# Can transfinite numbers be extended to more general algebraic structures?

Infinite ordinals and cardinals are beautiful mathematical objects, extending in a natural way $\mathbb{N}$ and sharing most of its properties. I have always wondered whether it would be possible to generalize the constructions of $\mathbb{Z}$ or $\mathbb{Q}$ to get transfinite integers or rationals. In general, putting a group structure on classes extending $\mathrm{Ord}$ or $\mathrm{Card}$ is not possible as I indicated some years ago. Basically, we would have $1+{\aleph}_{0}={\aleph}_{0}$ (and also $1+\omega =\omega $) which would imply $1=0$. However, we can consider a slightly modified problem with weaker constraints, as follows. First we assume we have a class of cardinals $C\subseteq \mathrm{Card}$ containing zero and stable by addition.

$$\text{(1)}\phantom{\rule{4em}{0ex}}\{\begin{array}{l}0\in C\\ \forall \kappa ,\lambda \in C,\kappa +\lambda \in C\end{array}$$

We define the class ${Z}_{C}=C\cup {C}^{-}$, where ${C}^{-}=\{-\kappa \mid \kappa \in C,\kappa \ne 0\}$ is a representation of "negative" cardinals. As we have seen, the class ${Z}_{C}$ need not be a group. Nevertheless, we can try to find an equivalence class $R$ over ${Z}_{C}$ such that ${G}_{C}=\raisebox{1ex}{${Z}_{C}$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$ is a group. Moreover, we want this equivalence class to be compatible with addition and opposite i.e.

$$\text{(2)}\phantom{\rule{4em}{0ex}}\begin{array}{c}\forall \kappa ,\lambda \in C,\phantom{\rule{mediummathspace}{0ex}}\overline{\kappa +\lambda}=\overline{\kappa}+\overline{\lambda}\\ \forall \kappa \in C,\phantom{\rule{mediummathspace}{0ex}}\overline{-\kappa}=-\overline{\kappa}\end{array}$$

Of course this implies that $\overline{0}$ is the identity element $0$ of the expected group. Similarly, we can settle the same problem for a class $C\subseteq \mathrm{Ord}$ and + is now understood as the ordinal addition. Can we follow this schema to contruct interesting infinite algebraic structures?

For $C\subseteq \mathrm{Card}$, we have for any infinite $\kappa \in C$ the relation $\overline{\kappa}=\overline{\kappa +\kappa}=\overline{\kappa}+\overline{\kappa}$ so $\overline{\kappa}=0$. Hence ${G}_{C\cap \omega}={G}_{C}$ and the initial problem is reduced to the case where $C\subseteq \omega $ contains only finite cardinals (= finite ordinals) and so the problem is still not really exciting. What about the case $C\subseteq \mathrm{Ord}$? In general it is possible to have ${\beta}_{1}+\gamma ={\beta}_{2}+\gamma $ without ${\beta}_{1},{\beta}_{2}$ being equal and so for any $\alpha $ such that ${\beta}_{1}\alpha ,{\beta}_{2}\alpha ,\mathrm{\gamma \alpha}\in C$, we get that $\overline{{\beta}_{1}\alpha}=\overline{{\beta}_{2}\alpha}$. In particular for any infinite ordinal $\gamma $ and $\alpha \in C$ such that $\mathrm{\gamma \alpha}\in C$ we obtain $\overline{\alpha}=0$ because $1+\gamma =\gamma $ (if this is not obvious to you, a more general statement is proved below). As a consequence, more general assumptions on the class $C$ strongly limit the structure of the group. For example if $C$ is closed under an infinite sum ($C=\mathrm{Ord}$ for example) then ${\text{G}}_{\text{C}}$ is trivial. Similarly, if we require $C$ to be stable by multiplication (in order to define a ring structure for example) then ${\text{G}}_{\text{C}}$ is trivial whenever $C$ contains an infinite ordinal $\gamma $ (hence the remaining case is again $C\subseteq \omega $). If ${G}_{C}$ is not trivial, we denote ${\alpha}_{0}\in C$ the least element such that $\overline{{\alpha}_{0}}\ne 0$ (in particular ${\alpha}_{0}>0$).

One additional natural hypothesis should be added. We know that for any ordinal $\alpha <\beta $ there exists a unique ordinal $\gamma $ such that $\alpha =\beta +\gamma $. We would very like $\overline{\gamma}$ to match the difference "$-\overline{\beta}+\overline{\alpha}$". For that purpose, we only require $\gamma $ to belong to $C$. Hence we assume that

$$\text{(3)}\phantom{\rule{4em}{0ex}}\forall \alpha ,\beta \in C,\forall \gamma \in \mathrm{Ord},\alpha =\beta +\gamma \Rightarrow \gamma \in C$$

Let's come back to the case $C\subseteq \omega $ with the new assumption (3) and suppose that ${G}_{C}$ is not trivial. Then ${\alpha}_{0}<\omega $ and any finite $m\in C$ and be written $m={\alpha}_{0}q+r$ with $r<{\alpha}_{0}$. By (3), $r\in C$ and because $C$ is stable by finite sums, $\overline{m}=\overline{{\alpha}_{0}}q+\overline{r}=\overline{{\alpha}_{0}}q$. Thus ${G}_{C}$ is the monogenic group generated by $\overline{{\alpha}_{0}}$.

