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Exercise 1.3.6
Show that is false for any . In particular, for any . This proves again that a “set of all sets” does not exist. [Hint: Let ; but .]
Answers
Proof. Following the hint, let . Clearly so that since any element in must also be in . Now suppose that also . We treat two cases.
Case: . Then and by the definition of . This is of course a contradiction.
Case: . Then, since we supposed that , the property in the definition of is satisfied so that . This is again a contradiction.
Since both exhaustive cases result in a contradiction, our supposition that cannot be true! So, since but , this suffices to show that is not a subset of 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 is such a set. Clearly then is a set that contains other sets. However, it was just shown that there is a such that . But is a set, so it must be in by definition, a contradiction.