AA Lovett

(1)

||

1.1 – Sets and Functions 1.1 – Sets and Functions

Exercise:

Exercise: 1 Section 1.1 1 Section 1.1 Question:

Question: LetLet U U ==

_{

nn

_∈

NN

_||

nn

_≤

1010

_}}

and consider the subsets and consider the subsets AA ==

_{

11,, 33,, 55,, 77,, 99

_}}

,, BB ==

_{

11,, 22,, 33,, 44,, 55

_}}

, and, and C

C = =

_{

11,, 22,, 55,, 77,, 88

_}}

. Calculate the following operations.. Calculate the following operations. a) a) AA

_∩∩

BB b) b) ((BB

_∪∪

C C ))

₋

AA c) c) ((AA

_∩∩

BB))

_∩∩

((AA

_∩∩

C C ))

_∩∩

((BB

_∩∩

C C )) d) (( d) ((AA

₋

BB))

₋

C C ))

_∩∩

((AA

₋

((BB

₋

C C )))) Solution:

Solution: WWe apply the deﬁnitions oe apply the deﬁnitions of set operations:f set operations: a) a) AA

_∩∩

BB = =

_{

11,, 33,, 55

_}}

b) b) ((BB

_∪∪

C C ))

₋

AA = =

_{{

11,, 22,, 33,, 44,, 55,, 77,, 88

_{}} −}

_{− {{}

11,, 33,, 55,, 77,, 99

_}}

= =

_{{

22,, 44,, 88

_}}

c) c) AA

_∩∩

BB

_∩∩

AA

_∩∩

C C

_∩∩

BB

_∩∩

C C ==

_{{

11,, 33,, 55

_}∩{

11,, 55,, 77

_}∩{

11,, 22,, 55

_}}

= =

_{{

22,, 44,, 66,, 77,, 88,, 99,, 1010

_}∩{

22,, 33,, 44,, 66,, 88,, 99,, 1010

_}∩{

44,, 66,, 88,, 99,, 1010

_}}

d) ((

d) ((AA

₋

BB))

₋

C C ))

_∩∩

((AA

₋

((BB

₋

C C )) = ()) = (

_{{

77,, 99

_{}} −}

₋

C C ))

_∩∩

((AA

₋

_{− {{}

33,, 44

_}}

) =) =

_{

99

_{}} ∩∩ {{}

11,, 55,, 77,, 99

_}}

= =

_{{

99

_}}

Exercise:

Question: LetLet U U ==

_{

a,b,c,d,e,f,ga,b,c,d,e,f,g

_}}

and consider the subsets and consider the subsets AA = =

_{

a,b,da,b,d

_}}

,, BB ==

_{

b,c,eb,c,e

_}}

, and C , and C ==

_{

c,d,f c,d,f

_}}

.. Calculate the following operations.

Calculate the following operations. a) a) C C

_∩

((AA

_∪∪

BB)) b) b) ((AA

_∪∪

C C ))

₋

BB c) c) ((AA

_∪∪

BB

_∪

C C ))

₋

((AA

_∩∩

BB

_∩∩

C C )) d) d) ((AA

₋

BB))

_∪∪

((BB

₋

C C )) Solution:

Solution: WWe apply the deﬁnitions oe apply the deﬁnitions of set operations:f set operations: a)

a) C C

_∩

((AA

_∪∪

BB) =) = C C

_{∩ {{}

_∩

a,b,c,d,ea,b,c,d,e

_}}

= =

_{

c,c, dd

_}}

b)

b) ((AA

_∪∪

C C ))

₋

BB = =

_{{

a,b,c,d,f a,b,c,d,f

_{}} −}

₋

BB = =

_{{

a,d,f a,d,f

_}}

c)

c) ((AA

_∪∪

BB

_∪

C C ))

₋

((AA

_∩∩

BB

_∩∩

C C ) =) =

_{{

a,b,c,d,e,f a,b,c,d,e,f

_{}} −}

_{− ∅∅}

= =

_{{

a,b,c,d,e,f a,b,c,d,e,f

_}}

d)

d) ((AA

₋

BB))

_∪∪

((BB

₋

C C ) =) =

_{{

a,a, dd

_{}} ∪∪ {{}

b,b, ee

_}}

= =

_{

a,b,d,ea,b,d,e

_}}

Exercise:

