Exercise 5.1.14

If A is infinite, then 𝒫 ( 𝒫 ( A ) ) is Dedekind infinite. [Hint: For each n 𝑵 , let S n = { X A | X | = n } . The set { S n n 𝑵 } is a countable subset of 𝒫 ( 𝒫 ( A ) ) .]

Answers

Lemma 1. If A is an infinite set then for any n 𝑵 there is a B A such that | B | = n .

Proof. Suppose that A is an infinite set and consider any n 𝑵 . Then 0 | A | so that there is an injective f : 𝑵 A . Now n 𝑵 but also n 𝑵 . So define the set

B = { f ( k ) k n } .

Clearly B A and we show that | B | = n by defining a mapping g : n B by

g ( k ) = f ( k )

for k n . Since f is injective clearly g is. Now consider any b B . By definition then there is a k n such that f ( k ) = b . Hence g ( k ) = f ( k ) = b so that g is surjective. Hence since g is bijective | B | = n as desired. □

Main Problem.

Proof. Suppose that A is infinite. Then for any n 𝑵 define

S n = { X 𝒫 ( A ) | X | = n } ,

noting that S n by Lemma 1. We also note that for n , m 𝑵 where n m we have S n S m since for any X S n and Y S m we have

| X | = n m = | Y |

so that X Y . From this it follows that

S = { S n n 𝑵 }

is a countable set. Also, for an S n S we have that each X S n is in 𝒫 ( A ) so that S n 𝒫 ( A ) . Hence each S n 𝒫 ( 𝒫 ( A ) ) . Thus S 𝒫 ( 𝒫 ( A ) ) . Since S is countable it then follows that 𝒫 ( 𝒫 ( A ) ) is Dedekind infinite by Exercise 5.1.10. □

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2024-07-15 11:42
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