AA Lovett

| Set Theory

1

1.1 – Sets and Functions Exercise: 1 Section 1.1 Question: Let U = n 10 and consider the subsets A = 1, 3, 5, 7, 9 , B = 1, 2, 3, 4, 5 , and N n C = = 1, 2, 5, 7, 8 . Calculate the following operations. a) A B b) (B C ) A c) (A B ) (A C ) (B C ) d) ((A ((A B ) C ) (A (B C )) ))

{

{ ∈ | ≤ }

}

{

}

{

}

∩ ∪ − ∩ ∩ ∩ ∩ ∩ − − ∩ − −

Solution: We apply the definitions of 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 d) ((A ((A B ) C ) (A (B C )) ) ) = ( 7, 9 3, 4 ) = 9 1, 5, 7, 9 = 9 C ) (A

∩ { } ∪ − { }−{ } { } ∩ ∩ ∩ ∩ ∩ { }∩{ }∩{ } { }∩{ − − ∩ − − { } − ∩ − { } { } ∩ {

} {}

}∩{ 4, 6, 8, 9, 10}

Exercise: 2 Section 1.1 Question: 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 ) B c) (A B C ) (A B C ) d) (A B ) (B C )

{

}

{

∩ ∩ ∪ ∪ −

∪ − ∪ − ∩ ∩ ∪ −

∩ ∩ ∪ ∪ −

∪ ∩{ ∩ { } { } − { }− { } ∪ − ∩ ∩ { }−∅ { ∪ − { } ∪ { } { }

}

{

}

{

}

Solution: We apply the definitions of 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 1.1 Question: As subsets of the reals, describe the differences differences between between the sets 3, 5 , [3, [3, 5] and (3, (3, 5). Solution: The set 3, 5 contains contains the integers integers 3 and 5. The closed interv interval al [3, [3 , 5] contains all real numbers between 3 and 5 including 3 and 5, while the open interval (3, (3 , 5) contains all real numbers between 3 and 5 not including 3 and 5.

{ }

{ }

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

∩ ⊆ ⊆ ∪

Solution: We use the definitions definitions of subsets and the intersection intersection and union of sets. a) Let x Let x A B. Then x Then x A and A and x B = B = x A, A, so A so A B A. A. b) Let x Let x A. A . We know that A B = y y A or y B , so x so x A = A = x A B . Hence A Hence A

∈ ∩ ∈

∈

∈ ⇒ ∈ ∩ ⊆ ∪ { | ∈ ∈ } ∈

Exercise: 5 Section 1.1 Question: Let A Let A and B be subsets of a set S . S . a) Prove Prove that that A B if and only if P(A) P (B ) b) Prove Prove that that P(A B ) = P (A) P(B ). c) Show Show that P(A B ) = P (A) P(B ) if and only if A A

⊆

Solution:

∩ ∪

∩ ∪

⊆

1

⊆ B or B or B ⊆ A. A.

⇒ ∈ ∪

⊆ A ∪ B.

2

CHAPTER CHAPTER 1. SET THEOR THEORY

⇒

⊆

∀ ∈

∈

a) (= ): Suppose A B . Then, a A, A , a B. B . Sinc Sincee P(B ) contains all the possible subsets of B , all the possible subsets subsets of A A must be in P(B ) because A because A B. B . Therefore, P(A) P (B ).

⊆ ⊆ (⇐=): =): Suppose P(A) ⊆ P(B ). Then Then ∀{a} ∈ P(A), {a} ∈ P(B ). Therefore, Therefore, there there must exist a subset C of P(B ) that contains every {a} from P(A). The subse subsett C leads C leads to the conclusion that every a ∈ A must also be in B in B . Therefore, A Therefore, A ⊆ B. B . b) By definition, P(A ∩ B ) = {{t1 , t2 ,...,tn } | ti ∈ A, ti ∈ B }. This This implies 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 and b ∈ B − A. A . Then Then the the set set { a, b} is in P(A ∪ B) B ) but not in P(A) or in P(B ). Theref Therefore ore by by the contrapositive, P(A ∪ B ) = P (A) ∪ P(B ) if A A ⊆ B or B or B ⊆ A. A. (⇐=): =): Suppose A Suppose A ⊆ B. B . Then, A Then, A ∪ B = B = B so P(A ∪ B ) = P (B ). Now suppose B suppose B ⊆ A. A . Then A Then A ∪ B = A = A so so P(A ∪ B ) = P (A). Either way, P(A ∪ B ) = P (A) ∪ P(B ). Exercise: 6 Section 1.1 Question: Give the the list descriptio description n of P( 1, 2, 3, 4 ).

{

}

Solution: Using the definition definition of a power 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 1.1 Question: Give the the list descriptio description n of



{{a1, a2, . . . , ak } ∈ P ({1, 2, 3, 4, 5}) a1 + a {{ + a2 + · · · + ak = 8}. Solution: We need to find all the subsets {1, 2, 3, 4, 5} whose subsets of { whose elements elements add to a total total of 8. Recall that no subset has repeated elements so {4, 4, } does not make sense. The set is {{1, 2, 5}, {1, 3, 4}, {3, 5}} .

Exercise: 8 Section 1.1 Question: Let A Let A,, B , and C and C be be subsets of a set S . S . a) Prove that (A ( A B ) C = = (A C ) (B C ). ). b) Find and prove prove a similar similar formula formula for A for A (B C ). ).

− −

− − − − −

Solution: S

S

B

A

B

C

A

C

a) In the first Venn diagram, the lighter shade represents ( A B ), and the darker shade, which overlaps some of (A B ), represents (A (A B ) C . In the second Venn Venn diagram, diagram, the lighter lighter shade represents represents (A ( A C ), ), while the darker shade represents (A ( A C ) (B C ). ). We observe from the diagrams that the darker regions are equal.

−

− −

−

− − −

−

2


⇒

⊆

∀ ∈

∈

a) (= ): Suppose A B . Then, a A, A , a B. B . Sinc Sincee P(B ) contains all the possible subsets of B , all the possible subsets subsets of A A must be in P(B ) because A because A B. B . Therefore, P(A) P (B ).

⊆ ⊆ (⇐=): =): Suppose P(A) ⊆ P(B ). Then Then ∀{a} ∈ P(A), {a} ∈ P(B ). Therefore, Therefore, there there must exist a subset C of P(B ) that contains every {a} from P(A). The subse subsett C leads C leads to the conclusion that every a ∈ A must also be in B in B . Therefore, A Therefore, A ⊆ B. B . b) By definition, P(A ∩ B ) = {{t1 , t2 ,...,tn } | ti ∈ A, ti ∈ B }. This This implies 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 and b ∈ B − A. A . Then Then the the set set { a, b} is in P(A ∪ B) B ) but not in P(A) or in P(B ). Theref Therefore ore by by the contrapositive, P(A ∪ B ) = P (A) ∪ P(B ) if A A ⊆ B or B or B ⊆ A. A. (⇐=): =): Suppose A Suppose A ⊆ B. B . Then, A Then, A ∪ B = B = B so P(A ∪ B ) = P (B ). Now suppose B suppose B ⊆ A. A . Then A Then A ∪ B = A = A so so P(A ∪ B ) = P (A). Either way, P(A ∪ B ) = P (A) ∪ P(B ). Exercise: 6 Section 1.1 Question: Give the the list descriptio description n of P( 1, 2, 3, 4 ).

{

}

Solution: Using the definition definition of a power 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 1.1 Question: Give the the list descriptio description n of



{{a1, a2, . . . , ak } ∈ P ({1, 2, 3, 4, 5}) a1 + a {{ + a2 + · · · + ak = 8}. Solution: We need to find all the subsets {1, 2, 3, 4, 5} whose subsets of { whose elements elements add to a total total of 8. Recall that no subset has repeated elements so {4, 4, } does not make sense. The set is {{1, 2, 5}, {1, 3, 4}, {3, 5}} .

Exercise: 8 Section 1.1 Question: Let A Let A,, B , and C and C be be subsets of a set S . S . a) Prove that (A ( A B ) C = = (A C ) (B C ). ). b) Find and prove prove a similar similar formula formula for A for A (B C ). ).

− −

− − − − −

Solution: S

S

B

A

B

C

A

C

a) In the first Venn diagram, the lighter shade represents ( A B ), and the darker shade, which overlaps some of (A B ), represents (A (A B ) C . In the second Venn Venn diagram, diagram, the lighter lighter shade represents represents (A ( A C ), ), while the darker shade represents (A ( A C ) (B C ). ). We observe from the diagrams that the darker regions are equal.

−

− −

−

− − −

−

3

1.1. SETS AND FUNCTION FUNCTIONS S

S

S

B

B

A

C

A

C

b) In the Venn diagram above, the lighter shade represents B C , and the darker shade represents A represents A (B C ). ). In the second diagram, the lighter region represents A B , and the darker region represents A represents A C , which overlaps some of A of A B . Thus, (A (A B ) (A C ) = A (B C ). ).

−

− ∪ −

− − − −

− − −

Exercise: 9 Section 1.1 Question: Let A Let A,, B , and C and C be be subsets of a set S . S . a) Prove Prove that that A B = if and only if A if A = = B B.. b) Prove Prove that that A (B C ) = (A B ) (A C ). ).

 ∅ ∩ 

∩  ∩

Solution: Let A Let A,, B , and C and C be be subsets of a set S . S . a) Suppose that A that A B = . 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 that A B = and B A = . Now for and two sets U sets U and T and T ,, the identity U identity U T = is equivalent to U to U T . T . Hence we deduce that A B and B and B A. A . Consequently, A Consequently, A = B = B.. The argument of the opposite direction is identical. Suppose that A = B = B.. Then A Then A B and B and B A. A . Thus A B = and B A = . We deduce that A B = (A B ) (B A) = . b) There are a variety ariety of ways to prove prove the identity identity A (B C ) = (A B ) (A C ). ). We could use use a well well designed designed Venn Venn diagram. We could also use a mem membership bership table which lists all p ossibilities ossibilities of an element element whether 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 the A and C column C column and nothing in the B column, that refers to the situations of an element in A in A,, not in B in B and in C in C ..

