5. 3.16 Solution

5. 3.16 Solution - INFS 501: H/W 5.3.16 - Complete Claim: A...

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INFS 501: H/W 5.3.16 - Complete Claim : ℘ A B =℘ A  ∩ ℘ B . Proof 1 : First we show P A B  ⊆ P A  ∩ P B by taking x P A B . Now, x P A B means x ⊆  A B . Thus, x ⊆  A B  ⊆ A,so x A, and x P A . Likewise, x P B . Thus, x P A  ∩ P B . Since x P A B was arbitrary, we have proven P A B  ⊆  P A  ∩ P B  . To go in the other direction, take an arbitrary x P A  ∩ P B . Now, x P A means x A. Likewise, x P B means x B. Thus, x A B. Therefore, x P A B . Since x P A  ∩ P B was arbitrary, we have proven P A  ∩ P B  ⊆ P A B . Since P A B  ⊆  P A  ∩ P B  and P A  ∩ P B  ⊆ P A B , P
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