Let $\mathbf{Ord}$ denote the class of ordinals. Equipped with its natural (commutative) sum and product, this is a semiring.

Say that a composition on $\mathbf{Ord}$ is a function $\circ: \mathbf{Ord} \times \mathbf{Ord}^{\geq \omega} \longrightarrow \mathbf{Ord}$, where $\mathbf{Ord}^{\geq \omega}$ is the class of transfinite ordinals, such that:

$(i)$: For $\varphi \in \mathbf{Ord}$, the map $_{\varphi}\circ: \mathbf{Ord}^{\geq \omega} \longrightarrow \mathbf{Ord}$ defined by $_{\varphi}\circ(\xi)=\varphi \circ \xi$ is strictly increasing, and $_{\omega}\circ=\operatorname{id_{\mathbf{Ord}^{\geq \omega}}}$.

$(ii)$: For $\alpha \in \mathbf{Ord}$ and $\beta,\gamma \in \mathbf{Ord}^{\geq \omega}$, we have $\beta \circ \gamma \in \mathbf{Ord}^{\geq \omega}$ and $\alpha \circ (\beta \circ \gamma)=(\alpha \circ \beta) \circ \gamma$.

$(iii$): For $\xi \in \mathbf{Ord}^{\geq \omega}$, the map $\circ_{\xi}: \mathbf{Ord}\longrightarrow \mathbf{Ord}$ defined by $\circ_{\xi}(\varphi)=\varphi \circ \xi$ is a strictly increasing semi-ring morphism with $\circ_{\omega}=\operatorname{id}_{\mathbf{Ord}}$.

Thus a composition is a way of seeing each ordinal number as an ordinal function with conditions of compatibility with the semi-ring and order structures.

Are there known compositions on the class of ordinal numbers?

  • $\begingroup$ What is $\bf Ord^{\geq\omega}$ for you? $\endgroup$
    – Asaf Karagila
    Nov 15, 2018 at 23:42
  • $\begingroup$ @AsafKaragila I just edited with the definition. $\endgroup$
    – nombre
    Nov 16, 2018 at 10:33
  • $\begingroup$ Ah. Because I thought you meant all the infinite sequences of ordinals. $\endgroup$
    – Asaf Karagila
    Nov 16, 2018 at 10:34
  • $\begingroup$ Yes I only realized now this was already a common notation for that. So, just ordinals! $\endgroup$
    – nombre
    Nov 16, 2018 at 10:36
  • $\begingroup$ Well, that makes (ii) much more reasonable now. $\endgroup$
    – Asaf Karagila
    Nov 16, 2018 at 10:49

1 Answer 1


So there is at least a somewhat natural way to define a composition of ordinals. I describe it briefly here. The proof that everything works is tedious.

First define $\mathbf{\Omega}_0=\omega^{\mathbf{Ord}}$, and for each ordinal $0<\alpha$, define the class $\mathbf{\Omega}_{\alpha}$ as the class of common fixed points of the functions $\Omega_{\beta},\beta<\alpha$ where $\Omega_{\beta}$ is the isomorphism $(\mathbf{Ord},<) \rightarrow (\mathbf{\Omega}_{\beta},<)$.

For $0<\alpha \in \mathbf{Ord}$, let $F_{\alpha}$ denote the isomorphism $(\mathbf{Ord}^{\geq \omega},<) \rightarrow (\mathbf{\Omega}_{\alpha} \setminus \mathbf{\Omega}_{\alpha+1},<)$. The function $F_0$ is defined by $F_0(\omega\otimes(1\oplus \beta)\oplus n)=2^n \omega^{\omega \otimes(1\oplus f(\beta))}$ for all $\beta \in \mathbf{Ord}$ and $n<\omega$, where $\oplus$ and $\otimes$ are the ordinal sum and product and $f$ is the isomorphism $(\mathbf{Ord},<) \rightarrow (\mathbf{Ord} \setminus \mathbf{\Omega}_1,<)$. All functions $F_{\alpha}$ are strictly increasing with $\forall \alpha<\beta \in \mathbf{Ord},\forall \xi \in \mathbf{Ord}^{\geq \omega}, \xi<F_{\alpha}(\xi)<F_{\beta}(\xi)$.

We also have $F_0(\xi+\psi)=F_0(\xi) F_0(\psi)$ and $F_0(\xi+n)=2^nF_0(\xi)$ for all ordinals $\xi,\psi \geq \omega$ and $n<\omega$.

Every ordinal $\alpha$ can be uniquely written as $\alpha=P_0(\omega) \alpha_0 +...+P_r(\omega) \alpha_r$ where $r \in \mathbb{N}$, $\alpha_0>...>\alpha_r$ are ordinals of the form $\alpha_i=\omega^{\beta_i}$ where $\beta_i$ is a limit ordinal, and $P_0,...,P_r \in \mathbb{N}[X]$. In this notation, each ordinal $\alpha_i$ for $0 \leq i \leq r$ with $\alpha_i>1$ can be written uniquely as $F_{\nu_{\alpha_i}}(\alpha_i')$ for certain $\nu_{\alpha_i} \in \mathbf{Ord}$ and $\alpha_i' \in \mathbf{Ord}^{\geq \omega}$. We define $\alpha \circ \beta$ inductively as $\alpha \circ \beta = \sum \limits_{\alpha_i>1} P_i(\beta)F_{\nu_{\alpha_i}}(\alpha_i' \circ \beta)+ \sum \limits_{\alpha_r=1} P_r(\beta)$.

This makes $\circ$ the unique composition of ordinals such that for $\alpha \in \mathbf{Ord}$ and $\xi \in \mathbf{Ord}^{\geq \omega}$, we have $F_{\alpha}(\omega) \circ \xi = F_{\alpha}(\xi)$.

For the general method to work, the function $F_0$ should transform sums into products and the ranges of the other functions $F_{\alpha}$ should be mutually disjoint subclasses of the class of ordinals $\omega^{\beta}$ where $0<\beta$ is limit. But there still is some freedom in the choice of functions $F_{\alpha}$. For instance one can make it so each function $F_{\alpha+1}$ is a solution of the Abel equation $\forall \omega\leq\xi,F_{\alpha+1}(\xi+1)=F_{\alpha}(F_{\alpha+1}(\xi))$.


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