For a sequence $(x_{\alpha})$ of surreal numbers indexed by the set of all ordinal numbers, we say that $\lim x_{\alpha}=l$ ($l$ is a surreal number) if for each surreal $\epsilon>0$, there exists an ordinal $\beta$ such that $|x_{\alpha}-l|<\epsilon$ for each ordinal $\alpha>\beta$.

Consider the sequence $x_{\alpha}=(1+1/\alpha)^\alpha$, where for each surreal $x,y$, $x^y$ is defined by $\exp(y\log x)$.

Q1: What is the normal form of $x_{\omega}$? Is its real part equals to $e$?

Q2: What is $\lim x_{\alpha}$?

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    $\begingroup$ What, $e$ is not surreal enough for you as it is? $\endgroup$
    – Asaf Karagila
    May 30, 2017 at 16:07
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    $\begingroup$ The set of all ordinal numbers? $\endgroup$
    – Nik Weaver
    May 30, 2017 at 16:34
  • $\begingroup$ I think you mean "standard part" rather than "real part". $\endgroup$ May 30, 2017 at 19:46
  • $\begingroup$ Of course many sequences indexed by the ordinal numbers do not converge at all in the surreals. That is, the left and/or right "set" has to be a proper class. Do you think this sequence is one? $\endgroup$ May 30, 2017 at 21:46
  • $\begingroup$ @JoelDavidHamkins : Any surreal $x$ can be written in a unique way as the sum $x_{\prec 1} + x_{\asymp 1} + x_{\succ 1}$ where $x_{\prec 1}$, the infinitesimal part, is infinitesimal, $x_{\asymp 1}$, the real part, is real, and $x_{\succ 1}$, the purely infinite part, has only strictly positive exponents in its normal form, or equivalently is a logarithm of some $\omega^y$. $\endgroup$
    – nombre
    May 30, 2017 at 22:01

1 Answer 1


$\DeclareMathOperator{\ee}{e}$If $\varepsilon$ is an infinitesimal surreal, the quantity $\log(1+\varepsilon)$ is actually equal to the formal sum à la Hahn series $\sum \limits_{n \in \mathbb{N}} \frac{(-1)^n\varepsilon^{n+1}}{n+1}$, and $\log(1+\varepsilon) - \varepsilon$ is negligeable with respect to $\varepsilon$. So the proof for real numbers can be applied here to show that the sequence converges to $\ee$.

As for $x_{\omega}$, this is $\exp(1 - \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ where $\exp(- \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ is infinitesimally close to $1$ because $a:= -\frac{1}{2\omega} + \frac{1}{3\omega^2} - ...$ is an infinitesimal. So the real part of $x_{\omega}$ is $\ee$.

For now I don't see what its normal form because some combinatorial cleverness seems to be required in unfolding $\sum \limits_{n \in \mathbb{N}} \frac{a^n}{n!}$. I'll edit this answer if I find something. In the meantime you can try to find it too: we know that exponents of $\omega$ in the $a^n$ and $\exp(a)$ will be negative integers. It might be easier to compute the exponential sum directly if you can find relations between the coeffifients $q_{n,k}$ of $\omega^{-k}$ in $a^n$ for different values of $n$.

  • 1
    $\begingroup$ @WillSawin That moment you forgot a dollar sign and the editing period timed out. $\endgroup$
    – user78249
    May 30, 2017 at 22:23
  • $\begingroup$ Doesn't this argument show that the limit is actually e? It seems to show that $(1+\epsilon)^{1/\epsilon}$ is equal to $e$ to within $O(\epsilon)$, so if $\epsilon$ gets smaller than the inverse of any ordinal, it should be equal ot $e$ within the inverse of every ordinal. $\endgroup$
    – Will Sawin
    May 30, 2017 at 22:46
  • $\begingroup$ @james.nixon Fixed. $\endgroup$
    – Will Sawin
    May 30, 2017 at 22:46
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    $\begingroup$ @WillSawin : $\DeclareMathOperator{\ee}{e}$$\DeclareMathOperator{\Ord}{Ord}$Yes, this shows that the limit is $e$, and for that matter, that $\exp$ can be defined as $x \mapsto \lim \limits_{\alpha \in \Ord} (1 + \frac{x}{\alpha})^{\alpha}$. What's missing in the anwser is the normal form of $x_{\omega}$. $\endgroup$
    – nombre
    May 31, 2017 at 7:11
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    $\begingroup$ @PaoloLipparini : I don't know if there is an easier way to define $s^{\alpha}$ for $\alpha \in \mathbf{Ord}$. There doesn't seem to be a strong connection between the standard surreal exponential function and ordinals. $\endgroup$
    – nombre
    Feb 14, 2020 at 20:52

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