Exercise 6.5.3

Simplify

(a) ( ω + 1 ) + ω .

(b) ω + ω 2 .

(c) ( ω + 1 ) ⋅ ω 2 .

Answers

(a) By Lemma 6.5.4c we have

( ω + 1 ) + ω = ω + ( 1 + ω ) = ω + ω = ω ⋅ 2 .

(b) We simply have

ω + ω 2 = ω ⋅ 1 + ω ⋅ ω = ω ⋅ ( 1 + ω ) = ω ⋅ ω = ω 2

(c)

Lemma 1. ( ω + 1 ) ⋅ n = ω ⋅ n + 1 for all n ∈ ω where n > 0 .

Proof. We show this by standard (as opposed to transfinite) induction on n . For n = 1 we have

( ω + 1 ) ⋅ n = ( ω + 1 ) ⋅ 1 = ω + 1 = ω ⋅ 1 + 1 = ω ⋅ n + 1 .

Now suppose that ( ω + 1 ) ⋅ n = ω ⋅ n + 1 so that we have

( ω + 1 ) ⋅ ( n + 1 ) = ( ω + 1 ) ⋅ n + ( ω + 1 ) (by Definition 6.5.6b) = ( ω ⋅ n + 1 ) + ( ω + 1 ) (by the induction hypothesis) = ω ⋅ n + ( 1 + ω ) + 1 (by the associativity of addition) = ω ⋅ n + ω + 1 = ω ⋅ ( n + 1 ) + 1 . (by Definition 6.5.6b)

This completes the inductive proof. □

Lemma 2. ( ω + 1 ) ⋅ ω = ω 2 .

Proof. Since ω is a limit ordinal we have by Definition 6.5.6c

( ω + 1 ) ⋅ ω = sup ⁡ { ( ω + 1 ) ⋅ n ∣ n ∈ ω } = sup ⁡ { ω ⋅ n + 1 ∣ n ∈ ω }

by Lemma 1. Now, since we have

ω ⋅ n + 1 < ( ω ⋅ n + 1 ) + ω = ω ⋅ n + ( 1 + ω ) = ω ⋅ n + ω = ω ⋅ ( n + 1 )

it is clear that we have

( ω + 1 ) ⋅ ω = sup ⁡ { ω ⋅ n + 1 ∣ n ∈ ω } = sup ⁡ { ω ⋅ n ∣ n ∈ ω } = ω ⋅ ω = ω 2

by Definition 6.5.6c. □

Main Problem.

We have

( ω + 1 ) ⋅ ω 2 = ( ω + 1 ) ⋅ ( ω ⋅ ω ) = [ ( ω + 1 ) ⋅ ω ] ⋅ w = ω 2 ⋅ ω = ω 2 + 1 = ω 3 ,

where we have used Lemma 2 and Definition 6.5.9b.

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2024-07-15 11:42
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