∪B)

b) (A ∪ C)−Bc) (A ∪B ∪ C)− (A ∩B ∩ C)

d) (A−B) ∪ (B − C)

Solution: We apply the definitions of set operations:

a) C ∩ (A ∪B) = C ∩ {a, b, c, d, e} = {c, d}b) (A ∪ C)−B = {a, b, c, d, f} −B = {a, d, f}c) (A ∪B ∪ C)− (A ∩B ∩ C) = {a, b, c, d, e, f} − ∅ = {a, b, c, d, e, f}d) (A−B) ∪ (B − C) = {a, d} ∪ {b, e} = {a, b, d, e}

Exercise: 3 Section 1.1Question: As subsets of the reals, describe the differences between the sets {3, 5}, [3, 5] and (3, 5).

Solution: The set {3, 5} contains the integers 3 and 5. The closed interval [3, 5] contains all real numbersbetween 3 and 5 including 3 and 5, while the open interval (3, 5) contains all real numbers between 3 and 5 notincluding 3 and 5.

Exercise: 4 Section 1.1Question: Prove that the following are true for all sets A and B.

a) A ∩B ⊆ A.

b) A ⊆ A ∪B.

Solution: We use the definitions of subsets and the intersection and union of sets.

a) Let x ∈ A ∩B. Then x ∈ A and x ∈ B =⇒ x ∈ A, so A ∩B ⊆ A.

b) Let x ∈ A. We know that A ∪B = {y | y ∈ A or y ∈ B}, so x ∈ A =⇒ x ∈ A ∪B. Hence A ⊆ A ∪B.

Exercise: 5 Section 1.1Question: Let A and B be subsets of a set S.

a) Prove that A ⊆ B if and only if P(A) ⊆ P(B)

b) Prove that P(A ∩B) = P(A) ∩ P(B).

c) Show that P(A ∪B) = P(A) ∪ P(B) if and only if A ⊆ B or B ⊆ A.

Solution:

1

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2 CHAPTER 1. SET THEORY

a) (=⇒): Suppose A ⊆ B. Then, ∀a ∈ A, a ∈ B. Since P(B) contains all the possible subsets of B, all thepossible subsets of A must be in P(B) because A ⊆ B. Therefore, P(A) ⊆ P(B).

(⇐=): Suppose P(A) ⊆ P(B). Then ∀{a} ∈ P(A), {a} ∈ P(B). Therefore, there must exist a subset Cof P(B) that contains every {a} from P(A). The subset C leads to the conclusion that every a ∈ Amust also be in B. Therefore, A ⊆ B.

b) By definition, P(A ∩ B) = {{t1, t2, ..., tn} | ti ∈ A, ti ∈ B}. This implies {ti} ∈ P(A) and {ti} ∈ P(B).Therefore, by definition of intersection, P(A ∩B) = P(A) ∩ P(B).

c) (=⇒): Suppose there are two sets A and B such that neither A ⊆ B nor B ⊆ A. Let a ∈ A − B andb ∈ B − A. Then the set {a, b} is in P(A ∪ B) but not in P(A) or in P(B). Therefore by thecontrapositive, P(A ∪B) = P(A) ∪ P(B) if A ⊆ B or B ⊆ A.

(⇐=): Suppose A ⊆ B. Then, A∪B = B so P(A∪B) = P(B). Now suppose B ⊆ A. Then A∪B = A soP(A ∪B) = P(A). Either way, P(A ∪B) = P(A) ∪ P(B).

Exercise: 6 Section 1.1Question: Give the list description of P({1, 2, 3, 4}).

Solution: Using the definition of a power set,

P({1, 2, 3, 4}) ={∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4},{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.

Exercise: 7 Section 1.1Question: Give the list description of {{a1, a2, . . . , ak} ∈ P({1, 2, 3, 4, 5})

∣∣∣ a1 + a2 + · · ·+ ak = 8}.

Solution: We need to find all the subsets of {1, 2, 3, 4, 5} whose elements add to a total of 8. Recall that nosubset has repeated elements so {4, 4, } does not make sense. The set is

{{1, 2, 5}, {1, 3, 4}, {3, 5}} .

Exercise: 8 Section 1.1Question: Let A, B, and C be subsets of a set S.

a) Prove that (A−B)− C = (A− C)− (B − C).

b) Find and prove a similar formula for A− (B − C).

Solution:

a)

S

A

B

C

S

A

B

C

In the first Venn diagram, the lighter shade represents (A−B), and the darker shade, which overlaps someof (A − B), represents (A − B) − C. In the second Venn diagram, the lighter shade represents (A − C),while the darker shade represents (A−C)−(B−C). We observe from the diagrams that the darker regionsare equal.


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1.1. SETS AND FUNCTIONS 3

b)

S

A

B

C

S

A

B

C

In the Venn diagram above, the lighter shade represents B−C, and the darker shade represents A−(B−C).In the second diagram, the lighter region represents A−B, and the darker region represents A−C, whichoverlaps some of A−B. Thus, (A−B) ∪ (A− C) = A− (B − C).

Exercise: 9 Section 1.1Question: Let A, B, and C be subsets of a set S.

a) Prove that A4B = ∅ if and only if A = B.

b) Prove that A ∩ (B4C) = (A ∩B)4(A ∩ C).

Solution: Let A, B, and C be subsets of a set S.

a) Suppose that A4B = ∅. Then by definition of the symmetric difference

(A−B) ∪ (B −A) = ∅.

If the union of two sets is the empty set, then each of the two sets must be empty. Hence we deduce thatA−B = ∅ and B −A = ∅. Now for and two sets U and T , the identity U − T = ∅ is equivalent to U ⊆ T .Hence we deduce that A ⊆ B and B ⊆ A. Consequently, A = B.

The argument of the opposite direction is identical. Suppose that A = B. Then A ⊆ B and B ⊆ A. ThusA−B = ∅ and B −A = ∅. We deduce that A4B = (A−B) ∪ (B −A) = ∅.

b) There are a variety of ways to prove the identity A ∩ (B4C) = (A ∩ B)4(A ∩ C). We could use a welldesigned Venn diagram. We could also use a membership table which lists all possibilities of an elementwhether it is in or not in one of the given three sets. Here is a membership table for both side of the equality.In this table, we put an

√in a column to indicate membership and nothing to indicate non-membership.

Hence if there is a√

in the A and C column and nothing in the B column, that refers to the situations ofan element in A, not in B and in C.

A B C (B4C) A ∩ (B4C) (A ∩B) (A ∩ C) (A ∩B)4(A ∩ C)√ √ √ √ √√ √ √ √ √ √√ √ √ √ √ √√√ √√ √√ √

Since the A ∩ (B4C) and column and the (A ∩B)4(A ∩ C) of this membership table are the same, thenthe sets are equal.

Exercise: 10 Section 1.1Question: Let S be a set and let {Ai}i∈I be a collection of subsets of S. Prove the following.

a)⋃i∈I

Ai =⋂i∈I

Ai.

b)⋂i∈I

Ai =⋃i∈I

Ai.

Solution: Let S be a set and let {Ai}i∈I be a collection of subsets of S.




a) We will prove the equation by proving set inclusion in both directions.

(=⇒) Let a ∈⋃i∈I

Ai. Then a /∈⋃i∈I

Ai or a /∈ Ai for every i ∈ I. And this implies that a ∈ Ai for every

i ∈ I. So then a ∈⋂i∈I

Ai. And so⋃i∈I

Ai ⊆⋂i∈I

Ai.

(⇐=) Let a ∈⋂i∈I

Ai. Then a ∈ Ai for every i ∈ I. Which implies that a /∈ Ai for eversy i ∈ I. And so

a /∈⋃i∈I

Ai. Which implies a ∈⋃i∈I

Ai. So⋂i∈I

Ai ⊆⋃i∈I

Ai.

And so⋃i∈I

Ai =⋂i∈I

Ai.

b) We will prove this equation by proving set inclusion in both directions.

(=⇒) Let a ∈⋂i∈I

Ai. So a /∈⋂i∈I

Ai. So a /∈ Ai for at least one i ∈ I. So a ∈ Ai for at least one i ∈ I. So

a ∈⋃i∈I

Ai. Which implies that⋂i∈I

Ai ⊆⋃i∈I

Ai.

(⇐=) Let a ∈⋃i∈I

Ai. Then a ∈ Ai for at least one i ∈ I. This implies that a /∈ Ai for at least one i ∈ I.

It follows that a /∈⋂i∈I

Ai. Which implies that a ∈⋂i∈I

Ai. So⋃i∈I

Ai ⊆⋂i∈I

Ai.

So⋂i∈I

Ai =⋃i∈I

Ai.

Exercise: 11 Section 1.1Question: Let P be the parabola in the plane whose equation is y = x2. Let {Aq}q∈P be the collection ofsubsets of R2 where Aq is the tangent line to P at q. Determine with proof

⋃q∈P Aq.

Solution: Sketching a picture of the standard parabola, we see that all the tangent lines are in some sensebeneath the parabola. Also, taking the (infinite, uncountable) union of all the tangent lines to the parabola, wemight guess that we would get all points (a, b) in the plane such that b ≤ a2. We need to prove this hypothesis.

Label the coordinates of a point q ∈ P as q = (x0, x20). From calculus, the tangent line to P at q has the

equation

y = x20 + 2x0(x− x0) =⇒ y = 2x0x− x2

0.

Now suppose that some point (a, b) in the plane is on a tangent line. Then, for some x0, we have b = 2x0a− x20.

Since a, b are given and x0 is unknown, this is an equation in x0. The quadratic formula gives for x0

x0 =2a±

√4a2 − 4b

2= a±

√a2 − b.

In particular, we note that there exists an x0 if and only if a2− b ≥ 0, confirming the hypothesis that b ≤ a2.

Exercise: 12 Section 1.1Question: Let {Ak}k∈N be the collection of subsets of R3 such that Ak = {(x, y, z) ∈ R3 | z ≥ ky}. Determineboth

⋃k∈NAk and

⋂k∈NAk.

Solution: Each subset Ak represents all points greater than the plane that is bound on the line z = ky. In thediagram below, imagine the x-axis is coming out of the page, and let k1, k2, and k3 represent the bounds of thesubsets A1, A2, and A3 respectively. We note that the larger k gets, the more steep the plane becomes. However,the condition does not hold true for z < 0 when y is 0. Therefore,

⋃k∈NAk = {(x, y, z) | z ≥ y}∪{(x, y, z) | x < 0},

and⋂k∈NAk = {(x, y, z) | z ≥ y} ∩ {(x, y, z) | x ≤ 0}.




z

y

k1k2k3

Exercise: 13 Section 1.1Question: In geometry of the plane, a subset S of the plane is called convex if for all p, q ∈ S the line segmentpq connecting p and q is a subset of S. Prove that the intersection of two convex sets is a convex set.

Solution: Let S and R both be convex sets. Consider S ∩R.

Case 1: If S ∩R is empty, than it is trivially a convex set.

Case 2: If it is non-empty, for any p, q ∈ S ∩ R, consider the line segment pq. Since p, q ∈ S and S is convex,we know that pq ⊆ S. By the same reasoning, since p, q ∈ R and R is convex, this implies that pq ⊆ R.Since pq is a subset of both S and R, we have pq ⊆ S ∩R. So S ∩R is convex.

This proves that S ∩R is convex.

Exercise: 14 Section 1.1Question: Inclusion-Exclusion Principle. Let A, B, and C be finite subsets of a set S.

a) Prove that |A ∪B| = |A|+ |B| − |A ∩B|.b) (*) Prove that |A ∪B ∪ C| = |A|+ |B|+ |C| − |A ∩B| − |A ∩ C| − |B ∩ C|+ |A ∩B ∩ C|.

Solution:

a) Let |A| = n and |B| = m. Now A ∪ B is all of the elements that are either in A or B. If we say that|A ∪ B| = |A| + |B| = n + m, for any element c that exists in both A and B, or equivalently A ∩ B, weaccount for that element twice, once in |A| and the second in |B|. So we must subtract 1 for the |A ∩ B|elements we account for twice. So |A ∪B| = |A|+ |B| − 1(|A ∩B|) = |A|+ |B| − |A ∩B|.

b) Let D = A∪B and consider |D∪C|. By applying the first result of this exercise, |D∪C| = |D|+|C|−|D∩C|.And if we put A ∪B back in for D and apply the first result again we get |A ∪B|+ |C| − |(A ∪B) ∩C| =|A|+ |B| − |A∩B|+ |C| − |(A∪B)∩C|. Now we will consider |(A∪B)∩C|. (A∪B)∩C contains all theelements that are in (A or B) and C. If A and C have n elements in common, or equivalently |A∩C| = n,and likewise |B∩C| = m, then |(A∪B)∩C| is certainly at most m+n. However, we should not double countwhatever elements are in both A∩C and B∩C, or A∩B∩C. So we must subtract out that amount, and wearrive at |(A∪B)∩C| = |A∩C|+ |B∩C|−|A∩B∩C|. Plugging this result back into our original equationwe get |A|+ |B|−|A∩B|+ |C|−|(A∪B)∩C| = |A|+ |B|+ |C|−|A∩B|−(|A∩C|+ |B∩C|−|A∩B∩C|) =|A|+ |B|+ |C| − |A ∩B| − |A ∩ C| − |B ∩ C|+ |A ∩B ∩ C|.

Exercise: 15 Section 1.1Question: Let U be a set and A,B ⊆ U .

a) Show by any means that A ∩B = A ∪B.

b) Show by any means that A ∪B = A ∩B.

Solution:




a) Observe the Venn diagrams below:

UUU

A B

UUU

A B

In the above diagrams, the left diagram represents A ∩B, and the right diagram represents A ∪ B. Inthe A ∪ B diagram, represents A, represents B, and represents where they overlap. A ∪ B isrepresented by the union of these shaded regions. We can observe the shaded regions from both diagramsdescribes the same set, thus A ∩B = A ∪B.

b) Observe the Venn diagrams below:

UUU

A B

UUU

A B

In the above diagrams, the left diagram represents A ∪B, and the right diagram represents A ∩B. In theA ∩ B diagram, represents A, represents B, and represents A ∩ B. Notice, the shadedregion describes the same set that is shaded in the A ∪B diagram, hence A ∪B = A ∩B.

Exercise: 16 Section 1.1Question: (*) Let n be a positive integer. Describe an algorithm (a finite list of well-defined instructions toaccomplish a task) to list all the subsets of {1, 2, 3, . . . , n}.

Solution:

Let S = {{∅}} and k = 1. For every set in S, add a new set to S representing the union of {k} and the setin S. Once this process is completed union {k} to S. Increase k by 1. Repeat this process until k = n. In otherwords, while k < n, S = S ∪ {s | t ∪ {k}, ∀t ∈ S} ∪ {k}.

