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This answer refers to $\omega_1$ in context of surreal numbers, and calls it "first uncountable ordinal".

But what exactly does it mean? How can it be represented in the $\{L|R\}$ form?

How do we know that "first uncountable ordinal" represents one surreal number and not a set of numbers? How is $\omega_1$ distinguished from $2\omega_1$?

I mean, the surreal $\omega$ can be considered equivalent to the germ at infinity of the function $f(x)=x$. But what about $\omega_1$? What is the representation?

According to my reasoning, if we take $\omega$ to be equal to the germ of $x$, the simplest constructed uncountable number is $\alpha=\int_0^1 \omega \, dx$. But it is the numerosity of the interval $[0,\pi)$, and not of the interval $[0,1)$.

On the other hand, we can simply say that the numerosity of the interval $[0,1)$ is the simplest uncountable number, but then it is constructed a bit more complicatedly: $N([0,1))=\frac1\pi\int_0^1 \omega \, dx$. Or, if we want to get rid of the $1/\pi$ coefficient, we should equate $\omega$ with the germ of $x/\pi$, which, I think, is not considered a canonical embedding of Hardy fields into surreals.

So, how is the canonical embedding determined? Could it be that the number $\omega_1$ could represent the numerosity of $[0,\pi)$?

Due to Euclid's principle, numerosities are totally ordered, so should be embedded into surreal numbers. On the other hand, what is the canonical embedding?

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  • $\begingroup$ Did you mean numerosity? $\endgroup$ Commented May 7 at 18:01
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    $\begingroup$ @J.W.Tanner thanks, fixed. $\endgroup$
    – Anixx
    Commented May 7 at 18:42

2 Answers 2

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There is nothing special about $\omega_1$ or indeed any infinite number in the surreals, and they cannot be defined purely from the field structure of the surreals. What I claim is that all infinite numbers in the surreals are automorphic. That is, all such numbers can be moved from one to the other by automorphisms of the surreal field, and consequently they look exactly the same with respect to the surreal algebraic structure.

For example, every statement in the language of fields that is true of $\omega$ is also true of $\omega+1$ and $\omega_1+\omega-5$ and $\omega_1$, and indeed there is an automorphism of the surreals $\pi:№\to№$ with $\pi(\omega_1)=\omega$.

The reason is that the surreal numbers are a saturated real-closed field. In any real-closed field, by a famous theorem of Tarski's, every assertion in the first-order language of fields is equivalent to a quantifier-free statement. From this it follows that the definable sets of surreals are finite unions of intervals with rational endpoints. Since all infinite numbers will sit the same way in those intervals, it follows that they all have the same $1$-type. By saturation it follows that there is an automorphism $\pi:№\to№$ moving any of them to another.

So there are automorphisms moving $\omega_1$ to any other infinite element, including to $\omega$ or $\sqrt{\omega}$ or $\omega_1+1$ or what have you.

What this argument shows is that the $\{L\mid R\}$ notation is not a field-theoretic concept — it is not respected by automorphisms of the field. The $\{L\mid R\}$ notation is not structural with respect to the surreal field as a field. The $\{L\mid R\}$ notation, in contrast, as well as the underlying tree structure of the surreals, are conceptions arising in a way of constructing the surreals. We are able to name particular numbers arising in that particular construction, and this is often extremely useful. But we cannot recover the name of a surreal number from the number itself in a purely algebraic manner.

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    $\begingroup$ @Anixx No, because the 1-type of $\omega_1$ with parameter $\omega$ is still the same as all the other large enough surreals. Anything above all the polynomials in $\omega$ will still be automorphic, because of the limited power to define things in real-closed fields. $\endgroup$ Commented May 7 at 13:30
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    $\begingroup$ There are wild automorphisms of the complex numbers. For that matter, there is an automorphism exchanging $i$ and $-i$. Yet, the working mathematician still treats each complex number, including $i$, as a definite thing; probably the mental picture of the “complex plane” helps here. Maybe things are not so different with the surreal numbers, where the “game” picture is part of what each surreal number is (even if it cannot be seen purely from the field structure). $\endgroup$ Commented May 7 at 14:27
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    $\begingroup$ @SamHopkins I agree very much with that. We generally understand the complex numbers as having the form $a+bi$, but this form is not deducible from the pure algebraic structure of the complex field. There are many different copies of $\mathbb{R}$ in $\mathbb{C}$, over which $\mathbb{C}$ arises as the algebraic closure, and one cannot define real-part, imaginary-part in $\mathbb{C}$ using only the field structure. This situation is something like the $\{L\mid R\}$ representation of surreal numbers being similarly not respected by the algebraic structure. $\endgroup$ Commented May 7 at 15:54
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    $\begingroup$ @MikeShulman The order is definable from the algebra, since $x\leq y$ if and only if $\exists z\ (x+z^2=y)$. $\endgroup$ Commented May 7 at 16:12
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    $\begingroup$ Absolutely agreed with the technical part of this answer, but I’d draw a slightly different conclusion from it, or at least a slightly different emphasis — not so much “the games are just one way of constructing the surreal field” (though that’s certainly true), but more “the field structure is not capturing the full structure we implictly consider on the surreals”. @SamHopkins’ analogy with the complexes is apt: the automorphisms show us that the embedding of the reals and the choice of $i$ are part of our implicit structure on $\mathbb{C}$. $\endgroup$ Commented May 7 at 21:46
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Assuming you already know how to define all the at most countable ordinals as surreal numbers, then $\omega_1 = \{ \text{those ordinals} \mid \}$. You can define ordinals in the context of surreal numbers as numbers whose right set is empty and the left set are other ordinal numbers. (This is not a vicious cycle because of regularity.) And a countable ordinal is just an ordinal whose left set is countable.

This number is different from $2 \omega_1$, if what you mean by that is the surreal number $2$ and the surreal number $\omega_1$ under surreal multiplication. In particular it is larger, as one can verify by looking up all the definitions.

I think there are too many questions in this post, not particularly well-connected. So I'll refrain from answering the rest, and recommend you split them off into a new question.

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    $\begingroup$ Thanks, so according to this answer mathoverflow.net/a/470718/10059, one can define $\omega_1$ with left set being the canonical Hardy field embedding into surreals, and the right-set empty. But I was always wondering how this construction works in surreal numbers? Is it presumed to be the simplest number that is greater than the left-set? How do we determine or choose the simplest? $\endgroup$
    – Anixx
    Commented May 7 at 10:47
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    $\begingroup$ My question was, how do we know that this definition of $\omega_1$ in the surreal numbers coincides with the definition of $\omega_1$ as the least upper bound of countable ordinals. $\endgroup$
    – Anixx
    Commented May 7 at 11:15
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    $\begingroup$ There is no LUB phenomenon in the surreals. No set of surreals has a LUB, except those sets with a largest element. $\endgroup$ Commented May 7 at 11:22
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    $\begingroup$ $\omega_1$ is the first-born upper bound of the countable ordinals, not the least upper bound. $\endgroup$ Commented May 7 at 11:24
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    $\begingroup$ @Anixx That is the least upper bound in the ordinals. Ordinals do have least upper bounds. This is indeed a bit confusing and I didn't make my phrasing as clear as I'd like. $\endgroup$
    – Trebor
    Commented May 7 at 11:38

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