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Am I correct that the hyperexponential $\exp_{\omega}$ is a bijection on positive infinite surreals?

An exponential level is an equivalence class for the relation $a \asymp_L b \Leftrightarrow \exists n,k \in \mathbb{N}, \exp_n(2^{-k} \log_n(a))\leqslant b \leqslant \exp_n(2^k \log_n(a))$ on the class of positive infinite numbers, where $\exp_n$ and $\log_n$ denote the $n$-fold iterates of $\exp$ and $\log$.

Am I correct that an exponential level is a convex interval?

If so, what is the relationship between them?

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  • $\begingroup$ Hi there, I'm probably best suited to answer your question, but I'm not sure what you mean by "an equivalence of <>E". Are you perhaps using notations from an article where you read about this? If so, this will be difficult to read by people here since these notions are niche and hardly consensual. The equivalence relation you seem to be alluding to is Berarducci and Mantova's relation $\asymp_L$ on positive infinite surreal numbers, where $a \asymp_L b \Leftrightarrow \exists n \in \mathbb{N}, |\log_n(a) -\log_n(b)|\leqslant 1$, $\log_n$ being the $n$-fold iterate of $\log$. $\endgroup$
    – nombre
    Commented Sep 19 at 12:05
  • $\begingroup$ The one I saw had an E, and it was defined as exp_n (2^-k log_n(a)) <= b <= exp_n (k log_n(a)) $\endgroup$
    – user23467
    Commented Sep 19 at 16:24
  • $\begingroup$ How do you get that math text? $\endgroup$
    – user23467
    Commented Sep 19 at 16:33
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    $\begingroup$ @user23467 Use mathjax. $\endgroup$
    – Noah Schweber
    Commented Sep 19 at 16:35
  • $\begingroup$ I edited accordingly. The two definitions are equivalent. For math text, the syntax is latex, and you have to put things between dollar signs. E.g. \log_n between two dollar signs is $\log_n$. $\endgroup$
    – nombre
    Commented Sep 19 at 16:36

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To partially answer the question, exponential levels are convex classes, and they are not intervals. This is because for instance an exponential level is closed under adding or substracting $1$, which cannot be the case for intervals with surreal bounds.

There is a relation between $\exp_{\omega}$ and exponential levels is that the exponential level of a positive infinite number $\exp_{\omega}(a)$ is the image under $\exp_{\omega}$ of the class of positive infinite elements $b$ for which there exists an $n \in \mathbb{N}$ with $a - \frac{1}{\log_n a}\leqslant b \leqslant a + \frac{1}{\log_n a}$.

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  • $\begingroup$ I don't understand. How do you go from the single number a to an entire class? Are you saying that exp_ω(b) is the same for every b in that class? $\endgroup$
    – user23467
    Commented Sep 20 at 6:03
  • $\begingroup$ The image of all such $b$ under $\exp_{\omega}$ is the whole exponential level of $\exp_{\omega}(a)$. $\endgroup$
    – nombre
    Commented Sep 20 at 7:44
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    $\begingroup$ I notice that the bounds you gave for b are infinitesimally different from a (because log_n(a) is infinite). So you're saying that exp_ω maps this infinitesimal neighborhood to an exponential level. Does that mean that if the infinitesimal support of a is empty, exp_ω(a) would be the log-atomic representative of that exponential level? $\endgroup$
    – user23467
    Commented Sep 20 at 9:24
  • $\begingroup$ Yes it is. In fact the hyperexponential function induces a bijection between positive numbers with purely infinite support and so-called $\kappa$-numbers, which are the simplest representatives for the wider equivalence classes $a \asymp_K b$ if $\exists n \in \mathbb[N},\log_n(a) \leqslant b \leqslant \exp_n(a)$. If you impose only empty infinitesimal support, then you remain $\log$-atomic, and more precisely you are a real-iterate of $\exp$ applied at a $\kappa$-number. You can find a list of relevant identities in Section 6.1 of the pre-print "The hyperserial field of surreal numbers". $\endgroup$
    – nombre
    Commented Sep 20 at 9:37

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