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Exercise 1.3.5
(a) Given , , and , there is a set such that if and only if or or .
(b) Generalize to four elements.
Answers
(a)
Proof. By the Axiom of Pair, the sets and uniquely exist. Then, by the Axiom of Union the set uniquely exists, which we claim is exactly the set we seek.
( ) Suppose that so that or . If then either or . On the other hand, if then it must be that . So in all cases it is true that or or as desired.
( ) Suppose that or or . In the first two cases clearly so that . In the last case so that clearly so that again . Note that in both cases.
This proves the results, and we can denote our set by . □
(b)
Proof. Suppose that the four elements are , , , and . That is, we want to prove the existence of a set such that if and only if , , , or . By part (a) just above, the set exists as does by the Axiom of Pair. Predictably, our set is then which exists by the Axiom of Union. We can denote this set in kind by . The proof that is the set we seek is directly analogous to the corresponding proof in part (a), so we do not repeat it here. □