Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

CHAP. 1] SET THEORY 13

(c) C={x∈N| 4 +x= 3 }

(a) Aconsists of the positive integers between 3 and 9; henceA={ 4 , 5 , 6 , 7 , 8 }.

(b) Bconsists of the even positive integers less than 11; henceB={ 2 , 4 , 6 , 8 , 10 }.

(c) No positive integer satisfies 4+x=3; henceC=∅, the empty set.

1.3 LetA={ 2 , 3 , 4 , 5 }.

(a) Show thatAis not a subset ofB={x∈N|xis even}.

(b) Show thatAis a proper subset ofC={ 1 , 2 , 3 ,..., 8 , 9 }.

(a) It is necessary to show that at least one element inAdoes not belong toB. Now 3∈Aand, sinceBconsists

of even numbers, 3∈/B; henceAis not a subset ofB.

(b) Each element ofAbelongs toCsoA⊆C. On the other hand, 1∈Cbut 1∈/A. HenceA=C. ThereforeA

is a proper subset ofC.

SET OPERATIONS

1.4 LetU={1,2, ..., 9} be the universal set, and let

A={ 1 , 2 , 3 , 4 , 5 },C={ 5 , 6 , 7 , 8 , 9 },E={ 2 , 4 , 6 , 8 },

B={ 4 , 5 , 6 , 7 },D={ 1 , 3 , 5 , 7 , 9 },F={ 1 , 5 , 9 }.

Find: (a)A∪BandA∩B;(b)A∪CandA∩C;(c)D∪FandD∩F.

Recall that the unionX∪Yconsists of those elements in eitherXorY(or both), and that the intersectionX∩Yconsists

of those elements in bothXandY.

(a) A∪B={ 1 , 2 , 3 , 4 , 5 , 6 , 7 }andA∩B={ 4 , 5 }

(b) A∪C={ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }=UandA∩C={ 5 }

(c) D∪F={ 1 , 3 , 5 , 7 , 9 }=DandD∩F=( 1 , 5 , 9 )=F

ObservethatF⊆D,so by Theorem 1.4 we must haveD∪F=DandD∩F=F.

1.5 Consider the sets in the preceding Problem 1.4. Find:

(a) AC,BC,DC,EC;(b)A\B, B\A, D\E; (c)A⊕B, C⊕D, E⊕F.

Recall that:

(1) The complementsXCconsists of those elements inUwhich do not belong toX.

(2) The differenceX\Yconsists of the elements inXwhich do not belong toY.

(3) The symmetric differenceX⊕Yconsists of the elements inXor inYbut not in both.

Therefore:

(a) AC={6, 7, 8, 9}; BC={ 1 , 2 , 3 , 8 , 9 }; DC={ 2 , 4 , 6 , 8 }=E; EC={ 1 , 3 , 5 , 7 , 9 }=D.

(b) A\B={ 1 , 2 , 3 }; B\A={ 6 , 7 }; D\E={ 1 , 3 , 5 , 7 , 9 }=D; F\D=∅.

(c) A⊕B={ 1 , 2 , 3 , 6 , 7 }; C⊕D={ 1 , 3 , 6 , 8 }; E⊕F={ 2 , 4 , 6 , 8 , 1 , 5 , 9 }=E∪F.

1.6 Show that we can have: (a) A∩B=A∩CwithoutB=C;(b)A∪B=A∪CwithoutB=C.

(a) LetA={1, 2},B={2, 3},C={2, 4}. ThenA∩B={ 2 }andA∩C={ 2 }; butB=C.

(b) LetA={1, 2},B={1, 3},C={2, 3}. ThenA∪B={ 1 , 2 , 3 }andA∪C={ 1 , 2 , 3 }butB=C.

1.7 Prove:B\A=B∩AC. Thus, the set operation of difference can be written in terms of the operations of

intersection and complement.

B\A={x|x∈B, x /∈A}={x|x∈B, x∈AC}=B∩AC.