I've been reading about the surreal numbers out of curiosity. Though I don't totally understand them, I wonder why I don't see them used more often. Surreal numbers extend the real line to also include infinitesimal and infinite numbers (transfinite ordinals like $\omega$ which is identified with the cardinal number $\aleph_0$, itself the cardinality of $\mathbb{N}$).

Often, there are interesting preferences which can't be represented by a real-valued utility function. The obvious example is for lexicographic preferences. But why not have $u:\mathbb{R}^n\rightarrow S^*$, like $u(a,b)=a+\frac{b}{\omega}$?

Lexicographic probability spaces are another place where these numbers could have application. If we want to differentiate between two probability zero events, let one have probability $\frac{1}{\omega}$ and the other $\frac{2}{\omega}$.

One might ask, sure you can use them, but what's the point? What do we gain? Fair question. I just kind of think the use would make things cleaner--making lexicographic prob spaces less of a tortured construction perhaps.

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    $\begingroup$ What would the surreal numbers buy you that an ordinary nonstandard model of the real numbers would not? $\endgroup$ Commented Nov 30, 2014 at 16:23
  • $\begingroup$ Probably nothing incredibly important, I concede. But isn't it more transparent to have a utility function $u(a,b,c)=a+\frac{bc}{\omega}$ than to explain this as the ordering generated by some binary relation $R$? $\endgroup$
    – Pburg
    Commented Nov 30, 2014 at 17:02
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    $\begingroup$ @Pburg: I asked why the surreals are better than a nonstandard model of $R$; you responded with a reason why the surreals might be better than a standard model of $R$. Are we failing to communicate? $\endgroup$ Commented Nov 30, 2014 at 17:28
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    $\begingroup$ @StevenLandsburg Sorry, I'm unfamiliar with the ordinary nonstandard approaches and their applications in economics, so I cannot answer your original question. $\endgroup$
    – Pburg
    Commented Nov 30, 2014 at 17:38
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    $\begingroup$ @StevenLandsburg Professor Landsbrurg, perhaps you could present summarily an ordinary nonstandard model as an answer here. I believe it would essentially answer the OP's question and also benefit more generally the users of this site. $\endgroup$ Commented Nov 30, 2014 at 17:47

2 Answers 2


The first thing that comes to mind is that some economic models have more than one agent.

If one agent has a utility function with values in ${\mathbb R}$, then two agents have a pair of utility functions with joint values in ${\mathbb R}^2$. If one agent has a utility function with values in $S^*$, then two agents have a pair of utility functions with joint values in ...what? We can't just form a product of $S^*$ with itself, because $S^*$ is not a set.

There are probably ways around these difficulties, but they're going to be clumsy and they're going to require lots of mucking around with foundations that are almost surely not interesting to most economists.

Set theory lets us form unions, products, coproducts and power sets. It lets us talk about subsets and supersets without having to worry about whether they exist. Economists use these constructions all the time without stopping to think about them. I doubt that we want to switch to a new foundation that forces us to think about them constantly.


This is about lexicographic probability and hyperreal-valued probability in the third paragraph. From the question and comment, it seems to me this question is more related to the usefulness of hyperreal numbers rather than surreal numbers.

One of central and controversial question in game theory was under what setting, common belief of rationality implies backward induction(see Aumann(1995), (1998) and Binmore(1996)).

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Consider a three-legged centipede game of incomplete information with Harsanyi's universal type space. Ann of type $t_a$ believes that Bob plays $``In"$, while Ann of type $u_a$ believes that Bob plays $“Out”$ . Bob, on the other hand believes Ann is type $u_a$ and plays $``Out"$. Then at the state $(t_a, (In,Out),Out)$ that is not the backward induction path, common belief of rationality still holds. What we need to verify are, at the given state:

  1. What Ann believes implies Ann is rational.
  2. What Ann believes Bob believes implies that Ann believes Bob is rational.
  3. What Ann believes Bob believes Ann believes implies Ann believes Bob believes Ann is rational.

    $\ldots \ \ldots$

Notice that we only need to check the first three steps, what Ann believes Bob believes Ann believes is the same as what Ann believes Bob believes Ann believes $\ldots$ Bob believes Ann believes. The same procedure applies to a similar situation of interchanging Ann and Bob in the above reasoning.

To make the baseline result that common belief implies backward induction, we can switch to hyperreal-valued belief system. Specifically, a player will not rule out any of her opponent's strategies. If she believes an opponent is rational, and strategy $s_1$ yields higher payoff compared with $s_2$ for this opponent, then she will believe that $s_1$ is infinitely more likely than $s_2$.

Now, if it's the case that Ann believes Bob is rational, then Bob should believe that then the chance Ann plays $“Out"$ at the third node conditional on that she plays $“In"$ at the first node is smaller than $\frac{2}{3}$. But if Ann believes Ann believes Bob believes Ann of type $t_a$ and $u_a$ is rational, then the standard part of the probability that Ann plays $“Out"$ at the third node conditional on that she plays $“In"$ at the first node equals $1$.

Of course, exactly the same argument works for lexicographic probability which is more parsimonious.

My guess is that one of drawback to prevent hyperreal probability form wide application may be that we can't define infinite product of infinitesimal number. For example, in an infinite extensive game, a player believe that the probability that the probability of reaching each node of a path equals $\epsilon$ which is the equivalence class of $(1, \frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}, \ldots,)$ up to a fixed ultrafilter on $\mathbb{N}$. Of course, we expect that the chance that this path is realized should also be infinitesimal. But notice that $\epsilon = \epsilon_n := \{a_i^n\}_{i \in \mathbb{Z_+}}$ in which

\begin{eqnarray}a_i^n= \begin{cases} 1, &i<n \cr mn^{n-1}, &i=n \cr \frac{1}{i}, &i>n\end{cases} \end{eqnarray}

Contrasted with finite product, if we define the inifinite product of hyperreal numbers as their componentwise product, then, in this case, we can make it equal any real number $m$.


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