Exercise: 17 Section 1.1Question: For each of these real-valued functions determine the largest possible domain D as a subset of Rand then prove whether f : D → R is an injection, surjection, both, or neither.

a) f(x) = −3x+ 4

b) f(x) = −3x2 + 7

c) f(x) = (x+ 1)/(x+ 2)

d) f(x) = x5 + 1

Solution: We decide if the function is an injection, surjection, both, or neither.

a) Let f(x) = −3x+ 4. This function can be defined over the whole domain R. Suppose that f(x1) = f(x2).Then −3x1 + 4 = −3x2 + 4. This implies that −3x1 = −3x2 so x1 = x2. This shows that f(x) is injective.To prove surjectivity, we attempt to solve for x in the expression y = f(x) for an arbitrary y. We getx = 4−y

3 . Since there is a solution in x for any y, then f is surjective. (f is bijective.)

b) Let f(x) = −3x2 + 7. The largest possible domain of definition is R. Note that f(−1) = 4 = f(1). Thisshows that f is not injective. To test for surjectivity, we attempt to solve for x in y = f(x). The equationy = −3x2 + 7 leads to x = ±

√(7− y)/3. However, if y > 7 there is no solution for x. Hence, since the

codomain of f is R, f is not surjective. (f is neither.)

c) Let f(x) = (x+1)/(x+2). This function is defined on R−{−2}. This is the domain. To test for injectivity,




suppose that f(x1) = f(x2). Then we have

x1 + 1

x1 + 2=x1 + 1

x1 + 2=⇒ (x1 + 1)(x2 + 2) = (x1 + 2)(x2 + 1)

=⇒ x1x2 + 2x1 + x2 + 2 = x1x2 + x1 + 2x2 + 2

=⇒ 2x1 + x2 = x1 + 2x2

=⇒ x1 = x2.

This shows that f is injective. To test for surjectivity, we attempt to solve y = f(x) for x given arbitraryy. We have

y =x+ 1

x+ 2=⇒ yx+ 2y = x+ 1 =⇒ xy − x = 1− 2y =⇒ x =

1− 2y

y − 1.

We see that there is no solution for x if y = 1. Hence f(x) is not surjective.

d) Consider the function f(x) = x5 + 1. This is defined over all R so this is the largest possible domain in R.To check for injectivity, consider the equality f(x1) = f(x2). This gives

x51 + 1 = x5

2 + 1 =⇒ x51 − x5

2 = 0 =⇒ (x1 − x2)(x41 + x3

1x2 + x21x

22 + x1x

32 + x4

2) = 0.

Obviously the equation holds if x1 = x2. Now we look for solutions of the second term. Note that ifx2 = 0, then the quartic equation implies that x1 = 0. But then x1 = x2, which we already know to be apossibility. Assuming that x1 6= x2, after division by x4

2 the quartic term implies(x1

x2

)4

+

(x1

x2

)3

+

(x1

x2

)2

+

(x1

x2

)+ 1 = 0

A graph of the function g(x) = x4 + x3 + x2 + x + 1 shows that g(x) has no solutions. Thus, the onlysolution to f(x1) = f(x2) is x1 = x2. Thus f is injective. For surjectivity, we see that y = f(x) impliesthat x = 5

√y − 1, which is defined for all y. Hence f is surjective. (f is bijective.)

Exercise: 18 Section 1.1Question: Given an explicit example of a function f : Z→ Z that is

a) bijective;

b) surjective but not injective;

c) injective but not surjective;

d) neither injective nor surjective.

Solution: Recall that bxc takes x and returns the nearest integer less than or equal to x. The solutionspresented here are not the only options!

a) f(x) = −x.

b) f(x) = bx/2c.c) f(x) = x ∗ 2.

d) f(x) = x2.

Exercise: 19 Section 1.1Question: Given an explicit example of a function f : N→ N that is

a) bijective;

b) surjective but not injective;

c) injective but not surjective;

d) neither injective nor surjective.

Solution: Recall that bxc takes x and returns the nearest integer less than or equal to x. The answers presentedhere are not the only options!

a) f(x) = x.

b) f(x) = bx/2c.c) f(x) = 2 ∗ x.




d) f(x) = 2 ∗ (bx/2c).

Exercise: 20 Section 1.1Question: Let f : A → B and g : B → C be functions. Prove that if f and g are bijective, then g ◦ f isbijective and

(g ◦ f)−1 = f−1 ◦ g−1.

Solution: Assume g ◦ f is not injective. Then there would exist a1, a2 ∈ A such that a1 6= a2 and g ◦ f(a1) =g ◦ f(a2). Let b1 = f(a1) and b2 = f(a2). We know that b1 6= b2 because f is bijective. Using substitution, wenotice that g(b1) = g(b2), but b1 6= b2. This creates a contradiction since g is bijective. Thus, g ◦ f is injective.Now assume that g ◦f is not surjective. Then there would exist at least one c ∈ C such that ∀a ∈ A, g(f(a)) 6= c.By substituting we observe ∀b ∈ B, g(b) 6= c. However, since g is bijective, this creates a contradiction. Thereforeg ◦ f is surjective. Hence, g ◦ f is bijective.

Let f(a) = b and g(b) = c. Then g(f(a)) = c so (g ◦ f)−1(c) = a. Therefore,

(g ◦ f)−1(c) = a = f−1(b) = f−1(g−1(c))

so (g ◦ f)−1 = f−1 ◦ g−1.

Exercise: 21 Section 1.1Question: Suppose that f and g are functions and that f ◦ g is injective.

a) Prove that g is injective.

b) Does it also follow that f is injective? Justify your answer (with a proof or counter-example).

Solution:

a) For the sake of contradiction, we assume g is not injective. Then for some a, b that exist in the domain ofg, g(a) = g(b). However, then f(g(a)) = f(g(b)) as well. This contradicts the injectivity of f ◦g. Thereforeg must be injective.

b) No it does not. Consider the following functions f, g : Z −→ Z:

g(x) =

{2x if x ≥ 0

2|x|+ 1 if x < 0.

f(x) = x2.

Now, g sends non-negative numbers to a unique, non-negative, even number and negative numbers to aunique, positive, odd number. However, f is injective on non-negative numbers so that f ◦ g is injective,but is not injective on all of Z. (Note: What is necessary is that f be injective on the range of g (in ourcase, the nonnegative numbers) but not necessarily the codomain of g(in our case, all of Z))

Exercise: 22 Section 1.1Question: Suppose that f and g are functions and that f ◦ g is surjective.

a) Prove that f is surjective.

b) Does it also follow that g is surjective? Justify your answer (with a proof or counter-example).

Solution:

a) Consider any a in the codomain of f . Since a is also in the codomain of f ◦ g and f ◦ g is a surjectivefunction, there must exist some b in the domain of f ◦ g so that a = (f ◦ g)(b). Set c = g(b). Then(f ◦ g)(b) = f(g(b)) = f(c) = a. So every element of the codomain is hit by f and f is surjective.

b) No it does not. Consider the functions f, g : N −→ N where f(x) = bx/2c and g(x) = 2x. Then for anyy ∈ N, (f ◦ g)(y) = f(g(y)) = f(2 ∗ y) = b(2 ∗ y)/2c = byc = y. So f ◦ g is surjective, but g is certainly notsince no odd, positive numbers will be hit.

Exercise: 23 Section 1.1Question: Restate the definition of (a) injective and (b) surjective as applied to a function f : A→ B in termsof properties of the sets f−1({b}).Solution:




a) For injectivity, |f−1({b})| ≤ 1.

b) For surjectivity, |f−1{b})| ≥ 1.

Exercise: 24 Section 1.1Question: For the following functions f , find the pre-image (or fiber) f−1(T ) of the given set T in the codomain.

a) f : R→ R with f(x) = sinx and T = {√

3/2}.b) f : R→ R with g(x) = x2 − 2 and T = [1, 2].

c) f : R→ R with h(x) = x3 − 2x and T = [−1, 0].

Solution: Note that the pre-image of f−1(T ) represents all of the domain values that map to the elements ofT .

a) f−1(T ) = sin−1(√

32 ) = {π3 ,

2π3 }

b) f−1(T ) = g−1([1, 2])

x2 − 2 = 1⇒ x = ±√

3

x2 − 2 = 2⇒ x = ±2

x

f(x)

−2 −1 1 2

1

2

−1

−2

x2 − 2

Therefore, the pre-image of g−1([1, 2]) is {[−2,−√

3], [√

3, 2]}.c) f−1(T ) = h−1([−1, 0]) Notice that 1 is a zero of the polynomial x3 − 2x + 1. This allows us to find the

x-values that satisfy x3 − 2x = −1.

x3 − 2x+ 1 = (x− 1)(x2 + x− 1)

x = 1,−1±

√5

2

x3 − 2x = 0⇒ x = 0,±√

2

x

f(x)

−2 −1 1 2

1

2

−1

−2

x3 − 2x

Therefore, the pre-image of h−1([−1, 0]) is {[−1−√

52 ,−

√2], [0, −1+

√5

2 ], [1,√

2]}.




Exercise: 25 Section 1.1Question: Let f : A −→ B be a function from the set A to the set B. Let S and T be subsets of the domainA.

1. Show that f(S ∪ T ) = f(S) ∪ f(T ).

2. Show that f(S ∩ T ) ⊆ f(S) ∩ f(T ).

3. Find an example of a function f : A→ B and subsets S and T in A such that f(S ∩ T ) 6= f(S) ∩ f(T ).

Solution:

a) We prove first that f(S ∪ T ) ⊆ f(S)∪ f(T ). Suppose that y ∈ f(S ∪ T ). Then there exists x ∈ S ∪ T suchthat f(x) = y. We have x ∈ S or x ∈ T , so y = f(x) ∈ f(S) or y = f(x) ∈ f(T ). Hence y ∈ f(S) ∪ f(T ),so we deduce that f(S ∪ T ) ⊆ f(S) ∪ f(T ).

Conversely, suppose that y ∈ f(S) ∪ f(T ). This y ∈ f(S) or y ∈ f(T ). We deduce that there exists x ∈ Swith f(x) = y or there exists x′ ∈ T with f(x′) = y. Thus, there exists an x ∈ S ∪ T such that x = f(y)and we deduce that f(S) ∪ f(T ) ⊆ f(S ∪ T ).

With these two set inclusions, we conclude that f(S ∪ T ) = f(S) ∪ f(T ).

b) Let y ∈ f(S ∩ T ). By definition, there exists x ∈ S ∩ T such that y = f(x). Then x ∈ S and x ∈ T soy = f(x) ∈ f(S) and y = f(x) ∈ f(T ). Thus y ∈ f(S) ∩ f(T ). We conclude that f(S ∩ T ) ⊆ f(S) ∩ f(T ).

c) Consider the function f : R→ R with f(x) = x2. Setting S = [−2,−1] and T = [1, 2], we find that

f(S ∩ T ) = f(∅) = ∅f(S) ∩ f(T ) = [1, 4] ∩ [1, 4] = [1, 4]

Obviously, these are not equal sets.

Note that if we attempted to prove that f(S) ∩ f(T ) ⊆ f(S ∩ T ), the reasoning would go as follows. Lety ∈ f(S) ∩ f(T ). Then y ∈ f(S) and y ∈ f(T ). Then there exists x ∈ S with f(x) = y and there existsx′ ∈ T with f(x′) = y. However, since x does not have to be equal to x′, there does not have to exist anelement x′′ ∈ S ∩ T such that f(x′′) = y.

Exercise: 26 Section 1.1Question: Let f : A −→ B be a function from the set A to the set B. Let V and W be subsets of the codomainB. Show the following.

a) f−1(V ∪W ) = f−1(V ) ∪ f−1(W ).

b) f−1(V ∩W ) = f−1(V ) ∩ f−1(W ).

Solution:

a) We will prove the equality by showing set inclusion in both directions.

(=⇒) Consider any element a ∈ f−1(V ∪W ). Then f(a) ∈ V ∪W so that f(a) exists in V or W . Thena ∈ f−1(V ) or f−1(W ). Which implies a ∈ f−1(V ) ∪ f−1(W ). This shows that f−1(V ∪ W ) ⊆f−1(V ) ∪ f−1(W ).

(⇐=) Consider any element b ∈ f−1(V )∪f−1(W ). So b exists in at least one of f−1(V ) or f−1(W ). Thenf(b) exists in V or W . Which implies f(b) ∈ V ∪W . So then b ∈ f−1(V ∪W ). This shows thatf−1(V ) ∪ f−1(W ) ⊆ f−1(V ∪W ).

This proves the equality f−1(V ∪W ) = f−1(V ) ∪ f−1(W ).

b) We will prove the equality by showing set inclusion in both directions.

(=⇒) Consider any element a ∈ f−1(V ∩W ). So f(a) exists in both V and W . Then a exists in bothf−1(V ) and f−1(W ). Which implies that a ∈ f−1(V ) ∩ f−1(W ). This shows that f−1(V ∩W ) ⊆f−1(V ) ∩ f−1(W ).

(⇐=) Consider any element a ∈ f−1(V ) ∩ f−1(W ). So a exists in both f−1(V ) and f−1(W ). Then f(a)exists in both V and W . Then, by definition, f(a) ∈ V ∩W . Which implies that a ∈ f−1(V ∩W ).This shows that f−1(V ) ∩ f−1(W ) ⊆ f−1(V ∩W ).

This proves the equality f−1(V ∩W ) = f−1(V ) ∩ f−1(W ).




Exercise: 27 Section 1.1Question: Let S and T be finite sets with |S| = |T |. Prove that a function f : S → T is injective if and onlyif it is surjective.

Solution:

(=⇒): Suppose f : S → T is injective. Then ∀s1, s2 ∈ S such that f(s1) = f(s2) implies s1 = s2. Therefore,every s ∈ S maps to a unique t ∈ T . Since |S| = |T |, there does not exist t ∈ T such that t is not associatedwith a unique s ∈ S. Thus f is surjective.

(⇐=): Suppose f : S → T is surjective. This means that there are at least |T | elements in S that map to T .Assume there exists f(s1) = f(s2) such that s1 = s2. Then f implies that |S| elements are mapped to atmost |S| − 1 elements in T . Thus there exists at least one t1 ∈ T where there does not exist s ∈ S suchthat f(s) = t1. This creates a contradiction because f is surjective. Therefore f is injective.

Exercise: 28 Section 1.1Question: Let S be a set. For each subset A ⊆ S we define the characteristic function of A as the functionχA : S → {0, 1} such that

χA(s) =

{1 if s ∈ A0 if s /∈ A.

Prove the following.

a) The association A 7→ χA is a bijection between P(S) and the set of functions from S to {0, 1}.b) χA∩B(s) = χA(s) · χB(s) for all s ∈ S.

c) χA∪B(s) = χA(s) + χB(s)− χA(s) · χB(s) for all s ∈ S.

d) χA(s) = 1− χA(s) for all s ∈ S.

e) χA−B(s) = χA(s)(1− χB(s)) for all s ∈ S.

Solution:

a) First, notice that each distinct characteristic function represents a unique subset of S. From proposition1.1.11 we know that there are 2n unique characteristic functions of S. We can write these subsets asn-tuples where each entry contains either a 1 or a 0. We can also picture any function from S to {0, 1} asa n-tuple (f(s1), f(s2), ..., f(sn)) where |S| = n. Note that each n-tuple will represent a distinct functionfrom S to {0, 1}. In comparing n-tuples, we observe that every function from S to {0, 1} represents aunique characteristic function of S implying that the association is injective. Since each entry has twooptions, and there are n entries, there are 2n distinct functions from S to {0, 1}. Thus the cardinalities ofthe functions of S to {0, 1} and P(S) are equal. Using the result in exercise 1.1.27, the association must besurjective as well. Hence, the association from A 7→ χA is a bijection between P(S) and the set of functionsfrom S to {0, 1}.

b) Suppose s ∈ A∩B. Then, s ∈ A and s ∈ B. This implies χA(s) = 1 and χB(s) = 1. From here we deducethat χA∩B(s) = 1 = χA(s) · χB(s). Now suppose s /∈ A ∩ B. Then s 6∈ A or s 6∈ B, so χA(s) = 0 orχB(s) = 0. Either way, χA∩B(s) = 0 = χA(s) · χB(s). Therefore, χA∩B(s) = χA(s) · χB(s) for all s ∈ S.

c) Suppose s ∈ A∪B. Then s ∈ A or s ∈ B. If s ∈ A∩B or if s ∈ B ∩A then we deduce that χA(s) +χB(s)works for both cases. If s ∈ A ∩B, then we need to account for counting an element twice so we subtractχA∩B(s) which we know to be χA(s) · χB(s) from above. Hence χA∪B(s) = χA(s) + χB(s)− χA(s) · χB(s)for all s ∈ S.

d) Suppose s ∈ A. Then s 6∈ A, so χA(s) = 0 and χA(s) = 1. If s ∈ A then s 6∈ A and we get the oppositeresult. Notice to reverse the results of χA(s) we subtract it from 1. Therefore χA(s) = 1− χA(s).

e) Note that A−B = A ∩B and χB(s) = 1− χB(s). Thus χA−B(s) = χA(s)(1− χB(s)) for all s ∈ S.

