Is possibile to define an exponentiation with respect an ordinal operation?

Question

It is well know the following resul holds.

Theorem

For any $(M,\bot,e)$ monoid there exists a unique esternal operation $\wedge_\ast$ from $X\times\omega$ into $X$ such that for any $x$ in $M$ the following results hold:

1. the equality $$ x\curlywedge_\bot 0=e $$
2. the equality holds $$ x\curlywedge_\bot(n+1)=(x\curlywedge_\bot n)\bot x $$ for any $n$ in $\omega$.

So let's we call $\curlywedge_\bot$ exponentiation of $\curlywedge_\bot$ so that let's we observe that the usual multiplication into $\omega$ is just the exponentiation of the usual summ. So observing the definition of ordinal multiplication I advanced the following conjecture: to follow I indicate the ordinal class with the symbol $\mathbf{Ord}$ whereas I indicate the limit ordinal class with the symbol $\mathbf{Lim}$.

Conjecture

Let $\bot$ be an operation on $\mathbf{Ord}$ with a neutral element $\mu$ in $\mathbf{Ord}$. Then there exists a unique operation $\curlywedge_\bot$ on $\mathbf{Ord}$ such that for every $\alpha$ in $\mathbf{Ord}$, the following statements hold:

1. $\alpha\curlywedge_\bot 0 = \mu$
2. $\alpha\curlywedge_\bot(\beta+1)=(\alpha\curlywedge_\bot\beta)\bot\alpha$ for every $\beta\in\mathbf{Ord}$
3. $\alpha\curlywedge\beta=\sup_{\gamma\in\beta}(\alpha\curlywedge_\bot\gamma)$ for every $\beta\in\mathbf{Lim}$ not empty.

So is the above conjecture true? if it is ture then how to prove it? can I prove it with recursion? Could someone help me, please?

Answer 1

By examining accurately the existence proof of sum I elaborated by my self what to follow.

Let us first recall the following theorem: to follow we will indicate with $\mathbf{Ord}$ the class of ordinals and with $\mathbf{Lim}$ the class of limit ordinals.

Theorem $0$

If $\mathbf F$, $\mathbf G$, and $\mathbf H$ are functionals, then there exists a unique functional $\Psi$ in $\mathbf{Ord}$ such that the following statements hold:

• $\Psi(\emptyset)$ equals $\mathbf F(\emptyset)$;
• $\Psi\big(S(\alpha)\big)$ equals $\mathbf G\big(\Psi(\alpha)\big)$ for every $\alpha$ in $\mathbf{Ord}$;
• $\Psi(\lambda)$ equals $\mathbf H(\Psi\restriction_\lambda)$ for every non-empty $\lambda$ in $\mathbf{Lim}$.

Now, armed with Theorem $0$, let's we prove the following important and useful result.

Theorem $1$

For every operation $\bot$ in $\mathbf{Ord}$ with a neutral element $\mu$, there exists a unique binary operation $\curlywedge_\bot$ in $\mathbf{Ord}$ such that the following statements hold for every $\alpha$ in $\mathbf{Ord}$:

1. $\alpha\curlywedge_\bot\emptyset=\mu$
2. $\alpha\curlywedge_\bot\operatorname{succ}\beta=(\alpha\curlywedge_\bot\beta)\bot\alpha$ for every $\beta$ in $\mathbf{Ord}$
3. $\alpha\curlywedge_\bot\beta=\sup_{\gamma\in\beta}(\alpha\curlywedge_\bot\gamma)$ for every non-empty $\beta$ in $\mathbf{Lim}$

Proof.

Let $\mathbf F$, $\mathbf G$, and $\mathbf H$ be predicates of one parameter $x$, one parameter $y$ and one parameter $z$ defined as follows

• $\mathbf F(x,y,z)$ is true if $z$ is $\mu$;
• $\mathbf G(x,y,z)$ is true if $z$ is $y\bot x$ whenever $x$ and $y$ are in $\mathbf{Ord}$, or if $z$ is the empty set otherwise;
• $\mathbf H(x,y,z)$ is true if $z$ is $\bigcup\operatorname{ran}y$ whenever $y$ is a function, or if $z$ is the empty set otherwise.

