Can the Surreals be Cauchy complete by allowing for ordinal length sequences?

Question

While it’s known that the surreal numbers aren’t complete ($1+1/2+1/4\dots$ approaches a gap instead of $2$), can it have a version of Cauchy completeness by allowing the sequences to be longer as is the case for the hyperreal internal sequences. That is to say, for a given function $s: On \to No$, if $\forall \epsilon \in No^+; \exists N \in On; \forall K>N; |s(K)-L|<\epsilon$, we have a Cauchy sequence which converges? The example give above would then converge to $2$ as expected (or at least I think it does). Is there a problem/improvement to this method and is this the mod natural way to “complete” the surreals or is there no way to do so?

After a comment by @Dave L. Renfro, the question would then be “Do the surreals form a radial space?”

Few people here know surreal numbers. Can you give at least one reference ? — Jean Marie, Oct 23 '22 at 19:20
I don't know enough about surreal numbers to say anything about your question, but it might be worth pointing out that transfinite sequences are NOT sufficient to obtain all limit points for topological spaces in general. See my comments to Intuition behind nets, and also see Sequential properties of topological spaces. — Dave L. Renfro, Oct 23 '22 at 20:26
@DaveL.Renfro It does in the case of the order topology on ordered fields, which is sequential in this sense (the sequences must be indexed by the cofinality of said ordered field). — nombre, Oct 24 '22 at 05:53

score 3 · Answer 1 · answered Oct 24 '22 at 06:37

For set-sized ordered fields $(F,+,\times,<)$, the following are equivalent:

$F$ has no proper dense ordered field extension
every Cauchy sequence $\operatorname{cof}(F) \rightarrow F$ converges, where $\operatorname{cof}(F)$ is the cofinality of $(F,<)$.
$F$ is complete as uniform space where the uniform structure is derived from the ordering.

We say that $F$ is Cauchy-complete if these conditions hold. There is a dense ordered field extension $\widetilde{F} \supseteq F$ which is Cauchy-complete, and this characterizes $\widetilde{F} / F$ up to unique isomorphism, because it is a final object of the category of dense extensions of $F$ (with commutative triangles as morphisms). It is also initial in the category of cofinal and Cauchy-complete extensions of $F$.

This extension is called the Cauchy-completion of $F$.

The equivalence above works for class-sized fields in NBG set theory, but there need not be a Cauchy-completion for all such fields. Indeed, for surreal numbers, the completion would be "too large to be a class". For instance, one can show in certain conservative extensions of NBG in which $\widetilde {\mathbf{No}}$ exists that $2^{\mathbf{On}}$ injects in it. This is because every strictly increasing and cofinal sequence $u:\mathbf{On} \longrightarrow \mathbf{On}$ induces a Cauchy sequence $C(u)=(\sum \limits_{\gamma<\alpha}\omega^{-u(\gamma)})_{\alpha \in \mathbf{On}}$ in $\mathbf{No}$, in such a way that no two distinct such sequences may have the same limit in an extension.

One can inject $2^{\mathbf{On}}$ into the class of such sequences $\mathbf{On} \longrightarrow \mathbf{On}$ by sending $v:\mathbf{On} \longrightarrow \{0,1\}$ to $u(v):\alpha \mapsto \alpha+v(\alpha)$ if $\alpha$ is a limit, and $\alpha+1 \mapsto u(v)(\alpha)+ 2^{v(\alpha)}$ for all $\alpha$.

If $\widetilde {\mathbf{No}}$ were a class in NBG, then it would inject in $\mathbf{On}$ (by the axiom of limitation of size): hence we would have an injection $2^{\mathbf{On}}\rightarrow\mathbf{On}$ as per the conservative extension of NBG, which cannot be.

In particular the surreal numbers themselves are not Cauchy-complete.

Can the Surreals be Cauchy complete by allowing for ordinal length sequences?

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