Lecture Notes, 29 Oct 2010

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§2.2: Set Operations (cont'd)

1. I didn't cover this in lecture, but it's useful information:

   |A ∪ B| ≠ |A| + |B|

   1. Why? Because possibly A ∩ B ≠ ∅!
   2. ∴ |A ∪ B| = |A| + |B| – |A ∩ B|

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2. Set Identities:
   1. Compare Table 1 (p. 124) to Table 6 (p. 24)
   2. Set identities are just like logical equivalences

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   3. Reminder:
      Let U = universe.
      Then ‾A =def U – A = {x | x ∈ U ∧ x ∉ A}
                                        = {x ∈ U | x ∉ A}
                                        = {x | x ∉ A}.

      "‾" is an operator on sets that corresponds to (or is defined in terms of) "¬", which is an operator on propositions

   4. Thm: (DeMorgan for sets)
      ‾(A ∪ B) = (‾A) ∩ (‾B)

      proof:

      ‾(A ∪ B)	= {x | x ∈ ‾(A ∪ B)}	: definition of set
      	= {x | x ∉ (A ∪ B)}	: definition of ‾
      	= {x | ¬[x ∈ (A ∪ B)]}	: definition of ∉
      	= {x | ¬[(x ∈ A) ∨ (x ∈ B)]}	: definition of ∪
      	= {x | (x ∉ A) ∧ (x ∉ B)}	: DeM for ∨ & def of ∉
      	= {x | (x ∈ ‾A) ∧ (x ∈ ‾B)}	: definition of ‾
      	= {x | x ∈ (‾A ∩ ‾B)}	: definition of ∩
      	= (‾A) ∩ (‾B)	: definition of set

      QED

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3. 1. (*) Lemma (based on def of ‾):
      Let S,T be sets.
      Then S – T = S ∩ ‾T

      proof:

      S – T = {x | (x ∈ S) ∧ (x ∉ T)}
                = {x | x ∈ S} ∩ {x | x ∉ T}
                = S ∩ ‾T
      QED

   2. Show (S – T) ∩ (T – S) = ∅:
      1. Proof #1:

         (S – T) ∩ (T – S)	= (S ∩ ‾T) ∩ (T ∩ ‾S)	: Lemma, based on def of ‾(*)
         	= (S ∩ ‾S) ∩ (T ∩ ‾T)	: commutativity of ∩
         	= ∅ ∩ ∅	: complement law (p. 124)
         	= ∅	: def of ∅ & ∩

         QED

      2. Proof #2:

         (S – T) ∩ (T – S)	= {x | x ∈ [(S – T) ∩ (T – S)]}	: def of set
         	= {x | [x ∈ (S – T)] ∧ [x ∈ (T – S)]}	: def of ∩
         	= {x | (x ∈ S ∧ x ∉ T) ∧ (x ∈ T ∧ x ∉ S)}	: def of –
         	= {x | x ∈ S ∧ x ∉ S} ∩ {x | x ∈ T ∧ x ∉ T}	: commutativity of ∧ & def of ∩
         	= ∅ ∩ ∅	: def of ∅
         	= ∅	: def of ∅ & ∩

         QED

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4. Show ‾[A ∪ (B ∩ C)] = ‾A ∩ (‾B ∪ ‾C):
   1. Proof #1:

      ‾[A ∪ (B ∩ C)]	= ‾A ∩ ‾(B ∩ C)	: DeM
      	= ‾A ∩ (‾B ∪ ‾C)	: DeM

      QED

   2. Proof #2:

      ‾[A ∪ (B ∩ C)]	= {x | x ∉ [A ∪ (B ∩ C)]}	: (justifications are left to you!)
      	= {x | ¬[x ∈ [A ∪ (B ∩ C)]]}
      	= {x | ¬[(x ∈ A) ∨ (x ∈ (B ∩ C))]}
      	= {x | ¬(x ∈ A) ∧ ¬(x ∈ (B ∩ C))}
      	= {x | (x ∈ ‾A) ∧ (x ∈ ‾(B ∩ C))}
      	= {x | (x ∈ ‾A) ∧ (x ∈ (‾B ∪ ‾C))}
      	= {x | x ∈ [‾A ∩ (‾B ∪ ‾C)]}
      	= ‾A ∩ (‾B ∪ ‾C)

      QED

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5. Can have ∪, ∩ over more than 2 sets:
   1. {0,1,2} ∪ {3,4} ∪ {x ∈ N | x ≥ 5} = N

   2. ({0,1,2} ∩ {0,2}) ∪ {3} = {0,2} ∪ {3} = {0,2,3}

   3. {0,1,2} ∩ ({0,2} ∪ {3}) = {0,1,2} ∩ {0,2,3} = {0,2}

   4. Note: I now have to introduce some notation that cannot easily be shown in HTML; I'll do my best.

      Notation:

      i = n
      ∪A_i =def A₁ ∪ A₂ ∪ … ∪ A_n
      i = 1

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