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Exercise 5.1.14
If is infinite, then is Dedekind infinite. [Hint: For each , let . The set is a countable subset of .]
Answers
Proof. Suppose that is an infinite set and consider any . Then so that there is an injective . Now but also . So define the set
Clearly and we show that by defining a mapping by
for . Since is injective clearly is. Now consider any . By definition then there is a such that . Hence so that is surjective. Hence since is bijective as desired. □
Main Problem.
Proof. Suppose that is infinite. Then for any define
noting that by Lemma 1. We also note that for where we have since for any and we have
so that . From this it follows that
is a countable set. Also, for an we have that each is in so that . Hence each . Thus . Since is countable it then follows that is Dedekind infinite by Exercise 5.1.10. □