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Exercise 7.1.5
for all . [Hint: Prove that by assigning to each the set of all well orderings for which the ordinal isomorphic to belongs to .]
Answers
Proof. First we show that by constructing an injective . So consider any so that . Then let be the set of well-orderings (so that ) of subsets such that is isomorphic to some . We then set , noting that clearly since for any we have that so that hence . Note also that because every is equipotent to some subset (by the definition of the Hartogs number) so that the well-ordering of according to will be in .
We claim that is injective. So consider and in (so that and ) such that . Then consider any . Then since there is a well-ordering of a subset of that is isomorphic to . Then since we have as well. It follows from this that . Thus since was arbitrary. A similar argument shows that so that we conclude that . This shows that is injective.
Hence we have shown that . We also have by Cantor’s Theorem (Theorem 5.1.8 in the text) that . It therefore follows from Exercise 4.1.2a that as desired. □