1. Let and ℛ be any nonempty collection of sets (distributivity, Parts A) For any set X, X⋂ (⋃) = ⋃{X ⋂S | S∈} and X⋃ (⋂) = ⋂{X⋃S | S∈}
Can you help prove this proof that it holds?

To prove two sets are equal, you have to show they are both subsets of one another.

• X ∩ (⋃ ) = ⋃ {X ∩ S | S ∈ }

Let x ∈ X ∩ (⋃ ). Then x ∈ X and x ∈ ⋃ . The latter means that x ∈ S for an arbitrary set S ∈ . So x ∈ X and x ∈ S, meaning x ∈ X ∩ S. That is enough to say that x ∈ ⋃ {X ∩ S | S ∈ }. So X ∩ (⋃ ) ⊆ ⋃ {X ∩ S | S ∈ }.

For the other direction, the proof is essentially the reverse. Let x ∈ ⋃ {X ∩ S | S ∈ }. Then x ∈ X ∩ S for some S ∈ , so that x ∈ X and x ∈ S. Because x ∈ S and S ∈ , we have that x ∈ ⋃ , and so x ∈ X ∩ (⋃ ). So ⋃ {X ∩ S | S ∈ } ⊆ X ∩ (⋃ ).

QED

• X ∪ (⋂ ) = ⋂ {X ∪ S | S ∈ }

Let x ∈ X ∪ (⋂ ). Then x ∈ X or x ∈ ⋂ . If x ∈ X, we're done because that would guarantee x ∈ X ∪ S for any set S, and hence x would belong to the intersection. If x ∈ ⋂ , then x ∈ S for all S ∈ , so that x ∈ X ∪ S for all S, and hence x is in the intersection. Therefore X ∪ (⋂ ) ⊆ ⋂ {X ∪ S | S ∈ }.

For the opposite direction, let x ∈ ⋂ {X ∪ S | S ∈ }. Then x ∈ X ∪ S for all S ∈ . So x ∈ X or x ∈ S for all S. If x ∈ X, we're done. If x ∈ S for all S ∈ , then x ∈ ⋂ , and we're done. So ⋂ {X ∪ S | S ∈ } ⊆ X ∪ (⋂ ).

QED

quit

step-by-step explanation:

answer: 0.64

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