I am attempting to define "the"/a limit for surreal functions:
Write for surreal numbers $x,p$ and $f$ a function: (Each surreal number $x$ is mapped to a unique surreal number $f(x)$.)
$$\lim_{x \rightarrow p} f(x) = L \iff \exists L \text{ surreal } \forall \epsilon > 0 \exists \hat{\delta}>0: \forall 0 < \delta \le \hat{\delta}: \text{ If } x = \{p-\delta|p+\delta\} \text{ then } L = \{ f(x)-\epsilon | f(x)+\epsilon \}$$
I could show - maybe I am being too optimistic - that $L$ is unique, if it has the property above.
The proof mimics essentially the same as in the definition of the limit in real case. My question is:
- Can this be independently verfied?
If it is not asked too much:
- If $\lim_{x \rightarrow p} f(x) = a$ and $\lim_{x \rightarrow p} b(x) = b$, do we have: $\lim_{x \rightarrow p} f(x)+g(x) = a+b$?
Thanks for your help!
In any ordered field, there is a notion of limit (hence also derivative) for functions using the usual "$\delta$-$\varepsilon$" definition. It is somewhat well-known that the corresponding topological and analytic notions retain many of the elementary properties of topology on real numbers (e.g. addition and multiplication are continuous), but none of the important ones (e.g. Rolle's theorem fails).
Your definition is not the same as that I am alluding to, and it is not too interesting for the following reason:
For each surreal number $p$, for all sufficiently small $\delta>0$, we have $p=\{p-\delta \ | \ p + \delta\}$. To see this, note that the function $\delta \mapsto p+\delta$ is injective, so for cardinality reasons, for small enough $\delta \neq 0$, the birth day (or length) of $p+\delta$ cannot be less than that of $p$. More precisely, assume for contradiction that for each positive ordinal $\alpha$, there is a non-zero number $\varepsilon_{\alpha}$ smaller in absolute value than $\frac{1}{\alpha}$, such that $p+\varepsilon_{\alpha}$ has birth day smaller than the birth day $\beta$ of $p$. Using this, we can define an injection $\mathbf{On}^{>0} \rightarrow \mathbf{No}(\beta+1)$ where $\mathbf{No}(\beta+1)$ is the set of numbers of birth day $\leq \beta$, and $\mathbf{On}^{>0}$ is the proper class of positive ordinals, which cannot be.
So if $L= \lim_p f$, then in particular we have $L=\{f(p) -\varepsilon \ | \ f(p) +\varepsilon\}$ for all $\varepsilon >0$. For sufficiently large $\varepsilon >0$, we have $0=\{f(p) -\varepsilon \ | \ f(p) +\varepsilon\}$, so $L=0$. But then $f(p)$ must be zero as well because $0=L$ should be as close as we want to it.
And we can see that conversely, if $f(p)=0$, then $0$ is indeed the limit of $f$ at $p$.
In other words, your definition is equivalent to $f(p)=0$, with $0$ being the limit if that occurs.