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# sols1 - M114S Solutions to HW#1 x1.1 Suppose that x ∈ A...

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Unformatted text preview: M114S, Solutions to HW #1 x1.1. Suppose that x ∈ A ∪ ( B ∩ C ). Then either x ∈ A (in which case x belongs to both A ∪ B and A ∪ C and hence to their intersection), or else x ∈ B ∩ C (so, again, x ∈ ( A ∪ B ) ∩ ( A ∪ C ). This proves half of what we want. For the opposite direction, suppose x ∈ ( A ∪ B ) ∩ ( A ∪ C ). Then we have two cases: if x ∈ A , then x ∈ A ∪ ( B ∩ C ); and if x / ∈ A , then x must belong to both B and to C . So again, x ∈ A ∪ ( B ∩ C ). We can also combine the two arguments in “symbolized” form: x ∈ A ∪ ( B ∩ C ) ⇐⇒ x ∈ A or x ∈ ( B ∩ C ) ⇐⇒ x ∈ A or [ x ∈ B and x ∈ C ] ⇐⇒ [ x ∈ ( A ∪ B ) and x ∈ ( A ∪ C )] or [ x ∈ ( A ∪ B ) and ( A ∪ C )] ⇐⇒ x ∈ ( A ∪ B ) ∩ ( A ∪ B ) . The second property A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) is proved similarly. Finally, we check that A \ ( A ∩ B ) = A \ B . If x ∈ A \ ( A ∩ B ), then x ∈ A , but x / ∈ A ∩ B , and so x cannot belong to B : therefore, x ∈ A \ B . For the opposite direction, if x ∈ A \ B , then x ∈ A and x / ∈ B , so, certainly, x / ∈ A ∩ B ; hence x ∈ A \ ( A ∩ B ). x1.2. First, let x ∈ C \ ( A ∪ B ). Then x belongs to C but not to A ∪ B ; so x belongs to neither A nor B ; thus, x belongs to C \ A and to C \ B , i.e., x ∈ ( C \ A ) ∩ ( C \ B ). Fortheoppositedirection,suppose x ∈ ( C \ A ) ∩ ( C \ B ); then x ∈ C , but also x is not in A and not in B ; so x / ∈ A ∪ B ; hence x ∈ C \ ( A ∪ B ). This verifies the first law. For the second law, we again argue both directions. Let x ∈ C \ ( A ∩ B ); then x belongs to C but not to A ∩ B ; so either x does not belong to A , or x does not belong to B , and in either case, we see that x ∈ ( C \ A ) ∪ ( C \ B )....
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sols1 - M114S Solutions to HW#1 x1.1 Suppose that x ∈ A...

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