# Cauchy completion of transfinite "rationals"

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 $$\text{No}$$?

• nombre's answer to your previous question proved that every Cauchy sequence $\mathrm{Ord}\to\mathbb{Q}_{\mathrm{Ord}}$ converges. Commented Jun 4, 2020 at 1:20
• Have you made any attempt to compute your "Hessenberg power" in any nontrivial cases? It seems entirely plausible to me that it ends up trivializing in some way such that your $f$ is totally uninteresting. Certainly, there is absolutely no reason to have any confidence that your $f$ will be "converging towards $e$" when you evaluate it on infinite ordinals. Commented Jun 4, 2020 at 1:28
• Are you saying that you are somehow modifying the definition of $\mathbb{Q}_{\mathrm{Ord}}$ so that the Hessenberg power operation is defined on all of it (not just on ordinals)? If so, you definitely need to say that explicitly and explain exactly how you are modifying the definition. None of this is clear from the question you have written. Commented Jun 4, 2020 at 1:46
• nombre's theorem says that any Cauchy sequence in the ordered field $(\mathbb{Q}_{\mathrm{Ord}},+,\times,\leq)$ converges. You have defined a sequence in that ordered field and are asking whether it is a Cauchy sequence that does not converge. nombre's theorem tells you the answer is no. It is completely irrelevant that you have defined some additional operation on this ordered field, because that additional operation has nothing to do with what it means for a sequence to be Cauchy or to converge. Commented Jun 4, 2020 at 2:17
• There are two additional problems. The biggest one is that you don't seem to have worked out the properties of your Hessenberg exponentiation for ordinal numbers, and it would make sense to start by that before looking at this sequence. The second is that your rules do not extend this definition to $\mathbb{Q}_{\mathbf{Ord}}$ (which is a subfield of $\mathbb{Q}[[x^{\mathbb{Z}_{\mathbf{Ord}}}]]$, not of $\mathbb{Q}[[x^{\mathbf{Ord}}]]$ that would only be a ring). How do you define $(\omega-1) \uparrow \omega$ for instance? Commented Jun 4, 2020 at 7:56

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 $$1$$.
• Interesting that the function jumps down to 1 at $\omega$ and thereafter. I wonder what's the most natural way to truly get an extension of the real numbers from $\mathbb{Q}_\text{Ord}$. Commented Jun 4, 2020 at 19:13