−

∅

−

−

∅

−

√

∅ ⊆ ∅

− −

⊆



− ∪ − ∅ ∩  ∩  ∩

∅

⊆ ⊆

⊆

√

√A B√ √C √ √ √ √ √ √ √ √ √

 √ √

(B C )

A

∩ (B C ) √ √

(A

√∩ B ) √

(A

√∩ C ) (A ∩ B)(A ∩ C ) √ √ √

√ √

∩ 

Since the A the A (B C ) and column and the (A (A the sets are equal.

∩ B)(A ∩ C ) of this membership table are the same, then

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

    Ai =

i

∈I

b)

Ai =

i∈I

⊆

{ }

Ai .

i

∈I Ai .

i∈I

{ }

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

4


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

⇒

∈



(= ) Let a

i

∈I

 ∈  ∈

∈ I . So then a then a

i

⇐

( =) Let a



∈

i

   ∈ Ai =

Ai

i



And this this implies implies that that a

∈ A i for every

Ai .

∈I

∈ A i for every i ∈ I .

∈

∈ I .

i

∈I

Ai . Then Then a

i

∈I

∈I

Ai . Which implies a implies a

And so

i

i

∈I

∈I

Ai or a or a / A i for every i

Ai . And so

i

a/

 ∈  ⊆ 

∈

Ai . Then Then a /

 ⊆ 

Ai . So

Ai

i

i

∈I

∈

∈ I .

Which Which implies implies that a / A i for eversy i

And And so

Ai .

i

∈I

∈I

Ai .

i

∈I

∈I

b) We will prove prove this equation by proving proving set inclusion inclusion in both directions. directions. (= ) Let a Let a

⇒

a

Ai . So a So a /

∈

i

 ∈

∈I

Ai . So a So a / A i for at least one i

i

∈I

Ai . Which implies that

i

Ai

i

∈I

( =) Let a Let a

⇐

∈



Ai . Then a Then a

i

∈I

∈

  Ai =

i



∈ I . So a So a ∈ A i for at least one i ∈ I . So

Ai .

i

∈I

It follows that a that a / So

 ∈  ⊆ 

∈I

∈ Ai for at least one i ∈ I . This implie impliess that a that a ∈ / A i for at least one i ∈ I .

Ai . Which implies that a that a

i

∈I

∈



Ai . So

i

 ⊆  Ai

i

∈I

Ai .

i

∈I

∈I

Ai .

i

∈I

∈I

Exercise: 11 Section 1.1 Question: Let P P be the parabola in the plane whose equation is y = x2 . Let Let Aq q∈P be the collection of subsets of R where A q is the tangent line to P at q at q .. Determine with proof q∈P A q . R 2 where A



{ }

Solution: Sketch Sketching ing a picture picture of the standard standard parabola, parabola, we see that all the tangent tangent lines are in some sense beneath the parabola. Also, taking the (infinite, uncountable) union of all the tangent lines to the parabola, we might guess that we would get all points ( a, b) in the plane such that b a 2 . We need to prove this hypothesis. Label the coordinates of a point q P as q = (x0 , x20 ). From calculus, calculus, the tangent tangent line to P at q has q has the equation

≤

∈ ∈

y = x = x20 + 2x 2 x0 (x

− x0) =⇒ y = y = 2x0 x − x20 .

Now suppose that some point (a, ( a, b) in the plane is on a tangent line. Then, for some x 0 , we have b have b = = 2x0 a Since a, Since a, b are given and x and x 0 is unknown, this is an equation in x 0 . The quadratic formula gives for x 0 x0 =

2a

± √ 4a2 − 4b = a ± 2

In particular, we note that there exists an x 0 if and only if a if a 2

 − a2

b.

− b ≥ 0, confirming the hypothesis that b ≤ a2.

Exercise: 12 Section 1.1 Question: Let Ak k∈N be the collection of subsets of R 3 such that A that A k = (x,y,z) x,y,z) both k∈N Ak and k∈N Ak .



− x20.

{ }

{



∈ R3 | z ≥ ky }. Determine

Solution: Each subset A subset A k represents all points greater than the plane that is bound on the line z = ky = ky . In the diagram below, imagine the x-axis is coming out of the page, and let k 1 , k 2 , and k and k 3 represent the bounds of the subsets A subsets A 1 , A 2 , and A and A3 respectively. We note that the larger k gets, k gets, the more steep the plane becomes. However, the condition does not hold true for z for z < 0 when y when y is is 0. Therefore, k∈N Ak = (x,y,z) x,y,z) z y (x,y,z) x,y,z) x < 0 , and k∈N Ak = (x,y,z) x,y,z) z y (x,y,z) x,y,z ) x 0 .



{

| ≥ } ∩ {

| ≤ }



{

| ≥ }∪{

|

}

5

1.1. SETS AND FUNCTIONS

k3 k2

z

k1

y

Exercise: 13 Section 1.1 Question: In geometry of the plane, a subset S of the plane is called convex if for all p, q S the line segment pq 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 1.1 Question: 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, we account 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 the elements 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 count whatever elements are in both A C and B C , or A B C . So we must subtract out that amount, and we arrive at (A B) C = A C + B C A B C . Plugging this result back into our original equation we 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 1.1 Question: 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:

⊆

∪ ∩

6

CHAPTER 1. SET THEORY

a) Observe the Venn diagrams below: U

A

U

B

A

B

∩

∪

In the above diagrams, the left diagram represents A B, and the right diagram represents A B. In the A B diagram, represents A, represents B, and represents where they overlap. A B is represented by the union of these shaded regions. We can observe the shaded regions from both diagrams describes the same set, thus A B = A B. b) Observe the Venn diagrams below:

∪

∩

∪

∪

U

A

U

B

A

B

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

∪ ∪

∩

∩ ∪

∩ B. In the

shaded

∩

Exercise: 16 Section 1.1 Question: (*) Let n be a positive integer. Describe an algorithm (a finite list of well-defined instructions to accomplish 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 set in S . Once this process is completed union k to S . Increase k by 1. Repeat this process until k = n. In other words, while k < n, S = S s t k , t S k .

{{∅}}

{ }

{ } ∪ { | ∪ { } ∀ ∈ } ∪ { }

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

f (x) = 3x + 4 f (x) = 3x2 + 7 f (x) = (x + 1)/(x + 2) f (x) = x 5 + 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 x 1 = 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 get x = 4−3 y . 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). This shows that f is not injective. To test for surjectivity, we attempt to solve for x in y = f (x). The equation y = 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,

−

−

−

−

−

−

−

−

±

 −

−{− }

7


suppose that f (x1 ) = f (x2 ). Then we have x1 + 1 x1 + 1 = = x1 + 2 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 arbitrary y. We have x + 1 1 2y y = = yx + 2y = x + 1 = xy x = 1 2y = x = . x + 2 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) = x 5 + 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 = x 52 + 1 = x51

⇒ − x52 = 0 =⇒ (x1 − x2)(x41 + x31x2 + x21x22 + x1x32 + x42) = 0.

Obviously the equation holds if x1 = x2 . Now we look for solutions of the second term. Note that if x2 = 0, then the quartic equation implies that x1 = 0. But then x 1 = x 2 , which we already know to be a possibility. Assuming that x 1 = x 2 , after division by x 42 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 only solution to f (x1 ) = f (x2 ) is x1 = x 2 . Thus f is injective. For surjectivity, we see that y = f (x) implies that x = 5 y 1, which is defined for all y. Hence f is surjective. (f is bijective.)

√ −

Exercise: 18 Section 1.1 Question: Given an explicit example of a function f : Z a) bijective; b) surjective but not injective; c) injective but not surjective; d) neither injective nor surjective.

→ Z that is

Solution: Recall that x takes x and returns the nearest integer less than or equal to x. The solutions presented here are not the only options! a) f (x) = x. b) f (x) = x/2 . c) f (x) = x 2. d) f (x) = x2 .

 

−   ∗

Exercise: 19 Section 1.1 Question: Given an explicit example of a function f : N a) bijective; b) surjective but not injective; c) injective but not surjective; d) neither injective nor surjective.

→ N that is

Solution: Recall that x takes x and returns the nearest integer less than or equal to x. The answers presented here are not the only options! a) f (x) = x. b) f (x) = x/2 . c) f (x) = 2 x.



  ∗

8


∗ 

d) f (x) = 2 ( x/2 ). Exercise: 20 Section 1.1 Question: Let f : A bijective and

→ B and g : B → C be functions. Prove that if f and g are bijective, then g ◦ f is (g ◦ f )−1 = f −1 ◦ g −1 .

◦

∈



◦

Solution: Assume g f is not injective. Then there would exist a 1 , a 2 A such that a 1 = a 2 and g f (a1 ) = g f (a2 ). Let b1 = f (a1 ) and b2 = f (a2 ). We know that b 1 = b 2 because f is bijective. Using substitution, we notice that g(b1 ) = g(b2 ), but b1 = b 2 . 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)) = c. By substituting we observe b B, g(b) = c. However, since g is bijective, this creates a contradiction. Therefore g 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 1.1 Question: 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 of g, g(a) = g(b). However, then f (g(a)) = f (g(b)) as well. This contradicts the injectivity of f g. Therefore g 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) = x 2 . Now, g sends non-negative numbers to a unique, non-negative, even number and negative numbers to a unique, 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 our case, the nonnegative numbers) but not necessarily the codomain of g(in our case, all of Z ))

◦

Exercise: 22 Section 1.1 Question: 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 surjective function, 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) = x/2 and g(x) = 2x. Then for any y N, (f g)(y) = f (g(y)) = f (2 y) = (2 y)/2 = y = y. So f g is surjective, but g is certainly not since no odd, positive numbers will be hit.

◦ ∈

◦ ◦

◦

◦

∗

 ∗

−→  

◦

  ◦

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

{}

→ B in terms

9


a) For injectivity, f −1 ( b ) b) For surjectivity, f −1 b )

| { } | ≤ 1. | { } | ≥ 1.

Exercise: 24 Section 1.1 Question: 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) = sin x and T = 3/2 . b) f : R R with g(x) = x2 2 and T = [1, 2]. c) f : R R with h(x) = x 3 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 of T . √ a) f −1 (T ) = sin−1 ( 23 ) = π3 , 2π 3 − − 1 1 b) f (T ) = g ([1, 2])

{

}

√ − 2 = 1 ⇒ x = ± 3 x2 − 2 = 2 ⇒ x = ±2 x2

f (x) x2

2

−2

1

−2 −1

1

2

x

−1 −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. x-values that satisfy x 3 − 2x = −1. x3 − 2x + 1 = (x − 1)(x2 + x − 1) √ − 1± 5 x = 1, 2 √ 3 x − 2x = 0 ⇒ x = 0, ± 2 f (x) x3

2

− 2x

1

−2 −1

1

2

x

−1 −2 Therefore, the pre-image of h −1 ([ 1, 0]) is [ −1−2

−

{

√ 5

,

√ −√ 2], [0, −1+2 5 ], [1, √ 2]}.

