Exercise 6.5.7

(a) If α 1 , α 2 , and β are ordinals and β 0 , then α 1 < α 2 if and only if β α 1 < β α 2 .

(b) For all ordinals α 1 , α 2 , and β 0 , β α 1 = β α 2 if and only if α 1 = α 2 .

Answers

(a)

Proof. ( ) We show this by transfinite induction on α 2 . So suppose and that α 1 < δ implies that β α 1 < β δ for all δ < α 2 and that α 1 < α 2 . If α 2 is a successor ordinal then α 2 = δ + 1 for some ordinal δ where α 1 δ since α 1 < α 2 . Then we have

β α 1 β δ (by the induction hypothesis if  α 1 < δ  and trivially if  α 1 = δ ) < β δ + β (by Lemma 6.5.4a since  0 < β ) = β ( δ + 1 ) (by Definition 6.5.6b) = β α 2 .

On the other hand if α 2 is a limit ordinal then α 1 + 1 < α 2 since α 1 < α 2 . Then we have that

β α 1 < β ( α 1 + 1 ) (by the induction hypothesis since  α 1 < α 1 + 1 < α 2 ) sup δ < α 2 β δ (since the supremum is an upper bound) = β α 2 . (by Definition 6.5.6c)

This completes the inductive proof.

( ) For this we assume that β α 1 < β α 2 . If it were the case that α 1 > α 2 than it would follow by the implication already shown that β α 1 > β α 2 , which is a contradiction. Similarly if α 1 = α 2 then β α 1 = β α 2 , another contradiction. Hence by the linearity of the order < it must be that α 1 < α 2 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 α 1 α 2 . Then either α 1 < α 2 or α 1 > α 2 . In the former case then part (a) implies that β α 1 < β α 2 so that β α 1 β α 2 . In the latter case part (a) similarly implies that β α 1 > β α 2 so that again β α 1 β α 2 .

( ) If α 1 = α 2 then we trivially have β α 1 = β α 2 . □

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2024-07-15 11:42
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