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Exercise 7.1.4
for all .
Answers
Proof. Suppose to the contrary that so that there is an injective from to . Then let so that clearly . But then considered as function from to is a bijection so that is equipotent to a subset of , which contradicts the definition of the Hartogs number. Hence it must be that as desired. □
Note that this does not imply that without using the Axiom of Choice.
Main Problem.
Proof. Since we are only interested in the cardinalities of and and their cardinal sum we can assume that they are disjoint. Clearly defined by for is an injective function so that by the definition of cardinal addition. So suppose that . Then there is a bijective function from into so that is an injective function from to . By definition this means that , but this contradicts Lemma 1. So it must be that as desired. □