Question: As subsets of the reals, describe the diﬀerencAs subsets of the reals, describe the diﬀerences betweees between the setsn the sets

_{

33,, 55

_}}

, , [3[3,, 5] and (35] and (3,, 5).5). Solution:

Solution: The The setset

_{

33,, 55

_}}

con contains the integtains the integers 3 and ers 3 and 5. 5. The closed inteThe closed intervrval [3al [3,, 5] contains all real numbers5] contains all real numbers between 3 and 5 including 3 and 5, while the open interval (3

between 3 and 5 including 3 and 5, while the open interval (3,, 5) contains all real numbers between 3 and 5 not5) contains all real numbers between 3 and 5 not including 3 and 5.

including 3 and 5. Exercise:

Question: ProProve that the follove that the following are true for all setswing are true for all sets A A and and B B .. a)

a) AA

_∩∩

BB

_⊆

A A.. b)

b) AA

_⊆

A A

_∪∪

BB.. Solution:

Solution: WWe use e use the deﬁnitthe deﬁnitions of subsets and the interseions of subsets and the intersection and union of sets.ction and union of sets. a) Let

a) Let x x

_∈

A A

_∩∩

B.B. Then Then x x

_∈

A A and and x x

_∈

B B = =

_⇒

x x

_∈

A, , so A so A A

_∩∩

BB

_⊆

A A.. b) Let

b) Let x x

_∈

A A. We know that. We know that A A

_∪∪

BB = =

_{{

yy

_| |

y y

_∈

A or y A or y

_∈

B B

_}}

, , soso x x

_∈

A A = =

_⇒

x x

_∈∈

A A

_∪∪

BB. Hence. Hence A A

_⊆

A A

_∪∪

BB.. Exercise:

Question: LetLet A A and and B B be subsets of a set be subsets of a set S S .. a)

a) ProProve thave thatt A A

_⊆

B B if and only if if and only if PP_((A_A))

_⊆

P P_((B_B))

b)

b) ProProve thave thatt PP_((A_A

_∩∩

_B_{B) =}_{) =} P P_((A_A))

_∩∩

PP_((B_B)._).

c)

c) ShoShow thatw thatPP_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_A))

_∪∪

PP_((B_{B) if and only if}_{) if and only if A}_A

_⊆

_B_{B or}_or B_B

_⊆

_A_A..

Solution: Solution:

(2)

22 CHACHAPTER PTER 1. 1. SET SET THEOTHEORRY Y

a) (=

_⇒

)):: Suppose Suppose AA

_⊆

B B . Then,. Then,

_∀

aa

_∈

A A,, aa

_∈

B B. . SiSincncee PP_((B_{B) contains all the possible subsets of}_{) contains all the possible subsets of B}_{B, all the}_{, all the}

possible subset

possible subsets s of of A A must be in must be in PP_((B_{B) because}_{) because A}_A

_⊆

_B_{B. Therefore,}_{. Therefore,} PP_((A_A))

_⊆

P P_((B_B)._).

((

_⇐

=)=):: Suppose Suppose PP_((A_A))

_⊆

PP_((B_B)._{). Th}_Then_en

_∀{

_a_a

_{} ∈}

PP_((A_A),_),

_{

_a_a

_{} ∈}

PP_((B_B)._{). There}_{Therefore, ther}_{fore, there must exist a subset}_{e must exist a subset C}_C

of

of PP_((B_{B) that contains every}_{) that contains every}

_{

_a_a

_}}

_from_from PP_((A_A)._{). The su}_{The subse}_bsett C_{C leads to the conclusion that every}_{leads to the conclusion that every a}_a

_∈

_A_A

must also be in

must also be in B B . Therefore,. Therefore, A A

_⊆

B B.. b) By deﬁnition,

b) By deﬁnition, PP_((A_A

_∩∩

_B_B)_{) =}₌

_{{

_tt₁₁_{, , tt}₂₂_,...,t_,...,t_n_n

_{} |}

_tt_ii

_∈

_A,_{A, tt}_ii

_∈

_B_B

_}}

_{. . Thi}_{This impl}_{s implies}_ies

_{

_tt_ii

_{} ∈}

PP_((A_{A) and}_{) and}

_{

_tt_ii

_{} ∈}

PP_((B_B)._).