Exercise: 29 Section 1.1Question: Provide all the details in the proof of Proposition 1.1.23.

Solution: We start with f(n) =⌊∣∣2n+ 1

2

∣∣⌋ where f : Z → N. Plugging in some points such as −2,−1, 0, 1, 2




we observe the following:

f(−2) = 3

f(−1) = 1

f(0) = 0

f(1) = 2

f(2) = 4

Noticing the negative integers point to odd natural numbers and the positive integers point to even naturalnumbers, we can describe any integer n in terms of any natural number m by using the definition of even andodd numbers

n =

{m2 if m is even

−(m+1

2

)if m is odd.

Since every integer n is represented by a different natural number m we can quickly deduce that the associationis surjective by definition. Assume that there are two integers a and b such that a 6= b but f(a) = f(b). Then,⌊∣∣2a+ 1

2

∣∣⌋ =⌊∣∣2b+ 1

2

∣∣⌋ which implies that 2a + 12 = 2b + 1

2 ⇒ a = b. This creates a contradiction, hence theassociation is injective as well. Thus since there exists a bijection between Z and N. Therefore, Z is countable.

Exercise: 30 Section 1.1Question: Let A, B, and C be sets. Prove that if |A| ≤ |B| and |B| ≤ |C|, then |A| ≤ |C|.Solution: Let A, B, and C be sets such that |A| ≤ |B| and |B| ≤ |C|. Also let a be an any element in A,f : A → B, and g : B → C. There exists a unique element in B such that f(a) = b, ∀a ∈ A. Also, there is aunique c ∈ C such that g(b) = c, ∀b ∈ B. By using substitution and composition we find g(f(a)) = c, ∀a ∈ A.This establishes a injective function f : A→ C therefore |A| ≤ |C|.

Exercise: 31 Section 1.1Question: Let A and B be finite sets. Prove that the number of distinct functions A→ B is |B||A|.Solution: Let the n-tuple, (f(a1), f(a2), f(a3), ..., f(an)), represent a distinct function f : A → B where|A| = n. Each entry in the n-tuple can be mapped to a different element in B resulting in m possibilities where|B| = m. Since every entry has m options and there are n entries, the total number of possible distinct n-tuplesis m1 ×m2 ×m3 × ...×mn = mn. By substituting we observe that the number of distinct functions is equal to|B||A|.

1.2 – The Cartesian Product; Operations; Relations

Exercise: 1 Section 1.2Question: Let A,B,C be sets. Explain why A×B × C is not the same set as A× (B × C).

Solution: The set A × B × C contains 3-tuples with the components (a, b, c), however, the set A × (B × C)contains pairs with the components (a, (b, c)).

Exercise: 2 Section 1.2Question: Let A,B,C,D be sets. Explain why A× (B × C)×D is not the same set as (A×B)× (C ×D).

Solution: The elements of the set A × (B × C) × D contains 3-tuples with three components, (a, (b, c), d),whereas the elements of the set (A×B)× (C ×D) contains pairs with two components, ((a, b), (c, d)).

Exercise: 3 Section 1.2Question: Let A = {1, 2, 3, 4} and B = {a, b}. Write out as a list a) A×B; b) A×A; c) B ×B ×A.

Solution:

a) {(1, a), (2, a), (3, a), (4, a), (1, b), (2, b), (3, b), (4, b)}b) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}



1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS 13

c) {(a, a, 1), (a, a, 2), (a, a, 3), (a, a, 4), (a, b, 1), (a, b, 2), (a, b, 3), (a, b, 4), (b, a, 1), (b, a, 2), (b, a, 3), (b, a, 4),(b, b, 1), (b, b, 2), (b, b, 3), (b, b, 4)}

Exercise: 4 Section 1.2Question: Write in list form {1, 3} × {2, 4} × {3, 5}.Solution: {(1, 2, 3), (1, 2, 5), (1, 4, 3), (1, 4, 5), (3, 2, 3), (3, 2, 5), (3, 4, 3), (3, 4, 5)}

Exercise: 5 Section 1.2Question: Write in list form {1} × {1, 2} × {1, 2, 3}.Solution: The elements in {1} × {1, 2} × {1, 2, 3} are triples. The Cartesian product set is

{(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 1), (1, 2, 2), (1, 2, 3)}.

Exercise: 6 Section 1.2Question: Justify the statement that A× ∅ = ∅ for all sets A.

Solution: By definition of the Cartesian product, A× ∅ = {(a, b)| a ∈ A and b ∈ ∅}. Since ∅ has no elements,A× ∅ has no elements and therefore is equal to ∅.

Exercise: 7 Section 1.2Question: (*) This exercise offers a proof that if A and B are countable sets, then A×B is countable.

a) Find a bijection between N∗ and N∗×N∗. [Hint: Count through the pairs (x, y) ∈ N∗×N∗ by successivelygoing through the lines of the form x+ y = n for n = 2, 3, 4, . . ..]

b) Use the bijection in the previous part to prove that if A and B are countable sets, then A×B is countable.

Solution:

a) Note that N∗ ×N∗ represents all the pairs of natural numbers which can be represented on a graph in thefirst quadrant. Using the hint, we establish a series of lines following the pattern x + y = n + 1 wheren = 1, 2, 3, .... Observe the intersections of the first couple of lines:

n = 1⇒ (1, 1)

n = 2⇒ (1, 2), (2, 1)

n = 3⇒ (1, 3), (2, 2), (3, 1)

As the pattern continues, we quickly observe that as n increases, every element in N∗ × N∗ is mapped toonce and only once. This establishes a bijection between N∗ × N∗ and N∗.

b) Suppose A and B are countable sets. Then both A and B have a bijection with N∗. It follows that thereexists a bijection between A × B and N∗ × N∗. Since we have already established a bijection betweenN∗ × N∗ and N∗, there exists a bijection between A×B and N∗. Therefore, A×B is countable.

Exercise: 8 Section 1.2Question: The operation ∗ on vectors of Rn defined by ~u ∗ ~v = proj~u ~v, i.e., projection of ~v onto ~u.

Solution:

Associative: Find any counter-example. Let ~a, ~b, and ~c be vectors in R2 such that ~a =< 1, 1 >, ~b =< 2, 1 >,and ~c =< 3, 1 >. Then (~a ∗~b) ∗ ~c = 4

3~a and ~a ∗ (~b ∗ ~c) = 2110~a. Therefore, ∗ is not associative.

Commutative: Find any counter-example. Using the same vectors above, observe that ~a∗~b = 32~a and~b∗~a = 3

5~b.

Since ~a ∗~b 6= ~b ∗ ~a, ∗ is not commutative.

Identity: Note that in order for ~a∗~e = ~a, proj~a ~e = ~a which is only possible when ~e = ~a since proj~a ~a = ~a. Thusit is impossible to have a unique identity.

Inverse: Since ∗ does not have an identity, it cannot be closed under inverses.




Idempotent: As shown earlier, proj~a ~a = ~a therefore, ∗ is idempotent.

Exercise: 9 Section 1.2Question: For the operation ? on the open interval [0, 1) described by a ? b = a+ b− ba+ bc where bxc is thegreatest integer less than or equal to x, determine whether it is associative, commutative, has an identity, hasinverses or is idempotent.

Solution:

Associative: Let a, b, c ∈ [0, 1). Then

(a ? b) ? c = (a+ b− ba+ bc) ? c= a+ b− ba+ bc+ c− ba+ b− ba+ bc+ cc

since bx+ nc = bxc+ n for all x ∈ R and all n ∈ Z

= a+ b+ c− ba+ bc+ ba+ bc − ba+ b+ cc= a+ b+ c− ba+ b+ cc.

Furthermore

a ? (b ? c) = a ? (b+ c− bb+ cc)= a+ b+ c− bb+ cc − ba+ b+ c− bb+ ccc= a+ b+ c− bb+ cc+ bb+ cc − ba+ b+ cc= a+ b+ c− ba+ b+ cc.

We conclude that (a ? b) ? c = a ? (b ? c). Hence ? is associative.

Commutative: For all a, b ∈ [0, 1) we do have

a ? b = a+ b− ba+ bc = b+ a− bb+ ac = b ? a.

So ? is commutative.

Identity: The element 0 ∈ [0, 1) serves as the identity.

Has Inverses: Let a ∈ [0, 1). If a = 0, then a is its own inverse. If a > 0, then the element b = 1− a, which isalso in [0, 1), satisfies

a ? b = a+ (1− a)− ba+ 1− ac = 1− b1c = 0.

Hence, every element in [0, 1) has an inverse.

Idempotent: Consider the element 0.5. We have 0.5?0.5 = 0, so 0.5 gives one counter example to idempotence.

Exercise: 10 Section 1.2Question: The operation 4 on nonnegative integers N defined by n4m = |m− n|.Solution:

Associative: Let a = 1, b = 2, and c = 3. Observe a4 (b4 c) = ||c−b|−a| = 0 while (a4 b)4 c = |c−|b−a|| =2. Therefore 4 is not associative.

Commutative: a4 b = |b− a| = |a− b| = b4 a. Therefore 4 is commutative.

Identity: a4 e = |e− a| = |a− e| = e4 a = a. Hence e = 0.

Inverse: Every element is it’s own inverse. a4 a = |a− a| = 0.

Idempotent: a4 a = 0 hence 4 is not idempotent.




Exercise: 11 Section 1.2Question: The operation ~ on points in the plane R2 where A~B is the midpoint of A and B.

Solution:

Associative: Let A = (a1, a2), B = (b1, b2), and C = (c1, c2). A ~ (B ~ C) = 2a1+b1+c14 and (A ~ B) ~ C =

a2+b2+2c24 . Thus, if A 6= C then A~ (B ~ C) 6= (A~B)~ C. Hence, ~ is not associative.

Commutative: The midpoint between A and B is the same as the midpoint between B and A, so ~ iscommutative.

Identity: Note that A~ E = A if A = E. Since E is dependent on A, there does not exist an identity for ~.

Inverse: Because there is no identity for ~, there cannot be any inverses.

Idempotent: A~A = (a1+a12 , a2+a2

2 ) = (a1, a2) = A. Therefore ~ is idempotent.

Exercise: 12 Section 1.2Question: The operation +× on C defined by a+×b = a+ b+ ab.

Solution:

Associative: Consider a, b, and c ∈ C.

a+×(b+×c) = a+ b+ c+ bc+ ab+ ac+ abc = a+ b+ ab+ c+ ac+ bc+ abc = (a+×b)+×c

Therefore +× is associative.

Commutative: a+×b = a+ b+ ab = b+ a+ ba = b+×a. Therefore +× is commutative.

Identity: Consider any q ∈ C. q+×0 = 0+×q = q so 0 is the identity.

Inverse: Let b be the inverse of any a ∈ C. Then a+×b = 0 which implies a + b(1 + a) = 0. We can quicklydeduce b = −a

1+a . From this observation we find that there exists an inverse for all values in C except −1.Thus, +× is not closed under inverses.

Idempotent: Let a = 1. Then a+×a = 1 + 1 + 1 = 3 which is not equal to a. Hence, +× is not idempotent.

Exercise: 13 Section 1.2Question: The operation 4 on P(S), where S is any set.

Solution:

Associative:

S

A

B

C

S

A

B

C

Note in the above diagrams, the darker shading represents the operation within the parenthesis. We observethrough the Venn diagrams that a4 (b4 c) = (a4 b)4 c. Thus, 4 is associative.

Commutative:




S

A B

Observing the Venn diagram, A4B = B4A therefore 4 is commutative.

Identity: A4∅ = ∅4A = A hence, ∅ is the identity of A.

Inverse: Each set is it’s own inverse as A4A = ∅.

Idempotent: A4A = ∅ implying 4 is not idempotent.

Exercise: 14 Section 1.2Question: The cross product on R3

Solution:

Associative: Find any counter-example. Let ~a =< 1, 2, 3 >, ~b =< 1, 1, 1 >, and ~c =< 1, 2, 1 >. Then(~a×~b)× ~c =< 4, 0,−4 > and ~a× (~b× ~c) =< 2,−4, 2 >. Since (~a×~b)× ~c 6= ~a× (~b× ~c), the cross productis not associative over R3.

Commutative: For any two vectors in R3, ~a×~b = −~b× ~a. Hence the cross product is not commutative.

Identity: The cross product is perpendicular to the plane created by the two vectors being multiplied bydefinition, therefore, there cannot exist a vector that satisfies ~a× ~e = ~a. Thus the cross product does nothave an identity.

Inverse: Because the cross product does not have an identity, it cannot be closed under inverses.

Idempotent: Note, ~b ×~b =< 0, 0, 0 > which does not equal ~b in every case therefore the cross product is notidempotent.

Exercise: 15 Section 1.2Question: For the power operator a∧b = ab on the set N∗ of positive integers determine whether it is associative,commutative, has an identity, has inverses or is idempotent.

Solution:

Associative: The following give a counter example

2∧(3∧4) = 2∧81 = 281 while (2∧3)∧4 = (23)∧4 = 212

Hence ∧ is not associative.

Commutative : Since 2∧3 = 8 and 3∧2 = 9, the operation ∧ is not commutative.

Identity : Assume that ∧ has an identity e. Then ae = a for all a ∈ N∗. Hence e = 1. However, by definition,we must also have ea = 1a = a for all a ∈ N∗, which leads to a contradiction. Hence ∧ does not have anidenity element.

Inverses: The operation cannot have an inverse since it does not have an identity.

Idempotent: Since 2∧2 = 4 6= 2, then ∧ is not idempotent.

Exercise: 16 Section 1.2Question: The composition operator ◦ on the set F(A,A) of functions from a set A to A (where A is any set).

Solution:




Associative: Let f , g, and h be functions in F(A,A). Then (h ◦ g) ◦ f = h(g(f(a))) = h ◦ (g ◦ f) where a ∈ A.Therefore, ◦ is associative.

Commutative: Consider the function j where j(a) = a1 or in other words, every input gives the same outputa1. Now let k be a function such that k(a1) 6= a1. Then j ◦ k(a) = a1, but k ◦ j(a) 6= a1. Therefore, ◦ isnot commutative.

Identity: Consider the function e where e(a) = a. Let q be any function in F(A,A). Then q ◦ e = q ande ◦ q = q. Therefore, e is the identity.

Inverse: Let q be any function in F(A,A). Then q ◦ q−1 = q(q−1(a)) = a and q−1 ◦ q = q−1(q(a)) = a wherea ∈ A. However, not every function from A to A is bijective, for example, the function j as describedabove. Therefore, ◦ is not closed under inverses.

Idempotent: Let l be a function such that l(a1) = a2 and l(a2) = a3. Then l(l(a1)) = a3 which is not equal tol(a1). Thus, ◦ is not idempotent.

Exercise: 17 Section 1.2Question: Prove that for all A,B,C ∈ P(S),

A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C).

