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# Lecture_10 - MA1100 Lecture 10 Sets Proving Set Identities...

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1 MA1100 Lecture 10 Sets Proving Set Identities Set Relations with Conditions Cartesian Product Power Sets

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MA1100 Lecture 10 2 Proving Set Identity U Set equality B A 1 2 3 4 Set region A B 2 A » B 1+2+3 A c 3+4 A – B 1 B – A 3 A c B c 4 A B c = A - B a formal proof should be given in terms of logical argument
MA1100 Lecture 10 3 To prove M = N, same as M Œ N and N Œ M . Prove that A – B = A B c . Proving Set Identity (Element-chasing) Proposition Let A and B be subsets of universal set U. (i) A – B Œ A B c (ii) A B c Œ A – B So x œ A and x B This implies x œ A B c Hence A – B Œ A B c Let x œ A – B i.e. x œ A and x œ B c Reverse the argument Proof

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MA1100 Lecture 10 4 Simple but Useful Arguments Let A and B be subsets of universal set U. 1. If x œ A , then x œ A » B 2. If x œ A and x œ B , then x œ A B 3. If x œ A B , then x œ A 4. If x œ A B , then x œ B 5. If x œ A » B , then x œ A or x œ B When using element-chasing method, the following arguments are useful.
MA1100 Lecture 10 5 Proving Set Identity (Algebra of Sets) Proposition Let A, B, C be subsets of universal set U. (A » B) – C = (A – C) » (B – C) Proof (A » B) – C = (A » B) C c = (A C c ) » (B C c ) = (A – C) » (B – C) We will use algebra of sets (i) A – B = A B c (ii) A (B » C) = (A B) » (A C)

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MA1100 Lecture 10 6 Set Relations with Conditions P Ø Q (i) M Œ N x œ M x œ N (i) M Œ N (ii) M = N (iii) M = « (ii) M = N (iii) M = « x œ M contradiction x œ M x œ N M = … = … = N
MA1100 Lecture 10 7 Set Relations with Conditions Proposition Let A and B be subsets of universal set U. Proof If A Œ B, then A B = A. Given A Œ B Try to show A B = A Let x œ A B. Then x œ A . Hence A B Œ A. This implies x œ A B. so x œ B. Let x œ A. Hence A Œ A B. Since A Œ B, So x œ A and x œ B.

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Lecture_10 - MA1100 Lecture 10 Sets Proving Set Identities...

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