Therefore, by deﬁnition of intersection,

Therefore, by deﬁnition of intersection, PP_((A_A

_∩∩

_B_{B) =}_{) =} P P_((A_A))

_∩∩

PP_((B_B)._).

c) (=

_⇒

)):: Suppose there are two sets Suppose there are two sets AA andand BB such that neither such that neither AA

_⊆

BB nornor BB

_⊆

AA. . LLeett aa

_∈

AA

₋

B B andand bb

_∈

BB

₋

A A. . ThThen ten the she setet

_{

a,a, bb

_}}

is in is in PP_((A_A

_∪

_B_{B) but not in}_{) but not in} PP_((A_{A) or in}_{) or in} PP_((B_B)._{). The}_Theref_{refore b}_{ore by the}_{y the}

contrapositive,

contrapositive,PP_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_A))

_∪∪

PP_((B_B)_{) if}_if A_A

_⊆

_B_{B or}_or B_B

_⊆

_A_A..

((

_⇐

=)=):: Suppose Suppose A A

_⊆

B B. Then,. Then, A A

_∪∪

BB = = B B soso PP_((A_A

_∪∪

_{B) =}_B_{) =} P P_((B_{B). Now suppose}_{). Now suppose B}_B

_⊆

_A_{A. Then}_{. Then A}_A

_∪∪

_B_B =_= A_A so_so P

P_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_{A). Either way,}_{). Either way,} PP_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_A))

_∪∪

PP_((B_B)._).

Exercise:

Question: Give tGive the list desche list descriptioription of n of PP₍₍

_{{

_{11,, 22,, 33,, 44}

_}}

_)._).

Solution:

Solution: Using the deﬁUsing the deﬁnition of a pownition of a power set,er set,

P

P₍₍

_{{

_{11,, 22,, 33,, 44}

_}}

_{) =}_{) =}

_{∅

_,,

_{{

₁₁

_}}

_,,

_{{

₂₂

_}}

_,,

_{{

₃₃

_}}

_,,

_{{

₄₄

_}}

_,,

_{{

_{11,, 22}

_}}

_,,

_{{

_{11,, 33}

_}}

_,,

_{{

_{11,, 44}

_}}

_,,

_{{

_{22,, 33}

_}}

_,,

_{{

_{22,, 44}

_}}

_,,

_{{

_{33,, 44}

_}}

_,,

{{

11,, 22,, 33

_}}

,,

_{{

11,, 22,, 44

_}}

,,

_{{

11,, 33,, 44

_}}

,,

_{{

22,, 33,, 44

_}}

,,

_{{

11,, 22,, 33,, 44

_}}

..

Exercise:

Question: Give tGive the list desche list descriptioription of n of

_{{

aa11,, aa22, . . . , a, . . . , akk

}

} ∈∈

P P((

{{

11,, 22,, 33,, 44,, 55

}}

))





aa11 + + a a22 + +

·· ·· ··

++ a akk = = 88

}}

..

Solution:

Solution: WWe need to ﬁnd all the sube need to ﬁnd all the subsets of sets of

_{

11,, 22,, 33,, 44,, 55

_}}

whos whose elemene elements add to a ts add to a totatotal of l of 8. 8. Recall that noRecall that no subset has repeated elements so

subset has repeated elements so

_{

44,, 44,,

_}}

does not make sense. The set is does not make sense. The set is

{{

11,, 22,, 55

_}}

,,

_{{

11,, 33,, 44

_}}

,,

_{{

33,, 55

_}}

..

Exercise:

Question: LetLet A A,, B B , and, and C C be subsets of a set be subsets of a set S S .. a) Prove that (

a) Prove that (AA

₋

BB))

₋

C C = = ((AA

₋

C C ))

₋

((BB

₋

C C ).). b)

b) Find and provFind and prove a e a similasimilar formur formula forla for A A

₋

((BB

₋

C C ).). Solution: Solution: a) a) S S A A B B C C S S A A B B C C

In the ﬁrst Venn diagram, the lighter shade represents (

In the ﬁrst Venn diagram, the lighter shade represents ( AA

₋

BB), and the darker shade, which overlaps some), and the darker shade, which overlaps some of

of ((AA

₋

BB), represents (), represents (AA

₋

BB))

₋

C C . . In the second VeIn the second Venn diagramnn diagram, , the lightthe lighter shade represeer shade represents (nts (AA

₋

C C ),), while the darker shade represents (

while the darker shade represents (AA

₋

C C ))

₋

((BB

₋

C C ). We observe from the diagrams that the darker regions). We observe from the diagrams that the darker regions are equal.

