For sake of well-definedness, here we consider only ordinals less than the first uncountable ordinal, $\Omega$. Just like $\infty$ in the notation $\lim_{n→\infty}$ is essentially $\omega$, $\Omega$ will be the new $\infty$. Likewise, surreal numbers will be capped by $\pm\Omega$.
I attempted to define the notion of limit of a sequence to arbitrary ordinal. It's just an extension of the usual $\epsilon$–$N$ definition:
$$
\lim_{n→O} f(n) = x \overset{\text{def.}}{\iff} \forall(\epsilon>0) \quad \exists(N<O) \quad \forall(n \text{ s.t. } N<n<O) \quad |f(n)-x|<\epsilon
$$
Where $\epsilon$ is real, $O$ is a limit ordinal, and $N$ and $n$ are ordinals.
Note that in this notion, the notion of uncountable summation follows. One might wonder whether the following limit converges:
$$
\lim_{n→\Omega}\sum_{k=0}^n\frac1{k^2}
$$
This makes sense only if $f$ is supposed to be a sequence of surreal numbers. But generally and unfortunately, limit doesn't uniquely exist amongst surreal numbers. For example, $\lim_{n→\omega}\frac1n$ converges to zero and every infinitesimal.
Furthermore, in this definition, $\lim_{n→\omega}n$ diverges. By the notion of a limit ordinal, $\lim_{n→\omega}n = \omega$ must satisfy.
To summarize, the definition above is flawed, and this question asks for a topology on surreal numbers such that:
-
Surreal numbers form a topological field
-
$\mathbb{R}$ as a subspace preserves its order topology
-
The countable ordinals as a subspace preserves its order topology
Best Answer
This is not an answer, but a comment: a field topology that would accomodate ordinal numbers would probably look wierd.
First note that in a topological field $F$, translations and non-zero homotheties and the multiplicative inversion are homeomorphisms (the latter on $F^{\times}$).
Consider a countable ordinal $\lambda$ of the form $\lambda = \omega^{\mu},\mu>0$. Every neighborhood $U$ of $\lambda$ in $\mathbf{No}$ (taken here to be the set of surreal numbers with length $<\Omega$) must contain some number $x$ with $x<\frac{\lambda}{2}$. Indeed $\Omega \cap U$ must contain some $\gamma$ for $\gamma<\lambda$. So any neighborhood $V=(V +\lambda)-\lambda$ of $0$ must contain a number $y$ with $y<-\frac{\lambda}{2}$. Since this is true for any $\lambda$ and any neighborhood $U$, we see by translations that each non-empty open set must be coinitial in $\mathbf{No}$. By symmetry, any non-empty open subset is also cofinal.
So every interval $(a,+\infty)$ is dense. Let $O$ be an open neighborhood of $1$ which does not contain $0$. For each $a\in \mathbf{No}^{>0}$, the set $O^{-1}$ intersects $(a,+\infty)$, so $O$ intersects $(0,a^{-1})$. By using homotheties, we see that any non-empty open subset which does not contain $0$ intersects any $(0,a^{-1})$, so any open interval containing $0$ is dense in $\mathbf{No}^{\times}$, hence in $\mathbf{No}$. So any non-empty open interval is dense.
So you'll have generalized sequences with values in $(-1,1)$ which tend to $\omega$.