Model theory 6. Ordinals and cardinals (correction)

Exercise 1 Claim 1.1 If α is an ordinal number, then so is α ∪ {α}. Proof. The set α ∪ {α} is transitive: An element x of α ∪ {α} is either α (hence x ⊂ α ∪ {α}) or an element of α (hence x ⊂ α ⊂ α ∪ {α} by transitivity of α). The relation ∈ is transitive on α ∪ {α}: assume that x ∈ y and y ∈ z for three elements x, y, z of α ∪ {α}. Either z ∈ α, and hence y and x are elements of α by transitivity of the set α, so x ∈ z by transitivity of the relation ∈ in α. Or z = α, and then x ∈ y ⊂ α by transitivity of the set α, so x ∈ z. The relation ∈ is a well-ordering of α ∪ {α}: as α is a maximal element of α ∪ {α} for , every non-empty set B of α ∪ {α} is either {α} or has a non-empty intersection A with the set α, and A has a least element, which is the least element of B. Claim 1.2 ω 2 and ω ω are countable. Proof. ω is countable. For every countable ordinal α and natural number n, the ordinal α + n is countable (it is in bijection with α). For every natural number n > 1, ω.n is coutable (by induction S on n. ω(n + 1) = ωn + ω = m∈ω (ωn + m), which is a countable union of coutable sets). As S ω 2 = n∈ω ωn, the ordinal ω 2 is a countable union of countable sets, hence countable. If α and β are countable ordinals, then so is α + β (by induction on β). One can show inductively on n that ω n is S countable (ω n+1 = ω n ω = m∈ω ω n m, which is a countable union of countable sets by the previous S remark). As ω ω = n∈ω ω n , it follows that ω ω is coutable. Claim 1.3 More generally, any ordinal number obtained by finitely many applications of ordinal operations (addition, multiplication and exponentiation) to ω or natural numbers is countable (by A is countable, we mean |A| 6 ω). Proof. We first claim that if α and β are countable ordinals, then so is α+β. We show that inductively on β. For β = 0, one has α + β = α, which is countable. If β = γ + 1, then α + β = (α + γ) + 1 = α + β ∪ {α + β}, which is countable by induction hypothesis. If β is a limit ordinal, one has S α + β = γ∈β α + γ, which is a countable union of countable sets by induction hypothesis, hence countable. We now claim that if α and β are countable ordinals, then so is αβ. We show that again by induction on β. For β = 0, one has αβ = 0. If β = γ + 1, then αβ = αγ + α, which is countable by S induction hypothesis and by the previous claim. If β is a limit ordinal, one has αβ = γ∈β αγ, which is a countable union of countable sets by induction hypothesis, hence countable. Finally, we claim that if α and β are countable ordinals, then so is αβ. We show that by induction on β. For β = 0, one has αβ = 1. If β = γ + 1, then αβ = αγ α, which is countable by induction S hypothesis and by the previous claim. If β is a limit ordinal, one has αβ = γ∈β αγ , which is a countable union of countable sets by induction hypothesis, hence countable. Claim 1.3 follows by induction on the number of operation performed. 1