(3)

a) (=

_⇒

)):: Suppose Suppose AA

_⊆

B B . Then,. Then,

_∀

aa

_∈

A A,, aa

_∈

B B. . SiSincncee PP_((B_{B) contains all the possible subsets of}_{) contains all the possible subsets of B}_{B, all the}_{, all the}

possible subset

possible subsets s of of A A must be in must be in PP_((B_{B) because}_{) because A}_A

_⊆

_B_{B. Therefore,}_{. Therefore,} PP_((A_A))

_⊆

P P_((B_B)._).

((

_⇐

=)=):: Suppose Suppose PP_((A_A))

_⊆

PP_((B_B)._{). Th}_Then_en

_∀{

_a_a

_{} ∈}

PP_((A_A),_),

_{

_a_a

_{} ∈}

PP_((B_B)._{). There}_{Therefore, ther}_{fore, there must exist a subset}_{e must exist a subset C}_C

of

of PP_((B_{B) that contains every}_{) that contains every}

_{

_a_a

_}}

_from_from PP_((A_A)._{). The su}_{The subse}_bsett C_{C leads to the conclusion that every}_{leads to the conclusion that every a}_a

_∈

_A_A

must also be in

must also be in B B . Therefore,. Therefore, A A

_⊆

B B.. b) By deﬁnition,

b) By deﬁnition, PP_((A_A

_∩∩

_B_B)_{) =}₌

_{{

_tt₁₁_{, , tt}₂₂_,...,t_,...,t_n_n

_{} |}

_tt_ii

_∈

_A,_{A, tt}_ii

_∈

_B_B

_}}

_{. . Thi}_{This impl}_{s implies}_ies

_{

_tt_ii

_{} ∈}

PP_((A_{A) and}_{) and}

_{

_tt_ii

_{} ∈}

PP_((B_B)._).

Therefore, by deﬁnition of intersection,

Therefore, by deﬁnition of intersection, PP_((A_A

_∩∩

_B_{B) =}_{) =} P P_((A_A))

_∩∩

PP_((B_B)._).

c) (=

_⇒

)):: Suppose there are two sets Suppose there are two sets AA andand BB such that neither such that neither AA

_⊆

BB nornor BB

_⊆

AA. . LLeett aa

_∈

AA

₋

B B andand bb

_∈

BB

₋

A A. . ThThen ten the she setet

_{

a,a, bb

_}}

is in is in PP_((A_A

_∪

_B_{B) but not in}_{) but not in} PP_((A_{A) or in}_{) or in} PP_((B_B)._{). The}_Theref_{refore b}_{ore by the}_{y the}

contrapositive,

contrapositive,PP_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_A))

_∪∪

PP_((B_B)_{) if}_if A_A

_⊆

_B_{B or}_or B_B

_⊆

_A_A..

((

_⇐

=)=):: Suppose Suppose A A

_⊆

B B. Then,. Then, A A

_∪∪

BB = = B B soso PP_((A_A

_∪∪

_{B) =}_B_{) =} P P_((B_{B). Now suppose}_{). Now suppose B}_B

_⊆

_A_{A. Then}_{. Then A}_A

_∪∪

_B_B =_= A_A so_so P

P_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_{A). Either way,}_{). Either way,} PP_((A_A

_∪∪

_B_{B) =}_{) =} P P_((A_A))

_∪∪

PP_((B_B)._).

Exercise:

Question: Give tGive the list desche list descriptioription of n of PP₍₍

_{{

_{11,, 22,, 33,, 44}

_}}

_)._).

Solution:

Solution: Using the deﬁUsing the deﬁnition of a pownition of a power set,er set,

P

P₍₍

_{{

_{11,, 22,, 33,, 44}

_}}

_{) =}_{) =}

_{∅

_,,

_{{

₁₁

_}}

_,,

_{{

₂₂

_}}

_,,

_{{

₃₃

_}}

_,,

_{{

₄₄

_}}

_,,

_{{

_{11,, 22}

_}}

_,,

_{{

_{11,, 33}

_}}

_,,

_{{

_{11,, 44}

_}}

_,,

_{{

_{22,, 33}

_}}

_,,

_{{

_{22,, 44}

_}}

_,,

_{{

_{33,, 44}

_}}

_,,

{{

11,, 22,, 33

_}}

,,

_{{

11,, 22,, 44

_}}

,,

_{{

11,, 33,, 44

_}}

,,

_{{

22,, 33,, 44

_}}

,,

_{{

11,, 22,, 33,, 44

_}}

..

