Abstract Algebra Theory and Applications - Computer Science ...

0.2 SETS AND EQUIVALENCE RELATIONS 7Proposition 0.1 Let A, B, and C be sets. Then1. A ∪ A = A, A ∩ A = A, and A \ A = ∅;2. A ∪ ∅ = A and A ∩ ∅ = ∅;3. A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C;4. A ∪ B = B ∪ A and A ∩ B = B ∩ A;5. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C);6. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).Proof. We will prove (1) and (3) and leave the remaining results to beproven in the exercises.(1) Observe thatandAlso, A \ A = A ∩ A ′ = ∅.(3) For sets A, B, and C,A ∪ A = {x : x ∈ A or x ∈ A}= {x : x ∈ A}= AA ∩ A = {x : x ∈ A and x ∈ A}= {x : x ∈ A}= A.A ∪ (B ∪ C) = A ∪ {x : x ∈ B or x ∈ C}= {x : x ∈ A or x ∈ B, or x ∈ C}= {x : x ∈ A or x ∈ B} ∪ C= (A ∪ B) ∪ C.A similar argument proves that A ∩ (B ∩ C) = (A ∩ B) ∩ C.□Theorem 0.2 (De Morgan’s Laws) Let A and B be sets. Then1. (A ∪ B) ′ = A ′ ∩ B ′ ;2. (A ∩ B) ′ = A ′ ∪ B ′ .