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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?

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    $\begingroup$ @NormalHuman: Do you have a bot to add those comments, or how did you manage to add it so quickly after posting? $\endgroup$ – celtschk Oct 25 '15 at 11:43
  • $\begingroup$ I hope that it’s a bot, since the comment is often added quite inappropriately. $\endgroup$ – Brian M. Scott Oct 25 '15 at 20:01
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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 $M = \inf M$.

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  • $\begingroup$ Great, thank you. Of course when you say "of unbounded birthdates" you mean "unbounded in $\mathbb N$", as all birth dates of dyadic fractions are bounded by $\omega$. $\endgroup$ – celtschk Oct 26 '15 at 10:40

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