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Exercise 1.3.3
(a) Prove that a “set of all sets” does not exist. [Hint: if is a set of all sets, consider .]
(b) Prove that for any set there is some .
Answers
(a)
Proof. Suppose that a set of all sets exists, and denote such a set by . Then, since the property defined by is a perfectly valid property, the set uniquely exists by the Axiom Schema of Comprehension and Lemma 1.3.3. Of course, it must be that either or , noting that since is a set. If then it must be that by the definition of since . However, means that , which is a contradiction. Hence, it must be that , but in this case is true so that by definition since , which is again a contradiction. Since a contradiction occurs no matter what, it must be that our original supposition that a set of all sets exists is untrue. □
(b)
Proof. Suppose that is a set. To avoid duplicating effort, we jump ahead a bit and utilize the result of Exercise 1.3.6 below. There it is proved that , which uniquely exists by the Axiom of Power Set, is not a subset of . Therefore, there must be some such that , which proves our result. This is because, by the definition of subset, we have the following logical equivalences:
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