# Definition of ordinal addition by transfinite induction (ZFC)

I work here in $$\mathrm{ZFC}$$. Transfinite induction is the following observation:

Let $$P(x)$$ be a property. If

$$(1) \ P(0)$$ is true

$$(2)$$ for any ordinal $$\alpha:$$ $$P(\alpha)$$ is true $$\Rightarrow P(\alpha + 1)$$ is true

$$(3)$$ for any limit ordinal $$\alpha$$: $$P(\beta)$$ is true for all $$\beta < \alpha \Rightarrow P(\alpha)$$ is true

Then $$P(\alpha)$$ is true for all ordinals $$\alpha$$

Then one definition of ordinal addition is the following:

Let $$\alpha$$ be an ordinal.

$$(1) \ \alpha + 0 = \alpha$$

$$(2) \ \forall$$ ordinals $$\beta: \alpha + (\beta + 1) = (\alpha + \beta) + 1$$

$$(3) \forall$$ limit ordinals $$\beta: \alpha + \beta = \mathrm{sup}\{ \alpha + \gamma \ | \ \gamma < \beta \}$$

How can we actually make sure that the preceding definition of ordinal addition is actually... a definition? That it makes sense? Probably, rigorously it means that we need to shows that $$\alpha_1 = \alpha_2, \beta_1 = \beta_2 \Rightarrow \alpha_1 + \beta_1 = \alpha_2 + \beta_2$$. Which can probably be attempted by transfinite induction. However, we also need to make sense of $$\{ \alpha + \gamma \ | \gamma < \beta \}$$. It should be proved that it is even a set.

Here lies the problem. If one tried to first prove that that definition for addition defines the unique "sum", then one would stumble on the fact that we don't know if $$\{ \alpha + \gamma \ | \gamma < \beta \}$$ is a set. But in other order one doesn't know if the sum is uniquely defined.

• You do not define things by induction but rather by recursion. You need an appropriate form of the transfinite recursion theorem to conclude the existence (and uniqueness) of $+$. Nov 3, 2018 at 0:03

We know $\{ \alpha + \gamma \ | \gamma < \beta \}$ is a set by the axiom of replacement. We replace each element of $\gamma$ by its addition with $\alpha$. For each element of $\gamma$ we have already defined that sum.

Actually, Instead of appealing to the primitive/usual form of Transfinite Recursion, you must appeal to the Parametric Version of it, quoted below. You can search the textbook Introduction to Set Theory by Karel Hrbacek and Thomas Jech to get more information about this version.

Let $$V$$ be the class of all sets, $$\operatorname{Ord}$$ be the class of all ordinals, and $$G_1,G_2,G_3$$ be class functions from $$V$$ to $$V$$.

There exists a class function $$F:V\times\operatorname{Ord}\to V$$ such that, for all $$z\in V$$

• $$F(z,0)=G_1(z,\emptyset)$$

• $$F(z,\alpha+1)=G_2(z,F_z(\alpha))$$ for all $$\alpha\in\operatorname{Ord}$$, with $$F_z(\alpha):=F(z,\alpha)$$

• $$F(z,\alpha)=G_3(z,F_z\restriction\alpha)$$ for all $$\alpha\neq 0$$ limit, with $$F_z\restriction \alpha:=\{\langle\beta,F(z,\beta)\rangle\mid\beta<\alpha\}$$

Here is my take to define not only addition but also multiplication and exponentiation:

We define $$G_1,G_2,G_3$$ as follows:

• $$G_1(\alpha,\beta)=\alpha$$

• $$G_2(\alpha,\beta)=\beta+1$$

• $$G_3(\alpha,\beta)=\begin{cases} \sup(\operatorname{ran}(\beta))&\text{if }\beta\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}$$

By Theorem, there is $$F:V\times\operatorname{Ord}\to V$$ such that

• $$F(\alpha,0)=G_1(\alpha,\emptyset)=\alpha$$

• $$F(\alpha,\beta+1)=G_2(\alpha,F_\alpha(\beta))= F_\alpha(\beta)+1= F(\alpha,\beta)+1$$ for all $$\alpha\in\operatorname{Ord}$$

• $$F(\alpha,\beta)=G_3(\alpha,F_\alpha\restriction\beta)=\sup(\operatorname{ran}(F_\alpha\restriction\beta))=\sup(\{F(\alpha,\gamma)\mid \gamma<\beta\})$$ for all $$\alpha\neq 0$$ limit

MILTIPLICATION

We define $$G_1,G_2,G_3$$ as follows:

• $$G_1(\alpha,\beta)=0$$

• $$G_2(\alpha,\beta)=\beta+\alpha$$

• $$G_3(\alpha,\beta)=\begin{cases} \sup(\operatorname{ran}(\beta))&\text{if }\beta\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}$$

By Theorem, there is $$F:V\times\operatorname{Ord}\to V$$ such that

• $$F(\alpha,0)=G_1(\alpha,\emptyset)=0$$

• $$F(\alpha,\beta+1)=G_2(\alpha,F_\alpha(\beta)) = F_\alpha(\beta)+\alpha = F(\alpha,\beta)+\alpha$$ for all $$\alpha\in\operatorname{Ord}$$

• $$F(\alpha,\beta)=G_3(\alpha,F_\alpha\restriction\beta)=\sup(\operatorname{ran}(F_\alpha\restriction\beta))=\sup(\{F(\alpha,\gamma)\mid \gamma<\beta\})$$ for all $$\alpha\neq 0$$ limit

EXPONENTIATION

We define $$G_1,G_2,G_3$$ as follows:

• $$G_1(\alpha,\beta)=1$$

• $$G_2(\alpha,\beta)=\beta\cdot\alpha$$

• $$G_3(\alpha,\beta)=\begin{cases} \sup(\operatorname{ran}(\beta))&\text{if }\beta\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}$$

By Theorem, there is $$F:V\times\operatorname{Ord}\to V$$ such that

• $$F(\alpha,0)=G_1(\alpha,\emptyset)=1$$

• $$F(\alpha,\beta+1) = G_2(\alpha,F_\alpha(\beta)) = F_\alpha(\beta)\cdot \alpha = F(\alpha,\beta)\cdot\alpha$$ for all $$\alpha\in\operatorname{Ord}$$

• $$F(\alpha,\beta)=G_3(\alpha,F_\alpha\restriction\beta)=\sup(\operatorname{ran}(F_\alpha\restriction\beta))=\sup(\{F(\alpha,\gamma)\mid \gamma<\beta\})$$ for all $$\alpha\neq 0$$ limit