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 ∈ ( ∩ α A α ) c , by the definition of complement x 6∈ ( ∩ α A α ). Therefore x 6∈ A α for some α ∈ Γ. Therefore x ∈ A c α for some α ∈ Γ. Thus x ∈ ∪ α A c α and, by the definition of union, x ∈ A c α for some α ∈ Γ. Therefore x 6∈ A α for some α ∈ Γ. Therefore x 6∈ ∩ α A α . Thus x ∈ ( ∩ α A α ) 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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