sols5 - 12 5. September 13th: Naive set theory. 5.1. Prove...

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12 5. September 13th: Naive set theory. 5.1 . Prove that ( A B ) C = A ( B C ) for all sets A , B , C . Answer: Let p , q , r respectively be the statements x A , x B , x C . By deFnition of intersection, x ( A B ) C means x ( A B ) x C , and again by deFnition of intersection this means ( x A x B ) x C . With our notation, this is ( )( p q ) r. Likewise, x A ( B C ) means ( ) p ( q r ) . So we have to verify that ( ) and ( ) are equivalent, and this is done by the following truth table: p q r p q q r ( p q ) r p ( q r ) T T T T T T T T T ± T ± ± ± T ± T ± ± ± ± T ± ± ± ± ± ± ± T T ± T ± ± ± T ± ± ± ± ± ± ± T ± ± ± ± ± ± ± ± ± ± ± Remark: The fact that is associative (the statement we just proved) implies that we can unambiguously write A 1 A 2 A 3 ∩ ··· ∩ A n without indicating parentheses, no matter how many sets A i we have. By contrast, A B C is ambiguous, because in general
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This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.

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sols5 - 12 5. September 13th: Naive set theory. 5.1. Prove...

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