Exercise 6.5.9

Show that the following rules to not hold for all ordinals α , β , and γ :

(a) If α + γ = β + γ then α = β .

(b) If γ > 0 and α ⋅ γ = β ⋅ γ , then α = β .

(c) ( β + γ ) ⋅ α = β ⋅ α + γ ⋅ α .

Answers

Lemma 1. If n > 0 is a natural number then n ⋅ ω = ω .

Proof. By Definition 6.5.6 we have that n ⋅ ω = sup ⁡ { n ⋅ k ∣ k < ω } but clearly n ⋅ k is a natural number for any k < ω so that n ⋅ ω = sup ⁡ { n ⋅ k ∣ k < ω } = sup ⁡ { k ∣ k < ω } = ω . □

Main Problem.

(a) Let α = 1 , β = 2 , and γ = ω so that

α + γ = 1 + ω = ω = 2 + ω = β + γ

by Lemma ?? but α = 1 < 2 = β so that α ≠ β .

(b) Again let α = 1 , β = 2 , and γ = ω so that γ = ω > 0 and

α ⋅ γ = 1 ⋅ ω = ω = 2 ⋅ ω = β ⋅ γ

by Lemma 1. Clearly though α = 1 < 2 = β so that α ≠ β .

(c) Here let α = ω , β = 1 , and γ = 2 . Then we have

( β + γ ) ⋅ α = ( 1 + 2 ) ⋅ ω = 3 ⋅ ω = ω

by Lemma 1, whereas

β ⋅ α + γ ⋅ α = 1 ⋅ ω + 2 ⋅ ω = ω + ω = ω ⋅ 2 ,

where we have used Lemma 1 here as well as Example 6.5.7b. That ( β + γ ) ⋅ α = ω = ω ⋅ 1 ≠ ω ⋅ 2 = β ⋅ α + γ ⋅ α follows from Exercise 6.5.7b.

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2024-07-15 11:42
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