Solution: Let x be in A ∩ (B ∪ C). Then, x ∈ A and x ∈ (B ∪ C) by the definition of the intersection ofsets. So, x ∈ A and x ∈ B, or x ∈ A and x ∈ C. By the definitions of the union and intersection of sets,x ∈ (A ∩B) ∪ (A ∩ C). Therefore, A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C).

Exercise: 18 Section 1.2Question: Let S be a set with a binary operation ∗. Assume that (a ∗ b) ∗ a = b for all a, b ∈ S. Prove thata ∗ (b ∗ a) = b for all a, b ∈ S.

Solution: Assume (a ∗ b) ∗ a = b for all a, b ∈ S.

(a ∗ b) ∗ a = b

(a ∗ b) ∗ a ∗ a = (b ∗ a)

a ∗ (a ∗ b) ∗ a ∗ a = a ∗ (b ∗ a)

(a ∗ b) ∗ a = a ∗ (b ∗ a)

b = a ∗ (b ∗ a)

Therefore, a ∗ (b ∗ a) = b for all a, b ∈ S.

Exercise: 19 Section 1.2Question: Consider the operations a∧b = ab and a × b on N∗. Prove that ∧ is right-distributive over × butnot left-distributive over ×.

Solution: Let a, b, c ∈ N∗. For right-distributivity, we observe that

(a× b)∧c = (ab)c = acbc = (a∧c)× (b∧c),

so ∧ is right-distributive over ×. In contrast, as a counter example to left-distributivity

2∧(3× 4) = 212 while 2∧3× 2∧4 = 23 · 24 = 27.

Exercise: 20 Section 1.2Question: Let S be a finite set with |S| = n. How many binary operations exist on S?

Solution: Let S be a finite set with |S| = n. By proposition 1.2.4 we know that the size of S × S is |S| · |S|.By definiton of a function, we know that for every tuple in S × S there exists an element in S that is mapped




to. Therefore, every element in S × S has |S| possible associations. Since there are |S| · |S| elements, each with|S| possibilities, then there are |S||S|·|S| distinct functions possible from S × S to S. By substitution, there are

nn2

possible binary operators on S.

Exercise: 21 Section 1.2Question: Let S = {1, 2}. How many binary operations on S are associative?

Solution: We observe that S × S = {(1, 1), (1, 2), (2, 1), (2, 2)}. Since each element maps to either 1 or 2,there are 24 binary operations on S. We can represent the binary operations as 4-tuples, (s1, s2, s3, s4), where(1, 1) = s1, (1, 2) = s2, (2, 1) = s3, and (2, 2) = s4. Therefore, by checking each of the sixteen binary operations,the following upheld associativity:

(1, 1, 1, 1), (1, 2, 2, 1), (2, 2, 2, 2), (2, 1, 1, 2), (1, 1, 1, 2), (1, 1, 2, 2), (1, 2, 1, 2), (1, 2, 2, 2)

Hence, there are eight binary operations on S that are associative.

Exercise: 22 Section 1.2Question: Let A and B be finite sets. Find the number of distinct relations from A to B.

Solution: Let A and B be finite sets. By definition, a distinct relation is a distinct subset of A × B. Recall,according to Proposition 1.2.4 that |A×B| = |A| · |B|. By Proposition 1.1.11, we know that |P(A×B)| = 2|A|·|B|.Therefore there are 2|A|·|B| distinct relations from A to B.

Exercise: 23 Section 1.2Question: Let A be a finite set with n elements. Prove that the number of reflexive relations on A is 2n

2−n

and that the number of symmetric relations on A is 2n(n+1)/2

Solution: Let A be a finite set with n elements. Define S to be the set of all possible reflexive elements inA2. Note, the smallest reflexive relation, which we will call B, is of size n such that B = {(a, a) | a ∈ A, ∀a}.Thus, we define S = {s | s ∈ A2 − B and B ∪ {s} is a reflexive relation}. We observe that S = A2 − B so|S| = n2 − n. With the union of P(S) and B we find all possible reflexive relations. By proposition 1.1.11 we

know that the size of P(S) is 2n2−n. Therefore, there are 2n

2−n possible reflexive relations on A. Similar tothe reflexive relations, if we can find the set of possible symmetric elements, then its power set will result in thenumber of symmetric relations on A.

A×A

n = 2

A×A

n = 3

A×A

n = 4

Each graph above represents the total number of symmetric elements for a set A of size n. By definition,every pair in the form of (a, a) is a distinct symmetric element. These elements, contained in B as defined before,are represented by the squares shown in the graphs above. Let T be the set of symmetric pairs in A2 −B suchthat T = {{(a, b), (b, a)} | ∀ a, b ∈ A2}.These pairs are represented by the circles on the graph. To illustrate,when n = 2, the only symmetric pair is {(a1, a2), (a2, a1)} represented by the single dot on the graph. Thuswhen n = 2, |T | = 1 and |B| = 2. Let X = T ∪ B. X contains all the symmetric elements of A2. Hence wesee when |A| = 2, that there are 3 symmetric elements. We observe that every time n increases, n symmetric

elements are added to X. It follows that |X| =∑ni=1 i. By substitution, |X| = n(n+1)

2 . Thus, the cardinality of

its power set results in 2n(n+1)

2 symmetric relations on A.

Exercise: 24 Section 1.2Question: For any set S, consider the relation G on P(S) defined by A G B to mean that A ∩B 6= ∅.Solution:




Reflexive: Choose any subset A. Then, A G A implies A ∩ A = A 6= ∅ which is true for all cases except theempty set. However, ∅ ∈ P(S) so G is not reflexive.

Symmetric: Let A and B be subsets of S. Then,

A G B ⇒ A ∩B 6= ∅ ⇒ B ∩A 6= ∅ ⇒ B G A

Therefore, G is symmetric.

Antisymmetric: Consider the set S = {1, 2, 3, 4}. Let A = {1, 2} and B = {3, 4}. We observe that both A G Band B G A exist, however it does not imply that they are equal. Hence, G is not antisymmetric.

Transitive: Consider the set S = {1, 2, 3, 4}. Let A = {1, 2}, B = {3}, and C = {2, 4}. Notice that while A G Band B G C satisfy the conditions of transitivity, A G C does not exist because A∩C = {2}. Hence, G is nottransitive.

Exercise: 25 Section 1.2Question: The relation % on S the set of people defined by p1 % p2 if p1 is taller than or the same height asp2.

Solution:

Reflexive: Any person is the same height as himself, therefore % is reflexive.

Symmetric: Consider the case where an individual p1 is taller than p2. Then p1 % p2 but p2 6% p1. Hence, %is not symmetric.

Antisymmetric: Assume (p1 % p2) and (p2 % p1), but p1 6= p2. This implies that either p1 is taller than p2 orvice versa. Without loss of generality, suppose p1 is taller than p2. Then p1 % p2 but p2 6% p1 which is acontradiction. Therefore, if p1 % p2 and p2 % p1, then p1 = p2. Thus % is antisymmetric.

Transitive: Assume that p1 6% p3, but p1 % p2 and p2 % p3 where p2 6= p3. This implies that p2 is taller thanp3 and p1 is taller than or equal to p2. Hence, p1 would have to be taller than p3 which is a contradiction.Therefore, if p1 % p2 and p2 % p3, then p1 % p3 implying % is transitive.

Exercise: 26 Section 1.2Question: The relation R on Z defined by nRm if n ≥ m2.

Solution:

Reflexive: Find any counter-example. Let n = 2. This creates a contradiction since 2 6≥ 4, therefore R is notreflexive.

Symmetric: Find any counter-example. Let n = 9 and m = 2. It is clear to see that nRm but the reverse isnot true. This creates a contradiction so R is not symmetric.

Antisymmetric: Let n and m be integers such that nRm and mRn. This implies that n ≥ m2 and m ≥ n2.By squaring both sides and substitution, we find that n ≥ n4 which is only true if n = 1. Applying thesame method the other direction, we find that m = 1 as well. Therefore, the only case where nRm andmRn is when n = m = 1. Thus, R is antisymmetric.

Transitive: Let a, b and c be integers. Assume a 6≥ c2, but aRb and bRc. Using some substitution, this impliesthat a ≥ b2 ≥ c2. This creates a contradiction, therefore R is transitive.

Exercise: 27 Section 1.2Question: Consider the relation � on S = R2 defined by (x1, y1) � (x2, y2) to mean x2

1 + y21 ≤ x2

2 + y22 . Prove

which of the properties reflexivity, symmetry, antisymmetry, and transitivity hold.

Solution:

Reflexivity: Let (x, y) ∈ R2. Then x2 + y2 ≤ x2 + y2, so (x, y) � (x, y). Hence � is reflexive.




Symmetry: Consider the elements (1, 1) and (1, 2). Then 12 + 12 ≤ 12 + 22 so (1, 1) � (1, 2). However12 + 22 6≤ 12 + 12 so (1, 2) 6� (1, 1). Hence � is not symmetric.

Antisymmetry: Note that (1, 2) � (2, 1) and (2, 1) � (1, 2) but since (1, 2) 6= (2, 1), then � is not antisymmetric.

Transitivity: Consider three points (x1, y1), (x2, y2), and (x3, y3). Suppose that (x1, y1) � (x2, y2) and (x2, y2) �(x3, y3). Then

x21 + y2

1 ≤ x22 + y2

2 and x22 + y2

2 ≤ x23 + y2

3 .

Hence x21 + y2

1 ≤ x22 + y2

2 so (x1, y1) � (x3, y3). Hence � is transitive.

Exercise: 28 Section 1.2Question: The relation $ on R defined by a $ b to mean ab = 0.

Solution:

Reflexive: Consider the case where a ∈ R and a 6= 0. Then a $ a ⇒ a2 = 0. This creates a contradiction asthis is not true in every case, thus $ is not reflexive.

Symmetric: Suppose a $ b. By definition, ab = 0 implying that either a = 0 or b = 0. In either case, ba = 0.Hence, b $ a therefore $ is symmetric.

Antisymmetric: Let a 6= 0 and b = 0. Then a $ b and b $ a, but a 6= b. Hence, $ is not antisymmetric.

Transitive: Let a 6= 0, b = 0, and c 6= 0. Then a $ b and b $ c, but a 6$ c since a 6= 0 and c 6= 0. Thus, $ isnot transitive.

Exercise: 29 Section 1.2Question: For any set S, consider the relation on P(S) defined by A B to mean that A ∪B = S.

Solution:

Reflexive: Let A ∈ P(S) and A 6= S. Then, A ∪A = A which is not equal to S, therefore is not reflexive.

Symmetric: Suppose A, B are in P(S) and A B. Then A ∪ B = B ∪ A = S which implies B A. Hence, is symmetric.

Antisymmetric: Let A = ∅ and B = S. Then, A B and B A, but A 6= B. Therefore, cannot beantisymmetric.

Transitive: Let A = ∅, B = S, and C 6= S such that A C. However, notice that A B and B C, butA C implies A ∪ C = S which creates a contradiction since C 6= S. Thus, is not transitive.

Exercise: 30 Section 1.2Question: The relation� on the set of pairs of points in the plane S = R2×R2 defined by (P1, Q1)� (P2, Q2)if the segment [P1, P2] intersects [Q1, Q2].

Solution:

Reflexive: Let P1, P2 be in S. It is not hard to see that (P1, P1) � (P2, P2) because any line intersects withitself. Hence, � is reflexive.

Symmetric: Suppose (P1, Q1) � (P2, Q2). Then [P1, P2] intersects [Q1, Q2] implying that [Q1, Q2] intersects[P1, P2] and (Q1, P1)� (Q2, P2). Thus, � is symmetric.

Antisymmetric: Let (P1, Q1)� (P2, Q2) such that [P1, P2] is perpendicular to [Q1, Q2]. It is not hard to showthat (P1, Q1)� (P2, Q2) and (Q1, P1)� (Q2, P2), but it is impossible for [P1, P2] to be equal to [Q1, Q2].Therefore, � is not antisymmetric.

Transitive: Let [P1, P2], [Q1, Q2], and [R1, R2] be lines in S and let [P1, P2] be parallel to [R1, R2] and per-pendicular to [Q1, Q2]. Then (P1, Q1) � (P2, Q2) and (Q1, R1) � (Q2, R2), but it is impossible for(P1, R1)� (P2, R2) since they are parallel to each other. Therefore, � is not transitive.




Exercise: 31 Section 1.2Question: Let S be a set and let R be a relation on S. Prove that if a relation is reflexive, symmetric, andanti-symmetric, then it is the = relation on S.

Solution: Let S be a set and let R be a relation on S such that R is reflexive, symmetric, and antisymmetric.By definition, R contains all elements of the form (s, s) ∀s ∈ S. Since R is symmetric, if sR t then tR s wheres, t ∈ S. Moreover, R is antisymmetric implying that if there is a symmetric pair (s, t), (t, s) ∈ R, then s = t.Hence, it is impossible for R to contain an element of the form (s, t) where s 6= t. Therefore, R = {(s, s), ∀s ∈ S}which is the = relation.

Exercise: 32 Section 1.2Question: Let P be the set of people who are living now. Let R be the relation on P defined by aRb if a andb are in the same nuclear family, i.e. if a is a self, child, parent, sibling, or spouse of b.

a) Decide whether R is reflexive, symmetric, antisymmetric, or transitive.

b) List all the family relations included in R(2) = R ◦R.

c) Give four commonly used family terms for relations in R(3) = R ◦R ◦R though not in R(2).

Solution:

a) For any person p, pR p since ’self’ is included in the nuclear family, therefore R is reflexive. Assumethere exists two people p1 and p2 such that p1Rp2 but p2 is not related to p1. It is not hard to see thatthis creates a contradiction as any two people who are in the same nuclear family satisfy the relation R.Therefore R is symmetric. In regards to antisymmetry, consider any p1Rp2 where p1 is the child of p2. Weknow that p1Rp2 and p2Rp1, however p1 and p2 are not the same person. Hence, R is not antisymmetric.Let p1 be the child of p2 and p2 be the sibling of p3. We can easily observe that p1Rp2 and p2Rp3, yet p1

is not in the same nuclear family as p3. Therefore R is not transitive.

b) R2 = {self, child, parent, sibling, spouse, grandchild, child-in-law, grandparent, uncle / aunt, niece /nephew, sibling-in-law, parent-in-law}

c) {cousin, great-grandchild, great-grandparent, great-uncle / aunt}

Exercise: 33 Section 1.2Question: We can define the graph of a relation R from R to itself as the subset of R2

{(x, y) ∈ R2 |xR y}.

a) Sketch the graph of the relation ≤.

b) Sketch the graph of the relation l defined by x l y if |x− y| = 1.

c) Provide defining geometric characteristics of a subset of R2 for a relations on R that are i) reflexive; ii)symmetric; iii) transitive; iv) antisymmetric.

Solution:

a)

y

x

y = x




b)

y

x

y = x− 1

y = x+ 1

c) i) A reflexive relation must contain the line y = x.ii) A symmetric relation must have a reflective mapping over the line y = x.iii) An antisymmetric relation cannot have a reflective mapping over the line y = x and must include pointsfrom the line y = x.iv) In a transitive relation, for any points (x, y) and (y, z) there exists a right triangle that contains thepoints (x, y), (y, z), and (x, z).

Exercise: 34 Section 1.2Question: Let S = {a, b, c, d, e} and consider the relation R on S described by

R = {(a, a), (a, c), (a, d), (b, c), (b, e), (c, b), (c, d), (e, a), (e, b)}.

Determine as a list in S × S, the composite relation R ◦R.

