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In Infinity and the Mind the author claims:

Conway derives the weird equation, $\infty=\sqrt[\Omega]{\omega}$, which almost magically ties together potential infinity $\infty$, the simplest actual infinity $\omega$, and the Absolute Infinite $\Omega$.

I have not seen this equation anywhere else & was wondering if it actually something Conway derived, & if so, where.

user784623
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    See Conway's On Numbers and Games for a lot of weird things, possibly including this one. – Angina Seng May 24 '20 at 05:17
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    Is that meant to be $\aleph_0$ instead of $\infty$ ? – jimjim May 24 '20 at 05:49
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    @Arjang: Most probably not. The context here is surreal numbers (see the tag), and in that context, $\infty$ denotes the gap between finite and positive infinite numbers. I'm not aware of $\aleph_0$ being assigned to any surreal number or gap (other than indirectly, through the common identification of $\aleph_0$ with $\omega$, but in that case, I don't see how it would make sense to use that in the context of this equation). – celtschk May 24 '20 at 06:04
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    If you take the $x$-th root of $\omega$ for sufficiently large $x$, (i.e. for $x$ close enough to the edge of the gap $\Omega$), then you get as close as you want (though always infinitely far) from the edge of the gap $+\infty$. I suspect this is what the author means. (but I don't think this is particularly "profound" or that it gives insight on surreal numbers) – nombre May 24 '20 at 15:51
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    @nombre You could basically rewrite that as an answer. – Mark S. May 24 '20 at 19:56

2 Answers2

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So there's a formalization of this relation. Given strictly monotonous function $f:X \rightarrow Y$ between dense linear orders $(X,<)$ and $(Y,<)$, there is a natural extension of $f$ to a strictly monotonous function $\widetilde{f}:\widetilde{X} \rightarrow \widetilde{Y}$ with same monotonicity as $f$. Here $\widetilde{X}$ denotes the set of gaps in $X$, i.e. the set of initial segments of $X$ without a maximum in $X$ (one other convention for gaps is that they be pairs $(L,R)$ where $L$ is a gap and $R$ is its set of strict upper bounds). Ordered by inclusion, it is a dense linear order, and the function $X \ni x \longmapsto (-\infty,x) \in \widetilde{X}$ is the canonical embedding of $X$ into $\widetilde{X}$.

Taking $f_0$ as the function $\mathbf{No}^{>0} \rightarrow \mathbf{No}:x \mapsto \omega^{\frac{1}{x}}$, and applying this to the gaps $\Omega:=\mathbf{No}$ and $\infty:=\{x \in \mathbf{No}^{>0} : \exists n \in \mathbb{N},x<n\}$, we get $\infty=\widetilde{f_0}(\Omega)$. This is just a fancy way of saying that $\omega^{\frac{1}{x}}$ gets as small as one wants among positive infinite elements when $x$ is large enough.

But the calculus on gaps that the sort of extension theorem I mention allows is an interesting tool. For instance in the case of fields, you can obtain a (not so well-behaved) sum and product of gaps. In the case of $\mathbf{No}$ which is a field of transseries, you can do much more and represent gaps as nested transseries. All of this is done in a slightly different context in Joris van der Hoeven's lecture notes.

The definition of $\widetilde{f}$ for strictly decreasing $f$ is as follows. The set $(\widetilde{Y},<)$ has the largest lower bound property (and the least upper bound property which we won't require for strictly decreasing functions). Indeed if $A \subseteq \widetilde{Y}$, then one can check that $\{y \in Y: \forall a(a \in A \Rightarrow y \in a)\}=\inf A$ (this is just the intersection of $A$). So you just define $\widetilde{f}(a):=\inf \{f(x) : x \in X,x\leq a\}$.

I let you check that the case of $f_0$, you get the desired result.

I should add that since $\mathbf{No}$ is a proper class, the collection $\widetilde{\mathbf{No}}$ does not make sense in ZFC or NBG, and one has to move into stronger set theories to talk about it. But the results then would carry over provided the set theory is nicely behaved with respect to ZFC (for instance Ackermann set theory).

nombre
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After contacting the author, I found out the formula he was referring to is in ONAG on page 38 & is given by Conway as $$\infty=\omega^{1/\mathbf{On}}$$ Letting $\Omega$ be equivalent to $\mathbf{On}$ leads to the derivation of the formula asked about $$\infty=\sqrt[\Omega]{\omega}$$

user784623
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