Ordinal Arithmetic (Non-Commutative)

What an ordinal is

An ordinal is the order type of a well-ordered set — a label for "the shape" of an increasing chain. The finite ordinals are $0,1,2,\dots$; the first infinite one is

$\omega = \text{order type of } (0 \prec 1 \prec 2 \prec \cdots),$

the natural numbers in their usual order. Ordinals are compared by length: $\alpha < \beta$ means a copy of $\alpha$ is a proper initial segment of $\beta$.

Sum: lay one order after the other

$\alpha + \beta$ is the order type you get by placing a copy of $\alpha$ first, then a copy of $\beta$ entirely above it. Formally it is defined by transfinite recursion:

$\alpha + 0 = \alpha,\qquad \alpha + (\beta+1) = (\alpha+\beta)+1,\qquad \alpha + \lambda = \sup_{\xi<\lambda}(\alpha+\xi)\ \ (\lambda\text{ limit}).$

The recursion runs on the right argument, which is exactly why left and right behave differently.

• $\omega + 2$: an $\omega$-chain, then two extra points on top — it has a new largest pair and a length strictly past $\omega$, so $\omega + 2 > \omega$.
• $2 + \omega$: two points, then a full $\omega$-chain on top. But "two points followed by $\omega$" is still just one increasing $\omega$-chain with no last element, so $2 + \omega = \omega$.

The finite head is absorbed: it sits below a copy of $\omega$ and gets swallowed by the limit. In general $n + \omega = \omega$ for every finite $n$, while $\omega + n \neq \omega$ for $n\ge 1$.

Product: replace each point by a block

$\alpha \cdot \beta$ is $\beta$ copies of $\alpha$, laid end to end (read it as "$\beta$ lots of $\alpha$"): replace every point of $\beta$ by a whole copy of $\alpha$.

• $\omega \cdot 2 = \omega + \omega$: two $\omega$-blocks in a row — length past $\omega$, so $\omega\cdot 2 > \omega$.
• $2 \cdot \omega$: $\omega$-many copies of the 2-block, i.e. $2+2+2+\cdots$, which is again a single $\omega$-chain, so $2 \cdot \omega = \omega$.

Same asymmetry as addition. Multiplication distributes only on the left:

$\alpha\cdot(\beta+\gamma) = \alpha\cdot\beta + \alpha\cdot\gamma\quad(\text{holds}),\qquad (\beta+\gamma)\cdot\alpha \ne \beta\cdot\alpha + \gamma\cdot\alpha\quad(\text{can fail}).$

For example $(1+1)\cdot\omega = 2\cdot\omega = \omega$, but $1\cdot\omega + 1\cdot\omega = \omega+\omega = \omega\cdot 2 \ne \omega$.

Side-by-side

Key fact used throughout: what is added/multiplied on the left, below a limit, gets absorbed; what is on the right survives.

Worked mini-example

Can $\mathbb{N}$ be re-ordered to have order type $\omega+2$? Yes — it is a countable ordinal, and a countably infinite set realizes any countable order type. Concretely re-order $\mathbb{N}$ as

$2 \prec 3 \prec 4 \prec \cdots \prec 0 \prec 1,$

i.e. $\{2,3,4,\dots\}$ as a copy of $\omega$, then $0,1$ on top: the result is well-ordered with order type exactly $\omega+2$. By contrast the usual order on $\mathbb{N}$ already has type $\omega$, and since $2+\omega=\omega$, that order also realizes $2+\omega$ — no re-ordering needed.

Exam trick: Ordinal $+$ and $\cdot$ are associative but not commutative, and the recursion runs on the right operand. So a finite ordinal on the left of $\omega$ vanishes ($n+\omega=\omega$, $n\cdot\omega=\omega$) while on the right it grows the ordinal ($\omega+n>\omega$, $\omega\cdot n>\omega$). Exams love to swap the operands ($\omega+2$ vs $2+\omega$) to see if you treat them as equal — they are not. Remember too that all of these are still countable, so each is separately realizable on $\mathbb{N}$.