This allows us to find the

10


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

−→

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 ) = 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 such that 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 S with 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 so y = 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. Let y f (S ) f (T ). Then y f (S ) and y f (T ). Then there exists x S with f (x) = y and there exists x T with f (x ) = y. However, since x does not have to be equal to x  , there does not have to exist an element x  S T such that f (x ) = y.

∈ ∈

∩ ∈ ∩

∈

∈

∩

⊆

∩

∈

Exercise: 26 Section 1.1 Question: Let f : A B be a function from the set A to the set B. Let V and W be subsets of the codomain B. 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 . Then a 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 ). Then f (b) exists in V or W . Which implies f (b) V W . So then b 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 ).

∪

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 both f −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 ).

∈

∩

11


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

| | | |

→ T is injective if and only

Solution:

⇒

→ ∀ ∈ ∈ ∈ | | | | ∈ ∈ (⇐=): 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 s 1 = s 2 . Then f implies that |S | elements are mapped to at most |S | − 1 elements in T . Thus there exists at least one t1 ∈ T where there does not exist s ∈ S such (= ): Suppose f : S T is injective. Then s1 , s2 S such that f (s1 ) = f (s2 ) implies s1 = s 2 . Therefore, every s S maps to a unique t T . Since S = T , there does not exist t T such that t is not associated with a unique s S . Thus f is surjective.

that f (s) = t 1 . This creates a contradiction because f is surjective. Therefore f is injective.

Exercise: 28 Section 1.1 Question: Let S be a set. For each subset A χA : S 0, 1 such that

→ { }

⊆ S we define the characteristic function of A as the function

χA (s) =



1 if s A 0 if s / A.

∈ ∈

Prove the following. a) The association A χ A 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 proposition 1.1.11 we know that there are 2 n unique characteristic functions of S . We can write these subsets as n-tuples where each entry contains either a 1 or a 0. We can also picture any function from S to 0, 1 as a n-tuple (f (s1 ), f (s2 ),...,f (sn )) where S = n. Note that each n-tuple will represent a distinct function from S to 0, 1 . In comparing n-tuples, we observe that every function from S to 0, 1 represents a unique characteristic function of S implying that the association is injective. Since each entry has two options, and there are n entries, there are 2 n distinct functions from S to 0, 1 . Thus the cardinalities of the functions of S to 0, 1 and P (S ) are equal. Using the result in exercise 1.1.27, the association must be surjective as well. Hence, the association from A χ A is a bijection between P (S ) and the set of functions from S to 0, 1 . b) Suppose s A B. Then, s A and s B. This implies χ A (s) = 1 and χ B (s) = 1. From here we deduce that χA∩B (s) = 1 = χA (s) χ B (s). Now suppose s / A B. Then s A or s B, so χA (s) = 0 or χB (s) = 0. Either way, χ A∩B (s) = 0 = χA (s) χB (s). Therefore, χ A∩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 χ A (s) + χB (s) works for both cases. If s A B, then we need to account for counting an element twice so we subtract χA∩B (s) which we know to be χ A (s) χB (s) from above. Hence χ A∪B (s) = χ A (s) + χB (s) χA (s) χB (s) for all s S . d) Suppose s A. Then s A, so χA (s) = 0 and χA (s) = 1. If s A then s A and we get the opposite result. Notice to reverse the results of χ A (s) we subtract it from 1. Therefore χ A (s) = 1 χA (s). e) Note that A B = A B and χ B (s) = 1 χB (s). Thus χ A−B (s) = χ A (s)(1 χB (s)) for all s S .

{ }

{ } ∈ ∩

→

∈ · ∈

∈ ∪

∈

∈ ∩

∈

{ }

{ }

{ }

∈

{ }

| |

· ∈ ∩

∈ ·

∈ ∩

∈

∈ ∩ ∈

∩

∈

−

∈

−

∈ ·

−

∈ −

−

·

∈

Exercise: 29 Section 1.1 Question: 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

12


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 natural numbers, we can describe any integer n in terms of any natural number m by using the definition of even and odd numbers m if m is even n = 2 m+1 if m is odd. 2

  −

Since every integer n is represented by a different natural number m we can quickly deduce that the association is surjective by definition. Assume that there are two integers a and b such that a = b but f (a) = f (b). Then, 2a + 21 = 2b + 21 which implies that 2a + 21 = 2b + 21 a = b. This creates a contradiction, hence the association is injective as well. Thus since there exists a bijection between Z and N . Therefore, Z is countable.

   



⇒

Exercise: 30 Section 1.1 Question: 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 a unique 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 1.1 Question: 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-tuples is m 1 m2 m3 ... mn = m n . 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 1.2 Question: 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 contains pairs with the components (a, (b, c)).

× ×

× ×

× ×

× (B × C )

Exercise: 2 Section 1.2 Question: 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 1.2 Question: 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)

{

{ {

}

{ }

×

}

×

× ×

}

13

1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS

{

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 1.2 Question: 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 1.2 Question: Write in list form 1 Solution: The elements in 1

{ } × {1, 2} × {1, 2, 3}. { } × {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 1.2 Question: Justify the statement that A = for all sets A. Solution: By definition of the Cartesian product, A = (a, b) a A has no elements and therefore is equal to .

× ∅ ∅

×∅

× ∅ {

∅

| ∈ A and b ∈ ∅}. Since ∅ has no elements,

Exercise: 7 Section 1.2 Question: (*) 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 successively going 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 the first quadrant. Using the hint, we establish a series of lines following the pattern x + y = n + 1 where n = 1, 2, 3,.... Observe the intersections of the first couple of lines:

×

⇒ (1, 1) ⇒ (1, 2), (2, 1) ⇒ (1, 3), (2, 2), (3, 1)

n = 1 n = 2 n = 3

As the pattern continues, we quickly observe that as n increases, every element in N∗ N∗ is mapped to once 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 there exists a bijection between A B and N∗ N∗ . Since we have already established a bijection between N∗ N∗ and N ∗ , there exists a bijection between A B and N ∗ . Therefore, A B is countable.

×

×

×

×

×

×

×

Exercise: 8 Section 1.2 Question: The operation on vectors of R n defined by  u v = proj v , i.e., projection of v onto u. u Solution:

∗

∗

Associative: Find any counter-example. Let a,  b, and c be vectors in R 2 such that a =< 1, 1 >,  b =< 2, 1 >, 21 and c =< 3, 1 >. Then (a  b) c = 34 a and a ( b c) = 10 a. Therefore, is not associative.

∗ ∗

∗ ∗

∗

Commutative: Find any counter-example. Using the same vectors above, observe that a  b = 23 a and  b a = 53 b. Since a  b =  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. Thus it is impossible to have a unique identity.

∗

∗

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

14


∗

Idempotent: As shown earlier, proja a = a therefore, is idempotent. Exercise: 9 Section 1.2 Question: For the operation  on the open interval [0, 1) described by a  b = a + b a + b where x is the greatest integer less than or equal to x, determine whether it is associative, commutative, has an identity, has inverses or is idempotent.

−



 

Solution:

∈

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

− a + b)  c − a + b + c − a + b − a + b + c since x + n = x + n for all x ∈ R and all n ∈ Z = a + b + c − a + b + a + b − a + b + c = a + b + c − a + b + c. (a  b)  c = (a + b = a + b

Furthermore a  (b  c) = a  (b + c = a + b + c = a + b + c = a + b + c

− b + c) − b + c − a + b + c − b + c − b + c + b + c − a + b + c − a + b + c.

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

− a + b = b + a − b + a = 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 also in [0, 1), satisfies a  b = a + (1 a) a + 1 a = 1 1 = 0.

∈

− −

− 

− a, which is

− 

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 1.2 Question: The operation Solution:

 on nonnegative integers N defined by n m = |m − n|.

Associative: Let a = 1, b = 2, and c = 3. Observe a 2. Therefore is not associative.

 (b  c) = ||c − b|− a| = 0 while (a  b)  c = |c −|b − a|| =

 Commutative: a  b = |b − a| = |a − b| = b  a. Therefore  is commutative. Identity: a  e = |e − a| = |a − e| = e  a = a. Hence e = 0. Inverse: Every element is it’s own inverse. a  a = |a − a| = 0. Idempotent: a  a = 0 hence  is not idempotent.

15


Exercise: 11 Section 1.2 Question: The operation  on points in the plane R 2 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 +b41 +c1 and (A  B)  C = a2 +b2 +2c2 . Thus, if A = C then A  (B  C ) = (A  B)  C . Hence,  is not associative. 4





Commutative: The midpoint between A and B is the same as the midpoint between B and A, so  is commutative. 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. 1 a2 +a2 Idempotent: A  A = ( a1 +a 2 , 2 ) = (a1 , a2 ) = A. Therefore  is idempotent.

Exercise: 12 Section 1.2 Question: 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. ×b = a + b + ab = b + a + ba = b+×a. Therefore +× is commutative. Commutative: a+ ×q = q so 0 is the identity. Identity: Consider any q ∈ C. q × +0 = 0+ Inverse: Let b be the inverse of any a ∈ C. Then a+ ×b = 0 which implies a + b(1 + a) = 0. We can quickly − a deduce b = 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 1.2 Question: The operation Solution:

 on P(S ), where S is any set.

Associative: S

S

B

A

B

C

A

C

Note in the above diagrams, the darker shading represents the operation within the parenthesis. We observe through the Venn diagrams that a (b c) = (a b) c. Thus, is associative.

 

Commutative:

 



16


S

A

B

Observing the Venn diagram, A

 B = B  A therefore  is commutative. Identity: A  ∅ = ∅ A = A hence, ∅ is the identity of A. Inverse: Each set is it’s own inverse as A  A = ∅. Idempotent: A  A = ∅ implying  is not idempotent. Exercise: 14 Section 1.2 Question: The cross product on R 3 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 = a ( b c), the cross product 3 is not associative over R .

× ×

−

× ×

Commutative: For any two vectors in R 3 , 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 by definition, therefore, there cannot exist a vector that satisfies a e = a. Thus the cross product does not have an identity.