Exercise:

Question: Give tGive the list desche list descriptioription of n of

_{{

aa11,, aa22, . . . , a, . . . , akk

}

} ∈∈

P P((

{{

11,, 22,, 33,, 44,, 55

}}

))





aa11 + + a a22 + +

·· ·· ··

++ a akk = = 88

}}

..

Solution:

Solution: WWe need to ﬁnd all the sube need to ﬁnd all the subsets of sets of

_{

11,, 22,, 33,, 44,, 55

_}}

whos whose elemene elements add to a ts add to a totatotal of l of 8. 8. Recall that noRecall that no subset has repeated elements so

subset has repeated elements so

_{

44,, 44,,

_}}

does not make sense. The set is does not make sense. The set is

{{

11,, 22,, 55

_}}

,,

_{{

11,, 33,, 44

_}}

,,

_{{

33,, 55

_}}

..

Exercise:

Question: LetLet A A,, B B , and, and C C be subsets of a set be subsets of a set S S .. a) Prove that (

a) Prove that (AA

₋

BB))

₋

C C = = ((AA

₋

C C ))

₋

((BB

₋

C C ).). b)

b) Find and provFind and prove a e a similasimilar formur formula forla for A A

₋

((BB

₋

C C ).). Solution: Solution: a) a) S S A A B B C C S S A A B B C C

In the ﬁrst Venn diagram, the lighter shade represents (

In the ﬁrst Venn diagram, the lighter shade represents ( AA

₋

BB), and the darker shade, which overlaps some), and the darker shade, which overlaps some of

of ((AA

₋

BB), represents (), represents (AA

₋

BB))

₋

C C . . In the second VeIn the second Venn diagramnn diagram, , the lightthe lighter shade represeer shade represents (nts (AA

₋

C C ),), while the darker shade represents (

while the darker shade represents (AA

₋

C C ))

₋

((BB

₋

C C ). We observe from the diagrams that the darker regions). We observe from the diagrams that the darker regions are equal.

(4)

1.1.

1.1. SETS SETS AND AND FUNCFUNCTIONTIONS S 33

b) b) S S A A B B C C S S A A B B C C

In the Venn diagram above, the lighter shade represents

In the Venn diagram above, the lighter shade represents B B

₋

C C , and the darker shade represents, and the darker shade represents A A

₋

((BB

₋

C C ).). In the second diagram, the lighter region represents

In the second diagram, the lighter region represents A A

₋

BB, and the darker region represents, and the darker region represents A A

₋

C C , which, which overlaps some of

overlaps some of A A

₋

BB. Thus, (. Thus, (AA

₋

BB))

_∪∪

((AA

₋

C C ) =) = A A

₋

((BB

₋

C C ).). Exercise:

Question: LetLet A A,, B B , and, and C C be subsets of a set be subsets of a set S S .. a)

a) ProProve thave thatt A A

_

BB = =

_∅∅

if and only if if and only if A A = = B B.. b)

b) ProProve thave thatt A A

_∩∩

((BB

_

C C ) = ) = ((AA

_∩∩

BB))

_

((AA

_∩∩

C C ).). Solution:

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

a) Suppose Suppose thatthat A A

_

BB = =

_∅∅

. Then by definition of the symmetric difference. Then by definition of the symmetric difference ((AA

₋

BB))

_∪∪

((BB

₋

AA) =) =

_∅∅

..

