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Exercise 1.4.4
Let be a set; show that a “complement” of does not exist. (The “complement” of is the set of all .)
Answers
Proof. Suppose that the complement of exist, and denote it by . Then the set exists by the Axiom of Union. Furthermore, the set then uniquely exists by the Axiom Schema of Comprehension.
Now consider any set so that either or . In the latter case of course by definition so that, either way, . Since is a set, it then follows that . Likewise, clearly any is also a set, so that would be exactly the set of all sets. However, it was proven in Exercise 1.3.3a that such a set cannot exist! Hence, it has to be that the complement also cannot exist. □