×

Inverse: Because the cross product does not have an identity, it cannot be closed under inverses. Idempotent: Note,  b idempotent.

×  b =< 0, 0, 0 > which does not equal  b in every case therefore the cross product is not

Exercise: 15 Section 1.2 Question: For the power operator a∧ b = a b 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 a e = a for all a N ∗ . Hence e = 1. However, by definition, we must also have e a = 1a = a for all a N ∗ , which leads to a contradiction. Hence ∧ does not have an idenity element.

∈

∈

Inverses: The operation cannot have an inverse since it does not have an identity. Idempotent: Since 2∧ 2 = 4 = 2, then ∧ is not idempotent.



Exercise: 16 Section 1.2 Question: The composition operator on the set

◦

Solution:

F (A, A) of functions from a set A to A (where A is any set).

17


Associative: Let f , g, and h be functions in Therefore, is associative.

◦

F (A, A). Then (h ◦ g) ◦ f = h(g(f (a))) = h ◦ (g ◦ f ) where a ∈ A.

Commutative: Consider the function j where j (a) = a 1 or in other words, every input gives the same output a1 . Now let k be a function such that k (a1 ) = a 1 . Then j k(a) = a 1 , but k j(a) = a 1 . Therefore, is not commutative.



◦

◦

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

◦

F (A, A).



◦

◦

Then q e = q and

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

F

∈

◦

◦

◦

Idempotent: Let l be a function such that l(a1 ) = a 2 and l(a2 ) = a 3 . Then l(l(a1 )) = a 3 which is not equal to l(a1 ). Thus, is not idempotent.

◦

Exercise: 17 Section 1.2 Question: 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 of sets. 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 1.2 Question: Let S be a set with a binary operation . Assume that (a b) a = b for all a, b a (b a) = b for all a, b S . Solution: Assume (a b) a = b for all a, b S .

∗ ∗

∗

∈ ∗ ∗

∗ ∗

∈ S .

Prove that

∈

(a (a b) a (a b) (a

∗ b) ∗ a = b ∗ ∗ a ∗ a = (b ∗ a) ∗ ∗ ∗ a ∗ a = a ∗ (b ∗ a) ∗ b) ∗ a = a ∗ (b ∗ a) b = a ∗ (b ∗ a)

∗ ∗

∈

Therefore, a (b a) = b for all a, b S . Exercise: 19 Section 1.2 Question: Consider the operations a∧ b = a b and a b on N∗ . Prove that ∧ is right-distributive over not left-distributive over . Solution: Let a, b, c N∗ . For right-distributivity, we observe that

∈

×

×

(a

× but

× 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 = 2 7 .

Exercise: 20 Section 1.2 Question: 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

| | | |

×

×

| | · | |

18


× | |

| |

| | · | |

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 2 nn possible binary operators on S .

||

×

Exercise: 21 Section 1.2 Question: 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) = s 1 , (1, 2) = s 2 , (2, 1) = s 3 , and (2, 2) = s 4 . 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 1.2 Question: 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 1.2 2 Question: Let A be a finite set with n elements. Prove that the number of reflexive relations on A is 2n −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 in A2 . 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 = n 2 n. With the union of P(S ) and B we find all possible reflexive relations. By proposition 1.1.11 we 2 2 know that the size of P(S ) is 2n −n . Therefore, there are 2 n −n possible reflexive relations on A. Similar to the reflexive relations, if we can find the set of possible symmetric elements, then its power set will result in the number of symmetric relations on A.

||

−

{ | ∈

−

n = 2

∪ { }

}

n = 3

A

{

| ∈ ∀ } −

n = 4

×A

A

×A

A

×A

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 A 2 B such that 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. Thus when n = 2, T = 1 and B = 2. Let X = T B. X contains all the symmetric elements of A2 . Hence we see 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) . Thus, the cardinality of 2

{{

| | | |

} | ∀ ∈ } { | | ∪ | |

its power set results in 2

(n+1) 2

n

}



symmetric relations on A.

Exercise: 24 Section 1.2 Question: For any set S , consider the relation  on Solution:

−

P(S )

| |

defined by A  B to mean that A

∩ B = ∅.

19


Reflexive: Choose any subset A. Then, A  A implies A P (S ) so  is not reflexive. empty set. However,

∅ ∈

∩ A = A = ∅ which is true for all cases except the

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

⇒ A ∩ B = ∅ ⇒ B ∩ A = ∅ ⇒ B  A

AB Therefore,  is symmetric.

Antisymmetric: Consider the set S = 1, 2, 3, 4 . Let A = 1, 2 and B = 3, 4 . We observe that both A  B and B  A exist, however it does not imply that they are equal. Hence,  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  B and B  C satisfy the conditions of transitivity, A  C does not exist because A C = 2 . Hence,  is not transitive.

{

}

{ }

{ }

{ } ∩

{ }

Exercise: 25 Section 1.2 Question: The relation  on S the set of people defined by p 1  p2 if p 1 is taller than or the same height as p2 . Solution: Reflexive: Any person is the same height as himself, therefore  is reflexive. Symmetric: Consider the case where an individual p 1 is taller than p 2 . Then p1  p 2 but p 2  p 1 . Hence,  is not symmetric.



Antisymmetric: Assume ( p1  p2 ) and ( p2  p1 ), but p 1 = p 2 . This implies that either p 1 is taller than p 2 or vice versa. Without loss of generality, suppose p1 is taller than p2 . Then p 1  p 2 but p2  p 1 which is a contradiction. Therefore, if p 1  p2 and p 2  p1 , then p 1 = p 2 . Thus  is antisymmetric.





Transitive: Assume that p 1  p 3 , but p1  p 2 and p2  p 3 where p 2 = p 3 . This implies that p2 is taller than p3 and p 1 is taller than or equal to p 2 . Hence, p 1 would have to be taller than p 3 which is a contradiction. Therefore, if p 1  p2 and p 2  p3 , then p 1  p3 implying  is transitive.





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

≥ m2.

Reflexive: Find any counter-example. Let n = 2. This creates a contradiction since 2 reflexive.

≥ 4, therefore R is not

Symmetric: Find any counter-example. Let n = 9 and m = 2. It is clear to see that nRm but the reverse is not 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 m 2 and m n 2 . By squaring both sides and substitution, we find that n n 4 which is only true if n = 1. Applying the same method the other direction, we find that m = 1 as well. Therefore, the only case where nRm and mRn is when n = m = 1. Thus, R is antisymmetric.

≥

≥

≥

Transitive: Let a, b and c be integers. Assume a c 2 , but aRb and bRc. Using some substitution, this implies that a b 2 c 2 . This creates a contradiction, therefore R is transitive.

≥ ≥

≥

Exercise: 27 Section 1.2 Question: Consider the relation  on S = R 2 defined by (x1 , y1 )  (x2 , y2 ) to mean x 21 + y12 which of the properties reflexivity, symmetry, antisymmetry, and transitivity hold. Solution:

≤ x22 + y22. Prove

Reflexivity: Let (x, y)

∈ R2. Then x 2 + y2 ≤ x2 + y 2, so (x, y)  (x, y). Hence  is reflexive.

20


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

≤



≤

12 + 22 so (1, 1)  (1, 2). However



Antisymmetry: Note that (1, 2)  (2, 1) and (2, 1)  (1, 2) but since (1, 2) = (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 + y12 x 22 + y22 and x22 + y22 x 23 + y32 .

≤

Hence x 21 + y12

≤

≤ x22 + y22 so (x1, y1)  (x3, y3). Hence  is transitive.

Exercise: 28 Section 1.2 Question: The relation  on R defined by a  b to mean ab = 0. Solution: Reflexive: Consider the case where a R and a = 0. Then a  a this is not true in every case, thus  is not reflexive.

∈



⇒ a 2 = 0.

This creates a contradiction as

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 = 0 and b = 0. Then a  b and b  a, but a = b. Hence,  is not antisymmetric.





Transitive: Let a = 0, b = 0, and c = 0. Then a  b and b  c, but a  c since a = 0 and c = 0. Thus,  is not transitive.







Exercise: 29 Section 1.2 Question: For any set S , consider the relation  on Solution:

P(S )





defined by A  B to mean that A

∪ B = S .

Reflexive: Let A

∈ P (S ) and A = 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 = antisymmetric.

∅ and B = S .

Then, A  B and B  A, but A = B. Therefore,  cannot be



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

∅ ∪





Exercise: 30 Section 1.2 Question: The relation  on the set of pairs of points in the plane S = R2 if the segment [P 1 , P 2 ] intersects [Q1 , Q2 ]. Solution:

× R2 defined by (P 1, Q1)  (P 2, Q2)

Reflexive: Let P 1 , P 2 be in S . It is not hard to see that (P 1 , P 1 )  (P 2 , P 2 ) because any line intersects with itself. Hence,  is reflexive. Symmetric: Suppose (P 1 , Q1 )  (P 2 , Q2 ). Then [P 1 , P 2 ] intersects [Q1 , Q2 ] implying that [Q1 , Q2 ] intersects [P 1 , P 2 ] and (Q1 , P 1 )  (Q2 , P 2 ). Thus,  is symmetric. Antisymmetric: Let (P 1 , Q1 )  (P 2 , Q2 ) such that [P 1 , P 2 ] is perpendicular to [Q1 , Q2 ]. It is not hard to show that (P 1 , Q1 )  (P 2 , Q2 ) and (Q1 , P 1 )  (Q2 , P 2 ), but it is impossible for [P 1 , P 2 ] to be equal to [Q1 , Q2 ]. Therefore,  is not antisymmetric. Transitive: Let [P 1 , P 2 ], [Q1 , Q2 ], and [R1 , R2 ] be lines in S and let [P 1 , P 2 ] be parallel to [R1 , R2 ] and perpendicular to [Q1 , Q2 ]. Then (P 1 , Q1 )  (P 2 , Q2 ) and (Q1 , R1 )  (Q2 , R2 ), but it is impossible for (P 1 , R1 )  (P 2 , R2 ) since they are parallel to each other. Therefore,  is not transitive.

21


Exercise: 31 Section 1.2 Question: Let S be a set and let R be a relation on S . Prove that if a relation is reflexive, symmetric, and anti-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 s R t then t R s where s, 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 = t. Therefore, R = (s, s), s S which is the = relation.

