Are there countably infinte surreal number?

Question

I was thinking about surreal numbers, and about how you can display them like a binary tree (like this) and that since they can be displayed as a binary tree shouldn't there be only countably infinite of them, and how if so, why aren't all real numbers countable infinite because there is a real number for every surreal number.

Answer 1

Note that that tree representation of the surreals is actually more complicated than it may first appear - it is infinitely tall! (Indeed, the surreal numbers constructed at a finite height are merely the dyadic fractions. You have to wait until height $\omega$ to get ${1\over 3}$.)

In fact, the tree has height $ON$, the ordertype of the ordinals. There are proper class many surreal numbers. In particular, there are way more surreal numbers than real numbers!

This may be easier to see if you think of the "dynamic" construction of the surreal numbers, where every surreal number has an ordinal "birthday" and we keep creating new surreal numbers at arbitrarily "late" ordinals.

I am not sure where you got the impression that there are as many real numbers as surreal numbers. I suspect you have slightly the wrong picture of how the reals sit inside the surreals. The surreals (similarly to the hyperreals) are a non-Archimedean ordered field: given any real number $r$, there are lots (= proper class many) of surreals $s$ which are "infinitesimally close" to $r$, in the sense that $\vert s-r\vert$ is nonzero but not greater than any positive rational number. So in fact, the real numbers are "far apart" from each other, as far as the surreal numbers are concerned! (And, of course, there are surreals not infinitesimally close to any real number. But such numbers' reciprocals will then be infinitesimally close to $0$, so we don't lose too much detail by focusing on the near-the-reals part of the surreal numbers.)

Answer 2

The binary tree you have given isn't a standard binary tree because it has nodes at $\omega$.