By W. Weiss

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**Example text**

Mk < ω, and α0 , . . , αk < β, then m 0 · ω α0 + · · · + m k · ω αk < ω β . Exercise 15. Prove this lemma and note that it implies that m · δ = δ for each positive integer m and each limit ordinal δ. There is an interesting application of ordinal arithmetic to Number Theory. Pick a number—say x = 54. We have 54 = 25 + 24 + 22 + 2 when it is written as the simplest sum of powers of 2. In fact, we can write out 54 using only the the arithmetic operations and the numbers 1 and 2. This will be the first step in a recursively defined sequence of natural numbers, {xn }.

But since l ∈ y, l ∈ / x and so l ⊆ n. Hence y = l ∪ {l} ⊆ n, contradicting that y ∈ x. We also prove (2) indirectly; suppose n ∈ N with {m : m ∈ n and m ∈ / N} = ∅. By Comprehension ∃x x = {m ∈ n : m ∈ / N} and so Foundation gives y ∈ x such that y ∩ x = ∅. Since y ∈ n, we have y = succ(l) for some l ∈ n. Since l ∈ y and y ∩ x = ∅ we must have l ∈ N. But then y = succ(l) ∈ N, contradicting that y ∈ x. Theorem 7. (Trichotomy of Natural Numbers) Let m, n ∈ N. Exactly one of three situations occurs: m ∈ n, n ∈ m, m = n.

1. 1 + ω = 2 + ω 2. 1 + ω = ω + 1 3. 1 · ω = 2 · ω 4. 2 · ω = ω · 2 5. 2ω = 4ω 6. (2 · 2)ω = 2ω · 2ω Lemma. If β is a non-zero ordinal then ω β is a limit ordinal. Exercise 11. Prove this lemma. Lemma. If α is a non-zero ordinal, then there is a largest ordinal β such that ω β ≤ α. Exercise 12. Prove this lemma. Show that the β ≤ α and that there are cases in which β = α. ) Lemma. γ ∈ ON α = β + γ. Exercise 13. Prove this lemma. Commonly, any function f with dom(f ) ⊆ ω is called a sequence. If dom(f ) ⊆ n + 1 for some n ∈ ω, we say that f is a finite sequence; otherwise f is an infinite sequence.

### An Introduction to Set Theory by W. Weiss

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