Exercise 5.2.5

Show that, for n > 0 , n 2 2 0 = 0 2 2 0 = 2 0 2 2 0 = 2 2 0 2 2 0 = ( 2 2 0 ) n = ( 2 2 0 ) 0 = ( 2 2 0 ) 2 0 = 2 2 0 .

Answers

Lemma 1. For any cardinal number κ

1 κ = κ .

Proof. Suppose that κ = | A | for a set A . We define f : A 1 × A by

f ( a ) = ( 0 , a )

for a A , noting that 1 = { 0 } . Clearly by simple inspection this is bijective so that

1 κ = | 1 × A | = | A | = κ

as desired. □

Main Problem.

Proof. First we note that clearly since 0 2 0 we have

2 0 2 2 0

by property (n) in section 5.1. So consider any cardinal n 𝑵 where n > 0 so that 1 n . We then have

2 2 0 = 1 2 2 0 (by Lemma  1 ) n 2 2 0 0 2 2 0 2 0 2 2 0 2 2 0 2 2 0 (repeated property (i) of 5.1) = ( 2 2 0 ) 2 (by property (o) of 5.1) = 2 2 2 0 (by Theorem 5.1.7b) = 2 2 0 . (by Theorem 5.2.2b)

We also have

2 2 0 = ( 2 2 0 ) 1 (by Exercise 5.1.2) ( 2 2 0 ) n ( 2 2 0 ) 0 ( 2 2 0 ) 2 0 (repeated property (n) of 5.1) = 2 2 0 2 0 (by Theorem 5.1.7b) = 2 2 0 (by Theorem 5.2.2b)

Clearly these together with the Cantor-Bernstein Theorem shows the desired result. □

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2024-07-15 11:42
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