∀ ∈

∈



∈ {

∀ ∈ }

Exercise: 32 Section 1.2 Question: Let P be the set of people who are living now. Let R be the relation on P defined by aRb if a and b 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, p R p since ’self’ is included in the nuclear family, therefore R is reflexive. Assume there exists two people p1 and p2 such that p1 R p2 but p 2 is not related to p1 . It is not hard to see that this 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 p 1 R p2 where p 1 is the child of p 2 . We know that p 1 R p2 and p 2 R p1 , however p 1 and p 2 are not the same person. Hence, R is not antisymmetric. Let p 1 be the child of p 2 and p 2 be the sibling of p 3 . We can easily observe that p 1 R p2 and p 2 R p3 , yet p 1 is not in the same nuclear family as p 3 . 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 1.2 Question: We can define the graph of a relation R from R to itself as the subset of R 2

{(x, y) ∈ R2 | x R y}. ≤

a) Sketch the graph of the relation . b) Sketch the graph of the relation  defined by x  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: y

y = x

x

a)

22


y

y = x + 1 y = x

−1

x

b) 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 points from the line y = x. iv) In a transitive relation, for any points (x, y) and (y, z) there exists a right triangle that contains the points (x, y), (y, z), and (x, z). Exercise: 34 Section 1.2 Question: 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 1.2 Question: Let R be a relation on a set A. Denote by R (n) the n-composite relation of R with itself: n times

 ◦ ◦ · · · ◦ 

(n) def

R

= 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) R. Then there exists some z A such that (x, z) and (z, y). This creates a contradiction by the definition of a transitive relation, thus (x, y) R(n) , (x, y) R. Therefore, R(n) R. ( =): Suppose R(n) R. Assume that there exists a R b and b R c such that a is not related to c. By definition of relation composition, a R c exists in R(n) . If a R c exists in R(n) , then it must also exist in R by definition of a subset causing a contradiction. Hence, if a R b and b R c, then a R c. Therefore, R is transitive.

⇒

⊆

∈ ∀ ∈

∈

∈

⊆

⇐

⊆

Exercise: 36 Section 1.2 Question: Let R be a relation that is reflexive and transitive. Prove that R n = 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 r1 R r1 and r1 R r2 , since R is reflexive, therefore r 1 R r2 must be in R 2 . We can easily observe that the same is true for R 3 in that r1 R r1 and r 1 R r1 and r1 R r2 , so (r1 , r2 ) must be in R3 . We can continue this process for any n to show that any (r1 , r2 ) in R must also be in R n . This implies that R R n . Therefore, R n = R for all n N∗ .

∈

⊆

⊆

∈

1.3 – Equivalence Relations Exercise: 1 Section 1.3 Question: 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

×

×

23

1.3. EQUIVALENCE RELATIONS

(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 is symmetric. Suppose (a, b) R (c, d) and (c, d) R (e, f ). Then a + d = b + c and c + f = d + e. Using subtraction and substitution 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 1.3 Question: Let C be the set of people in your abstract algebra class. Describe a “natural” relation satisfying each 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 a R b if a and b live within one kilometer (or one mile) of each other. b) Reflexive and transitive, but not symmetric. R, where a R b if a earns a final grade that is less than or equal to the final grade that b earns. c) Symmetric and transitive, but not reflexive. The simplest relation that satisfies these conditions is the empty relation. Symmetry and transitivity are satisfied trivially. (The hypothesis is always false so the conditional statement is always true.) Note that if a R b, where a = b, then by symmetry b R a and then by transitivity a R a. So for this combinations of properties to hold, there must be an element that is not in relation to any other element. d) An equivalence relation. R, where a R b if a and b entered college the same semester.



Exercise: 3 Section 1.3 Question: Let P be the set of living people. For all a, b P , define the relation a R 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 c be people such that a R b and b R c. It does not follow that person a has met person c in every case implying R is not transitive. Therefore, R is not an equivalence relation.

∈

Exercise: 4 Section 1.3 Question: Let P be the set of living people. For all a, b P , define the relation a R b if a and b live in a common town. Solution: Obviously, a R a exists for all people within the town. It quickly follows that if a R b then b R a and therefore is symmetric as well. Suppose a R b and b R c. Then a and b live in a common town, and b and c live in the same town. Hence, a and c live in the same town meaning a R c and R is transitive. Therefore, R is an equivalence relation.

∈

Exercise: 5 Section 1.3 Question: Let be the set of circles in R2 and let R be the relation of concentric on . Prove or disprove whether the described relation is an equivalence relation. If the relation is not an equivalence relation, determine which properties it lacks. Solution: Two circles are concentric if and only if they have the same center.

C

C

C

Reflexivity: If C is a circle in , then it has the same center as itself. Symmetry: If C 1 , C 2 symmetry holds.

∈ C , then if C 1 has the same center as C 2, then C 2 has the same center as C 1. ∈ C

So

Transitivity: Let C 1 , C 2 , C 3 . Suppose that C 1 is concentric with C 2 and that C 2 is concentric with C 3 . Then C 1 and C 2 have the same center and C 2 and C 3 have the same center. Hence C 1 and C 3 have the same center, so C 1 is concentric with C 3 . Hence concentric is transitive.

24


So the concentric relation is an equivalence relation. Exercise: 6 Section 1.3 Question: Let S = Z Z and define the relation R on S by (m1 , m2 ) R (n1 , n2 ) if m 1 m2 = n1 n2 .

×

∈

Solution: Consider any (m1 , m2 ) S . We observe (m1 , m2 ) R (m1 , m2 ) implies m1 m2 = m1 m2 which is always true. Hence, R is reflexive. Suppose (m1 , m2 ) R (n1 , n2 ). We know m 1 m2 = n 1 n2 so n1 n2 = m 1 m2 and therefore (n1 , n2 ) R (m1 , m2 ). Hence, R is symmetric. Suppose (m1 , m2 ) R (n1 , n2 ) and (n1 , n2 ) R ( p1 , p2 ). Then, m1 m2 = n 1 n2 and n 1 n2 = p 1 p2 . By substitution, we find m 1 m2 = p 1 p2 implying (m1 , m2 ) R ( p1 , p2 ). Hence, R is transitive and therefore is an equivalence relation. Exercise: 7 Section 1.3 Question: Let S = Z Z and define the relation R on S by (m1 , m2 ) R (n1 , n2 ) if m 1 n1 = m 2 n2 .

×

Solution: Consider the relation (1, 2) R (1, 2). Since 1 an equivalence relation.

× 1 = 2 × 2, then R is not reflexive. Therefore, R is not

Exercise: 8 Section 1.3 Question: Let S = Z Z and define the relation R on S by (m1 , m2 ) R (n1 , n2 ) if m 1 n2 = m 2 n1 .

×

Solution: Suppose m1 , m2 S , Z and (m1 , m2 ) R (m1 , m2 ). Then m1 m2 = m2 m1 for any (m1 , m2 ) thus R is reflexive. Suppose (m1 , m2 ) R (n1 , n2 ) where n1 , n2 Z. Then, m1 n2 = m2 n1 which can be easily rearranged to show n1 m2 = n2 m1 . This implies (n1 , n2 ) R (m1 , m2 ), and therefore R is symmetric. Suppose r1 , r2 , s1 , s2 , t1 , t2 Z such that (r1 , r2 ) R (s1 , s2 ) and (s1 , s2 ) R (t1 , t2 ). Then, r1 s2 = r2 s1 and s1 t2 = s2 t1 . Using some substitution we can arrive at the desired outcome as seen below,

∈

∈

∈

∈

r1 s2 = s 1 r2 r1 s2 t1 = s 1 r2 t1 r1 s1 t2 = s 1 r2 t1 r1 t2 = t 1 r2 Therefore, since R is reflexive, symmetric, and transitive, R is an equivalence relation. Exercise: 9 Section 1.3 Question: Let P 3 be the set of polynomials with real coefficients and of degree 3 or less. Define the relation R on P 3 by p(x) R q (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 12 and q (x) = 2x 7. We observe that p(x) R q (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 an equivalence relation.

−

−

−

−

−

−

− − −

Exercise: 10 Section 1.3 Question: Consider the set C 0 (R) of continuous functions over R . Define the relation R on C 0 (R) by f R g if there 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 equivalence relation, determine which properties it lacks. Solution: We check the defining properties for an equivalence relation. Reflexivity: Let f be any function in C 0 (R). Setting a = b = 0 we have f (x) = f (x + a) + b for all x. Hence R is reflexive. Symmetry: Suppose that f R g. Then there exists some a, b R such that g(x) = f (x + a) + b. Then g(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.

−

∈

∈

− −

∈

25


Transitivity: Let f , g , h C 0 (R) and suppose that f R g and g R h. There there exist a1 , a2 , b1 , b2 that g(x) = f (x + a1 ) + b1 and h(x) = g(x + a2 ) + b2 for all x R.

∈

∈ R such

∈

Then h(x) = f (x + a + 2 + a1 ) + b1 + b2 for all x

∈ R. Hence f R h. Thus R is transitive.

Exercise: 11 Section 1.3 Question: Let Pfin (R) be the set of finite subsets of R and define the relation on Pfin (R) by A B if the sum 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 of elements 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 of elements in A is equal to the sum of elements in C by substitution. Hence, is transitive and therefore is an equivalence relation.

∼

∈

∼

∼ ∼

∈

∼

∼

∼

∼

∼

∼

∼

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

n

→∞

− an) = 0.

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

n

→∞

− an) = nlim →∞(0) = 0

Also, notice lim (bn

n

→∞

− an) = − nlim →∞(an − bn ) = 0

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

n

→∞

− cn ) = 0

and

lim (an

n

→∞

− bn) = 0.

Thus lim (an

n

→∞

− cn) = nlim − − − − →∞((an bn ) + (bn cn )) = nlim →∞(an bn ) + nlim →∞(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 rule here because we know that each of the sequences involved in the sums converge (to 0). Exercise: 13 Section 1.3 Question: 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 counterexample for reflexivity. Let an = 2n . Then, 2 limn→∞ (an ) = implying R cannot be reflexive. Therefore, R cannot be an equivalence relation. It should be noted, however, that R is both symmetric and transitive.

