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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

1.1

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].

L1L2L3L4

1.2

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

1.3

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.

L1L2L4

2.1

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 $α$ , $β$ , $γ$ .

step 1.1step 1.2step 1.3L5L6

3.1

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 $γ$ .

step 2.1L1L4L5L6

4.1

Claims (a) and (b) are established.

step 2.1step 3.1∎

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.

Depends on

Used by

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ordinal arithmetic (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 2 (Ordinal numbers) (standard reference, not scraped)
Open Logic Project, Open Logic Text (standard reference, not scraped)

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