If $k$ is an ordered field, the least ordinal $s(k)$ which doesn't embed in $(k,<)$ is regular. This is because every interval of an ordered field embeds in every infinite interval so given a strictly increasing map $f: \alpha < s(k) \rightarrow s(k)$ you can define an embedding $(\sup(f(\alpha)),\in) \rightarrow (k,<)$. Moreover $s(k)$ is always uncountable since $s(\mathbb{Q}) = \omega_1$.

Note that these results make use of the field structure of $k$. They fail even for ordered rings such as the ring obtained from $({\omega}^{\omega},\underline{+},\underline{\times})$ (natural operations), by adding additive inverses and extending the operations like one usually does.

Let $\lambda$ be an $\varepsilon$-number. Let $No(\lambda)$ denote the field of surreal numbers of length $<\lambda$. I am trying to figure out what $s(No(\lambda))$ is.

Clearly, $s(No(\lambda))$ is at least equal to ${\lambda}^+$ and at most equal to ${|No(\lambda)|}^+$. Assuming GCH, this yields $s(No(\lambda)) = {\lambda}^+$ when $\lambda$ is itself a cardinal, and ${\lambda}^+ \leq s(No(\lambda)) \leq ({\lambda}^+)^+$ otherwise. This is just a computation of $|No(\lambda)|$ using elemenraty cardinal arithmetic and the ordered-set-focused definition of $No(\lambda)$.

Can these results (or their "equivalent" modulo GCH where $2^{\alpha}$ replaces ${\alpha}^+$) be proven in ZFC by taking advantage of the structure of $No(\lambda)$?

  • $\begingroup$ Can you clarify what you mean by "length"? Since $\lambda$ is an $\epsilon$-number, I guess what you mean is equivalent to saying that the number is born before $\lambda$? $\endgroup$ – Joel David Hamkins Dec 18 '15 at 1:05
  • $\begingroup$ Yes, "length" in Gonshor's approach is the same as birthdate in Conway's. $\endgroup$ – nombre Dec 18 '15 at 1:39

I claim that $s(\text{No}(\lambda))=\lambda^+$.

To see this, first let me mention that it is a standard exercise in elementary set theory to prove that there is no increasing or decreasing $\lambda^+$-sequence in the set ${}^\lambda 2$ of all binary $\lambda$-sequences, ordered in the lexical order, so that $s<t$ just in case $s$ is below $t$ at the first point of difference. I'll let you prove that on your own, or I can post later if you want.

From this, I claim, it follows that $\lambda^+$ does not embed into the set of surreal numbers born before $\lambda$. The reason is that every surreal number born before $\lambda$ is characterized uniquely by a $<\lambda$-sequence from $\{-1,1\}$, depending on which side of the cut in the previously born surreals the new number has landed. The surreal number order agrees exactly with the lexical order on the these sequences. (A shorter sequence is below a longer one, if the next digit is $1$, and above, if the next digit is $-1$; alternatively, imagine adding a tail of $0$s to compare sequences of difference lengths.) Since $\lambda^+$ does not embed into ${}^\lambda 2$, it cannot embed into ${}^{<\lambda}\{-1,1\}$.

The argument shows that you could even allow the surreals born on day $\lambda$ itself, continuing up to any particular day $\beta<\lambda^+$, and still $\lambda^+$ would not embed in.

  • $\begingroup$ The first fact is basically the result I am looking for. Since you say this is a standard exercice, I'll try to do it! Thank you. $\endgroup$ – nombre Dec 18 '15 at 1:40
  • $\begingroup$ Hint: if $\lambda^+$ embeds into ${}^\lambda 2$, then fix an embedding; eventually, the first digits of the targets stabilize, and then the next, and so on. But $\lambda^+$ is regular, so every $\lambda$ sequence is bounded. $\endgroup$ – Joel David Hamkins Dec 18 '15 at 1:44
  • $\begingroup$ Same argument works with ${}^\lambda\lambda$. $\endgroup$ – Joel David Hamkins Dec 18 '15 at 2:08
  • $\begingroup$ Okay this was pretty natural indeed, thanks. $\endgroup$ – nombre Dec 18 '15 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.