If the union of two sets is the empty set, then each of the two sets must be empty. Hence we deduce that If the union of two sets is the empty set, then each of the two sets must be empty. Hence we deduce that A

A

₋

BB = =

_∅

and and B B

₋

AA = =

_∅∅

. Now for and two sets U . Now for and two sets U andand T T , the identity, the identity U U

₋

T =T =

_∅∅

is equivalent to is equivalent to U U

_⊆

T T .. Hence we deduce that

Hence we deduce that A A

_⊆

B B andand B B

_⊆

A A. Consequently,. Consequently, A A = = B B.. The argument of the opposite direction is identical. Suppose that

The argument of the opposite direction is identical. Suppose that A A = = B B. Then. Then A A

_⊆

B B andand B B

_⊆

A A. Thus. Thus A

A

₋

BB = =

_∅

and and B B

₋

AA = =

_∅

. We deduce that. We deduce that A A

_

BB = = ((AA

₋

BB))

_∪∪

((BB

₋

AA) =) =

_∅∅

.. b)

b) There are a There are a vvarietariety y of ways to of ways to proprove the identitve the identityy AA

_∩∩

((BB

_

C C ) ) = = ((AA

_∩∩

BB))

_

((AA

_∩∩

C C ). ). WWe could use could use a e a wewellll desig

designed Vened Venn diagram. nn diagram. WWe could e could also use a also use a memmembership table which lists all pbership table which lists all p ossibilossibilities of an ities of an elemeelementnt whether it is in or not in one of the given three sets. Here is a membership table for both side of the equality. 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 this table, we put an

√

in a column to indicate membership and nothing to indicate non-membership. in a column to indicate membership and nothing to indicate non-membership. Hence if there is a

Hence if there is a

√

in the in the A A and and C C column and nothing in the column and nothing in the B B column, that refers to the situations of column, that refers to the situations of an element in

an element in A A, not in, not in B B and in and in C C .. A

A B B C C ((BB

_

C ))C AA

_∩∩

((BB

_

C ) C ) ((AA

_∩∩

BB) ) ((AA

_∩∩

C C ) ) ((AA

_∩∩

BB))

_

((AA

_∩∩

C C ))

√

√ √

√

Since the

Since the A A

_∩∩

((BB

_

C ) and column and the (C ) and column and the (AA

_∩∩

BB))

_

((AA

_∩∩

C C ) of this membership table are the same, then) of this membership table are the same, then the sets are equal.

the sets are equal. Exercise:

Question: LetLet S S be a set and let be a set and let

_{

AAii

}}

ii

_∈I

be a collection of subsets of be a collection of subsets of S S . Prove the following.. Prove the following.

a) a)



ii

_∈I

A Aii = =



ii

_∈I

A Aii.. b) b)



ii

_∈I

A Aii = =



ii

_∈I

A Aii.. Solution:

(5)

a)

a) WWe will e will proprove the equation by provinve the equation by proving g set inclusioset inclusion n in both in both direcdirectionstions.. (= (=

_⇒

) Let) Let aa

_∈



ii

_∈I

A Aii. . ThThenen a a //

∈∈



ii

_∈I

A

Aii oror a a //

∈∈

A Aii for every for every i i

∈ I

. . And thiAnd this implis implies thaes thatt aa

∈

A Aii for every for every

ii

_∈

_{∈ I}

_I

. So then. So then a a

_∈



ii

_∈I

A Aii. And so. And so



ii

_∈I

A Aii

⊆



ii

_∈I

A Aii.. ((

_⇐

=) Let=) Let aa

_∈



ii

_∈I

A

Aii. . ThThenen aa

∈

A Aii for every for every ii

∈ I

. . WhicWhich implieh implies thats that a a //

∈∈

A Aii for eversy for eversy ii

∈ I

. . AnAnd sod so

a a //

_∈∈



ii

_∈I

A

Aii. Which implies. Which implies a a

∈∈



ii

_∈I

A Aii. . SoSo



ii

_∈I

A Aii

⊆



ii

_∈I

A Aii.. And so And so



ii

_∈I

A Aii = =



ii

_∈I

A Aii.. b)

b) WWe will e will proprove this equation by provinve this equation by proving set g set inclusinclusion in ion in both directioboth directions.ns. (=

(=

_⇒

) Let) Let a a

_∈



ii

_∈I

A

Aii. . SoSo a a //

∈∈



ii

_∈I

A

Aii. . SoSo a a //

∈∈

A Aii for at least one i for at least one i

∈

∈ I

I

. . SoSo a a

∈

A Aii for at least one for at least one i i

∈

∈ I

I

. . SoSo

a a

_∈∈



ii

_∈I

A

Aii. Which implies that. Which implies that



ii

_∈I

A Aii

⊆



ii

_∈I

A Aii..

((

_⇐

=) Let=) Let a a

_∈



ii

_∈I

A

Aii. Then. Then a a

∈

A Aii for at least one for at least one i i

∈

∈ I

I

. . This impThis implielies thats that a a //

∈∈

A Aii for at least one for at least one ii

∈

∈ I

I

..

