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If one encodes the real numbers as the surreal numbers with countable birthdays, it seems that the tree representation can be mapped to the naturals by a simple breadth first traversal.

What am I missing?

JPER
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    Traversing the tree breadth first you will reach the surreal numbers with finite birthdays, which correspond to the dyadic rational numbers, not the real numbers. – jjagmath Sep 07 '23 at 00:29
  • countable=finite? – JPER Sep 07 '23 at 00:54
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    No, countable means either finite or of the size of the natural numbers. – jjagmath Sep 07 '23 at 01:03
  • Weirdly enough, a very similar question has been asked before: https://math.stackexchange.com/questions/2728504/do-surreals-prove-reals-are-countable. (That question looks at multiples of $1\over \omega$ instead of nodes on the countable birthday tree, but the fundamental issue is the same in each case - a countable-looking set is not actually countable.) – Noah Schweber Sep 07 '23 at 01:29
  • And this one is a dupe I think: https://math.stackexchange.com/questions/1956014/are-there-countably-infinte-surreal-number/1956020#1956020 – Noah Schweber Sep 07 '23 at 01:32
  • I am assuming $ \forall n \in \mathbb{N}, \frac{\omega}{n}>|\mathbb{N}| $. That is why I am puzzled. – JPER Sep 08 '23 at 00:26

1 Answers1

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Think about how the surreals form:

  1. On Day $0$: we start with $0$.
  2. On Day $1$: $-1, 1$ are added. (Note that the day number is the highest number added on that day.)
  3. On Day $2$: $\frac 12, 2$ and their opposites are added.
  4. On Day $3$: $\frac14, \frac34, \frac32, 3$ and their opposites are added.
  5. On day $4$: $\frac18, \frac38, \frac58, \frac78, \frac54, \frac74, \frac52, 4$ and their opposites are added.

etc. Note that all the numbers are rationals with powers of $2$ in the denominator (including $2^0$ for the integers). If you understand how the surreals develop, this will continue for every finite day. These are called "dyadic rationals".

Once you go through all the finite days, the surreals will consist of exactly all the dyadic rationals. It is only on the first infinite day, $\omega$, that the remaining real numbers are added, But not only them, $-\omega$ and $\omega$ are also both added on this day. So you cannot define the reals as all surreals defined by day $\omega$. That would be the extended reals instead.

Further, if you define the reals as you did as all surreals with countable birthdays, then you are not done. $\omega$ is infinite, but still countable, as are $\omega + 1, \omega + 2, \dots$ and $2\omega, 3\omega, \dots$, and even $\omega^2, \omega^3, \dots$. They are all countable ordinals. That is, the sets of all ordinals $\le$ to them can all be put in one-to-one correspondence with a subset of the natural numbers. By the time $\omega^\omega$ is reached, we are finally dealing with uncountable ordinals. (Exactly where this transition occurs is the subject of the Continuum Hypothesis.)

By the time you get through all the countable ordinals, you have gone far, far beyond the real numbers.

Paul Sinclair
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  • I appreciate the clarifications! If one limits the birthdays to those $ \le \omega $ and limit those nodes whose birthday is $\omega$ to those who have children, does traversal of these nodes preclude a map to naturals? – JPER Sep 08 '23 at 03:37
  • What do you mean by "those who have children"? If you limit to birthdays $\le \omega$, no number with birthday $\omega$ has children. But throwing out $-\omega, \omega$ does leave just the reals. By the time you get to day $\omega$ in a breadth-first traversal, you've already gone through the infinitely many numbers. Then on day $\omega$, there are uncountably many new numbers in that breadth. – Paul Sinclair Sep 08 '23 at 11:55
  • Some nodes whose birthday is $\omega$ will have children which are (albeit not new) dyadic rationals (e.g. 1/3). One could even imagine an embedded cone of duplicates. think I mean these. – JPER Sep 09 '23 at 21:33
  • No. That is not how the surreals are built. Any such "duplicates" do not have $\omega$ as their birthday. Their birthday is the day they first get added. You do not add them again. And by the way, if you only consider representations ${L\mid R}$ where $L \cup R$ consists of all older surreals, you get every new surreal for that birthday, and only new surreals. You only get older surreals when you use representations that do not include them in either set. Finally, "dyadic" means the denominator is a power of $2$. $\frac 13$ is not a dyadic rational. – Paul Sinclair Sep 09 '23 at 21:41
  • Thanks, that hepls – JPER Sep 17 '23 at 00:51