Exercise 1.4.4

Let A be a set; show that a “complement” of A does not exist. (The “complement” of A is the set of all x A .)

Answers

Proof. Suppose that the complement of A exist, and denote it by A ¯ . Then the set A A ¯ exists by the Axiom of Union. Furthermore, the set V = { X A A ¯ X  is a set } then uniquely exists by the Axiom Schema of Comprehension.

Now consider any set X so that either X A or X A . In the latter case of course X A ¯ by definition so that, either way, X A A ¯ . Since X is a set, it then follows that X V . Likewise, clearly any X V is also a set, so that V 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 A ¯ also cannot exist. □

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2024-07-15 11:42
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