It follows that

It follows that a a //

_∈∈



ii

_∈I

A

Aii. Which implies that. Which implies that a a

∈



ii

_∈I

A Aii. . SoSo



ii

_∈I

A Aii

⊆



ii

_∈I

A Aii.. So So



ii

_∈I

A Aii = =



ii

_∈I

A Aii.. Exercise:

Question: LetLet P P be the parabola in the plane whose equation is be the parabola in the plane whose equation is yy == xx22. . LeLett

_{

AAqq

}}

qq

_∈

P P be the collection of be the collection of

subsets of

subsets of R R22 wherewhere A Aqq is the tangent line to is the tangent line to P P atat q q . Determine with proof . Determine with proof



_q_q

_∈

_P_P A Aqq..

Solution:

Solution: SkeSketchtching a ing a pictupicture of the re of the standstandard parabolaard parabola, , we see that all the we see that all the tangtangent lines are in some senseent lines are in some sense beneath the parabola. Also, taking the (inﬁnite, uncountable) union of all the tangent lines to the parabola, we 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 (

might guess that we would get all points ( a,a, bb) in the plane such that) in the plane such that b b

_≤

a a22_{. We need to prove this hypothesis.}_{. We need to prove this hypothesis.}

Label the coordinates of a point

Label the coordinates of a point q q

_∈

P P asas q q = = ((xx00,, xx0202). ). FFrom calcurom calculus, the tangenlus, the tangent line tot line to P P atat q q has the has the

equation equation

yy = = x x22₀₀ + 2 + 2xx00((xx

−

xx00) =) =

⇒

y y = = 22xx00xx

−

xx2200..

Now suppose that some point (

Now suppose that some point (a,a, bb) in the plane is on a tangent line. Then, for some) in the plane is on a tangent line. Then, for some x x00, we have, we have b b = = 22xx00aa

−

xx2200..

Since

Since a, a, bb are given and are given and x x00 is unknown, this is an equation in is unknown, this is an equation in x x00. The quadratic formula gives for. The quadratic formula gives for x x00

x x00 = = 22aa

±

√

_44a_a22

₋

_44bb 22 == a a

±



aa 2 2

−

b.b. In particular, we note that there exists an

In particular, we note that there exists an x x00 if and only if if and only if a a22

−

bb

≥

0, conﬁrming the hypothesis that 0, conﬁrming the hypothesis that b b

≤

a a22..

Exercise:

Question: LetLet

_{

AAkk

}}

kk

_∈

NN be the collection of subsets of be the collection of subsets of R R33 such thatsuch that A A_k_k = =

{

((x,y,zx,y,z))

∈∈

RR33

||

zz

≥

ky ky

}}

. Determine. Determine

both



_k_k

_∈

NN A Akk andand



_k_k

_∈

NN A Akk..

Solution:

Solution: Each Each subsetsubset A Akk represents all points greater than the plane that is bound on the line represents all points greater than the plane that is bound on the line z z = = ky ky . In the. In the

diagram below, imagine the x-axis is coming out of the page, and let

diagram below, imagine the x-axis is coming out of the page, and let k k11,, k k22, and, and k k33 represent the bounds of the represent the bounds of the

subsets

subsets A A11,, A A22, and, and A A33 respectively. We note that the larger respectively. We note that the larger k k gets, the more steep the plane becomes. However, gets, the more steep the plane becomes. However,

the condition does not hold true for

the condition does not hold true for z z << 0 when 0 when y y is 0. Therefore, is 0. Therefore,



_k_k

_∈

NN A Akk = =

{{

((x,y,zx,y,z))

||

zz

≥

y y

}∪{

((x,y,zx,y,z))

||

x x << 0 0

}}

,,

and and



_k_k

∈

(6)

1.1. SETS AND FUNCTIONS 5 z y k1 k2 k3 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:

(7)

a) Observe the Venn diagrams below: U U U A B U U U 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 U U A B U U U 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 A

_∩

B. Notice, the shaded region describes the same set that is shaded in the A

_∪

B diagram, hence A

_∪

B = A

_∩

B.

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

_}

.

