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Let $ZF^\times$ denote the set of axioms of Zermelo-Fraenkel set theory without the axiom of infinity. The set $V_\omega$ of all hereditarily finite sets is a model of $ZF^\times$, and $Ord^{V_\omega}=\omega$.

The surreal numbers, in $ZFC$ form a proper class which is an ordered field that contains all the ordinals and the order is Dedekind-complete (in the sense that every cut is realized).

Does that make any sense to define the surreal numbers in $ZF^\times$, and if so - is the result just the rational numbers (which realizes every cut defined by a finite type, and contains "all the ordinals" of $V_\omega$)?

To add on the comment of Zhen Lin below, suppose we work in $NBG$ set theory, again without axioms of infinity, however now we can require that class-cuts will be realized as well. If in the $ZF^\times$ we cannot get the entire rational numbers, is it possible in this case?

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Can you actually get all the rationals? I think you can only get the diadic rationals $m/2^n$. –  Alon Amit Nov 1 '11 at 22:18
    
@Alon: If I had any idea how to approach this on my own (or the time to actually do it properly nowadays) I would probably not post this question to begin with :-) –  Asaf Karagila Nov 1 '11 at 22:24
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@Asaf: Well, if I remember correctly, in the usual construction of the surreal numbers, rational numbers with denominators not $2^n$ have birthday $\omega$, together with all the other real numbers. As such, if we only permit sets in our Conway cuts then we can never obtain all the rational numbers... –  Zhen Lin Nov 1 '11 at 22:53
    
You can definitely get each dyadic rational, since they’re all born on finite days. I’m pretty sure that if you could get each rational, you could get $\{\mathbb{Z}|\;\}=\omega$, since to get the non-dyadic rationals you need to be able to collect infinite sets of dyadics. –  Brian M. Scott Nov 1 '11 at 22:59
    
I guess that I do not understand the surreal numbers enough to answer this on my own. I do get the gist of what you guys say. @Zhen: I thought that the usual surreal construction allowed only set cuts. Am I wrong about this? –  Asaf Karagila Nov 1 '11 at 23:09
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First of all, for the base case of $ZF^{-\infty}$: as was noted, the set of numbers that can be produced here is the same as the set of all numbers 'born' on finite days in Conway's usual construction. I finally got around to digging through my references and it turns out that none of On Numbers And Games, Winning Ways or Surreal Numbers actually give a full proof that the only numbers born on day $n$ are dyadic rationals. Here's a rough outline of the process (omitting the very basic proofs of the usual properties for addition, inequality, etc. and the obvious restriction to positive numbers, since negatives follow easily) that would go into a proof:

  1. Prove the Simplicity Theorem: if both $X_L$ and $X_R$ are non-empty, the value of $x = \{X_L|X_R\}$ is the simplest (first-created) value strictly between them. Another version of this is that if $x=\{X_L|X_R\}$ and $Y_L\lt x, Y_R \gt x$ then $x=\{X_L,Y_L|X_R,Y_R\}$. This implies that if the numbers born by day $n$ are $0, x_1, \ldots, x_m$ then the numbers born on day $n+1$ will be just $\{0|x_1\}, \{x_1|x_2\}, \ldots, \{x_{m-1}|x_m\}, \{x_m|\}$ - they fill in the gaps.
  2. Use this to prove the simplest form of the translation theorem: if $x$ is positive, then $1+x = \{1+X_L|1+X_R\}$. This then implies that the numbers greater than $1$ born on day $n+1$ are $1$ more than the numbers born on day $n$ and lets us restrict attention to the numbers between $0$ and $1$.
  3. By induction, assume that the numbers between $0$ and $1$ born by day $n$ are the dyadic rationals $t/2^{n-1}, 0\lt t\lt 2^{n-1}$. Consider the gap $(a,a+2^{-(n-1)})$ on day $n$ and the number $x_{new} = \{a|a+2^{-(n-1)}\}$ born on day $n+1$; we know that $x_{new} > a$ and $x_{new} < a+2^{-(n-1)}$. Then $x_{new}+x_{new} = \{a+x_{new}|a+2^{-(n-1)}+x_{new}\}$; the left side is thus greater than $a+a$ and less than $a+a+2^{-(n-1)}$; similarly, the right side is greater than $a+a+2^{(n-1)}$ and less than $a+a+2^{-(n-2)}$. The simplest number in this span is $a+a+2^{-(n-1)}$, and so by the simplicity theorem $x_{new}+x_{new} = a+a+2^{-(n-1)}$ and thus $x_{new} = a+2^{-n}$.

For the $NBG^{-\infty}$ case, things are a bit more subtle. It seems clear that you can't get numbers born on any day $\gt\omega$, because those numbers will involve 'class members' - numbers born on at least day $\omega$, with a proper class on either their left or right side, and proper classes aren't available as members. At first I thought you could get all the numbers born on day $\omega$, but the catch is that class comprehension doesn't give you arbitrary countable sets of hereditarily-finite (surreal) numbers (i.e. dyadics), but only definable sets, and what's definable depends on what your language is. If you restrict to $\{\leq, +\}$ then the (traditional) numbers that you can get are all the rationals; if your language is $\{\leq, +, \times\}$ then you should get all algebraic numbers. Over either language, you'll also get $\omega, -\omega$, and all numbers of the form $d\pm 1/\omega$ for dyadic $d$, since these are defined by cuts of the form $\{d|d+{1\over2}, d+{1\over4}, \ldots\}$ and similar. Proving that these are all the defined numbers seems relatively straightforward: if $X_L$ has a largest member $x_L$ then we must have $\inf(X_R) = x_L$ (otherwise there will be some dyadic rational in the range $(x_L, \inf(X_R)$) and so the value is $x_L+{1\over\omega}$ (as this is the simplest number larger than $x_L$ and smaller than all the $X_R$); the dual statement obviously holds if $X_R$ has a smallest member $x_R$. If neither $X_L$ nor $X_R$ has an extremal member in the appropriate direction, then they must have a shared infimum/supremum; at this point it comes down to showing that the supremum of all the values satisfying some system of linear (respectively polynomial) integer inequalities is a rational, respectively algebraic, number.

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Wonderful!! Many thanks! –  Asaf Karagila Nov 3 '11 at 20:57
    
Hm, not that I've thought very long about it, but oughtn't $\{\le,+,\times\}$ to be definable classes in $NBG^{-\infty}$? –  Henning Makholm Nov 3 '11 at 23:21
    
@HenningMakholm: the point is that these operations have no inherent definitions; they're defined as operations on surreal numbers, and they turn out to be equivalent to the 'classical' definitions, but that's something that actually needs proof. I should probably use $\oplus$ and $\otimes$ for them. –  Steven Stadnicki Nov 3 '11 at 23:30
    
@Steven, oh I somehow thought we were just working in $\omega$ there, but of course not. Never mind, then. But how can the class comprehension axiom care about language choices we make as users of the system? The language of NBG is just the language of NBG, right? –  Henning Makholm Nov 3 '11 at 23:33
    
@HenningMakholm Hmm - I see your point here. I was taking class comprehension ('For any formula $\phi$ with no class quantifiers there's a class $A$ s.t. $\forall x (x\in A \leftrightarrow \phi(x))$') to mean 'formulas in the language of the surreals' as opposed to 'formulas in the language of set theory'. Of course you're right that $\oplus$ and $\otimes$ should be definable themselves (as should $\leq$), although that may not be possible in $NBG^{-\infty}$ with class cuts because of the need to quantify over class parameters, I think... –  Steven Stadnicki Nov 3 '11 at 23:49
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