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Exercise 6.5.7
(a) If , , and are ordinals and , then if and only if .
(b) For all ordinals , , and , if and only if .
Answers
(a)
Proof. ( ) We show this by transfinite induction on . So suppose and that implies that for all and that . If is a successor ordinal then for some ordinal where since . Then we have
On the other hand if is a limit ordinal then since . Then we have that
This completes the inductive proof.
( ) For this we assume that . If it were the case that than it would follow by the implication already shown that , which is a contradiction. Similarly if then , another contradiction. Hence by the linearity of the order it must be that as desired. □
(b)
Proof. This follows from part (a) in the same way as the proof of Lemma 6.5.4b but is repeated for completeness.
( ) We prove this part by contrapositive, so suppose that . Then either or . In the former case then part (a) implies that so that . In the latter case part (a) similarly implies that so that again .
( ) If then we trivially have . □