Surreal Numbers and Set Theory

Question

Hello,

I looked through MathOverflow's existing entries but couldn't find a satisfactory answer to the following question:

What is the relationship between No, Conway's class of surreal numbers, and V, the Von Neumann set-theoretical universe?

In particular, does V contain all the surreal numbers? If so, then is there a characterization of the surreal numbers as sets in V? And does No contain large cardinals?

I came across surreal numbers recently, but was surprised by the seeming lack of discussion of their relationship to traditional set theory. If they are a subclass of V, then I suppose that could explain why so few people are studying them.

Thank you, Alex

Answer 1

Yes, every surreal number is also an element of $V$ (at least, once you choose some method of encoding ordered pairs of sets as sets). The (highly recursive) characterization of which elements of $V$ are surreal numbers is precisely the definition of the surreal numbers: an element of $V$ is a surreal number just in case it is an ordered pair $(L, R)$ of sets of surreal numbers, with every element of $L$ being less than every element of $R$, with the definition of the ordering being, etc., etc. (Note, however, that this particular embedding of the surreal numbers into $V$ doesn't respect surreal number equivalence; it depends on the particular selection of $L$ and $R$ sets defining the surreal number.)

Conversely, while not every element of $V$ is a surreal number, there is at least a natural way of encoding every ordinal (and thus every cardinal, under the usual Axiom of Choice-based identification of cardinals with their initial ordinals) as a surreal number: the encoding of the ordinal $\alpha$ is the surreal number specified with an empty right set and a left set consisting of the encodings of all the ordinals less than $\alpha$. Thus, if there are large cardinals in $V$, there are corresponding elements in $No$.

Answer 2

Using the wikipedia definition, each surreal number is an equivalence class, but there is a standard trick of turning equivalence classes into sets, namely by only considering the elements of the equivalence class of minimal ordinal rank. Using this trick, each surreal is a set, so the class of surreal numbers is a subclass of $V$, as should be the case.

Everything else was already said by Sridhar Ramesh.

Answer 3

I'm probably very late in answering this, but, in addition to what Stefan and Sridar have already said, I wanted to point out that there is another equivalent definition for a surreal number. You can think of a surreal number as a function from some ordinal $\alpha$ into your favourite two-element set (whose elements are usually denoted $-,+$). Then the order among surreal numbers defined this way is the "lexicographical" one, namely, given two surreal numbers, the bigger will be the one with bigger domain, and if the domains are equal, then look at the first place where they differ: the bigger will be the one that has a "$+$" in that place. With this definition, you don't need to worry about equivalence classes, and I personally find it much easier to work with. You can find all of this in Harry Gonshor's book, "An Introduction to the Theory of Surreal Numbers", volume 110 of the London Mathematical Society Lecture Note Series.

Edit: I made a mistake in describing the order, so let me describe the actual order with some more detail: given a surreal number $s:\alpha\rightarrow${$+,-$}`, we abuse notation by saying that $s(\beta)=0$ for all $\beta\geq\alpha$. Now given two surreal numbers $s,t$, look at the first ordinal $\alpha$ such that $s(\alpha)\neq t(\alpha)$ (we consider the possibility that $s(\alpha)$ or $t(\alpha)$ equal $0$). Then we will say that $s < t$ iff $s(\alpha) < t(\alpha)$, with the convention that $- < 0 < +$.