Exercise 1.3.6

Show that 𝒫 ( X ) X is false for any X . In particular, 𝒫 ( X ) X for any X . This proves again that a “set of all sets” does not exist. [Hint: Let Y = { u X u u } ; Y 𝒫 ( X ) but Y X .]

Answers

Proof. Following the hint, let Y = { u X u u } . Clearly Y X so that Y 𝒫 ( X ) since any element in Y must also be in X . Now suppose that also Y X . We treat two cases.

Case: Y Y . Then Y X and Y Y by the definition of Y . This is of course a contradiction.

Case: Y Y . Then, since we supposed that Y X , the property in the definition of Y is satisfied so that Y Y . This is again a contradiction.

Since both exhaustive cases result in a contradiction, our supposition that Y X cannot be true! So, since Y 𝒫 ( X ) but Y X , this suffices to show that 𝒫 ( X ) is not a subset of X as desired by the definition of inclusion. □

This can also serve as a lemma to prove (differently than Exercise 1.3.3a) that a “set of all sets” cannot exist. To see this, suppose that V is such a set. Clearly then 𝒫 ( V ) is a set that contains other sets. However, it was just shown that there is a Y 𝒫 ( V ) such that Y V . But Y is a set, so it must be in V by definition, a contradiction.

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2024-07-15 11:42
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