hw1sol - Math 280A Fall 2011 Homework 1 Solutions 1...

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Unformatted text preview: Math 280A, Fall 2011 Homework 1 Solutions 1. According to the definition on page 3, A △ B := ( A \ B ) ∪ ( B \ A ) . But A \ B = A ∩ B c and B \ A = B ∩ A c , so the first equality in the first display to be proved is clear. For the second equality, observe that A ∪ B can be expressed as a disjoint union: A ∪ B = ( A \ B ) ∪ ( B \ A ) ∪ ( A ∩ B ) . Consequently, ( A ∪ B ) \ ( A ∩ B ) = ( A \ B ) ∪ ( B \ A ) = A △ B, as desired. Next, A c △ B c = ( A c \ B c ) ∪ ( B c \ A c ) = ( A c ∩ B ) ∪ ( B c ∩ A ) = ( B \ A ) ∪ ( A \ B ) = B △ A = A △ B. Finally, let x be an arbitrary element of ( A 1 ∪ A 2 ) △ ( B 1 ∪ B 2 ). Then x ∈ A 1 ∪ A 2 or x ∈ B 1 ∪ B 2 but not both. In the former case, x ∈ B c 1 ∩ B c 2 and either x ∈ A 1 or x ∈ A 2 . In particular, either x ∈ A 1 ∩ B c 1 ⊂ A 1 △ B 1 or x ∈ A 2 ∩ B c 2 ⊂ A 2 △ B 2 , so x ∈ ( A 1 △ B 1 ) ∪ ( A 2 △ B 2 ) in the former case. The same is also true in the latter case ( x ∈ B 1 ∪ B 2 but x / ∈ A 1 ∪ A 2 ). Thus in either case, x ∈ ( A 1 △ B 1 ) ∪ ( A 2 △ B 2 ). Since x was an arbitrary element of...
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hw1sol - Math 280A Fall 2011 Homework 1 Solutions 1...

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