Exercise 7.1.5

| h ( A ) | < | 𝒫 ( 𝒫 ( A × A ) ) | for all A . [Hint: Prove that | 𝒫 ( h ( A ) ) | | 𝒫 ( 𝒫 ( A × A ) ) | by assigning to each X 𝒫 ( h ( A ) ) the set of all well orderings R A × A for which the ordinal isomorphic to R belongs to X .]

Answers

Proof. First we show that | 𝒫 ( h ( A ) ) | | 𝒫 ( 𝒫 ( A × A ) ) | by constructing an injective f : 𝒫 ( h ( A ) ) 𝒫 ( 𝒫 ( A × A ) ) . So consider any X 𝒫 ( h ( A ) ) so that X h ( A ) . Then let Y be the set of well-orderings R A × A (so that R 𝒫 ( A × A ) ) of subsets B A such that ( B , R ) is isomorphic to some α X . We then set f ( X ) = Y , noting that clearly f ( X ) = Y 𝒫 ( 𝒫 ( A × A ) ) since for any R Y we have that R 𝒫 ( A × A ) so that Y 𝒫 ( A × A ) hence Y 𝒫 ( 𝒫 ( A × A ) ) . Note also that Y because every α h ( A ) is equipotent to some subset B A (by the definition of the Hartogs number) so that the well-ordering of B according to α will be in Y .

We claim that f is injective. So consider X 1 and X 2 in 𝒫 ( h ( A ) ) (so that X 1 h ( A ) and X 2 h ( A ) ) such that f ( X 1 ) = f ( X 2 ) . Then consider any α X 1 . Then since f ( X 1 ) there is a well-ordering R f ( X 1 ) of a subset of A that is isomorphic to α . Then since f ( X 1 ) = f ( X 2 ) we have R f ( X 2 ) as well. It follows from this that α X 2 . Thus X 1 X 2 since α was arbitrary. A similar argument shows that X 2 X 1 so that we conclude that X 1 = X 2 . This shows that f is injective.

Hence we have shown that | 𝒫 ( h ( A ) ) | | 𝒫 ( 𝒫 ( A × A ) ) | . We also have by Cantor’s Theorem (Theorem 5.1.8 in the text) that | h ( A ) | < | 𝒫 ( h ( A ) ) | . It therefore follows from Exercise 4.1.2a that | h ( A ) | < | 𝒫 ( 𝒫 ( A × A ) ) | as desired. □

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2024-07-15 11:42
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