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


3 Answers 3


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 $No$.

  • $\begingroup$ Thank you for your answer! So $V$ also contains $\omega/2$, $\omega-1$, $\sqrt{\omega}$, etc. but these numbers don't appear as stages in the cumulative hierarchy $V_\alpha$? $\endgroup$ Jul 23, 2011 at 8:34
  • $\begingroup$ The stages in the cumulative hierarchy are the ordinal numbers; as indicated in the second paragraph, these are the surreal numbers you can get if you restrict right sets to always be empty. The surreal numbers $\omega/2$, $\omega - 1$, $\sqrt{\omega}$, etc., cannot be created without invoking a non-empty right set at some point. $\endgroup$ Jul 23, 2011 at 19:33
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    $\begingroup$ (The other word for "ordinal numbers" that is sometimes used in the context of surreal number theory is "birthdays") $\endgroup$ Jul 23, 2011 at 19:33

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.

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    $\begingroup$ Thank god for von Neumann and the axiom of foundation, allowing Scott's trick. :-) $\endgroup$
    – Asaf Karagila
    Jul 21, 2011 at 21:35
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    $\begingroup$ It is interesting to note that one also needs the power set axiom for Scott's trick, since otherwise the set of least-rank representatives of a class may not be a set. In particular, Scott's trick does not work properly in the theory ZFC-, which is often used with in connection with ultrapowers, where one truly wants to use it. The fact of the matter is that there are models of ZFC- having measurable cardinals, whose ultrapowers have height taller than the original universe, and therefore have no internal Mostowski collapse. In these models, Scott's trick fails and there is no way around it. $\endgroup$ Oct 21, 2012 at 2:30

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 $- < 0 < +$.

  • $\begingroup$ The order you give is not lexicographic order. I think you must have made a typo. $\endgroup$ Apr 14, 2012 at 15:16
  • $\begingroup$ Well, in some sense, the order I gave was the lexicographical one, although the wrong one. I edited the answer to describe the actual order (I wasn't able to surround the $+,-$ with curly brackets). $\endgroup$ Apr 14, 2012 at 18:13
  • $\begingroup$ I'm learning some new MathJax tricks, now I could fix the curly brackets thing. $\endgroup$ May 2, 2012 at 12:05

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