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Ordinal multiplication is associative, and $α \cdot (β + γ) = α \cdot β + α \cdot γ$

Statement

For all ordinals $α$ , $β$ , $γ$ (Ordinal (von Neumann)), with $+$ and $\cdot$ as in Ordinal addition $α + β$ and Ordinal multiplication $α \cdot β$ :

(a) Left distributivity. $α \cdot (β + γ) = α \cdot β + α \cdot γ$ .

(b) Associativity. $(α \cdot β) \cdot γ = α \cdot (β \cdot γ)$ .

Distributivity holds on the left only. The right-hand law $(β + γ) \cdot α = β \cdot α + γ \cdot α$ is false, and so is commutativity of $\cdot$ ; both are refuted among this page's false statements, and both refutations are named in the Remarks below.

No choice principle is used.

Facts & Assumptions

Given: Ordinals $α$ , $β$ , $γ$ . For a set $A$ of ordinals, $\sup A = ⋃ A$ is its least upper bound (Basic closure properties of ordinals, claim (e)).

[L1]

$α \cdot 0 = 0$ , $α \cdot δ^{+} = α \cdot δ + α$ , and $α \cdot λ = \sup {α \cdot ξ : ξ \in λ}$ for limit $λ$ (Ordinal multiplication $α \cdot β$ ).

[L2]

$α + 0 = α$ , $α + δ^{+} = (α + δ)^{+}$ , and $α + λ = \sup {α + ξ : ξ \in λ}$ for limit $λ$ (Ordinal addition $α + β$ ).

[L3]

Ordinal addition is associative (Ordinal addition is associative).

[L4]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + β = β$ and $1 \cdot β = β$ : $β \cdot 0 = 0 \cdot β = 0$ and $β + 0 = 0 + β = β$ (claim (a)); $β < γ$ implies $α + β < α + γ$ (claim (b)); for $α > 0$ , $β < γ$ implies $α β < α γ$ , and $α \cdot β = 0$ exactly when $α = 0$ or $β = 0$ (claim (d)); if $μ$ is a limit ordinal and $D \subseteq μ$ is nonempty with $\sup D = μ$ , then $α + μ = \sup {α + η : η \in D}$ and, for $α > 0$ , $α \cdot μ = \sup {α \cdot η : η \in D}$ (claim (f)); and $β + λ$ and, for $β > 0$ , $β \cdot λ$ are limit ordinals whenever $λ$ is (claim (g)).

[L5]

Every ordinal is exactly one of $0$ , a successor, or a limit (Successor and limit ordinals).

[L6]

Transfinite induction over the ordinals: if a property $P$ of ordinals fails at some $β_{0}$ , apply Transfinite induction to the well-order $(β_{0}^{+}, \in)$ and to $S = {ξ \in β_{0}^{+} : P (ξ)}$ ; since every nonempty set of ordinals has an $\in$ -least element (Trichotomy and well-ordering of the ordinals), it follows that if $P$ holds at $ξ$ whenever it holds at every ordinal in $ξ$ , then $P$ holds at every ordinal.

Proof

technique · direct

1.1L1L2L3L4

Claim (a) at $γ = 0$ and at a successor: $α \cdot (β + 0) = α \cdot β = α \cdot β + 0 = α \cdot β + α \cdot 0$ ; and assuming $α (β + δ) = α β + α δ$ , the successor clauses give $α (β + δ^{+}) = α ((β + δ)^{+}) = α (β + δ) + α = (α β + α δ) + α = α β + (α δ + α) = α β + α δ^{+}$ , the middle equality by [L3].

1.2L4

Claim (a) at a limit $γ = λ$ when $α = 0$ : both sides are $0$ , since $0 \cdot μ = 0$ for every $μ$ by [L4].

