Cauchy completion of transfinite "rationals"

Question

Let the Hessenberg power $\alpha^\beta$ be the supremum of ordinals that are order-isomorphic to some well-order on the set of finite-support functions $\beta \rightarrow \alpha$ that extends the following partial order:

$$f \leq g \leftrightarrow \forall x \in \beta : f(x) \leq g(x)$$

Start with the construction given in this question. Add the power function $\uparrow : \mathbb{Q}^\text{Ord} \times \text{Ord} \rightarrow \mathbb{Q}_\text{Ord}$ that extends the Hessenberg power $\text{Ord} \times \text{Ord} \rightarrow \text{Ord}$ through the following equalities: \begin{align} \left(\frac{a}{b}\right)^c &= \frac{a^c}{b^c} \\ (a - b)^c &= (-1)^c (b - a)^c \\ (-1)^c &= \begin{cases} +1 & \text{$c$ is an even ordinal} \\ -1 & \text{$c$ is an odd ordinal} \end{cases} \end{align}

Let $f : \text{Ord} \rightarrow \mathbb{Q}_\text{Ord}$, where

\begin{align} f(n) &= \left(1 + \frac{1}{n}\right)^n \\ &= \frac{(n+1)^n}{n^n} \end{align}

and we use the Hessenberg power for transfinite arguments. Is $f$ Cauchy but not convergent in $\mathbb{Q}_\text{Ord}$? If so, what is the relationship between the Cauchy completion of $\mathbb{Q}_\text{Ord}$ and the field of surreal numbers $\text{No}$?

Answer 1

By nombre's answer to your previous question, $f$ cannot be a Cauchy sequence that does not converge, since every Cauchy sequence $\mathrm{Ord}\to\mathbb{Q}_{\mathrm{Ord}}$ converges.

However, that argument is overkill, since we can compute $f$ quite explicitly and see that it does converge. Suppose $\alpha$ and $\beta$ are ordinals such that $\beta$ is infinite and $1<|\alpha|\leq|\beta|$. Then I claim that your Hessenberg power $\alpha^\beta$ is always equal to the cardinal $|\beta|^+$.

First, clearly $\alpha^\beta\leq|\beta|^+$, since the set of finite-support functions $\beta\to\alpha$ has cardinality $|\beta|$. On the other hand, for any well-ordering on $\beta$ (say of order-type $\gamma$), we can well-order the finite-support functions $\beta\to\alpha$ using the reverse lexicographic order with respect to this well-ordering on $\beta$. This gives an ordering whose order-type is the usual ordinal exponentiation $\alpha^\gamma$, and in particular its order type is greater than or equal to $\gamma$. The supremum of all possible $\gamma$ is just $|\beta|^+$, so your Hessenberg power satisfies $\alpha^\beta\geq|\beta|^+$ and hence $\alpha^\beta=|\beta|^+$.

(More generally, this shows that for arbitrary $\alpha$ and infinite $\beta$, your Hessenberg power $\alpha^\beta$ is at least as large as the usual ordinal power $\alpha^{|\beta|^+}$. I don't know whether they are actually equal when $|\alpha|>|\beta|$. Note that you could see ahead of time that your exponentiation operation is not likely to be very natural, since it depends on $\beta$ only as a set, and does not use the ordering of $\beta$ at all.)

In particular, then, for any infinite $\alpha$, your $f(\alpha)$ is just $\frac{|\alpha|^+}{|\alpha|^+}=1$, so your $f$ converges to $1$.