Exercise 7.1.4

| A | < | A | + h ( A ) for all A .

Answers

Lemma 1. | h ( A ) | | A | for any set A .

Proof. Suppose to the contrary that | h ( A ) | | A | so that there is an injective f from h ( A ) to A . Then let X = f [ h ( A ) ] so that clearly X A . But then f considered as function from h ( A ) to X is a bijection so that h ( A ) is equipotent to a subset of A , which contradicts the definition of the Hartogs number. Hence it must be that | h ( A ) | | A | as desired. □

Note that this does not imply that h ( a ) > | A | without using the Axiom of Choice.

Main Problem.

Proof. Since we are only interested in the cardinalities of A and h ( A ) and their cardinal sum we can assume that they are disjoint. Clearly f : A A h ( A ) defined by f ( x ) = x for x A is an injective function so that | A | | A | + h ( A ) by the definition of cardinal addition. So suppose that | A | = | A | + h ( A ) . Then there is a bijective function f from A h ( A ) into A so that f h ( A ) is an injective function from h ( A ) to A . By definition this means that h ( A ) | A | , but this contradicts Lemma 1. So it must be that | A | < | A | + h ( A ) as desired. □

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2024-07-15 11:42
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