Homepage › Solution manuals › Karel Hrbáček › Introduction to Set Theory › Exercise 8.1.16
Exercise 8.1.16
If for any sets and either or , then the Axiom of Choice holds. [Hint: Compare and .]
Answers
Proof. Consider any set . Then we have that either or (where denotes the Hartogs number of ). Now, it cannot be that . For, if it were, then there would be an injection from into . Then clearly and is a bijection from to . Thus is equipotent to , which violates the definition of the Hartogs number.
Therefore it must be that . Hence there is an injection from into . Since is an ordinal and , it follows that is a set of ordinals, which is well-ordered by Theorem 6.2.6d. So, ordering according to (considered as a bijection from to ) results in a well-ordering of . Since was an arbitrary set, this shows the Well-Ordering Principle, from which the Axiom of Choice follows by Theorem 8.1.13. □