set operations. when sets are equal a equals b iff for all x, x is in a iff x is in b or … and...

Set Operations

When sets are equal

][ BxAxxBA

A equals B iff for all x, x is in A iff x is in B

)]()[( AxBxBxAxxBA

or

ABBABA

… and this is what we do to prove sets equal

Note: remember all that stuff about implication?

tivecontraposi

inverse

converse

nalbiconditio

qppq

qp

pq

pqqpqp

}|{ BxAxxBA

Union of two sets

Give me the set of elements, x where x is in A or x is in B

}5,4,3,2,1{A }8,7,6,5,4{B

}8,7,6,5,4,3,2,1{BA

Example

BABA0 0 00 1 11 0 11 1 1 OR

A membership table

Union of two sets

}|{ BxAxxBA

Note: we are using set builder notation and the

laws of logical equivalence (propositional equivalence)

}|{ BxAxxBA

Intersection of two sets

Give me the set of elements, x where x is in A and x is in B

}5,4,3,2,1{A }8,7,6,5,4{B

}5,4{ BA

Example

BABA0 0 00 1 01 0 01 1 1

AND

}|{ BxAxxBA

),(disjoint{} BABA

Disjoint sets

][ BxAxx

}|{ BxAxxBA

Difference of two sets

Give me the set of elements, x where x is in A and x is not in B

}5,4,3,2,1{A }8,7,6,5,4{B

}3,2,1{ BA

Example

BA BA0 0 00 1 01 0 11 1 0 BA

BABA

}|{ BxAxxBA

Note: Compliment of a set

}|{ AxUxxA

)()( ABBABA

Symmetric Difference of two sets

Give me the set of elements, x where x is in A and x is not in B OR x is in B and x is not in A

}5,4,3,2,1{A }8,7,6,5,4{B

}8,7,6,3,2,1{BA

Example

BABA0 0 00 1 11 0 11 1 0

XOR

)()( ABBABA

}|{ AxxA

Complement of a set

Give me the set of elements, x where x is not in A

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

}10,9,8,7,6,5,4,3,2,1,0{U

}10,9,8,7,6,0{A

Example

Not

U is the “universal set”

AUA

Cardinality of a Set

In claire

• A = {1,3,5,7}• B = {2,4,6}• C = {5,6,7,8}

• |A u B| ?• |A u C|• |B u C|• |A u B u C|

Cardinality of a Set

|BA| |B| |A| || BA

The principle of inclusion-exclusion

Cardinality of a Set

|BA| |B| |A| || BA

The principle of inclusion-exclusion

U

Potentially counted twice (“over counted”)

Set Identities

{} AA

AUA Identity

UUA {}{} A

Domination

•Think of• U as true (universal)• {} as false (empty)• Union as OR• Intersection as AND• Complement as NOT

AAA AAA

Indempotent

Note similarity to logical equivalences!

Set Identities

ABBA ABBA

Commutative

CBACBA )()(CBACBA )()(

Associative

)()()( CABACBA )()()( CABACBA

Distributive

BABA

BABA De Morgan

Note similarity to logical equivalences!

lawsnegation

laws absorption)(

)(

laws sMorgan' De)(

)(

law vedistributi)()()(

)()()(

laws eassociativ)()(

)()(

laws ecommutativ

lawnegation double)(

lawst indempoten

law domination

lawidentity

NameeEquivalenc

Fpp

Tpppqpp

pqppqpqp

qpqprpqprqp

rpqprqprqprqp

rqprqppqqp

pqqp

pp

ppp

pppFFp

TTppFp

pTp

Four ways to prove two sets A and B equal

• a membership table• a containment proof

• show that A is a subset of B• show that B is a subset of A

• set builder notation and logical equivalences• Venn diagrams

B BA :RTP A

Prove lhs is a subset of rhs

Prove rhs is a subset of lhs

B BA :RTP A

… set builder notation and logical equivalences

}|{ BAxx

)}(|{ BAxx

)}(|{ BxAxx

)}()(|{ BxAxx

}|{ BxAxx }|{ BAxx

Defn of complement

Defn of intersection

De Morgan law

Defn of complement

Defn of union

B BA :RTP A

prove using membership table

Class

B BA :RTP AMe

0000111

1101001

1011010

1111000

BABABABABA

They are the same

B BA :RTP A

prove using set builder and logical equivalence

Class

B BA :RTP AMe

notationbuilder set of Meaning

ion)(intersect ofDefn }|{

complement ofDefn )}()(|{

ofDefn )}()(|{

lawMorgan De ))}()(|{

)( ofDefn }(|{

ofDefn )}(|{

}|{

BA

BAxx

BxAxx

BxAxx

BxAxx

unionBxAxx

BAxx

BAxxBA

{})( :RTP ABA

)}(|{)(

ABxAxxABA

)}(|{ AxBxAxx

}|{ BxAxAxx

}{}|{ Bxxx

{}

Prove using set builder and logical equivalences

A containment proof

See the text book

That’s a cop out if ever I saw one!

A containment proof Guilt kicks in

To do a containment proof of A = B do as follows

1. Argue that an arbitrary element in A is in B i.e. that A is an improper subset of B

2. Argue that an arbitrary element in B is in A i.e. that B is an improper subset of A

3. Conclude by saying that since A is a subset of B, and vice versa then the two sets must be equal

ni

n

i

AAAA

...211

Collections of sets

ni

n

i

AAAA

...211