Exercise 9.1.14

Evaluate the cardinality of 0 < α < ω 1 α . [Answer: 2 1 .]

Answers

Lemma 1. If α and β are ordinals and α β , then α β = 2 β .

Proof. First we have that

2 β α β

by property (n) after Lemma 5.6.1 since clearly 2 α . We also have

α β ( 2 α ) β (again by property (n), and Theorems 5.1.8 and 5.1.9) = 2 α β (by Theorem 5.1.7b) = 2 β . (by Corollary 7.2.2 since  α β )

Hence it follows from the Cantor-Bernstein Theorem that α β = 2 β as desired. □

Main Problem.

We claim that | 0 < α < ω 1 α | = 2 1 .

Proof. First, let I = { α 0 < α < ω 1 } so that clearly 0 < α < ω 1 α = α . It is trivial to show that the mapping

f ( α ) = { α + 1 0 α < ω 0 α ω 0 α < ω 1

for α ω 1 is a bijection from ω 1 to I , and hence | I | = | ω 1 | = 1 . Also, since α < ω 1 for any α I , it follows from Lemma ?? that | α | 0 . Therefore, we first have

| α | = | α | (by the definition of the cardinal product) 0 (by Exercise 9.1.8 since  | α | 0  for all  α I ) = 0 | I | (by Exercise 9.1.10) = 0 1 (since we have shown that  | I | = 1 ) = 2 1 . (by Lemma  1 )

Now, let J = { α 1 < α < ω 1 } . For any a = a 2 , a 3 , , a ω , a ω + 1 , = a α α J α , it is easy to show that the function that maps this to b = 0 , a 2 , a 3 , , a ω , a ω + 1 , α is bijective so that | α | = | α | . It should also be clear that | J | = | I | = 1 . Lastly, we clearly have 2 | a | for all a J since 1 < α . We then have

2 1 = 2 | J | (since  | J | = 1 ) = 2 (by Exercise 9.1.10) | α | (by Exercise 9.1.8 since  2 | α |  for all  α J ) = | α | (by the definition of the cardinal product) = | α | (by what was shown above)

It therefore follows from the Cantor-Bernstein Theorem that | 0 < α < ω 1 α | = | α | = 2 1 as desired. □

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2024-07-15 11:42
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