1.3L1L2L4

Claim (a) at a limit $γ = λ$ when $α > 0$ , assuming $α (β + ξ) = α β + α ξ$ for every $ξ \in λ$ : the set $D = {β + ξ : ξ \in λ}$ is a nonempty subset of the limit ordinal $β + λ$ with $\sup D = β + λ$ by [L2] and [L4], so $α (β + λ) = \sup {α (β + ξ) : ξ \in λ} = \sup {α β + α ξ : ξ \in λ}$ by [L4]; and $E = {α ξ : ξ \in λ}$ is a nonempty subset of the limit ordinal $α \cdot λ$ with $\sup E = α \cdot λ$ by [L1] and [L4], so $α β + α λ = \sup {α β + α ξ : ξ \in λ}$ by [L4]; the two right-hand sides are the same set's supremum.

2.1step 1.1step 1.2step 1.3L5L6

The three cases of [L5] are exhaustive and steps 1.1, 1.2 and 1.3 derive claim (a) at $γ$ from claim (a) at every ordinal in $γ$ , so by [L6] claim (a) holds for all ordinals $α$ , $β$ , $γ$ .

3.1step 2.1L1L4L5L6

Claim (b), by induction on $γ$ . At $γ = 0$ both sides are $0$ by [L1] and [L4]. At $γ = δ^{+}$ , assuming $(α β) δ = α (β δ)$ : $(α β) δ^{+} = (α β) δ + α β = α (β δ) + α β = α (β δ + β) = α (β δ^{+})$ , the third equality being step 2.1. At $γ = λ$ a limit: if $α = 0$ or $β = 0$ then both sides are $0$ by [L4], since $α β = 0$ in that case and $α \cdot (β \cdot λ)$ is $0$ either because $α = 0$ or because $β \cdot λ = 0$ ; otherwise $α > 0$ and $β > 0$ , so $α β > 0$ by [L4], and assuming $(α β) ξ = α (β ξ)$ for every $ξ \in λ$ one gets $(α β) λ = \sup {(α β) ξ : ξ \in λ} = \sup {α (β ξ) : ξ \in λ}$ by [L1], while $D = {β ξ : ξ \in λ}$ is a nonempty subset of the limit ordinal $β \cdot λ$ with $\sup D = β \cdot λ$ , so $α (β λ) = \sup {α (β ξ) : ξ \in λ}$ by [L4]; the three cases of [L5] are exhaustive, so [L6] gives claim (b) for all $γ$ .

4.1step 2.1step 3.1∎

Claims (a) and (b) are established.

Remarks

Where the limit cases really need the continuity clause. In both inductions the limit step is the assertion that multiplication on the left commutes with a supremum taken over any set unbounded in a limit ordinal. That is claim (f) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + β = β$ and $1 \cdot β = β$ in its refined form, and it is used twice in step 1.3 and once in step 3.1. Without it one is left comparing $\sup {α (β + ξ)}$ with $\sup {α β + η : η < α λ}$ , which are indexed by different sets.

The degenerate cases are not decoration. At $α = 0$ the ordinal $α \cdot λ$ is $0$ , not a limit, so the continuity clause does not apply and the case has to be handled separately; the same happens in claim (b) at $β = 0$ . Both are one line, and both are wrong to skip.

Right distributivity is false, so the two laws are not a package. $(1 + 1) \cdot ω = 2 \cdot ω = ω$ , while $1 \cdot ω + 1 \cdot ω = ω + ω$ , which is strictly larger. That computation is FALSE: $(β + γ) \cdot α = β \cdot α + γ \cdot α$ for all ordinals.

Commutativity fails too, and separately. $2 \cdot ω = ω$ while $ω \cdot 2 = ω + ω$ , so associativity and left distributivity are the whole of what survives; the computation is FALSE: ordinal multiplication is commutative. These are the two refutations the Statement above points at.

Ordinal multiplication is associative, and $α \cdot (β + γ) = α \cdot β + α \cdot γ$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Ordinal multiplication is associative, and α⋅(β+γ)=α⋅β+α⋅γα⋅(β+γ)=α⋅β+α⋅γ

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Ordinal multiplication is associative, and $α \cdot (β + γ) = α \cdot β + α \cdot γ$