Lecture 2.2: Set Theory*
CS 250, Discrete Structures, Fall 2011
Nitesh Saxena
*Adopted from previous lectures by Cinda Heeren
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HW1 Posted
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Due at 11am 09/09/11
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Outline
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Set Theory, Operations and Laws
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Set Theory - Operators
like
“exclusive
or”
The symmetric difference, A  B, is:
A  B = { x : (x  A  x  B) v (x  B  x  A)}
= (A - B) U (B - A)
U
B
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A
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Set Theory - Operators
A  B = { x : (x  A  x  B) v (x  B  x  A)}
= (A - B) U (B - A)
Proof: { x : (x  A  x  B) v (x  B  x  A)}
= { x : (x  A - B) v (x  B - A)}
= { x : x  ((A - B) U (B - A))}
= (A - B) U (B - A)
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Set Theory - Famous Laws
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Two pages of (almost) obvious.
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One page of HS algebra.
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One page of new.
Don’t
memorize
them,
understand
them!
They’re in
Rosen, p. 130
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Set Theory - Famous Laws
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Identity
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Domination A U U = U
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AU=A
AU=A
A=
AUA=A
Idempotent A  A = A
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Set Theory - Famous Laws
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Excluded Middle
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Uniqueness
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Double complement
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AUA=U
AA=
A=A
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Set Theory – Famous Laws
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Commutativity A U B = B U A
AB= BA
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Associativity
(A U B) U C = A U (B U C)
(A  B)  C = A  (B  C)
A U (B  C) = (A U B)  (A U C)
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Distributivity A  (B U C) = (A  B) U (A  C)
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Set Theory – Famous Laws
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DeMorgan’s I
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DeMorgan’s II (A  B) = A U B
(A U B) = A  B
p
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q
Venn Diagrams
are good for
intuition, but
we aim for a
more formal
proof.
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3 Ways to prove Laws or set equalities
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Show that A  B and that A  B.
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Use a membership table.
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New & important
Like truth tables
Use logical equivalences to prove
equivalent set definitions.
Not hard, a little tedious
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Example – the first way
Prove that (A U B) = A  B
1.
() (x  A U B)  (x  A U B) 
(x  A and x  B)  (x  A  B)
2. () (x  A  B)  (x  A and x  B) 
(x  A U B)  (x  A U B)
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Example – the second way
Prove that (A U B) = A  B using a
membership table.
0 : x is not in the specified set
1 : otherwise
A B A
B
AB
AUB
AUB
1
1
0
0
0
1
0
1
0
0
0
1
1
1
1
0
0
0
0
1
1
0
1
0
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0
0
1
1
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Example – the third way
Prove that (A U B) = A  B using
logically equivalent set definitions.
(A U B) = {x : (x  A v x  B)}
= {x : (x  A)  (x  B)}
= {x : (x  A)  (x  B)}
=AB
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Another example: applying the laws
X  (Y - Z) = (X  Y) - (X  Z). True or
False?
Prove your response.
(X  Y) - (X  Z) = (X  Y)  (X  Z)’
= (X  Y)  (X’ U Z’)
= (X  Y  X’) U (X  Y  Z’)
=
U (X  Y  Z’)
= (X  Y  Z’)
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A Proof (direct and indirect)
AB=
Pv that if (A - B) U (B - A) = (A U B) then
Suppose to the contrary, that A  B  , and that x  A  B.
Then x cannot be in A-B and x cannot be in B-A.
Then x is not in (A - B) U (B - A).
a)
b)
c)
d)
AUB=
A=B
AB=
A-B = B-A = 
But x is in A U B since (A  B)  (A U B).
Thus, A  B = .
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Today’s Reading
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Rosen 2.1 and 2.2
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