Exercise 9.1.3

Find some cardinals κ n , λ n ( n 𝑵 ) such that k n < λ n for all n , but κ n = λ n .

Answers

Let κ n = 1 and λ n = 2 for all n 𝑵 . We claim that these satisfy the required properties.

Proof. Clearly we have κ n = 1 < 2 = λ n for all n 𝑵 . It then follows from Exercise 9.1.4 (and also the more general Theorem 9.1.3) that

κ n = 1 = n 𝑵 1 = n < 0 1 = 0 1 = 0 = 0 2 = n < 0 2 = n 𝑵 2 = 2 = λ n

as desired. □

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2024-07-15 11:42
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