Exercise 2 Let α, β, γ be ordinal numbers. Claim 2.1

1. If β < γ, then α + β < α + γ, but β + α < γ + α does not hold in general.

2. If β 6 γ, then β + α 6 γ + α. 3. If β < γ and α > 0, then αβ < αγ, but βα < γα does not hold in general. 4. If β < γ and α > 0, then αβ < αγ , but β α < γ α does not hold in general. Proof. 1. By induction on γ. If γ = 0, there is nothing to show. If γ = δ + 1, then β 6 δ, so that S α + β 6 α + δ < (α + δ) + 1 = α + γ. If γ is a limit ordinal, then α + β < δ β1 > · · · > βk . If one has γ 6 β1 , then ω γ 6 ω β1 6 ρ, a contradiction, so one has α = ω γ · δ + ω β1 · n 1 + · · · + ω βk · n k , with γ > β1 > · · · > βk (and one also has α > γ for otherwise one would have α 6 ω α < ω γ 6 α). We show uniqueness by induction on α. This is obvious for any finite α. Let α = ω α1 · n1 + · · · + ω αk · nk = ω β1 · m1 + · · · + ω β` · m` . If α1 < β1 , then ω α1 < ω β1 , so α 6 ω α1 · (n1 + · · · + nk ) < ω α1 +1 6 ω β1 6 α, a contradiction. It follows that α1 = β1 . If n1 < m1 , one has α 6 ω α1 · n1 + ω α2 (n2 + · · · + nk ) < ω α1 · n1 + ω α2 +1 6 ω α1 · n1 + ω α1 6 ω α1 · m1 6 α, a contradiction, from which one has n1 = m1 . By Claim 2.2, either α = ω α1 · n1 , and we are done, or α > ω α1 · n1 , and there is a unique δ such that α = ω α1 · n1 + δ. It follows that δ = ω α2 · n2 + · · · + ω αk · nk = ω β2 · m2 + · · · + ω β` · m` . As δ < α, one has k = `, ni = mi and αi = βi for every i ∈ {2, . . . , k} by induction hypothesis.

Exercise 3 Claim 3.1 1. The cardinal addition κ + λ, multiplication κ · λ and exponentiation κλ are welldefined. + and · are associative and · is distributive over +. 2. For any set X, the power set P(X) has cardinality 2|X| . 3. For any cardinal numbers λ, κ and µ, one has κλ+µ = κλ κµ . Proof. Claim 3.2 Let α, β be ordinal numbers, and λ an infinite cardinal number. 1. The ordinal α + β is isomorphic to the disjoint union α In particular, one has |α + β| = |α| + |β|. `

`

β (defined to be α × {0} ∪ β × {1}).

[note that α + β and α β are even order-isomorphic if one equipes the disjoint union with the contenation ordering 6 defined by x 6 y iff x, y ∈ α × {0} and x ⊂ y, or x, y ∈ β × {1} and x ⊂ y, or x ∈ α × {0} and y ∈ β × {1}]. 2. λ + λ = λ. 3

3. λ · λ = λ. Proof. 1. We show by induction on β that there is an isomorphism fβ : α + β −→ α β such that for all γ < β, the restriction of fβ to α + γ is fγ . If β = 0, then β is the empty set, and there is nothing ` ` to show. If β = γ + 1 and fγ : α + γ −→ α γ is an isomorphism sets, we define fβ : α + β −→ α β by putting g(x) = f (x) if x ∈ α + γ or g(x) = (β, 1) if x = α + β. This is clearly an isomorphism whose restriction to α + γ is fγ . If β is a limit ordinal, and for every γ < β one has an isomophism ` ` S fγ : α + γ −→ α γ and defines fβ : α + β −→ α γ to be γ · · · > αk of ordinals, allowing one of ni or mi to be zero (but not both of them) for every i. We define f (α, β) = ω α1 · 2n1 3m1 + · · · + ω αk · 2nk 3mk . We first claim that f is well-defined, i.e. that f (α, β) is indeed an element of λ. As n1 or m1 is non-zero (n1 say), one has α > ω α1 so ω α1 ∈ λ, and in particular |ω α1 | < λ. As f (α, β) < ω α1 · n for some natural number n, by 1, one has |f (α, β)| 6 |ω α1 · n| = |ω α1 | + · · · + |ω α1 | = |ω α1 | < λ. We claim that f is injective. If f (α, β) = f (γ, δ), where γ = ω γ1 · p1 + · · · + ω γk · p`

and

δ = ω γ1 · q1 + · · · + ω γk · q` ,

then ω α1 · 2n1 3m1 + · · · + ω αk · 2nk 3mk = ω γ1 · 2p1 3q1 + · · · + ω γk · 2p` 3q` . By uniqueness of the Cantor normal form, one must have k = `, (α1 , . . . , αk ) = (γ1 , . . . , γk ) and (2n1 3m1 , . . . , 2nk 3mk ) = (2p1 3q1 , . . . , 2pk 3qk ). By uniqueness of the prime decomposition in N, one must have (n1 , . . . , nk ) = (p1 , . . . , pk ) and (m1 , . . . , mk ) = (q1 , . . . , qk ). It follows that α = γ and β = δ. Claim 3.3 The set [λ]