Exercise 1.3.3

(a) Prove that a “set of all sets” does not exist. [Hint: if V is a set of all sets, consider { x V x x } .]

(b) Prove that for any set A there is some x A .

Answers

(a)

Proof. Suppose that a set of all sets exists, and denote such a set by V . Then, since the property 𝐏 ( x ) defined by x x is a perfectly valid property, the set W = { x V 𝐏 ( x ) } uniquely exists by the Axiom Schema of Comprehension and Lemma 1.3.3. Of course, it must be that either W W or W W , noting that W V since W is a set. If W W then it must be that 𝐏 ( W ) by the definition of W since W V . However, 𝐏 ( W ) means that W W , which is a contradiction. Hence, it must be that W W , but in this case 𝐏 ( W ) is true so that W W by definition since W V , 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 A 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 𝒫 ( A ) , which uniquely exists by the Axiom of Power Set, is not a subset of A . Therefore, there must be some x 𝒫 ( A ) such that x A , which proves our result. This is because, by the definition of subset, we have the following logical equivalences:

𝒫 ( A ) ⊈ A ¬ x [ x 𝒫 ( A ) x A ] x ¬ [ x 𝒫 ( A ) x A ] x ¬ [ x 𝒫 ( A ) x A ] x [ x 𝒫 ( A ) x A ] .
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2024-07-15 11:42
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