Solution: R ◦R = {(a, a), (a, b), (a, c), (a, d), (b, a), (b, b), (b, d), (c, c), (c, e), (e, a), (e, c), (e, d), (e, e)}

Exercise: 35 Section 1.2Question: Let R be a relation on a set A. Denote by R(n) the n-composite relation of R with itself:

R(n) def=

n times︷︸︸︷R ◦R ◦ · · · ◦R.

Prove that the relation R is transitive if and only if R(n) ⊆ R for all n = 1, 2, 3, . . ..

Solution: (=⇒): Suppose R is transitive. Assume that R(n) contains an element (x, y) such that (x, y) 6∈ R.Then there exists some z ∈ A such that (x, z) and (z, y). This creates a contradiction by the definition of atransitive relation, thus ∀(x, y) ∈ R(n), (x, y) ∈ R. Therefore, R(n) ⊆ R. (⇐=): Suppose R(n) ⊆ R. Assumethat there exists aR b and bR c such that a is not related to c. By definition of relation composition, aR c existsin R(n). If aR c exists in R(n), then it must also exist in R by definition of a subset causing a contradiction.Hence, if aR b and bR c, then aR c. Therefore, R is transitive.

Exercise: 36 Section 1.2Question: Let R be a relation that is reflexive and transitive. Prove that Rn = R for all n ∈ N∗.Solution: Let R be a relation that is reflexive and transitive. From exercise 1.2.35 we know that Rn ⊆ R.Consider the case when n = 2 and let (r1, r2) be any pair in R. Then r1Rr1 and r1Rr2, since R is reflexive,therefore r1Rr2 must be in R2. We can easily observe that the same is true for R3 in that r1Rr1 and r1Rr1

and r1Rr2, so (r1, r2) must be in R3. We can continue this process for any n to show that any (r1, r2) in Rmust also be in Rn. This implies that R ⊆ Rn. Therefore, Rn = R for all n ∈ N∗.

1.3 – Equivalence Relations

Exercise: 1 Section 1.3Question: Let S = Z × Z and let R be the relation on S defined by (a, b)R(c, d) means that a + d = b + c.Show that R is an equivalence relation. Concisely describe the equivalence classes of R.

Solution: Let S = Z × Z and let R be the relation on S defined by (a, b)R(c, d) means that a + d = b + c.For any (a, b) in S it is not hard to see that (a, b)R (a, b) means a+ b = a+ b therefore R is reflexive. Suppose



1.3. EQUIVALENCE RELATIONS 23

(a, b)R (c, d). Then, a+ d = b+ c which is equivalent to c+ b = d+ a. This implies (c, d)R (a, b) therefore R issymmetric. Suppose (a, b)R (c, d) and (c, d)R (e, f). Then a+d = b+c and c+f = d+e. Using subtraction andsubstitution we find that a− b = e− f . With some arranging we observe a+ f = b+ e implying (a, b)R (e, f).Hence, R is transitive. Therefore, since R is reflexive, symmetric, and transitive it is an equivalence relation.Each distinct equivalence class describes the solutions to a function f : Z −→ Z where f(x) = x+ c for c ∈ Z.

Exercise: 2 Section 1.3Question: Let C be the set of people in your abstract algebra class. Describe a “natural” relation satisfyingeach of the combination of properties listed below.

1. Reflexive and symmetric, but not transitive.

2. Reflexive and transitive, but not symmetric.

3. Symmetric and transitive, but not reflexive.

4. An equivalence relation.

Solution: We describe relations for each of the following situations.

a) Reflexive and symmetric, but not transitive. R, where aR b if a and b live within one kilometer (or onemile) of each other.

b) Reflexive and transitive, but not symmetric. R, where aR b if a earns a final grade that is less than orequal to the final grade that b earns.

c) Symmetric and transitive, but not reflexive. The simplest relation that satisfies these conditions is theempty relation. Symmetry and transitivity are satisfied trivially. (The hypothesis is always false so theconditional statement is always true.) Note that if aR b, where a 6= b, then by symmetry bR a and thenby transitivity aRa. So for this combinations of properties to hold, there must be an element that is notin relation to any other element.

d) An equivalence relation. R, where aR b if a and b entered college the same semester.

Exercise: 3 Section 1.3Question: Let P be the set of living people. For all a, b ∈ P , define the relation aR b if a and b have met.

Solution: It is not hard to see that R is both reflexive and symmetric by definition. However, let a, b, and cbe people such that aR b and bR c. It does not follow that person a has met person c in every case implying Ris not transitive. Therefore, R is not an equivalence relation.

Exercise: 4 Section 1.3Question: Let P be the set of living people. For all a, b ∈ P , define the relation aR b if a and b live in acommon town.

Solution: Obviously, aRa exists for all people within the town. It quickly follows that if aR b then bR a andtherefore is symmetric as well. Suppose aR b and bR c. Then a and b live in a common town, and b and c livein the same town. Hence, a and c live in the same town meaning aR c and R is transitive. Therefore, R is anequivalence relation.

Exercise: 5 Section 1.3Question: Let C be the set of circles in R2 and let R be the relation of concentric on C. Prove or disprovewhether the described relation is an equivalence relation. If the relation is not an equivalence relation, determinewhich properties it lacks.

Solution: Two circles are concentric if and only if they have the same center.

Reflexivity: If C is a circle in C, then it has the same center as itself.

Symmetry: If C1, C2 ∈ C, then if C1 has the same center as C2, then C2 has the same center as C1. Sosymmetry holds.

Transitivity: Let C1, C2, C3 ∈ C. Suppose that C1 is concentric with C2 and that C2 is concentric with C3.Then C1 and C2 have the same center and C2 and C3 have the same center. Hence C1 and C3 have thesame center, so C1 is concentric with C3. Hence concentric is transitive.




So the concentric relation is an equivalence relation.

Exercise: 6 Section 1.3Question: Let S = Z× Z and define the relation R on S by (m1,m2)R (n1, n2) if m1m2 = n1n2.

Solution: Consider any (m1,m2) ∈ S. We observe (m1,m2)R (m1,m2) implies m1m2 = m1m2 which isalways true. Hence, R is reflexive. Suppose (m1,m2)R (n1, n2). We know m1m2 = n1n2 so n1n2 = m1m2 andtherefore (n1, n2)R (m1,m2). Hence, R is symmetric. Suppose (m1,m2)R (n1, n2) and (n1, n2)R (p1, p2). Then,m1m2 = n1n2 and n1n2 = p1p2. By substitution, we find m1m2 = p1p2 implying (m1,m2)R (p1, p2). Hence, Ris transitive and therefore is an equivalence relation.

Exercise: 7 Section 1.3Question: Let S = Z× Z and define the relation R on S by (m1,m2)R (n1, n2) if m1n1 = m2n2.

Solution: Consider the relation (1, 2)R (1, 2). Since 1× 1 6= 2× 2, then R is not reflexive. Therefore, R is notan equivalence relation.

Exercise: 8 Section 1.3Question: Let S = Z× Z and define the relation R on S by (m1,m2)R (n1, n2) if m1n2 = m2n1.

Solution: Suppose m1,m2 ∈ Z and (m1,m2)R (m1,m2). Then m1m2 = m2m1 for any (m1,m2) ∈ S,thus R is reflexive. Suppose (m1,m2)R (n1, n2) where n1, n2 ∈ Z. Then, m1n2 = m2n1 which can be easilyrearranged to show n1m2 = n2m1. This implies (n1, n2)R (m1,m2), and therefore R is symmetric. Supposer1, r2, s1, s2, t1, t2 ∈ Z such that (r1, r2)R (s1, s2) and (s1, s2)R (t1, t2). Then, r1s2 = r2s1 and s1t2 = s2t1.Using some substitution we can arrive at the desired outcome as seen below,

r1s2 = s1r2

r1s2t1 = s1r2t1

r1s1t2 = s1r2t1

r1t2 = t1r2

Therefore, since R is reflexive, symmetric, and transitive, R is an equivalence relation.

Exercise: 9 Section 1.3Question: Let P3 be the set of polynomials with real coefficients and of degree 3 or less. Define the relationR on P3 by p(x)Rq(x) to mean that q(x)− p(x) has 5 as a root.

Solution: Notice, 5 will always be a root of f(x)−f(x) so R is reflexive. Consider the case when p(x) = x−12and q(x) = 2x−7. We observe that p(x)Rq(x) since q(x)−p(x) = x−5. However, because p(x)−q(x) = −x−5,5 is not a root of p(x) − q(x) therefore q(x) is not related to p(x). Since R is not symmetric it cannot be anequivalence relation.

Exercise: 10 Section 1.3Question: Consider the set C0(R) of continuous functions over R. Define the relation R on C0(R) by f R g ifthere exist some a, b ∈ R such that

g(x) = f(x+ a) + b for all x ∈ R.

Prove or disprove whether the described relation is an equivalence relation. If the relation is not an equivalencerelation, determine which properties it lacks.

Solution: We check the defining properties for an equivalence relation.

Reflexivity: Let f be any function in C0(R). Setting a = b = 0 we have f(x) = f(x + a) + b for all x. HenceR is reflexive.

Symmetry: Suppose that f R g. Then there exists some a, b ∈ R such that g(x) = f(x + a) + b. Theng(x)− b = f(x+ a) for all x ∈ R. Setting y = x+ a, we have g(y− a)− b = f(y). This holds for all y ∈ R.Hence g R f . Thus, R is symmetric.




Transitivity: Let f, g, h ∈ C0(R) and suppose that f R g and g Rh. There there exist a1, a2, b1, b2 ∈ R suchthat

g(x) = f(x+ a1) + b1 and h(x) = g(x+ a2) + b2 for all x ∈ R.

Then h(x) = f(x+ a+ 2 + a1) + b1 + b2 for all x ∈ R. Hence f Rh. Thus R is transitive.

Exercise: 11 Section 1.3Question: Let Pfin(R) be the set of finite subsets of R and define the relation ∼ on Pfin(R) by A ∼ B if thesum of elements in A is equal to the sum of elements in B. Prove that ∼ is an equivalence relation.

Solution: It is not hard to see A ∼ A for all A ∈ R, so ∼ is reflexive. Similarly, if A ∼ B, then the sum ofelements in A is equal to the sum of elements in B. Thus, we observe that this implies that B ∼ A. Therefore,∼ is symmetric. Moreover, given any A,B,C ∈ R where A ∼ B and B ∼ C, it quickly follows that the sum ofelements in A is equal to the sum of elements in C by substitution. Hence, ∼ is transitive and therefore is anequivalence relation.

Exercise: 12 Section 1.3Question: Let `∞(R) be the set of sequences of real numbers. Define the relation R on `∞(R) by (an)R (bn) if

limn→∞

(bn − an) = 0.

Solution: It is not hard to see that R is reflexive since

limn→∞

(an − an) = limn→∞

(0) = 0

Also, noticelimn→∞

(bn − an) = − limn→∞

(an − bn) = 0

therefore R is symmetric. For transitivity, suppose that (an)R(bn) and (bn)R(cn). Then

limn→∞

(bn − cn) = 0 and limn→∞

(an − bn) = 0.

Thuslimn→∞

(an − cn) = limn→∞

((an − bn) + (bn − cn)) = limn→∞

(an − bn) + limn→∞

(bn − cn) = 0 + 0 = 0,

where the second equality holds by virtue of the addition law of limits. Note that we can apply the addition rulehere because we know that each of the sequences involved in the sums converge (to 0).

Exercise: 13 Section 1.3Question: Let `∞(R) be the set of sequences of real numbers. Define the relation R on `∞(R) by (an)R (bn)if the sequence (an + bn)∞n=1 converges.

Solution: Using the properties of limits we can show limn→∞(an + an) = 2 limn→∞(an). Find any counter-example for reflexivity. Let an = 2n. Then, 2 limn→∞(an) = ∞ implying R cannot be reflexive. Therefore, Rcannot be an equivalence relation. It should be noted, however, that R is both symmetric and transitive.

Exercise: 14 Section 1.3Question: Let S be the set of lines in R2 and let R be the relation of perpendicular.

Solution: We observe that any given line cannot be perpendicular to itself, therefore R cannot be reflexive.Thus R is not an equivalence relation.

Exercise: 15 Section 1.3Question: Let W be the words in the English language (i.e., have an entry in the Oxford English Dictionary).Define the relation R on W by w1Rw2 is w1 comes before w2 in alphabetical order.

Solution: This relation is not symmetric. Find any counter-example. Consider the two words, w1 =”ball” andw2 =”stick”. We observe that w1Rw2, but the reverse is not true. Hence, R cannot be an equivalence relation.




Exercise: 16 Section 1.3Question: Let C0([0, 1]) be the set of continuous real-valued functions on [0, 1]. Define the relation ∼ onC0([0, 1]) by

f ∼ g ⇐⇒∫ 1

0

f(x) dx =

∫ 1

0

g(x) dx.

Show that ∼ is an equivalence relation and describe (with a precise rule) a complete set of distinct representativesof ∼.

Solution: First show that ∼ is an equivalence relation.

Reflexivity: For all f ∈ C0([0, 1]), we have∫ 1

0f(x) dx =

∫ 1

0f(x) dx so f ∼ f .

Symmetry: Suppose that f, g ∈ C0([0, 1]) with f ∼ g. Then∫ 1

0

f(x) dx =

∫ 1

0

g(x) dx.

Equality is reversible so g ∼ f . Hence ∼ is symmetric.

Transitivity: Suppose that f, g, h ∈ C0([0, 1]) with f ∼ g and g ∼ h. Then∫ 1

0

f(x) dx =

∫ 1

0

g(x) dx and

∫ 1

0

g(x) dx =

∫ 1

0

h(x) dx.

Hence∫ 1

0f(x) dx =

∫ 1

0h(x) dx. Thus f ∼ h. Hence ∼ is an equivalence relation.

Every function f ∈ C0([0, 1]) is in relation with the constant function g(x) = c, where c =∫ 1

0f(x) dx.

However, constant functions that are not equal are not in relation with each other. Hence the constantfunctions are a complete set of distinct representatives.

Exercise: 17 Section 1.3Question: Let C∞(R) be the set of all real-value functions on R such that all its derivatives exist and arecontinuous. Define the relation R on C∞(R) by f R g if f (n)(0) = g(n)(0) for all positive, even integers n.

a) Prove that R is an equivalence relation.

b) Describe concisely all the elements in the equivalence class [sinx].

Solution:

a) It is not hard to see that for any f ∈ C∞(R), f R f implies f (n)(0) = f (n)(0). We conclude that R isreflexive. Similarly, for any f, g ∈ C∞(R) if f R g, then we observe g R f quickly follows since f (n)(0) =g(n)(0). Thus R is symmetric as well. Finally, for any given f, g, h ∈ C∞(R) such that f R g and g Rh, wefind f Rh because f (n)(0) = g(n)(0) = h(n)(0) for all positive even integers n. Consequently, R is reflexive,symmetric, and transitive implying that R is an equivalence relation.

b) If f(x) = sinx then for all positive and even integers n, we have f (n)(0) = 0, so this is the definingcharacteristic of functions in [sinx]. (If a function in [sinx] is equal to its power series in a neighborhoodof 0, then [sinx] consists of functions that are a constant plus an odd function.)

Exercise: 18 Section 1.3Question: Let S = {1, 2, 3, 4} and the relation ∼ on P(S), defined by A ∼ B if and only if the sum of elementsin A is equal to the sum of elements in B, is an equivalence relation. List the equivalence classes of ∼.