∞

Exercise: 14 Section 1.3 Question: Let S be the set of lines in R 2 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 1.3 Question: Let be the words in the English language (i.e., have an entry in the Oxford English Dictionary). Define the relation R on by w 1 R w2 is w 1 comes before w 2 in alphabetical order. Solution: This relation is not symmetric. Find any counter-example. Consider the two words, w 1 =”ball” and w2 =”stick”. We observe that w 1 R w2 , but the reverse is not true. Hence, R cannot be an equivalence relation.

W

W

26


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



1

∼ g ⇐⇒

f



∼ on

1

f (x) dx =

0

g(x) dx.

0

Show that is an equivalence relation and describe (with a precise rule) a complete set of distinct representatives of . Solution: First show that is an equivalence relation.

∼

∼

∼

  ∼   1 0

Reflexivity: For all f C 0 ([0, 1]), we have

∈

∈ C 0([0, 1]) with f

Symmetry: Suppose that f , g

1 0 f (x) dx so f

f (x) dx =

∼ f .

g. Then

1

1

f (x) dx =

0

g(x) dx.

0

Equality is reversible so g

∼ f . Hence ∼ is symmetric. Transitivity: Suppose that f , g , h ∈ C 0 ([0, 1]) with f ∼ g and g ∼ h. Then

 

1

1

f (x) dx =

0

Hence



1 0 f (x) dx =

1 0





1

g(x) dx

and

0



1

g(x) dx =

0

h(x) dx.

0

∼ h. Hence ∼ is an equivalence relation.

h(x) dx. Thus f

 1

Every function f C 0 ([0, 1]) is in relation with the constant function g(x) = c, where c = 0 f (x) dx. However, constant functions that are not equal are not in relation with each other. Hence the constant functions are a complete set of distinct representatives.

∈

Exercise: 17 Section 1.3 Question: Let C ∞ (R) be the set of all real-value functions on R such that all its derivatives exist and are continuous. 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 [sin x]. 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 is reflexive. 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 R h, we find f R h 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) = sin x then for all positive and even integers n, we have f (n) (0) = 0, so this is the defining characteristic of functions in [sin x]. (If a function in [sin x] is equal to its power series in a neighborhood of 0, then [sin x] consists of functions that are a constant plus an odd function.)

∈

∈

∈

Exercise: 18 Section 1.3 Question: Let S = 1, 2, 3, 4 and the relation on P (S ), defined by A B if and only if the sum of elements in 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 1.3 Question: Let be the set of (non-degenerate) triangles in the plane. a) Prove that the relation of similarity on triangles in is an equivalence relation. b) Concisely describe a complete set of distinct representatives of .

T

Solution:

∼

T

∼

∼ }} { }

27


∈ T ∼

∼

a) Suppose you have any 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 to see that the corresponding angles of triangle t are equal to the corresponding angles of triangle s implying that 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 are also equal to the corresponding angles of t. Therefore the corresponding angles of triangle r are equal to those 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 any triangle such that x, y,z R>0 and x + y + z = 180. For example, [30, 60, 90] is the distinct representative for all triangles with 30, 60, and 90 as their respective angles.

∈

∼

∈

∼ ∈

∼

∼

∼

∼

∼

Exercise: 20 Section 1.3 Question: 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 under multiplication. 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 ) . Then ad = 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 1.3 Question: Let S = 1, 2, 3, 4, 5, 6 . For the partitions of S given below, write out the equivalence relation as a 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 1.3 Question: Let S = a,b,c,d,e . For the partitions of S given below, write out the equivalence relation as a subset 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)

{ {

}

}

28


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 1.3 Question: Let C 1 ([a, b]) be the set of continuously differentiable functions on the interval [ a, b]. Define the relation on C 1 ([a, b]) as f g if and only if f  (x) = g  (x) for all x (a, b). Prove that is an equivalence relation on C 1 ([a, b]). Describe the elements in the equivalence class for a given f C 1 ([a, b]). Solution: First, we observe for any f C 1 ([a, b]) that f  (x) = f  (x) for all x (a, b). Hence, is reflexive. Similarly, given any f, g C 1 ([a, b]) we observe that if f g then f  (x) = g  (x) for all x (a, b). Thus we know g (x) = f  (x) which establishes that is also symmetric. For transivity, consider any f , g , h C 1 ([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 is transitive. Therefore, is an equivalence relation. The elements of the equivalence class for a given f are the vertical translations of f . In other words, [f (x)] = f (x) + k k is some constant in R .

∼

∼

∈

∈

∈

∼

∼

∼

∼ ∼

∼

∈ ∈

∼ ∈ ∈

∈

{

|

}

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

×

∼

×

∈ M n×n(R),

∼ is an equivalence relation on M n×n(R). 2. Prove that the function f : M n×n (R)/ ∼ → R defined by f ([A]) = det A is a well-defined function on the quotient set M n×n (R)/ ∼. 3. Determine with proof or counter-example whether the function g : M n×n (R)/ ∼ → R defined by g([A]) = 1. Prove that similarity

Tr A, the trace of A, is a well-defined function.

Solution: Let M n×n (R) be the set of n n matrices with real coefficients and define the relation between matrices. a) We prove that is an equivalence relation.

×

∼ as similarity

∼

∈ M n×n(R), since I AI −1 = A then A ∼ A. Symmetry: Let A, B ∈ M n×n (R) such that A ∼ B . Then there exists an invertible matrix S such that B = SAS −1 . Hence A = S −1 BS . Thus B ∼ A so ∼ is symmetric. Transitivity: Let A,B, C ∈ M n×n (R) such that A ∼ B and B ∼ C . Thus there exist invertible matrices −1 −1 Reflexivity: For all matrices A

A and T such that B = SAS

and C = T BT

. Then

C = T (SAS −1 )T −1 = (T S )A(T S )−1 .

∼ C so ∼ is transitive. b) Suppose that B ∼ A. Then B = SAS −1 for some invertible matrix S . Then Hence A

det B = (det S )(det A)(det S )−1 = det A. Therefore, the function det is well-defined on the set of -equivalence classes on M n×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 1.3 Question: 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.

∈ ∼

∼

| − |

∼ ∈

− ∈

29


∼ has an uncountable number of equivalence classes.

b) (*) Prove that

Solution: a) Choose any q Q such that q < . Then, if x we find,

∈

| − y| = q , y ∼ x and |x − y| < . Rearranging our equation y = x + q y = x q

−

Notice that the domain of these functions spans R . Therefore for any x y x and x y < . A graphical representation is shown below. y y = x +  y = x + q y = x q y = x 

∼

| − |

−

∈ R there exists y ∈ R such that

−

x

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

∼

Exercise: 26 Section 1.3 Question: 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) R 1 R2 ; (b) R 1 R2 ; (c) R 1 R2 . [Note that R 1 R2 , and similarly for the others, is a relation as a subset of S S .] Solution: Let R 1 and R 2 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) R 1 R2 . Thus R1 R2 is reflexive. Suppose that (a, b) R1 R2 . Then since R1 is an equivalence relation, (b, a) R 1 and similarly for R 2 . Hence (b, a) R1 R2 . Thus R 1 R2 is symmetric. Finally, suppose that (a, b) and (b, c) are pairs in R 1 R2 . Then since R 1 is an equivalence relation (a, c) R1 and the same holds for R 2 , so (a, c) R 2 . Thus (a, c) R1 R2 and hence R 1 R2 is transitive. b) R1 R2 is not an equivalence relation. Since R1 is reflexive, for all a S , the pair (a, a) R 1 and thus (a, a R1 R2 . Hence, R 1 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, R 1 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) / R 1 and (a, b) / R2 . For example, let S = P ( 1, 2, 3, 4, 5 ) and let R 1 be the equivalence relation on S of same cardinality and let R 2 be the equivalence relation of elements summing to the same value. Then 1, 4 R1 R2 5 because the elements in the set both add to the same value and 5 R1 R 2 2 because the sets are the same cardinality. However, 1, 4 is not in relation to 2 under R1 R2 . Hence, the union of two transitive relations is not necessarily transitive. c) R1 R2 cannot be an equivalence relation since it is not reflexive. For all a S , the pair (a, a) is in both R1 and in R 2 . Hence, the pair (a, a) does not occur in R 1 R2 .

∩ ×

∪

∩

∩

∈

∈ ∩

∩

∈ ∩ ∪

∪ ∈ ∪ ∈

{

{ }

∈ ∈ ∩

∪

∈ ∩ ∈

∩

∈

∩

∈ ∈ ∪ ∈ ∪

∈

}

{ }





∈ ∈ ∪

∈

∈

∈

∈ ∪

{ } ∪ { }

{ } ∪ { } ∪



∈

∈

Exercise: 27 Section 1.3 Question: 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 k Z means all the multiples of k.

{ | } 2. {{3n, 3n + 1, 3n + 2 }| n ∈ Z}. 3. {{k | n2 ≤ k ≤ (n + 1)2 }|n ∈ N}.

30


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 2 Z 3Z, so the subsets are 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 defined by f (m) = m/3 . 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 map via 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 and these distinct sets are not mutually disjoint. The collection of subsets is not a partition of Z for the additional reason that it does not cover all of Z . In fact,

{

}

{

{

}

n2 , n2 + 1, . . . , (n + 1)2 = N.

}

n

∈N

d) This collection of subsets is a partition. Consider the equivalence relation on Z defined by a b if and only if a = b . This is indeed an equivalence relation and the equivalence classes are precisely subsets of n, n . Z of the form

∼

| | | | {− }

Exercise: 28 Section 1.3 Question: Let S be a set. Prove that there is a bijection between the set of partitions of S and the set of equivalence 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 that f ( ) = R. Let R be an equivalence relation on S and assume there does not exist over S such that f ( ) = R. However, by proposition 1.3.12 we know all the distinct equivalence classes of R are disjoint and their union is equal to S . This creates a contradiction because it satisfies the definition of a partition. Hence, for every equivalence relation R on S , there must exist where f ( ) = R. Therefore, f is surjective. Now assume there exists partitions = Ai i∈I and = Bj j ∈J such that = and f ( ) = f ( ) = R. This would imply every A i and B j represents a distinct equivalence class of R by proposition 1.3.14. However, since = there must exist some Ai and Bj where Ai Bj = A i and A i Bj = . This creates a contradiction by proposition 1.3.12 since the distinct equivalence classes of any equivalence relation are disjoint. Therefore, f must also be injective which establishes a bijection between the set of partitions of S and the set of equivalence classes on S .

