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Exercise 9.1.14
Evaluate the cardinality of . [Answer: .]
Answers
Proof. First we have that
by property (n) after Lemma 5.6.1 since clearly . We also have
Hence it follows from the Cantor-Bernstein Theorem that as desired. □
Main Problem.
We claim that .
Proof. First, let so that clearly . It is trivial to show that the mapping
for is a bijection from to , and hence . Also, since for any , it follows from Lemma ?? that . Therefore, we first have
Now, let . For any , it is easy to show that the function that maps this to is bijective so that . It should also be clear that . Lastly, we clearly have for all since . We then have
It therefore follows from the Cantor-Bernstein Theorem that as desired. □