Dr. Hackney STA Solutions pg 6

Dr. Hackney STA Solutions pg 6 - Second Edition 1-3 b....

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Unformatted text preview: Second Edition 1-3 b. Suppose x ( A ) c , by the definition of complement x 6 ( A ). Therefore x 6 A for some . Therefore x A c for some . Thus x A c and, by the definition of union, x A c for some . Therefore x 6 A for some . Therefore x 6 A . Thus x ( A ) c . 1.10 For A 1 ,...,A n ( i ) n [ i =1 A i ! c = n i =1 A c i ( ii ) n i =1 A i ! c = n [ i =1 A c i Proof of ( i ): If x ( A i ) c , then x / A i . That implies x / A i for any i , so x A c i for every i and x A i . Proof of ( ii ): If x ( A i ) c , then x / A i . That implies x A c i for some i , so x A c i . 1.11 We must verify each of the three properties in Definition 1.2.1. a. (1) The empty set { ,S } . Thus B . (2) c = S B and S c = B . (3) S = S B ....
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