A

A

A B { } ∩ 

A { }

A

A  B ∩  ∅

A

A

B

A  B

Exercise: 29 Section 1.3 Question: Call p(n) the number of equivalence relations (equivalently, by Exercise 1.3.28, partitions) on a set of cardinality n. (The numbers p(n) are called the Bell numbers after the Scottish-born mathematician E. T. Bell.) a) (*) Prove that p(0) = 1 and that for all n 1, p(n) satisfies the condition

≥

n 1

p(n) =

−

 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 number of 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 equivalence relation on the empty set itself. Suppose that we know the value of p(k) for 0 k n 1. To determine p(n), we count up the number of possible partitions based on how many elements are in the equivalence class 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 equivalence 1 class [n] = 1, 2, . . . , n . Now for any given j, there are n− ways to choose the remaining elements in the j

{

}

≤ ≤ −

{

}

{ }

| −{ }| −

 

31


− −

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, for 1 each j , there are p(n 1 j) n− partitions. Summing over j = 0 to n 1 gives the number of possible j partitions (equivalence classes) on 1, 2, . . . , n . b) Using this recursive formula, we get

{

}

 

− −

{

−

}

n p(n)

1 2 3 4 5 6 7 1 2 5 15 52 203 877

Exercise: 30 Section 1.3 Question: 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 which establishes that is symmetric. For transivity, suppose that a, b, c R such that a b and b c. Then b a = x and c b = y where x, y Z. Setting c = b + y and solving we find,

− ∼

∈ ∼ −

−

−

∈

∈

∈ ∼

∈

∈

∼

Since x + y is always an integer, we conclude is transitive. Therefore, b) Notice, b a = x and d c = y where x, y Z.

−

∈

− ∼

− a = x − a = x + y − a = x + y

b b + y c

−

∼

∼ −

∼ is an equivalence relation.

−

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 Since π Z, ac bd.

∈

∈

∼

∼

∼ b and c ∼ d, yet ac ∼ bd = π.

Exercise: 31 Section1.3 Question: Let S be a set and let refinement of if

A = {Ai}i∈I be a partition of S . Another partition B = {Bj }j∈J is called a ∀ 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 s 1 ∼B s 2 =⇒ s 1 ∼A s 2 . Solution: Suppose that B is a refinement of the partition A. Suppose that s 1 ∼B s 2 . This is equivalent to the statement that s 1 and s 2 are both in B j for some index j ∈ J . Since B is a refinement of A, then B j ⊆ A i for some i ∈ I , and therefore, s 1 and s 2 are both in this A i . Hence s 1 ∼A s 2 . Conversely, suppose that ∀s1 , s2 ∈ S, s1 ∼ B s2 −→ s1 ∼ A s2 . Consider a given subset Bj in the partition B and let s ∈ Bj . Since A is a partition of S , then s ∈ A i for some index i ∈ I  . For all s ∈ Bj , we have s ∼B s and so by our hypothesis, s ∼A s . Consequently, s ∈ A i and this for all s ∈ B j . Thus B j ⊆ A i . This establishes that B is a refinement of A. A

Exercise: 32 Section 1.3 Question: Let S be a set and let collection of sets

A = {Ai }i∈I and B = {Bj }j∈J be two partitions of S . {Ai ∩ Bj | i ∈ I and j ∈ J} − {∅}

Prove that the

32


is a partition of S . Solution: Call the collection of subsets of S

C

{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 i 0 ∈ I and a unique j 0 ∈ J such that s ∈ Ai and s ∈ Bj . Then s ∈ A i ∩ Bj and hence the union of subsets in C is all of S . Now consider A i ∩ Bj and A i ∩ Bj sets in C and suppose that (Ai ∩ Bj ) ∩ (Ai ∩ Bj )  = ∅. 0

0

1

0

1

2

0

2

1

1

2

2

Then by associativity

∩ Ai ) ∩ (Bj ∩ Bj ) = ∅. However, Ai ∩ Ai  = ∅ if and only if i1 = i2 and similarly Bj ∩ Bj  = ∅ if and only if j1 = j2 . Hence, we have i 1 = i 2 and j 1 = j 2 . Thus sets in C are either equal or disjoint. Consequently, since C also covers S , C is a (Ai1

1

2

1

2

2

1

2

partition of S .

Exercise: 33 Section 1.3 Question: Let S be a set and let R be any relation on S . Design an algorithm that determines the smallest equivalence 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 computationally 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 contains S exists by referring to Exercise 1.3.26 and generalizing. It is not hard to prove that the intersection of any collection of equivalence relations on S is again an equivalence relation on S . Hence, we can define the smallest equivalence relation on S containing R as Re



R : R R e

⊆

e

where the intersection is over all equivalence relations Re that contain R. We can call this relation the equivalence closure of R. From an algorithmic perspective, one of the challenges of this exercise is to realize that if we begin with the relation R as a subset of S S , and adjoin pairs to force symmetry and then adjoin more pairs to force transitivity, 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 of ˜ as m nonnegative integers, we define M ˜ ij = min(1, mij ). Hence, if mij 1, then m ˜ ij = 1 but if m ij = 0, then m ˜ ij = 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 ) R 0 otherwise.

∈

• Replace A with A + I . • Replace A with A.˜ [These first two steps ensure reflexivity.] • While A = A2  + (A2 ) , replace A with A2  + (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 both nonzero. This process will add a nonzero entry corresponding to a pair that is required by transitivity. The operation 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 A2  + (A2 ) will again be a matrix of 0s and 1s with new entries turned to 1 for transitivity or symmetry reasons.

→

33

1.4. PARTIAL ORDERS

The algorithm stops because if A = A2  + (A2 ) then no transitivity requirement or symmetry requirement will add a new pair to the relation (nonzero entry to A). Hence, A will then correspond to an equivalence relation. Furthermore, the algorithm will terminate because before the while loop, A contains at most n2 n entries 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 n 2 n times.

−

−

1.4 – Partial Orders Exercise: 1 Section 1.4 Question: Let S = a,b,c,d,e (where we consider all the labels unique elements). In the following relations on S determine with explanation whether or not the relation is a partial order. If it fails antisymmetry then remove 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 of the form (x, x) is (a, c). The pair (c, a) is not in the relation so the relation is antisymmetric. For the same reason, 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 relation is 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 and antisymmetric but it is not transitive because of the pairs (d, a) and (a, c). In the modified relation R = 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 antisymmetry 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 1.4 Question: In microeconomics (the study of consumer behavior), one considers consumer’s utility (preference) in regards to pairs of commodities. Let (q 1 , q 2 ) N2 be a pair of nonnegative integers representing quantities of two 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 (q 1 , q 2 ) (or these are equal), we write (q 1 , q 2 )  (q 1 , q 2 ). Note that this is a partial order on N 2 , 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 is in people’s psychology that if (q 1 , q 2 )  (q 1 , q 2 ) and (q 1 , q 2 )  (q 1 , q 2 ), so the concept of preference is transitive. We do not have (q 1 , q 2 )  (q 1 , q 2 ) and (q 1 , q 2 )  (q 1 , q 2 ) 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. Note that in the concept of preference, given two pairs (q 1 , q 2 ) and (q 1 , q 2 ) it is perfectly possible to not prefer one over the other. Such pairs would be incomparable.

∈

Exercise: 3 Section 1.4 Question: Let S = R >0 on S defined by

× R>0 be the positive first quadrant in the Cartesian plane. Consider the relation R (x1 , y1 ) R (x2 , y2 ) =⇒ x 1 y1 ≥ x2 y2

Prove or disprove that R is a partial order. Solution: We check the three axioms for a partial order.

34


∈

≤ 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)  = (2, 1) So the relation is not antisymmetric. Reflexivity: For all (x, y) S , we do have xy

Transitivity: We have already shown that the relation is not a partial order since it fails antisymmetry. The relation is transitive. Not a partial order. Exercise: 4 Section 1.4 Question: Prove that for any real x >

√ 2, the inequality √ 2 < 1 x + 2 2 x √ √ 1 1 2 √ Solution: Since 2 < x, then < , which implies < 2 < x. So x

 

x

2

1 2

< x holds.

  x +

2 1 < (x + x) = x. x 2

 √ 2, we have (x − √ 2)2 > 0, which is equivalent to √ √ 1 x2 − 2 2x + 2 > 0 ⇐⇒ x2 + 2 > 2 2x ⇐⇒ 2

Also, from x =

  x +

2 > x

√

2.

The result follows. Exercise: 5 Section 1.4 Question: Let (S, ) be a partial order in which every element has an immediate successor. Prove that it is not necessarily true that for any two elements a  b that any chain between a and b has finite length. Solution:

Consider the subset of (Q, ) with the set S =

≤

 −  ∈  ∪   ∈  − − 1

1 n

n

N∗

1+

1 n

n

N∗ . In the poset

1 (S, ), every element has an immediate successor: the immediate successor of 1 n1 is 1 n+1 , and the immediate 1 1 successor of an element of the form 1 + n is 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 immediate successor.

≤

∈

Exercise: 6 Section 1.4 Question: Prove the three claims about properties of  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 collection of sets An is countable. Furthermore, we claim that the sets An partition Q>0 . Indeed for all fractions xy , expressed in reduced form, we have xy A x+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 m with k < m k+1 , define f (m) as the (m k )th element (ordered by ) in Ak . This function is injective since the A k 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

∈

≤

−

→

→

| | · · · | | { }

−

m

F (m) = ( 1) f

  m + 1 2

≤

.

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

A { } ⊆

Solution: Let S be a set. Recall that a refinement of a partition = Ai i∈I of S is another partition = Bj j ∈J such that for all j J , there exists an i I such that Bj Ai . We will write  if is a refinement of .

B { }

∈

A

∈

A = {Ai }i∈I of S , for each i ∈ I we have Ai ⊆ Ai so A  A.

1. For a partition

B A B

35

1.4. PARTIAL ORDERS

B A

A B ∈

∈ ⊆

∈ ⊆ ⊆

⊆

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

B B A

B

A



C { } ⊆ B ⊆



C B

3. Finally, let be a refinement of and let = C k k∈K be a partition of S such that  . Then for all k K , there exists j J such that C k B j . Since is a refinement of , then there exists i I such that Bj Ai . Hence, since is transitive, C k Ai . Thus  . Hence, the relation of refinement is transitive.