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) f (x) =

₋

3x + 4 b) f (x) =

₋

3x2_{+ 7}

c) f (x) = (x + 1)/(x + 2) d) f (x) = x5_{+ 1}

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

a) Let f (x) =

₋

3x + 4. This function can be deﬁned over the whole domain R. Suppose that f (x1) = f (x2).

Then

₋

3x1 + 4 =

−

3x2 + 4. This implies that

−

3x1 =

−

3x2 so x1 = x2. This shows that f (x) is injective.

To prove surjectivity, we attempt to solve for x in the expression y = f (x) for an arbitrary y. We get x = 4

−

₃y . 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.)

(8)

1.1. SETS AND FUNCTIONS 7 suppose that f (x1) = f (x2). Then we have

x1 + 1 x1 + 2 = x1 + 1 x1 + 2 =

_⇒

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

_⇒

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

_⇒

2x1 + x2 = x1 + 2x2 =

_⇒

x1 = x2.

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

y = x + 1

x + 2 =

⇒

yx + 2y = x + 1 =

⇒

xy

−

x = 1

−

2y =

⇒

x =

1

₋

2y y

₋

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

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

x5₁ + 1 = x5₂ + 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 x1 = x2, which we already know to be a

possibility. Assuming that x1



= x2, after division by x42 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 = x2. 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

_→

Z that is a) bijective;

b) surjective but not injective; c) injective but not surjective; d) neither injective nor surjective.

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_.

Question: Given an explicit example of a function f :N

_→

N that is a) bijective;

b) surjective but not injective; c) injective but not surjective; d) neither injective nor surjective.

Solution: Recall that

_

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

_

.

(9)

d) f (x) = 2

_∗

(

_

x/2

_

). Exercise: 20 Section 1.1

Question: Let f : A

_→

B and g : B

_→

C be functions. Prove that if f and g are bijective, then g

_◦

f is bijective and

(g

_◦

f )

−

1 = f

−

1

_◦

g

−

1.

Solution: Assume g

_◦

f is not injective. Then there would exist a1, a2

∈

A such that a1



= a2 and g

◦

f (a1) =

g

_◦

f (a2). Let b1 = f (a1) and b2 = f (a2). We know that b1



= b2 because f is bijective. Using substitution, we

notice that g(b1) = g(b2), but b1



= b2. This creates a contradiction since g is bijective. Thus, g

◦

f is injective.

Now assume that g

_◦

f is not surjective. Then there would exist at least one c

_∈

C such that

_∀

a

_∈

A, g(f (a))

_

= 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_.

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) = x2.

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))

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.

Question: Restate the deﬁnition of (a) injective and (b) surjective as applied to a function f : A

_→

B in terms of properties of the sets f

−

1₍

{

b

_}

). Solution:

(10)

1.1. SETS AND FUNCTIONS 9 a) For injectivity,

_|

f

−

1₍

{

b

_}

)

_{| ≤}

1. b) For surjectivity,

_|

f

−

1

{

b

_}

)

_{| ≥}

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) = x3

−

2x and T = [

₋

1, 0].

Solution: Note that the pre-image of f

−

1_{(T ) represents all of the domain values that map to the elements of}

T . a) f

−

1(T ) = sin

−

1(

√

₂3) =

_{

π₃,2π₃

_}

b) f

−

1_{(T ) = g}

₋

1_{([1, 2])} x2

₋

2 = 1

_⇒

x =

_±

√

3 x2

₋

2 = 2

_⇒

x =

_±

2 x f (x)

−

2

₋

1 1 2 1 2

−

1

−

2 x2

−

2

Therefore, the pre-image of g

−

1([1, 2]) is

_{

[

₋

2,

₋

√

3], [

√

3, 2]

_}

. c) f

−

1_{(T ) = h}

₋

1_([

−

1, 0]) Notice that 1 is a zero of the polynomial x3

−

2x + 1. This allows us to ﬁnd the x-values that satisfy x3

−

2x =

₋

1. x3

₋

2x + 1 = (x

₋

1)(x2 + x

₋

1) x = 1,

−

1

±

√

₅ 2 x3

₋

2x = 0

_⇒

x = 0,

_±

√

2 x f (x)

−

2

₋

1 1 2 1 2

−

1

−

2 x3

₋

2x

Therefore, the pre-image of h

−

1_([

−

1, 0]) is

_{

[

−

1

₋

√

5

2 ,

−

√

2], [0,

−

1+

√

5

(11)

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.}

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 )