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On the sound of sounding ridiculous, but in the line of "There are no stupid quetsions": Is there a way to express $\omega_1$ (and in general $\omega_k$ with $k >= 1$ as a Conway game (that is $<L|R>$, with L and R the left and right options). And is there a way to do it such that addition, multiplication, etc. makes sense ?

I can express $\omega_1 = < {i}_{i\in\mathbb{R}} | >$, which is essential the same as $< f: \mathbb{R} \to \mathbb{R} |>$ where $f$ is increasing (${i}_{i\in\mathbb{R}} = f(i)$), which isn't exactly a Conway game notation. I think this also borders on an idea of Gonshor, to express the surreals as maps from initial segments of ordinals to a two-element set). In general, I have replaced the sequence $<0,1,2, ...|>$ from one of the forms of $\omega_0$ by an increasing "sequence" where the index is a positive real number.

Questions are:

  • The above question(s) ?
  • Is my idea correct or at leat in the right direction ?
  • Are there any other ways to do this ? (of course this is an open question if my idea is wrong in the first place).

And yes, I realize my question is little bit broad and I haven't an idea which model you should be work in (ZF, ZFC, NBG, etc), nor do I know how the answer varies with the choice of the model).

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  • $\begingroup$ I've made an attempt to fix the LaTeX in your post, but I am not sure if I did it 100% to what you meant to write. I hope the changes I've made at least is enough for you to make whatever modifications you feel necessary. $\endgroup$ Jan 25, 2011 at 23:13
  • $\begingroup$ I just cleaned this up with a LaTeX editor on my PC, Thans $\endgroup$ Jan 25, 2011 at 23:24
  • $\begingroup$ Since you're speaking of LaTeX: it's better to use \langle and \rangle for angle brackets (not "less than" and "greater than", which are relational operators). $\endgroup$ Jan 26, 2011 at 7:07

2 Answers 2

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Every ordinal number $\alpha$ has a canonical copy as a surreal number $\hat\alpha$, whose left set has as members $\hat\beta$ for every $\beta\lt\alpha$, and whose right set is empty. Succinctly, $$\hat\alpha=\{\ \{\hat\beta\mathrel{:} \beta\lt\alpha\ \}\mathrel{|} \}.$$

One can prove by induction that $\alpha\mapsto\hat\alpha$ is isomorphism of the ordinals with their usual order and arithmetic to the corresponding suborder of the surreals.

For example, for the successor case, $\alpha+1$ is the next ordinal after $\alpha$, and $\hat(\alpha+1)=\{\hat\alpha\mid\}$ is the next surreal after $\hat\alpha$, or $\hat\alpha+1$ in the surreals.

In particular, $\omega_1$ would be represented by the surreal number $\hat\omega_1$, which I think is very different from the surreal numbers that you propose.

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I should note that your expression for $\omega_1$ isn't actually correct; in part because the cardinality of the reals isn't necessarily $\aleph_1$ (this is just Cantor's Hypothesis, which of course Paul Cohen proved independent of ZF), but mostly because the surreal number you've given — $\langle i_{i\in\mathbb{R}}|\rangle$ — is actually equal to $\omega\stackrel{\mathrm{def}}{=}\omega_0$ (exercise: show that whoever plays first in $\langle i_{i\in\mathbb{R}}|\rangle - \omega$ loses). As JDH notes above, the standard way of defining the ordinals copies over into the surreals to give surreal definitions of $\omega_1$, $\omega_2,$ etc; On Numbers And Games even mentions this in passing, IIRC. That said, AFAIK the canonical arithmetic operations ($\sqrt{}$, etc) are relatively boring on ordinals of higher cardinality; in effect they suffer from (as I understand it - this isn't really my field!) predicativity issues. You might want to have a look at the Wikipedia page on (an) 'Ordinal collapsing function' - there's a pretty close tie between surreal definitions of various ordinal-related 'numbers' and the canonical ordinal notations discussed there.

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