Solution: [{1}], [{2}], [{3}] = {{1, 2}}, [{4}] = {{1, 3}}, [{1, 4}] = {{2, 3}}, [{2, 4}] = {{1, 2, 3}}, [{3, 4}] ={{1, 2, 4}}, [{1, 3, 4}], [{2, 3, 4}], [{1, 2, 3, 4}], [∅]

Exercise: 19 Section 1.3Question: Let T be the set of (non-degenerate) triangles in the plane.

a) Prove that the relation ∼ of similarity on triangles in T is an equivalence relation.

b) Concisely describe a complete set of distinct representatives of ∼.

Solution:




a) Suppose you have any t ∈ T . Then, t ∼ t implies that triangle t has equal corresponding angles with itself.This is always true, thus ∼ is reflexive. Now suppose you have to triangles s, t ∈ mathcalT such that s ∼ t.Then the corresponding angles of triangle s are equal to the corresponding angles of t. It is not hard tosee that the corresponding angles of triangle t are equal to the corresponding angles of triangle s implyingthat ∼ is symmetric. For transivity, suppose you have any r, s, t ∈ mathcalT such that r ∼ s and s ∼ t.Then the corresponding angles of triangle r are equal to the corresponding angles of triangle s which arealso equal to the corresponding angles of t. Therefore the corresponding angles of triangle r are equal tothose in triangle t. Hence, ∼ is transitive. Therefore, ∼ is an equivalence relation.

b) The set of distinct representatives of ∼ will be equal to the unique combinations of the angles x, y, z of anytriangle such that x, y, z ∈ R>0 and x+ y+ z = 180. For example, [30, 60, 90] is the distinct representativefor all triangles with 30, 60, and 90 as their respective angles.

Exercise: 20 Section 1.3Question: Prove that the relation defined in Example 1.3.10 is an equivalence relation.

Solution: Choose any (a, b) in S. We quickly observe that ab = ba which holds true for all a, b undermultiplication. Thus, R is reflexive. Now suppose that R is symmetric and choose any (a, b), (c, d) ∈ S such that(a, b) ∼ (c, d). Then, ad = bc so we can show that cb = da implying that (c, d) ∼ (a, b). Hence, R is symmetric.Suppose R is also transitive. Let (a, b), (c, d), (e, f) be in S such that (a, b) ∼ (c, d) and (c, d) ∼ (e, f). Thenad = bc and df = ec. By multiplying both sides by ab we observe,

abdf = abec

afbd = beac

afbd = bebd

af = be

Consequently, (a, b) ∼ (e, f). Thus ∼ is transitive and therefore an equivalence relation.

Exercise: 21 Section 1.3Question: Let S = {1, 2, 3, 4, 5, 6}. For the partitions of S given below, write out the equivalence relation asa subset of S × S.

a) {{1, 2}, {3, 4}, {5, 6}}b) {{1}, {2}, {3, 4, 5, 6}}c) {{1, 2}, {3}, {4, 5}, {6}}

Solution:

a) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 2), (2, 1), (3, 4), (4, 3), (5, 6), (6, 5)}b)

{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 4), (3, 5), (3, 6), (4, 3), (4, 5), (4, 6), (5, 3),

(5, 4), (5, 6), (6, 3), (6, 4), (6, 5)}

c) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 2), (2, 1), (4, 5), (5, 4)}

Exercise: 22 Section 1.3Question: Let S = {a, b, c, d, e}. For the partitions of S given below, write out the equivalence relation as asubset of S × S.

a) {{a, d, e}, {b, c}}b) {{a}, {b}, {c}, {d}, {e}}c) {{a, b, d, e}, {c}}

Solution:

a) {(a, a), (b, b), (c, c), (d, d), (e, e), (a, d), (a, e), (d, a), (d, e), (e, a), (e, d), (b, c), (c, b)}b) {(a, a), (b, b), (c, c), (d, d), (e, e)}




c)

{(a, a), (b, b), (c, c), (d, d), (e, e), (a, b), (a, d), (a, e), (b, a), (b, d), (b, e), (d, a), (d, b),

(d, e), (e, a), (e, b), (e, d)}

Exercise: 23 Section 1.3Question: Let C1([a, b]) be the set of continuously differentiable functions on the interval [a, b]. Define therelation ∼ on C1([a, b]) as f ∼ g if and only if f ′(x) = g′(x) for all x ∈ (a, b). Prove that ∼ is an equivalencerelation on C1([a, b]). Describe the elements in the equivalence class for a given f ∈ C1([a, b]).

Solution: First, we observe for any f ∈ C1([a, b]) that f ′(x) = f ′(x) for all x ∈ (a, b). Hence, ∼ is reflexive.Similarly, given any f, g ∈ C1([a, b]) we observe that if f ∼ g then f ′(x) = g′(x) for all x ∈ (a, b). Thus weknow g′(x) = f ′(x) which establishes that ∼ is also symmetric. For transivity, consider any f, g, h ∈ C1([a, b])where f ∼ g and g ∼ h. Then f ′(x) = g′(x) = h′(x) for all x ∈ (a, b). Thus f ′(x) = h′(x) implying that ∼ istransitive. Therefore, ∼ is an equivalence relation. The elements of the equivalence class for a given f are thevertical translations of f . In other words, [f(x)] = {f(x) + k | k is some constant in R}.

Exercise: 24 Section 1.3Question: Let Mn×n(R) be the set of n×n matrices with real coefficients. For two matrices A,B ∈Mn×n(R),we say that B is similar to A if there exists and invertible n× n matrix S such that B = SAS−1.

1. Prove that similarity ∼ is an equivalence relation on Mn×n(R).

2. Prove that the function f : Mn×n(R)/∼ → R defined by f([A]) = detA is a well-defined function on thequotient set Mn×n(R)/∼.

3. Determine with proof or counter-example whether the function g : Mn×n(R)/∼ → R defined by g([A]) =TrA, the trace of A, is a well-defined function.

Solution: Let Mn×n(R) be the set of n×n matrices with real coefficients and define the relation ∼ as similaritybetween matrices.

a) We prove that ∼ is an equivalence relation.

Reflexivity: For all matrices A ∈Mn×n(R), since IAI−1 = A then A ∼ A.

Symmetry: Let A,B ∈ Mn×n(R) such that A ∼ B. Then there exists an invertible matrix S such thatB = SAS−1. Hence A = S−1BS. Thus B ∼ A so ∼ is symmetric.

Transitivity: Let A,B,C ∈Mn×n(R) such that A ∼ B and B ∼ C. Thus there exist invertible matricesA and T such that B = SAS−1 and C = TBT−1. Then

C = T (SAS−1)T−1 = (TS)A(TS)−1.

Hence A ∼ C so ∼ is transitive.

b) Suppose that B ∼ A. Then B = SAS−1 for some invertible matrix S. Then

detB = (detS)(detA)(detS)−1 = detA.

Therefore, the function det is well-defined on the set of ∼-equivalence classes on Mn×n(R).

c) There is a property in the algebra of matrices that Tr(AB) = Tr(BA), even if the matrices do not commute.Suppose that B ∼ A with B = SAS−1 for some invertible matrix S. Then

Tr(B) = Tr(SAS−1) = Tr(ASS−1) = Tr(A).

Hence the function g is a well-defined function.

Exercise: 25 Section 1.3Question: Define the relation ∼ on R by a ∼ b if and only if b− a ∈ Q.

a) Prove that all real x ∈ R, there exists y ∈ [x]∼ that is arbitrarily close to x. (In other words, for all ε > 0,there exists y with y ∼ x and |x− y| < ε.




b) (*) Prove that ∼ has an uncountable number of equivalence classes.

Solution:

a) Choose any q ∈ Q such that q < ε. Then, if |x− y| = q, y ∼ x and |x− y| < ε. Rearranging our equationwe find,

y = x+ q

y = x− q

Notice that the domain of these functions spans R. Therefore for any x ∈ R there exists y ∈ R such thaty ∼ x and |x− y| < ε. A graphical representation is shown below.

y

x

y = x− ε

y = x+ εy = x+ qy = x− q

b) Since ∼ is reflexive, every value in R has its own equivalence class. Since R is uncountable, it follows thatthere are an uncountable amount of equivalence classes.

Exercise: 26 Section 1.3Question: Let R1 and R2 be equivalence relations on a set S. Determine (with a proof or counterexample)which of the following relations are also equivalence classes on S. (a) R1 ∩ R2; (b) R1 ∪ R2; (c) R14R2. [Notethat R1 ∪R2, and similarly for the others, is a relation as a subset of S × S.]

Solution: Let R1 and R2 be equivalence relations on a set S.

a) R1∩R2 is an equivalence relation. For all a ∈ S, the pair (a, a) is in both R1 and R2. Hence (a, a) ∈ R1∩R2.Thus R1∩R2 is reflexive. Suppose that (a, b) ∈ R1∩R2. Then since R1 is an equivalence relation, (b, a) ∈ R1

and similarly for R2. Hence (b, a) ∈ R1 ∩R2. Thus R1 ∩R2 is symmetric. Finally, suppose that (a, b) and(b, c) are pairs in R1 ∩R2. Then since R1 is an equivalence relation (a, c) ∈ R1 and the same holds for R2,so (a, c) ∈ R2. Thus (a, c) ∈ R1 ∩R2 and hence R1 ∩R2 is transitive.

b) R1 ∪ R2 is not an equivalence relation. Since R1 is reflexive, for all a ∈ S, the pair (a, a) ∈ R1 and thus(a, a ∈ R1∪R2. Hence, R1∪R2 is reflexive. Suppose that (a, b) ∈ R1∪R2. Thus (a, b) ∈ R1 or (a, b) ∈ R2.Both R1 and R2 are equivalence relations. If (a, b) ∈ Ri, then (b, a) ∈ Ri so (b, a) ∈ R1∪R2. Thus, R1∪R2

is symmetric. Finally, suppose that (a, b) and (b, c) are in R1 ∪ R2. In the case where (a, b) ∈ R1 and(b, c) ∈ R2, it does not appear that transitivity would need to hold if (b, c) /∈ R1 and (a, b) /∈ R2.

For example, let S = P({1, 2, 3, 4, 5}) and let R1 be the equivalence relation on S of same cardinality andlet R2 be the equivalence relation of elements summing to the same value. Then {1, 4}R1∪R2 {5} becausethe elements in the set both add to the same value and {5}R1 ∪ R2{2} because the sets are the samecardinality. However, {1, 4} is not in relation to {2} under R1 ∪ R2. Hence, the union of two transitiverelations is not necessarily transitive.

c) R14R2 cannot be an equivalence relation since it is not reflexive. For all a ∈ S, the pair (a, a) is in bothR1 and in R2. Hence, the pair (a, a) does not occur in R14R2.

Exercise: 27 Section 1.3Question: Which of the following collections of subsets of the integers for partitions? If it is not a partition,explain which properties fail.

1. {pZ | p is prime}, where kZ means all the multiples of k.

2. {{3n, 3n+ 1, 3n+ 2} |n ∈ Z}.

3. {{k |n2 ≤ k ≤ (n+ 1)2} |n ∈ N}.




4. {{n,−n} |n ∈ N}.

Solution: Testing to satisfy the properties of a partition.

a) The set of subsets {pZ | p is prime} is not a partition of Z since for example 6 ∈ 2Z ∩ 3Z, so the subsetsare not disjoint. Also, the union of all these sets is not all of Z but Z− {−1, 1}.

b) The set of subsets {{3n, 3n+1, 3n+2} |n ∈ Z} is a partition of Z. Consider the function f : Z→ Z definedby f(m) = bm/3c. It is easy to see that

{3n, 3n+ 1, 3n+ 2} = f−1(n).

Since f is a function, the union of all f−1(n) gives Z. Furthermore, since an element in Z does not mapvia f to distinct n values, these pre-image sets are disjoint.

c) This is not a partition because for example {1, 2, 3, 4} is one set in the partition as is {4, 5, 6, 7, 8, 9} andthese distinct sets are not mutually disjoint. The collection of subsets is not a partition of Z for theadditional reason that it does not cover all of Z. In fact,⋃

n∈N{n2, n2 + 1, . . . , (n+ 1)2} = N.

d) This collection of subsets is a partition. Consider the equivalence relation on Z defined by a ∼ b if andonly if |a| = |b|. This is indeed an equivalence relation and the equivalence classes are precisely subsets ofZ of the form {−n, n}.

Exercise: 28 Section 1.3Question: Let S be a set. Prove that there is a bijection between the set of partitions of S and the set ofequivalence classes on S.

Solution: Let f be a function from the set of partitions of S to the set of equivalence classes on S such thatf(A) = R. Let R be an equivalence relation on S and assume there does not exist A over S such that f(A) = R.However, by proposition 1.3.12 we know all the distinct equivalence classes of R are disjoint and their unionis equal to S. This creates a contradiction because it satisfies the definition of a partition. Hence, for everyequivalence relation R on S, there must exist A where f(A) = R. Therefore, f is surjective. Now assume thereexists partitions A = {Ai}i∈I and B = {Bj}j∈J such that A 6= B and f(A) = f(B) = R. This would implyevery Ai and Bj represents a distinct equivalence class of R by proposition 1.3.14. However, since A 6= B theremust exist some Ai and Bj where Ai ∩ Bj 6= Ai and Ai ∩ Bj 6= ∅. This creates a contradiction by proposition1.3.12 since the distinct equivalence classes of any equivalence relation are disjoint. Therefore, f must also beinjective which establishes a bijection between the set of partitions of S and the set of equivalence classes on S.

Exercise: 29 Section 1.3Question: Call p(n) the number of equivalence relations (equivalently, by Exercise 1.3.28, partitions) on aset of cardinality n. (The numbers p(n) are called the Bell numbers after the Scottish-born mathematicianE. T. Bell.)

a) (*) Prove that p(0) = 1 and that for all n ≥ 1, p(n) satisfies the condition

p(n) =

n−1∑j=0

p(n− j − 1)

(n− 1

j

).

b) Use the previous part to calculate p(n) for n = 1, 2, 3, 4, 5, 6, 7.

Solution: Call p(n) the number of partitions that exist on the set {1, 2, . . . , n}. This will be the same numberof partitions on any set of size n.

a) The value p(0) = 1 comes from the comment that the empty set satisfies all the conditions for an equivalencerelation on the empty set itself. Suppose that we know the value of p(k) for 0 ≤ k ≤ n− 1. To determinep(n), we count up the number of possible partitions based on how many elements are in the equivalenceclass of n besides n. In other words, let j = |[n]−{n}|. The index j = 0 corresponds to the equivalence class[n] being the singleton set {n} and the index j = n− 1, corresponds to the situation where the equivalenceclass [n] = {1, 2, . . . , n}. Now for any given j, there are

(n−1j

)ways to choose the remaining elements in the




equivalence class of [n]. Furthermore, for each of those choices, there are n− 1− j elements remaining in{1, 2, . . . , n} from which to create the remaining equivalence classes that make up the partition. Thus, foreach j, there are p(n− 1− j)

(n−1j

)partitions. Summing over j = 0 to n− 1 gives the number of possible

partitions (equivalence classes) on {1, 2, . . . , n}.b) Using this recursive formula, we get

n 1 2 3 4 5 6 7p(n) 1 2 5 15 52 203 877

Exercise: 30 Section 1.3Question: Consider the relation ∼ on R defined by x ∼ y if y − x ∈ Z.

a) Prove that ∼ is an equivalence relation.

b) Prove that if a ∼ b and c ∼ d, then (a+ c) ∼ (b+ d).

c) Decide with a proof or counter-example whether ac ∼ bd, whenever a ∼ b and c ∼ d.