What about the general case? Unfortunately, it turns out that if ${G}_{C}$ is not trivial it is still a monogenic group generated by $\overline{{\alpha}_{0}}$. To prove this, we need to recall some equalities on ordinals. First, for any $\alpha >0$, we have $1+{\omega}^{\alpha}={\omega}^{\alpha}$. This is clearly true for $\alpha =1$ and we prove the general case by induction on $\alpha $: $1+{\omega}^{\alpha +1}=1+\omega {\omega}^{\alpha}=1+\left(1+\omega \right){\omega}^{\alpha}=\left(1+{\omega}^{\alpha}\right){+\omega}^{\alpha +1}={\omega}^{\alpha}+{\omega}^{\alpha +1}=\left(1+\omega \right){\omega}^{\alpha}=\omega {\omega}^{\alpha}={\omega}^{\alpha +1}$ and $1+{\omega}^{\lambda}=\underset{0<\alpha <\lambda}{sup}\left({1+\omega}^{\alpha}\right)=\underset{0<\alpha <\lambda}{sup}{\omega}^{\alpha}={\omega}^{\lambda}$. Now, if we have two ordinals $\alpha <\beta $ and if $\gamma >0$ is such that $\beta =\alpha +\gamma $ we have ${\omega}^{\alpha}+{\omega}^{\beta}={\omega}^{\alpha}\left(1+{\omega}^{\gamma}\right)={\omega}^{\alpha}{\omega}^{\gamma}={\omega}^{\beta}$. Finally, if we have two ordinals ${\gamma}_{1}<{\gamma}_{2}$ we can write their Cantor Normal Forms:

$$\text{(4)}\phantom{\rule{4em}{0ex}}\begin{array}{c}{\gamma}_{1}={\omega}^{{\beta}_{1}^{1}}{k}_{1}^{1}+{\omega}^{{\beta}_{2}^{1}}{k}_{2}^{1}+\dots +{\omega}^{{\beta}_{n}^{1}}{k}_{n}^{1}\\ {\gamma}_{2}={\omega}^{{\beta}_{1}^{2}}{k}_{1}^{2}+{\omega}^{{\beta}_{2}^{2}}{k}_{2}^{2}+\dots +{\omega}^{{\beta}_{n}^{2}}{k}_{m}^{2}\end{array}$$

where ${\beta}_{1}^{2}\ge {\beta}_{1}^{i}>{\beta}_{2}^{i}>\dots >{\beta}_{2}^{i}$ and ${k}_{j}^{i}$ are positive integers. Using the equality ${\omega}^{\alpha}+{\omega}^{\beta}={\omega}^{\beta}$ for $\alpha <\beta $ it is clear that ${\gamma}_{1}+{\gamma}_{2}={\gamma}_{2}$ if ${\beta}_{1}^{2}>{\beta}_{1}^{1}$.

Now our element ${\alpha}_{0}\in C$ can be written ${\alpha}_{0}={\omega}^{\beta}k+\gamma $ where ${\omega}^{\beta}k$ is the first term of its Cantor Normal Form. For any $\delta \in C$, we can write its euclidean division by ${\alpha}_{0}$: $\delta ={\alpha}_{0}l+\epsilon $ where $\epsilon <{\alpha}_{0}$. By the previous discussion, if $l\ge \omega $ then we can write the Cantor Normal Forms of the two elements ${\alpha}_{0}<\delta $ as in (4) and see that ${\beta}_{1}^{2}>{\beta}_{1}^{1}$. Hence ${\alpha}_{0}+\delta =\delta $ and so $\overline{{\alpha}_{0}}=0$, which is a contradiction. So $l<\omega $ and because $C$ is stable by finite sums, ${\alpha}_{0}l\in C$ and by the property (3) we get $\epsilon \in C$. This means that $\overline{\delta}=\overline{{\alpha}_{0}}l+\overline{\epsilon}=\overline{{\alpha}_{0}}l$ and hence ${G}_{C}$ is generated by $\overline{{\alpha}_{0}}$ as claimed above.

As a conclusion, if $C\subseteq \mathrm{Card}$ satisfies the properties (1), (2) above and ${G}_{C}=\raisebox{1ex}{${Z}_{C}$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$ is a group then $C\subseteq \omega \subseteq \mathrm{Ord}$. If $C\subseteq \mathrm{Ord}$ satisfies properties (1), (2), (3) above and ${G}_{C}=\raisebox{1ex}{${Z}_{C}$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$ is a group then ${G}_{C}$ is isomorphic to $\mathbb{Z}$ or to $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$n\mathbb{Z}$}\right.$. Conversely, we can build these groups from $C=\omega $ by defining the relation $R$ as the equality (respectively the equality modulo $n$). But we do not get any new groups...