For every ordinal $\alpha$, define $a_{\alpha} = \{0\} \ | \ \{a_{\beta} \ | \ \beta < \alpha\}$.

In Harry Gonshor's approach of surreals where they are $(+,-)$ sequences of ordinal domain, $a_{\alpha}$ is a plus followed by $\alpha$ minuses. And $a_{\alpha}$ is the least stricly positive surreal of length less than or equal to $1+\alpha$.

I am trying to prove that for a positive surreal $x \geq 0$, if there is an ordinal $\alpha < l(x)$ such that the final segment $x|^{\alpha}$ * in $x$ starting at $\alpha$ is a (strictly) positive omnific integer, then $x = \{x-a_{\omega_0.\alpha}\} \ | \ \{x+a_{\omega_0.\alpha}\}$

*Let $\gamma$ be the unique ordinal such that $l(x) = \alpha + \gamma$. $l(x|^{\alpha}) = \gamma$ and $\forall \beta < \gamma, x|^{\alpha}(\beta) = x(\alpha + \beta)$

Does someone know how to prove such an implication? Is it even true?

(one can prove that for any ordinal $\alpha, a_{\omega_0.\alpha} = {\omega}^{-\alpha}$)

For now, all I could come up with is an inductive proof of the following:

For any ordinal $\alpha$, if $x < y$ are two surreal numbers such that $x|^{\alpha},y|^{\alpha}$ are Conway integers, then $x + {\omega}^{-\alpha} \leq y$.

edit: I just found out my proof of the above statement was flawed in the treatment of two cases, and that the statement is false. (for instance, $1$ and $1 + {\omega}^{-3}$ don't satisfy the statement for $\alpha = 2$.) It could be that the result holds if $l(x),l(y) \geq \alpha$, but my inductive method to prove it doesn't apply as easily here.



You must log in to answer this question.

Browse other questions tagged .