∈

∈

⊆

⊆

A

C A

∈

These three results show that refinement is a partial order on the set of partition of S . Exercise: 8 Section 1.4 Question: Draw the Hasse diagram of the partial order Solution: The Hasse diagram of (P( 1, 2, 3, 4 ), ) is

{

{1, 2}

⊆ on P({1, 2, 3, 4}).

} ⊆ {1, 2, 3, 4}

{1, 2, 3}

{1, 2, 4}

{1, 3, 4}

{2, 3, 4}

{1, 3}

{1, 4}

{2, 3}

{2, 4}

{1}

{2}

{3 }

{ 4}

{3, 4}

∅ Exercise: 9 Section 1.4 Question: 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

{ }≤

{

}≤

6 5 4 3 2 1 Exercise: 10 Section 1.4 Question: Let A = a,b,c,d,e,f,g . Draw the Hasse diagram for the partial order  given as a subset of A A as

×

{

}

 = (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)

{

}

36


Solution: The Hasse diagram of (A, ) is e

h g

c a

b

f

d

Exercise: 11 Section 1.4 Question: 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 → t 2 holds if the letter portion of the blood type donates according to the way described in the examples for this section and if someone with a + designation can 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

→

ab+

a+

−

a

−

ab

o+

o

{ −}

b+

−

b

−

b) Under the partial order on +, is described completely by + in the lexicographic order on B  , we have

→ +, − → −, and − → +. So for example,

(a, +)

→lex ( ab, −) because the pairs already differ on the first entry and a → ab in B. donor relation on B  , a+ cannot donate to ab−.

However, we see that in the actual

Hence the actual donor relation on B  is not the lexicographic partial order on B

× {+, −}.

Exercise: 12 Section 1.4 Question: Consider the set of triples of integers Z 3 . Define the relation  on Z 3 by (a1 , a2 , a3 )  (b1 , b2 , b3 )

⇐⇒



a1 + a2 + a3 < b1 + b2 + b3 a1 + a2 + a3  lex b 1 + b2 + b3

if a 1 + a2 + a3 = b 1 + b2 + b3 ; if a 1 + a2 + a3 = b 1 + b2 + b3 ,



where lex is the lexicographic order on Z 3 (with each copy of Z equipped with the partial order  is a partial order on Z 3 . Prove also that  is a total order. Solution: We first prove that  is a partial order on Z 3 .

≤). Prove that

Reflexivity : Let (a1 , a2 , a3 ) Z 3 . Then since a 1 + a2 + a3 = a 1 + a2 + a3 , we use the lexicographic order to compare the entries but all entries are equal and this satisfies the lexicographic order. Hence (a1 , a2 , a3 )  (a1 , a2 , a3 ).

∈

37

1.4. PARTIAL ORDERS

Antisymmetry : Suppose that (a1 , a2 , a3 )  (b1 , b2 , b3 ) and (b1 , b2 , b3 )  (a1 , a2 , a3 ). Assume that a1 + a2 + a 3 = b1 + b 2 + b 3 , then a1 + a 2 + a 3 < b1 + b 2 + b 3 and b1 + b 2 + b 3 < a1 + a 2 + a 3 , which is a contradiction. Hence we must have a1 + a 2 + a 3 = b1 + b 2 + b 3 . Thus, we compare the triples by lexicographic order. However, the lexicographic order is a partial order, which is antisymmetric so we can conclude that (a1 , a2 , a3 ) = (b1 , b2 , b3 ).



Transitivity : Suppose that (a1 , a2 , a3 )  (b1 , b2 , b3 ) and (b1 , b2 , b3 )  (c1 , c2 , c3 ). We can break the situation into three cases. If a 1 + a2 + a3 = b 1 + b2 + b3 = c 1 + c2 + c3 , then we compare the triples by lexicographic order, which is transitive so (a1 , a2 , a3 )  (c1 , c2 , c3 ). If a 1 + a2 + a3 = b 1 + b2 + b3 = c 1 + c2 + c3 (resp. a1 + a 2 + a3 = b1 + b 2 + b3 = c 1 + c 2 + c3 ) then a1 + a2 + a 3 < c1 + c2 + c 3 so (a1 , a2 , a3  (c1 , c2 , c3 ). Finally, if all thee of the sums are different, then we deduce that a 1 + a2 + a3 < b1 + b2 + b3 < c1 + c2 + c3 so we can conclude that (a1 , a2 , a3  (c1 , c2 , c3 ). We conclude that  is transitive.





Finally, to show that  is a total order, let (a1 , a2 , a3 ), (b1 , b2 , b3 ) Z3 be arbitrary, distinct elements. There are two cases. If a1 + a 2 + a 3 = b1 + b 2 + b 3 , then either a1 + a 2 + a 3 < b1 + b 2 + b 3 , which implies that (a1 , a2 , a3 )  (b1 , b2 , b3 ), or the reverse is true. If a1 + a2 + a3 = b1 + b2 + b 3 , then we compare the triples by the lexicographic order. In this case, let i be the least index for which a i = b i . Then (a1 , a2 , a3 )  (b1 , b2 , b3 ) if ai < bi and (b1 , b2 , b3 )  (a1 , a2 , a3 ) if b i < ai . Hence, every pair of elements is comparable and the relation is a total order.

∈





Exercise: 13 Section 1.4 Question: Let (Ai , i ) be posets for i = 1, 2, . . . , n and define lex as the lexicographic order on A1 A2 An . Prove that  lex is a total order if and only if  i is a total order on A i for all i. Solution: First suppose that i is a total order on Ai for all i. Let (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) be elements in A1 A2 An . If ai = bi for i = 1, . . . , n, then (a1 , a2 , . . . , an ) lex (b1 , b2 , . . . , bn ). Otherwise, let j be the least index for which ai = b i . Since all i are total orders, then j is a total order. Thus, either aj j bj or b j j aj . Hence, (a1 , a2 , . . . , an ) lex (b1 , b2 , . . . , bn ) or the reverse is true. Thus lex is a total order. Now suppose that lex is a total order. Let j be any index with 1 j n and let aj , bj Aj . Consider a pair of n-tuples (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) such that ai = b i for all i < j. Since (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) are comparable in  lex , then a j j bj or b j j aj . Since a j , bj were arbitrary, then  j is a total order. Thus, we conclude that all the partial orders  i are partial orders.

× ×···

× × · · ·

≤ ≤

∈

Exercise: 14 Section 1.4 Question: Let  be the lexicographic order on R3 , where each R is equipped with the usual . Prove or  then a + c   disprove the following statement: For all vectors a,  b, c,  d, if a   b and c  d, b +  d. Solution: We first prove that if a   b, then a + c   b + c. With the lexicographic order, a   b if and only 3 if the first index j 1, 2, 3 for which aj = b j has aj < bj . For any c R , for all indices a i = b i if and only ai + ci = b i + ci . Hence the first index for with a i + ci = b i + ci is j and then we have aj + cj < bj + cj . Thus a + c   b + c.  Since the above result was arbitrary, a   b implies a +c   b+c and c  d implies  c + b  d + b. By transitivity of , we deduce that a + c   b + c   b +  d.

≤

∈ {

}





∈

Exercise: 15 Section 1.4 Question: 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.

38


j i

h g

f

e c

d a

b

{ }

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 of greater 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 1.4 Question: Consider the partial order on R 2 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)  (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 2 x y on the interior of A. We can parametrize the boundary of A by (cos t, sin t) with t [0, 2π]. We can optimize 2x yon the boundary by optimizing 2 cos t sin t. This is optimized with tan t = 12 . With the result that sin t = 12 cos t, we deduce that the optimum values occur at

−

− −

 √ − √  2 , 5

1 5

and

−

−

∈

−

−

−

∈

−

− √ √  2 1 , 5 5

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

− y > √ 55 = √ 5 as well as the point √ 25 , − √ 15 √ 2 √ bounds of A are all the points (x, y) with 2x − y < − 5 as well as the point − √ , 15 . 5

b) The upper bounds to A are any (x, y) with 2x









. The lower

Exercise: 17 Section 1.4 Question: Consider the lexicographic order on R 2 coming from the standard ( R, ). Let A be the closed disk of 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.

39

1.4. PARTIAL ORDERS

Solution: The lexicographic order on R 2 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 them with 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) R 2 x > 6 (6, y) y 2 . The set of lower bounds of A is 2 (x, y) R 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 the unique greatest lower bound.

≤

− {

−

∈ |

− }∪{ −

{ ∈ | | ≤ }

}∪{

| ≥ }

Exercise: 18 Section 1.4 Question: Prove that in a finite lattice, there exists exactly one maximal element and one minimal element. Solution: Let (S, ) be a finite lattice. We first prove that a maximal element exists. Since S is finite every chain in S is finite. Consequently, there exists a chain C of maximal length. We can consider the set of positive integers defined by y C x  y x C .

{|{ ∈ |

}| | ∈ }

By the well-ordering of the integers, this set must have a least element and this least element must be 1. The element M that gives this least element satisfies y C M  y . Hence, the only element y such that M  y is M itself, which means that M is maximal. Let M 1 and M 2 be two maximal elements. Let M be a least upper bound to M 1 and M 2 . In particular, M 1  M and M 2  M . Since M 1 and M 2 are maximal, then M 1 = M = M 2 . Hence, the maximal element is unique. A similar reasoning holds with minimal elements and greatest lower bounds. Hence, S contains a unique minimal element.

|{ ∈ |

}|

Exercise: 19 Section 1.4 Question: 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 monotonic property on unequal elements x and y that satisfy x y. The five unequal pairs of donating relation give the following output by f . in B in 1, 2, 3 true of false o a 1 2 T o b 1 2 T o ab 1 3 T a a b 2 3 T b 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 that h(x) = a and h(y) = (b). The elements x and y are distinct since h is a bijection. Now since is a total a. Thus, no bijection can be monotonic. order, x y or y x. However, neither a b nor b

{

≤

}≤

≤

→

→

{

→

} ≤

Exercise: 20 Section 1.4 Question: Let (S, ), (T,  ), and (U,  ) be three posets. Let f : S T and g : T U be monotonic functions. Prove that the composition g f : S U is monotonic. Solution: Let x, y S with x  y. Since f is monotonic, then f (x)  f (y). Now since g is monotonic, then g(f (x))  g(f (y)). Thus we have proved that

∈

◦

→

→

x  y = (g f )(x)  (g f )(y).

⇒ ◦

Hence, the composition g

◦ f is monotonic.

◦

→

AA Lovett

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