Solution:

a) First, we observe that x − x = 0 for any x ∈ R. Since 0 ∈ Z, ∼ is reflexive. Suppose ∼ is symmetric.Then for any x, y ∈ R if x ∼ y, then y − x = z where z ∈ Z. Rearranging we find that x− y = −z whichestablishes that ∼ is symmetric. For transivity, suppose that a, b, c ∈ R such that a ∼ b and b ∼ c. Thenb− a = x and c− b = y where x, y ∈ Z. Setting c = b+ y and solving we find,

b− a = x

b+ y − a = x+ y

c− a = x+ y

Since x+ y is always an integer, we conclude ∼ is transitive. Therefore, ∼ is an equivalence relation.

b) Notice, b− a = x and d− c = y where x, y ∈ Z.

b− a = x

b− a+ y = x+ y

b− a+ d− c = x+ y

(b+ d)− (a+ c) = x+ y

Therefore, since x+ y ∈ Z, (a+ c) ∼ (b+ d).

c) Find any counter-example. Let a = π, b = π, c = 1, and d = 2. Then, a ∼ b and c ∼ d, yet ac ∼ bd = π.Since π 6∈ Z, ac 6∼ bd.

Exercise: 31 Section1.3Question: Let S be a set and let A = {Ai}i∈I be a partition of S. Another partition B = {Bj}j∈J is called arefinement of A if

∀j ∈ J , ∃i ∈ I, Bj ⊆ Ai.Let A and B be two partitions of a set S and let ∼A (resp. ∼B) as the equivalence relation corresponding to A(resp. B). Prove that B is a refinement of A if and only if s1 ∼B s2 =⇒ s1 ∼A s2.

Solution: Suppose that B is a refinement of the partition A. Suppose that s1 ∼B s2. This is equivalent to thestatement that s1 and s2 are both in Bj for some index j ∈ J . Since B is a refinement of A, then Bj ⊆ Ai forsome i ∈ I, and therefore, s1 and s2 are both in this Ai. Hence s1 ∼A s2.

Conversely, suppose that ∀s1, s2 ∈ S, s1 ∼B s2 −→ s1 ∼A s2. Consider a given subset Bj in the partitionB and let s ∈ Bj . Since A is a partition of S, then s ∈ Ai for some index i ∈ I. For all s′ ∈ Bj , we haves ∼B s′ and so by our hypothesis, s ∼A s′. Consequently, s′ ∈ Ai and this for all s′ ∈ Bj . Thus Bj ⊆ Ai. Thisestablishes that B is a refinement of A.

Exercise: 32 Section 1.3Question: Let S be a set and let A = {Ai}i∈I and B = {Bj}j∈J be two partitions of S. Prove that thecollection of sets

{Ai ∩Bj | i ∈ I and j ∈ J } − {∅}




is a partition of S.

Solution: Call C the collection of subsets of S

{Ai ∩Bj | i ∈ I and j ∈ J } − {∅}

Let s be any element in S. Since A and B are partitions of S, there exists a unique i0 ∈ I and a unique j0 ∈ Jsuch that s ∈ Ai0 and s ∈ Bj0 . Then s ∈ Ai0 ∩Bj0 and hence the union of subsets in C is all of S.

Now consider Ai1 ∩Bj1 and Ai2 ∩Bj2 sets in C and suppose that

(Ai1 ∩Bj1) ∩ (Ai2 ∩Bj2) 6= ∅.

Then by associativity(Ai1 ∩Ai2) ∩ (Bj1 ∩Bj2) 6= ∅.

However, Ai1 ∩ Ai2 6= ∅ if and only if i1 = i2 and similarly Bj1 ∩ Bj2 6= ∅ if and only if j1 = j2. Hence, wehave i1 = i2 and j1 = j2. Thus sets in C are either equal or disjoint. Consequently, since C also covers S, C is apartition of S.

Exercise: 33 Section 1.3Question: Let S be a set and let R be any relation on S. Design an algorithm that determines the smallestequivalence relation on S that contains the relation R.

Solution: There are a variety of ways to accomplish this depending on the context and, if working computa-tionally with a finite set S, how we store the equivalence relation.

From a purely theoretical standpoint, we can deduce that a smallest equivalence relation on S that containsS exists by referring to Exercise 1.3.26 and generalizing. It is not hard to prove that the intersection of anycollection of equivalence relations on S is again an equivalence relation on S. Hence, we can define the smallestequivalence relation on S containing R as ⋂

Re :R⊆Re

Re

where the intersection is over all equivalence relations Re that contain R. We can call this relation the equivalenceclosure of R.

From an algorithmic perspective, one of the challenges of this exercise is to realize that if we begin withthe relation R as a subset of S × S, and adjoin pairs to force symmetry and then adjoin more pairs to forcetransitivity, we may need to go back and adjoin new pairs that are required for symmetry again and vice versa.It is not clear when this process will terminate.

Suppose that |S| = n and label the elements of S by S = {s1, s2, . . . , sn}. For an n × n matrix M ofnonnegative integers, we define M as mij = min(1,mij). Hence, if mij ≥ 1, then mij = 1 but if mij = 0, thenmij = 0.

Here is an algorithm for determining the equivalence closure of R.

• Let A = (aij) be the n× n matrix defined by

aij =

{1 if (si, sj) ∈ R0 otherwise.

• Replace A with A+ I.

• Replace A with A. [These first two steps ensure reflexivity.]

• While A 6= Ã2 + (A2)>, replace A with Ã2 + (A2)>.

• Return A.

The operation A2 will have a nonzero entry position (i, j), if there is some k such that aik and bkj are bothnonzero. This process will add a nonzero entry corresponding to a pair that is required by transitivity. Theoperation A → A + A> adds a matrix to its transpose, which adds a nonzero entry to complete for symmetry

reasons. Now if A is a matrix of 0s and 1s, then Ã2 + (A2)> will again be a matrix of 0s and 1s with new entriesturned to 1 for transitivity or symmetry reasons.



1.4. PARTIAL ORDERS 33

The algorithm stops because if A = ˜A2 + (A2)> then no transitivity requirement or symmetry requirementwill add a new pair to the relation (nonzero entry to A). Hence, A will then correspond to an equivalencerelation. Furthermore, the algorithm will terminate because before the while loop, A contains at most n2 − nentries that are 0 and at each stage of the while loop that changes A will change at least one 0 entry to a 1.Hence, the while loop can repeat at most n2 − n times.

1.4 – Partial Orders

Exercise: 1 Section 1.4Question: Let S = {a, b, c, d, e} (where we consider all the labels unique elements). In the following relationson S determine with explanation whether or not the relation is a partial order. If it fails antisymmetry thenremove a least number of pairs and if it fails transitivity then add some pairs to make the relation a partial order.

1. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c)}.

2. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (a, d)}.

3. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (d, a)}.

4. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (b, c), (c, d), (d, e), (a, e)}.

Solution: Consider each partial order.

1. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c)}. This relation is reflexive. The only pair in the relation not ofthe form (x, x) is (a, c). The pair (c, a) is not in the relation so the relation is antisymmetric. For the samereason, the relation is also transitive.

2. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (a, d)}. For the same reasons as the previous part, this relationis a partial order.

3. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (d, a)}. In this example, the relation is again reflexive andantisymmetric but it is not transitive because of the pairs (d, a) and (a, c). In the modified relationR′ = R ∪ {(d, c)}, we now have transitivity and R′ still has antisymmetry.

4. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (b, c), (c, d), (d, e), (a, e)}. This relation satisfies reflexivity and anti-symmetry but fails transitivity. In order to get transitivity, we need add the pairs (b, d), (c, e), and (b, e).Then R ∪ {(b, d), (b, e), (c, e)} is a partial order.

Exercise: 2 Section 1.4Question: In microeconomics (the study of consumer behavior), one considers consumer’s utility (preference)in regards to pairs of commodities. Let (q1, q2) ∈ N2 be a pair of nonnegative integers representing quantities oftwo commodities. Explain why, given two specific commodities and a given consumer, the relation of preferable(or equal) is a partial order.

Solution: When a consumer prefers the pair of commodities in the quantities (q′1, q′2) over (q1, q2) (or these are

equal), we write (q1, q2) 4 (q′1, q′2). Note that this is a partial order on N2, where the ith entry represents the

quantity of the ith commodity. By including the equality of pairs into the relation, the relation is reflexive. It isin people’s psychology that if (q1, q2) 4 (q′1, q

′2) and (q′1, q

′2) 4 (q′′1 , q

′′2 ), so the concept of preference is transitive.

We do not have (q1, q2) 4 (q′1, q′2) and (q′1, q

′2) 4 (q1, q2) at the same time unless forced by the requirement that

the pairs are equal. So preference, with the additional assumption of including equality, is a partial order. Notethat in the concept of preference, given two pairs (q1, q2) and (q′1, q

′2) it is perfectly possible to not prefer one

over the other. Such pairs would be incomparable.

Exercise: 3 Section 1.4Question: Let S = R>0 × R>0 be the positive first quadrant in the Cartesian plane. Consider the relation Ron S defined by

(x1, y1)R (x2, y2) =⇒ x1y1 ≥ x2y2

Prove or disprove that R is a partial order.

Solution: We check the three axioms for a partial order.




Reflexivity: For all (x, y) ∈ S, we do have xy ≤ xy so the relation is reflexive.

Antisymmetry: Consider the points (2, 1) and (1, 2). Then 1 × 2 ≤ 2 × 1 and 2 × 1 ≤ 1 × 2 so (1, 2)R (2, 1)and (2, 1)R (1, 2) even through (1, 2) 6= (2, 1) So the relation is not antisymmetric.

Transitivity: We have already shown that the relation is not a partial order since it fails antisymmetry. Therelation is transitive.

Not a partial order.

Exercise: 4 Section 1.4Question: Prove that for any real x >

√2, the inequality

√2 < 1

2

(x+ 2

x

)< x holds.

Solution: Since√

2 < x, then 1x <

1√2, which implies 2

x <√

2 < x. So

1

2

(x+

2

x

)<

1

2(x+ x) = x.

Also, from x 6=√

2, we have (x−√

2)2 > 0, which is equivalent to

x2 − 2√

2x+ 2 > 0⇐⇒ x2 + 2 > 2√

2x⇐⇒ 1

2

(x+

2

x

)>√

2.

The result follows.

Exercise: 5 Section 1.4Question: Let (S,4) be a partial order in which every element has an immediate successor. Prove that it isnot necessarily true that for any two elements a 4 b that any chain between a and b has finite length.

Solution: Consider the subset of (Q,≤) with the set S ={

1− 1n

∣∣∣n ∈ N∗} ∪ {1 + 1n

∣∣∣n ∈ N∗}. In the poset

(S,≤), every element has an immediate successor: the immediate successor of 1− 1n is 1− 1

n+1 , and the immediate

successor of an element of the form 1 + 1n is 1 + 1

n−1 . By construction S is a chain since it is a total order.Furthermore, 0, 2 ∈ S, so it is an infinite chain between 0 and 2, such that every element has an immediatesuccessor.

Exercise: 6 Section 1.4Question: Prove the three claims about properties of 4 in Example 1.4.8. Prove that Q≥0 is countable.Conclude that Q is countable.

Solution: Consider the sets described in Example 1.4.8. Since |An| ≤ n, each set is finite. Also, the collectionof sets An is countable. Furthermore, we claim that the sets An partition Q>0. Indeed for all fractions x

y ,

expressed in reduced form, we have xy ∈ Ax+y−1. We can set up a bijection f : N∗ → Q>0 as follows. Define

`1 = 1 and then for all positive integers k, set `k = |A1| + · · · + |Ak|. Set f(1) = 1 and then for all integers mwith `k < m ≤ `k+1, define f(m) as the (m − `k)th element (ordered by ≤) in Ak. This function is injectivesince the Ak are mutually disjoint and is surjective since {Ak} cover Q>0.

We can now define a bijection between F : N→ Q by F (0) = 0 and

F (m) = (−1)mf

(⌊m+ 1

2

⌋).

Exercise: 7 Section 1.4Question: Let S be a set. Show that the relation of refinement is a partial order on the set of partitions of S.

Solution: Let S be a set. Recall that a refinement of a partition A = {Ai}i∈I of S is another partitionB = {Bj}j∈J such that for all j ∈ J , there exists an i ∈ I such that Bj ⊆ Ai. We will write B 4 A if B is arefinement of A.

1. For a partition A = {Ai}i∈I of S, for each i ∈ I we have Ai ⊆ Ai so A 4 A.




2. Suppose that B 4 A and A 4 B. Then for all j ∈ J , there exists an i ∈ I such that Bj ⊆ Ai. However,for this index i, there exists j′ ∈ J such that Ai ⊆ Bj′ . Hence Bj ⊆ Ai ⊆ Bj′ . However, since the sets inB are mutually disjoint, we conclude that j = j′. Hence Bj = Ai. Since j was arbitrary, we conclude thatB = A. Thus the relation of refinement is antisymmetric.

3. Finally, let B be a refinement of A and let C = {Ck}k∈K be a partition of S such that C 4 B. Then for allk ∈ K, there exists j ∈ J such that Ck ⊆ Bj . Since B is a refinement of A, then there exists i ∈ I suchthat Bj ⊆ Ai. Hence, since ⊆ is transitive, Ck ⊆ Ai. Thus C 4 A. Hence, the relation of refinement istransitive.

These three results show that refinement is a partial order on the set of partition of S.

Exercise: 8 Section 1.4Question: Draw the Hasse diagram of the partial order ⊆ on P({1, 2, 3, 4}).Solution: The Hasse diagram of (P({1, 2, 3, 4}),⊆) is

∅

{1} {2} {3} {4}

{1, 2} {1, 3} {1, 4} {2, 3} {2, 4} {3, 4}

{1, 2, 3} {1, 2, 4} {1, 3, 4} {2, 3, 4}

{1, 2, 3, 4}

Exercise: 9 Section 1.4Question: Draw the Hasse diagram for the poset ({1, 2, 3, 4, 5, 6},≤).

Solution: The Hasse diagram of ({1, 2, 3, 4, 5, 6},≤) is

1

2

3

4

5

6

Exercise: 10 Section 1.4Question: Let A = {a, b, c, d, e, f, g}. Draw the Hasse diagram for the partial order 4 given as a subset ofA×A as

4 = {(a, a), (b, b), (c, c), (d, d), (e, e), (f, f), (g, g), (a, c),

(b, c), (d, g), (a, e), (b, e), (c, e), (d, h), (g, h)}




Solution: The Hasse diagram of (A,4) is

a b d f

c g

e h

Exercise: 11 Section 1.4Question: A person’s blood type is usually listed as one of the eight elements in the set

B′ = {o+, o−, a+, a−, b+, b−, ab+, ab−}.

We define the donor relation → on B′ as follows. The relation t1 → t2 holds if the letter portion of the bloodtype donates according to the way described in the examples for this section and if someone with a + designationcan only give to someone else with +, while someone with − can give to anybody.

1. Draw the Hasse diagram for (B′,→).

2. Show that the (B′,→) poset does not have the lexicographic order on B × {+,−}.

Solution: We study the partial order of blood type B′ including the rhesus.

a) The Hasse diagram for the poset (B′,→) is

o−

a−

a+

o+

ab−

b−

b+

ab+

b) Under the partial order on {+,−} is described completely by +→ +, − → −, and − → +. So for example,in the lexicographic order on B′, we have

(a,+)→lex (ab,−)

because the pairs already differ on the first entry and a → ab in B. However, we see that in the actualdonor relation on B′, a+ cannot donate to ab−.

Hence the actual donor relation on B′ is not the lexicographic partial order on B × {+,−}.

