Exercise 8.1.16

If for any sets A and B either | A | | B | or | B | | A | , then the Axiom of Choice holds. [Hint: Compare A and B = h ( A ) .]

Answers

Proof. Consider any set A . Then we have that either | A | | h ( A ) | or | h ( A ) | | A | (where h ( A ) denotes the Hartogs number of A ). Now, it cannot be that | h ( A ) | | A | . For, if it were, then there would be an injection f from h ( A ) into A . Then clearly ran f A and f is a bijection from h ( A ) to ran f . Thus h ( A ) is equipotent to ran f A , which violates the definition of the Hartogs number.

Therefore it must be that | A | | h ( A ) | . Hence there is an injection g from A into h ( A ) . Since h ( A ) is an ordinal and ran g h ( A ) , it follows that ran g is a set of ordinals, which is well-ordered by Theorem 6.2.6d. So, ordering A according to g (considered as a bijection from A to ran g ) results in a well-ordering of A . Since A was an arbitrary set, this shows the Well-Ordering Principle, from which the Axiom of Choice follows by Theorem 8.1.13. □

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2024-07-15 11:42
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