Now, if $\Phi$ is a property of one parameter $x$, one parameter $y$ and one parameter $z$ then for some entity $a$ let be $\Phi_a$ the property of one parameter $b$ and one parameter $c$ such that $\Phi_a$ holds if and only if $\Phi(a,b,c)$ holds: so for some entity $x$ let's we observe that the predicates $\mathbf F_x$, $\mathbf G_x$, and $\mathbf H_x$ are functionals. Thus, by Theorem $0$, for every $\theta$ in $\mathbf{Ord}$ there exists a function $\Psi_\theta$ in $\mathbf{Ord}$ such that the following statements hold:

a. $\Psi_\theta(\emptyset)=\mathbf F_\theta(\emptyset)$;

b. $\Psi_\theta(\operatorname{succ}\delta)=\mathbf G_\theta\big(\Psi_\theta(\delta)\big)$ for every $\delta$ in $\mathbf{Ord}$

c. $\Psi_\theta(\delta)=\mathbf H_\theta(\Psi_\theta\restriction_\delta)$ for every non-empty $\delta$ in $\mathbf{Lim}$.

Now, let us prove that the definition $$ \Psi(\theta,\delta):=\Psi_\theta(\delta) \tag{$\alpha$} \label{def: esponenziazione ordinale} $$ with $\theta$ and $\delta$ in $\mathbf{Ord}$ defines a function in $\mathbf{Ord}$ such that statements $1$-$3$ hold.

Since $\Psi_\theta(\delta)$ is unique for $\theta$ and $\delta$ in $\mathbf{Ord}$, clearly, def. \eqref{def: esponenziazione ordinale} defines a function in $\mathbf{Ord}$. Thus, by statement (a), we observe that $$ \Psi(\theta,\emptyset)=\Psi_\theta(\emptyset)=\mathbf F_\theta(\emptyset)=\mu \tag{$\beta$} \label{eq: /Psi(/theta,/emptyset)=/emptyset} $$ for every $\theta$ in $\mathbf{Ord}$; furthermore, by statement (b), we observe that $$ \Psi(\theta,\operatorname{succ}\delta)=\Psi_\theta(\operatorname{succ}\delta)=\mathbf G_\theta\big(\Psi_\theta(\delta)\big)=\Psi_\theta(\delta)\bot\theta=\Psi(\theta,\delta)\bot\theta \tag{$\gamma$} \label{eq: /Psi(/theta,/succ/delta)=/Psi(/theta,/delta)/oplus/theta} $$ for every ordinal $\theta$ and $\delta$ in $\mathbf{Ord}$; finally, if $\theta$ is an ordinal, then by the Replacement Axiom, $\Psi_\theta\restriction_\delta$ is a function for every $\delta$ in $\mathbf{Ord}$ and thus, by st. (c), we conclude that $$ \Psi(\theta,\delta)=\Psi_\theta(\delta)=\mathbf H_\theta(\Psi_\theta\restriction_\delta)=\bigcup(\operatorname{ran}\Psi_\theta\restriction_\delta)=\bigcup_{\lambda\in\delta}\Psi_\theta(\lambda)=\bigcup_{\lambda\in\delta}\Psi(\theta,\lambda) \tag{$\delta$} \label{eq: /Psi(/theta,/succ/delta)=/Psi(/theta,/delta)/oplus/theta esponenziazione} $$ for every non-empty $\delta$ in $\mathbf{Lim}$.

Well, by equations \eqref{eq: /Psi(/theta,/emptyset)=/emptyset} - \eqref{eq: /Psi(/theta,/succ/delta)=/Psi(/theta,/delta)/oplus/theta esponenziazione}, we conclude that statements $1$-$3$ are true with respect to $\Psi$.

Now, if $\Phi$ is a functional for which statements $1$-$3$ are true then it is not difficult to conclude (it is sufficient to distinguish the case where $\beta$ is empty, successor and limit) the equality $$ \Psi(\alpha,\beta)=\Phi(\alpha,\beta) $$ holds for every $\alpha$ and $\beta$ in $\mathbf{Ord}$. Therefore, by setting $$ \Psi:=\curlywedge_\bot $$ the statement follows.