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Ordinal multiplication is associative, and $\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gamma$

Statement

For all ordinals $\alpha$ , $\beta$ , $\gamma$ (Ordinal (von Neumann)), with $+$ and $\cdot$ as in Ordinal addition $\alpha + \beta$ and Ordinal multiplication $\alpha \cdot \beta$ :

(a) Left distributivity. $\alpha \cdot (\beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$ .

(b) Associativity. $(\alpha \cdot \beta) \cdot \gamma = \alpha \cdot (\beta \cdot \gamma)$ .

Distributivity holds on the left only. The right-hand law $(\beta + \gamma) \cdot \alpha = \beta \cdot \alpha + \gamma \cdot \alpha$ 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 $\alpha$ , $\beta$ , $\gamma$ . For a set $A$ of ordinals, $\sup A = \bigcup A$ is its least upper bound (Basic closure properties of ordinals, claim (e)).

[L1]

$\alpha \cdot 0 = 0$ , $\alpha \cdot \delta^{+} = \alpha \cdot \delta + \alpha$ , and $\alpha \cdot \lambda = \sup\{\alpha \cdot \xi : \xi \in \lambda\}$ for limit $\lambda$ (Ordinal multiplication $\alpha \cdot \beta$ ).

[L2]

$\alpha + 0 = \alpha$ , $\alpha + \delta^{+} = (\alpha + \delta)^{+}$ , and $\alpha + \lambda = \sup\{\alpha + \xi : \xi \in \lambda\}$ for limit $\lambda$ (Ordinal addition $\alpha + \beta$ ).

[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 $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ : $\beta \cdot 0 = 0 \cdot \beta = 0$ and $\beta + 0 = 0 + \beta = \beta$ (claim (a)); $\beta < \gamma$ implies $\alpha + \beta < \alpha + \gamma$ (claim (b)); for $\alpha > 0$ , $\beta < \gamma$ implies $\alpha\beta < \alpha\gamma$ , and $\alpha \cdot \beta = 0$ exactly when $\alpha = 0$ or $\beta = 0$ (claim (d)); if $\mu$ is a limit ordinal and $D \subseteq \mu$ is nonempty with $\sup D = \mu$ , then $\alpha + \mu = \sup\{\alpha + \eta : \eta \in D\}$ and, for $\alpha > 0$ , $\alpha \cdot \mu = \sup\{\alpha \cdot \eta : \eta \in D\}$ (claim (f)); and $\beta + \lambda$ and, for $\beta > 0$ , $\beta \cdot \lambda$ are limit ordinals whenever $\lambda$ 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 $\beta_0$ , apply Transfinite induction to the well-order $(\beta_0^{+}, \in)$ and to $S = \{\xi \in \beta_0^{+} : P(\xi)\}$ ; 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 $\xi$ whenever it holds at every ordinal in $\xi$ , then $P$ holds at every ordinal.

Proof

technique · direct

1.1

Claim (a) at $\gamma = 0$ and at a successor: $\alpha \cdot (\beta + 0) = \alpha \cdot \beta = \alpha \cdot \beta + 0 = \alpha \cdot \beta + \alpha \cdot 0$ ; and assuming $\alpha(\beta + \delta) = \alpha\beta + \alpha\delta$ , the successor clauses give $\alpha(\beta + \delta^{+}) = \alpha((\beta + \delta)^{+}) = \alpha(\beta + \delta) + \alpha = (\alpha\beta + \alpha\delta) + \alpha = \alpha\beta + (\alpha\delta + \alpha) = \alpha\beta + \alpha\delta^{+}$ , the middle equality by [L3].

L1L2L3L4

1.2

Claim (a) at a limit $\gamma = \lambda$ when $\alpha = 0$ : both sides are $0$ , since $0 \cdot \mu = 0$ for every $\mu$ by [L4].

1.3

Claim (a) at a limit $\gamma = \lambda$ when $\alpha > 0$ , assuming $\alpha(\beta + \xi) = \alpha\beta + \alpha\xi$ for every $\xi \in \lambda$ : the set $D = \{\beta + \xi : \xi \in \lambda\}$ is a nonempty subset of the limit ordinal $\beta + \lambda$ with $\sup D = \beta + \lambda$ by [L2] and [L4], so $\alpha(\beta + \lambda) = \sup\{\alpha(\beta + \xi) : \xi \in \lambda\} = \sup\{\alpha\beta + \alpha\xi : \xi \in \lambda\}$ by [L4]; and $E = \{\alpha\xi : \xi \in \lambda\}$ is a nonempty subset of the limit ordinal $\alpha \cdot \lambda$ with $\sup E = \alpha \cdot \lambda$ by [L1] and [L4], so $\alpha\beta + \alpha\lambda = \sup\{\alpha\beta + \alpha\xi : \xi \in \lambda\}$ 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 $\gamma$ from claim (a) at every ordinal in $\gamma$ , so by [L6] claim (a) holds for all ordinals $\alpha$ , $\beta$ , $\gamma$ .

step 1.1step 1.2step 1.3L5L6

3.1

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

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 $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ 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 $\sup\{\alpha(\beta + \xi)\}$ with $\sup\{\alpha\beta + \eta : \eta < \alpha\lambda\}$ , which are indexed by different sets.

The degenerate cases are not decoration. At $\alpha = 0$ the ordinal $\alpha \cdot \lambda$ 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 $\beta = 0$ . Both are one line, and both are wrong to skip.

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

Commutativity fails too, and separately. $2 \cdot \omega = \omega$ while $\omega \cdot 2 = \omega + \omega$ , 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

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Direct dependencies and their dependencies through the next three levels: 41 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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⋅γ\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gammaα⋅(β+γ)=α⋅β+α⋅γ