Exercise: 12 Section 1.4Question: Consider the set of triples of integers Z3. Define the relation 4 on Z3 by

(a1, a2, a3) 4 (b1, b2, b3)⇐⇒

{a1 + a2 + a3 < b1 + b2 + b3 if a1 + a2 + a3 6= b1 + b2 + b3;

a1 + a2 + a3 4lex b1 + b2 + b3 if a1 + a2 + a3 = b1 + b2 + b3,

where 4lex is the lexicographic order on Z3 (with each copy of Z equipped with the partial order ≤). Prove that4 is a partial order on Z3. Prove also that 4 is a total order.

Solution: We first prove that 4 is a partial order on Z3.

Reflexivity : Let (a1, a2, a3) ∈ Z3. Then since a1 + a2 + a3 = a1 + a2 + a3, we use the lexicographic order tocompare the entries but all entries are equal and this satisfies the lexicographic order. Hence (a1, a2, a3) 4(a1, a2, a3).




Antisymmetry : Suppose that (a1, a2, a3) 4 (b1, b2, b3) and (b1, b2, b3) 4 (a1, a2, a3). Assume that a1 +a2 + a3 6= b1 + b2 + b3, then a1 + a2 + a3 < b1 + b2 + b3 and b1 + b2 + b3 < a1 + a2 + a3, which isa contradiction. Hence we must have a1 + a2 + a3 = b1 + b2 + b3. Thus, we compare the triples bylexicographic order. However, the lexicographic order is a partial order, which is antisymmetric so we canconclude that (a1, a2, a3) = (b1, b2, b3).

Transitivity : Suppose that (a1, a2, a3) 4 (b1, b2, b3) and (b1, b2, b3) 4 (c1, c2, c3). We can break the situationinto three cases. If a1 + a2 + a3 = b1 + b2 + b3 = c1 + c2 + c3, then we compare the triples by lexicographicorder, which is transitive so (a1, a2, a3) 4 (c1, c2, c3). If a1 + a2 + a3 = b1 + b2 + b3 6= c1 + c2 + c3 (resp.a1 + a2 + a3 6= b1 + b2 + b3 = c1 + c2 + c3) then a1 + a2 + a3 < c1 + c2 + c3 so (a1, a2, a3 4 (c1, c2, c3).Finally, if all thee of the sums are different, then we deduce that a1 + a2 + a3 < b1 + b2 + b3 < c1 + c2 + c3so we can conclude that (a1, a2, a3 4 (c1, c2, c3). We conclude that 4 is transitive.

Finally, to show that 4 is a total order, let (a1, a2, a3), (b1, b2, b3) ∈ Z3 be arbitrary, distinct elements. Thereare two cases. If a1 + a2 + a3 6= b1 + b2 + b3, then either a1 + a2 + a3 < b1 + b2 + b3, which implies that(a1, a2, a3) 4 (b1, b2, b3), or the reverse is true. If a1 + a2 + a3 = b1 + b2 + b3, then we compare the triples bythe lexicographic order. In this case, let i be the least index for which ai 6= bi. Then (a1, a2, a3) 4 (b1, b2, b3) ifai < bi and (b1, b2, b3) 4 (a1, a2, a3) if bi < ai. Hence, every pair of elements is comparable and the relation is atotal order.

Exercise: 13 Section 1.4Question: Let (Ai,4i) be posets for i = 1, 2, . . . , n and define4lex as the lexicographic order on A1×A2×· · ·An.Prove that 4lex is a total order if and only if 4i is a total order on Ai for all i.

Solution: First suppose that 4i is a total order on Ai for all i. Let (a1, a2, . . . , an) and (b1, b2, . . . , bn) beelements in A1 × A2 × · · ·An. If ai = bi for i = 1, . . . , n, then (a1, a2, . . . , an) 4lex (b1, b2, . . . , bn). Otherwise,let j be the least index for which ai = bi. Since all 4i are total orders, then 4j is a total order. Thus, eitheraj 4j bj or bj 4j aj . Hence, (a1, a2, . . . , an) 4lex (b1, b2, . . . , bn) or the reverse is true. Thus 4lex is a total order.

Now suppose that 4lex is a total order. Let j be any index with 1 ≤ j ≤ n and let aj , bj ∈ Aj . Considera pair of n-tuples (a1, a2, . . . , an) and (b1, b2, . . . , bn) such that ai = bi for all i < j. Since (a1, a2, . . . , an) and(b1, b2, . . . , bn) are comparable in 4lex, then aj 4j bj or bj 4j aj . Since aj , bj were arbitrary, then 4j is a totalorder. Thus, we conclude that all the partial orders 4i are partial orders.

Exercise: 14 Section 1.4Question: Let 4 be the lexicographic order on R3, where each R is equipped with the usual ≤. Prove ordisprove the following statement: For all vectors ~a,~b,~c, ~d, if ~a 4 ~b and ~c 4 ~d, then ~a+ ~c 4 ~b+ ~d.

Solution: We first prove that if ~a 4 ~b, then ~a + ~c 4 ~b + ~c. With the lexicographic order, ~a 4 ~b if and onlyif the first index j ∈ {1, 2, 3} for which aj 6= bj has aj < bj . For any ~c ∈ R3, for all indices ai = bi if and onlyai + ci = bi + ci. Hence the first index for with ai + ci 6= bi + ci is j and then we have aj + cj < bj + cj . Thus

~a+ ~c 4 ~b+ ~c.Since the above result was arbitrary, ~a 4 ~b implies ~a+~c 4 ~b+~c and ~c 4 ~d implies ~c+~b 4 ~d+~b. By transitivity

of 4, we deduce that~a+ ~c 4 ~b+ ~c 4 ~b+ ~d.

Exercise: 15 Section 1.4Question: Answer the following questions pertaining to the poset described by the Hasse diagram below.

1. List all the minimal elements.

2. List all the maximal elements.

3. List all the maximal elements in the subposet with {a, b, c, d, e, f, g}.

4. Determine the length of the longest chain and find all chains of that length.

5. Find the least upper bound of {a, b}, if it exists.

6. Find the greatest lower bound of {b, c}, if it exists.




a b

cd

e

f g

h i

j

7. List all the upper bounds of {f, d}.

Solution: In the given Hasse diagram, we have the following named elements.

a) The minimal elements are a and b.

b) The maximal element is j.

c) The maximal elements in {a, b, c, d, e, f, g} are e, f , and g.

d) There are 3 chains of length 5. They are {a, d, e, i, j}, {a, d, g, h, j}, and {a, c, f, h, j}. There is no chain ofgreater length.

e) The subposet {a, b} has only one upper bound, namely e, and this is the least upper bound.

f) The subposet {b, c} does not have a lower bound.

g) The set of upper bounds of {f, d} is {h, j}.

Exercise: 16 Section 1.4Question: Consider the partial order on R2 given in Example 1.4.6. Let A be the unit disk

A = {(x, y) ∈ R2 |x2 + y2 ≤ 1}.

a) Show that A has both a maximal and minimal element. Find all of them.

b) Find all the upper bounds and all the lower bounds of A.

Solution: The partial order on R2 is defined by (x1, y1) 4 (x2, y2) if and only if 2x1 − y1 < 2x2 − y2 or(x1, y1) = (x2, y2).

a) The solution to this problem is directly related to the calculus problem of optimizing 2x− y for (x, y) ∈ A.The gradient of 2x − y is never 0 so there are no optimal values to 2x − y on the interior of A. We canparametrize the boundary of A by (cos t, sin t) with t ∈ [0, 2π]. We can optimize 2x − yon the boundaryby optimizing 2 cos t − sin t. This is optimized with tan t = − 1

2 . With the result that sin t = − 12 cos t, we

deduce that the optimum values occur at(2√5,− 1√

5

)and

(− 2√

5,

1√5

)with the former corresponding to a maximum and the latter being a minimum element with respect to 4.

b) The upper bounds to A are any (x, y) with 2x− y > 5√5

=√

5 as well as the point(

2√5,− 1√

5

). The lower

bounds of A are all the points (x, y) with 2x− y < −√

5 as well as the point(− 2√

5, 1√

5

).

Exercise: 17 Section 1.4Question: Consider the lexicographic order on R2 coming from the standard (R,≤). Let A be the closed diskof center (1, 2) and radius 5.

1. Show that A has both a maximal and minimal element. Find all of them.

2. Find all the upper bounds and all the lower bounds of A.

3. Show that A has both a least upper bound and a greatest lower bound.




Solution: The lexicographic order on R2 coming from the standard (R,≤) is a total order.

a) A maximal element of A will have a greatest x value and if there are any ties on the x value, we break themwith the y value. The maximal element is the point (6, 2). No point in A has a greater x-value. Similarly,the point with the least x value is (−4, 2) and there is no other point with the same x component. Hence(−4, 2) is minimal.

b) The set of upper bounds to A is {(x, y) ∈ R2 |x > 6} ∪ {(6, y) | y ≥ 2}. The set of lower bounds of A is{(x, y) ∈ R2 |x < −4} ∪ {(−4, y) | y ≤ 2}.

c) The single maximal element is the unique least upper bound to A and the single minimal element is theunique greatest lower bound.

Exercise: 18 Section 1.4Question: Prove that in a finite lattice, there exists exactly one maximal element and one minimal element.

Solution: Let (S,4) be a finite lattice. We first prove that a maximal element exists. Since S is finite everychain in S is finite. Consequently, there exists a chain C of maximal length. We can consider the set of positiveintegers defined by

{|{y ∈ C |x 4 y}| |x ∈ C}.

By the well-ordering of the integers, this set must have a least element and this least element must be 1. Theelement M that gives this least element satisfies |{y ∈ C |M 4 y}|. Hence, the only element y such that M 4 yis M itself, which means that M is maximal.

Let M1 and M2 be two maximal elements. Let M be a least upper bound to M1 and M2. In particular,M1 4 M and M2 4 M . Since M1 and M2 are maximal, then M1 = M = M2. Hence, the maximal element isunique.

A similar reasoning holds with minimal elements and greatest lower bounds. Hence, S contains a uniqueminimal element.

Exercise: 19 Section 1.4Question: Let (B,→) be the poset of blood types equipped with the donor relation. (See Example 1.4.4.)

a) Consider the poset ({1, 2, 3},≤). Show that the function f : B −→ {1, 2, 3} defined by f(o) = 1, f(a) = 2,f(b) = 2 and f(ab) = 3 is a monotonic function.

b) Show that there exists no isomorphism between (B,→) and ({1, 2, 3, 4},≤).

Solution: Let (B,→) be the poset of blood types.

a) It is obvious that for all x ∈ B, we have f(x) ≤ f(x). Consequently, we only need to check the monotonicproperty on unequal elements x and y that satisfy x→ y. The five unequal pairs of donating relation givethe following output by f .

in B in {1, 2, 3} true of falseo→ a 1 ≤ 2 To→ b 1 ≤ 2 To→ ab 1 ≤ 3 Ta→ ab 2 ≤ 3 Tb→ ab 2 ≤ 3 T

Thus, we have exhaustively checked that f is monotonic.

b) Let h be a bijection from ({1, 2, 3, 4},≤) to (B,→). Let x and y be the elements in {1, 2, 3, 4} such thath(x) = a and h(y) = (b). The elements x and y are distinct since h is a bijection. Now since ≤ is a totalorder, x ≤ y or y ≤ x. However, neither a→ b nor b→ a. Thus, no bijection can be monotonic.

Exercise: 20 Section 1.4Question: Let (S,4), (T,4′), and (U,4′′) be three posets. Let f : S → T and g : T → U be monotonicfunctions. Prove that the composition g ◦ f : S → U is monotonic.

Solution: Let x, y ∈ S with x 4 y. Since f is monotonic, then f(x) 4′ f(y). Now since g is monotonic, theng(f(x)) 4′′ g(f(y)). Thus we have proved that

x 4 y =⇒ (g ◦ f)(x) 4′′ (g ◦ f)(y).

Hence, the composition g ◦ f is monotonic.




Exercise: 21 Section 1.4Question: Prove that the poset (R,≤) is not isomorphic to (R− {0},≤).

Solution: Suppose that f : R→ R−{0} is an isomorphism from the poset (R,≤) to (R−{0},≤). Let a, b ∈ R.For all x such that a ≤ x ≤ b, we have f(a) ≤ f(x) ≤ f(b). Furthermore, for all y with f(a) ≤ y ≤ f(b), we alsohave a ≤ f−1(y) ≤ b. Hence, the interval [a, b] is mapped to the interval [f(a), f(b)].

Let a and b be such that f(a) = −1 and f(b) = 1. Then f([a, b]) is the interval [−1, 1]. However, [−1, 1] isnot a subset of R− {0}.

Exercise:Question: Prove that the poset of integers greater that a fixed number k (with partial order ≤) is isomorphicto the (N,≤).

Solution: It is obvious that f : N→ {k+ 1, k+ 2, . . .} defined by f(n) = n+ k+ 1 is a bijection. Furthermore,for all m,n ∈ N, we have m ≤ n implies m+ k + 1 ≤ n+ k + 1. The result follows.

Exercise: 23 Section 1.4Question: Let (S,41) and (T,42) be two partially ordered sets and let f : S → T be a monotonic function.

a) Prove that if A is a subset of S with an upper bound u, then f(u) is an upper bound of f(A).

b) Prove with a counterexample that f(lub(A)) is not necessarily equal to lub(f(A)).

c) Prove that if f is an isomorphism, then f(lub(A)) = lub(f(A)).

Solution: Let (S,41) and (T,42) be two partially ordered sets and let f : S → T be a monotonic function.

a) Since u is an upper bound of A, then x 41 u for all x ∈ A. Hence, f(x) 42 f(u) for all x ∈ A, or in otherwords, y 42 f(u) for all y ∈ f(A). Thus f(u) is an upper bound of f(A).

b) Consider the function tan−1 : R → R and A = R. Then lub(f(A)) = π2 whereas lub(A) does not exist so

f(lub(A)) does not exist either.

c) Suppose now that f is surjective and let u = lub(A). We know that f(u) is an upper bound of f(A). Lety′ be another upper bound of f(A). Since f is surjective, there exists u′ ∈ S such that f(u′) = y′. Sinceb 42 y

′ for all b ∈ f(A), then since f is an isomorphism, a 41 u′ for all a ∈ A. Thus u′ is also an upper

bound of A. Hence u 41 u′. Thus, f(u) 42 y

′. We conclude that f(u) is a least upper bound of f(A).

Exercise: 24 Section 1.4Question: Prove or disprove that (Z,≤) and (Q,≤) are isomorphic as posets.

Solution: Suppose that h is an isomorphism from a poset (S,41) to (T,42). Suppose that y is an immediatesuccessor to an element x in S. Then x 41 y and x 6= y and if z is any element such that x 41 x 41 y, then z = xor z = y. Then h(x) 6= h(y) and h(x) 42 h(y). Furthermore, for all z′ such that h(x) 42 z

′ 42 h(y), then sinceh−1 is also monotonic we deduce that x 41 h

−1(z′) 41 y. Thus h−1(z′) = x, which implies that z′ = h(x) orh−1(z′) = y, which implies that z′ = h(y). Hence, if y is an immediate successor to x, then h(y) is an immediatesuccessors to h(x).

In this particular example, every element in (Z,≤) has an immediate successor but no element in the poset(Q,≤) has an immediate successor. Hence, they cannot be isomorphic as posets.

Exercise: 25 Section 1.4Question: Determine whether the posets corresponding to the following Hasse diagrams are lattices. If theyare not, explain why.

a

b c

d e

f

g

(a)

a

b c

d e

f g

h(b)

a

b c

d e

f g

h(c)

b c

d e

f g

h(d)



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