As is well, known, the surreal numbers have gaps. As far as I understand, this means that a set of surreal numbers will not always have a supremum or infimum in the surreal numbers.
So I thought about the "next best" thing, which is an upper and lower bound that is almost tight. This is what I've come up with:
Be $M$ a non-empty set of surreal numbers. Then I define two surreal numbers $\operatorname{up} M$ and $\operatorname{low} M$ as follows:
Let $b(x)$ denote the birthday of the surreal number $x$. Then we can define the set $$S_M = \{x \in \mathrm{No}| \exists y\in M: b(x) < b(y)\},$$ that is the set of all ordinals born earler than some member of $x$.
Now for each member $x\in M$ we can define the sets $$L_M(x) = \{y\in S_M: y < x\}\quad\text{and}\quad R_M(x) = \{y\in S_M: y > x\}.$$ Unless I'm mistaken, this implies $x = \{L_M(x)|R_M(x)\}$.
Next, I define the sets $$L^\cap = \bigcap_{x\in M} L_M(x)\quad\text{and}\quad L^\cup = \bigcup_{x \in M} L_M(x)$$ and analogously $R^\cap$ and $R^\cup$.
Finally, I define $$\operatorname{low} M = \{L^\cap|R^\cup\}, \quad \operatorname{up} M = \{L^\cup|R^\cap\}.$$
Now what I can prove is the following:
- $\operatorname{low} M$ is a lower bound of $M$.
- $\operatorname{up} M$ is an upper bound of $M$.
- If $M$ has a minimal element, then $\operatorname{low} M = \min M$.
- If $M$ has a maximal element, then $\operatorname{up} M = \max M$.
What I believe to be true, but don't know how to prove/disprove is:
- If $M\subset\mathbb R$ is bounded in $\mathbb R$, then $$\operatorname{low} M = \inf M + \epsilon$$ where $\inf M$ is the infimum in $\mathbb R$ and $\epsilon$ is an infinitesimal number.
- If $M\subset N$, then $\operatorname{low} M\gtrsim\operatorname{low} N$ where $x\gtrsim y$ means either $x\ge y$ or $(x-y)/x$ is infinitesimal (I suspect $\ge$ generally not to hold, as $S_N$ can be a proper superset of $S_M$, thus allowing for a tighter bound).
- And of course the corresponding statements for the upper bound.
Now my question is: Are those last points indeed true, and if so, how could they be proved?
Here is an idea of proof for your first belief, if you don't understand why things are this way I can elaborate:
-If $M$ has a least element then like you said, low $M = \inf M$ holds.
-Else $M$ contains dyadic numbers of unbounded birthdates or a non dyadic number because otherwise it would be finite and have a least element. Then $L^{\cap}$ is the set of dyadic numbers that are strict lower bounds of $M$. $\inf M$ is stricly lower than any element of $R^{\cup}$ so either $\inf M$ is dyadic, and it is in $L^{\cap}$ so low $M = \inf M + \frac{1}{\omega_0}$, or $\inf M$ is not dyadic, then it is not in $L^{\cap}$ and this implies low $M = \inf M$.