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I was just reading through the construction of the surreal numbers on wikipedia, and I read through some of the examples. I noticed that all of the examples were how certain types of already existing numbers (such as reals or hyperreals) could be constructed. However, I'm curious about examples of numbers that are unique to the set of surreal numbers. I've heard that they are the largest possible ordered field, so I would imagine that many if not most of surreal numbers are of this type. So what are some examples?

RothX
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    I'm not sufficiently conversant with hyperreals to know whether $\sqrt{\infty}$ is one. – Gerry Myerson Apr 22 '18 at 01:06
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    @MJD $+_2$ and $*$ are games, not surreal numbers. Every number is a game but not vice versa. – Mike Earnest Apr 22 '18 at 01:15
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    God kills a kitten every time someone mentions surreal numbers and hyperreal numbers in the same sentence. They have nothing to do with each other (beyond some very superficial things). – Eric Wofsey Apr 22 '18 at 01:17
  • @MikeEarnest I don't see why $+_2$ isn't a number? As far as I can tell, $0 < $ {$0 | 2$} is true. – RothX Apr 22 '18 at 02:27
  • You are right, ${0|{0|2}} = \frac12$ is a number. I got confused because $+_2$ is usually defined as ${0|{0|-2}}$, which is not a number. – Mike Earnest Apr 22 '18 at 02:53
  • I meant ${0\mid{0\mid-2}}$ and Mike Earnest was quite correct. – MJD Apr 22 '18 at 12:26
  • @EricWofsey All I said is that hyperreals are a subset of the surreals. Is that not correct? – RothX Apr 22 '18 at 15:03
  • It is incorrect. It is possible to embed the hyperreals in the surreals, but not in any canonical way. (For that matter, there is not really any such thing as "the" hyperreals, but rather many different objects that can be called "hyperreal fields".) – Eric Wofsey Apr 22 '18 at 15:06
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    How about {$1,2,3,...|\omega$}? This number is smaller than $\omega$ and larger than any finite number (as far as I can tell). Not sure if any number that is not unique to the surreals acts like that. – RothX Apr 22 '18 at 15:15
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    @RothX That's just $\omega-1$. You could instead try ${1,2,3,\ldots | \omega-1,\omega-2,\omega-3,\ldots}$, but that turns out to be $\omega/2$ (which you can prove by adding it to itself and using the definition of surreal addition). I'd be interested in an answer to this myself. – BlueRaja - Danny Pflughoeft May 01 '20 at 15:58
  • @GerryMyerson $\sqrt{\infty}$ is not a surreal number (and in most contexts means the same as $\infty$). – Anixx Jun 15 '24 at 20:22
  • @Ani, perhaps I should have written $\sqrt{\omega}$ which, according to the answer posted four years ago by user mjqxxxx, is a surreal number. – Gerry Myerson Jun 15 '24 at 22:35

3 Answers3

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A few remarks:

$\DeclareMathOperator{\Noo}{\mathbf{No}}$ -$\Noo$ is not "the largest possible ordered field", but rather "a universal ordered field" with nice properties. Universality is to be understood in the weak sense: every ordered field embeds in it, not in a unique way in general.

-The differents fields (or frameworks whithin which one can talk) of hyperreals do not compare in a canonical way with surreals. That $\Noo$ is universal requires some choice, and for instance, no explicit embedding of $^*\mathbb{R}$ into $\Noo$ is known. Thus it makes only little sense to compare which number lies in which field, and it is more fruitful to understand global properties of the different fields.

-Although in hyperreal fields constructed by ultrafilters on $\mathbb{N}$ one can embed ordinals below $\varepsilon_0$ in a somewhat natural way preserving natural aritmetic (send $\omega$ to the class of $id_{\mathbb{N}}$ modulo the ultrafilter and embeds other ordinals using operations on ordinals and on hyperreals), it is not usual to consider that hyperreals contain distinguished infinite elements such as "$\omega$".

-However, for any infinite element $x$ in $^*\mathbb{R}$, $x-1 \in ^*\mathbb{R}$ is strictly below $x$ but infinite.

-While an hyperreal field does not canonically contain all ordinals, it is closed under extensions of real functions, most of which aren't explicitly known on $\Noo$. IN this sense many expressions such as $\sin(H)$ make sense in $^*\mathbb{R}$ but not in $\Noo$.

-To finish on a positive note, a somewhat similar question is asked here.

nombre
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  • What does it mean when you say "canonically"? – RothX Apr 22 '18 at 20:52
  • I meant it in an informal way. There is no single order embedding of $\omega_1$ into $^*\mathbb{R}$ which is "more natural", or "more simple" than another. – nombre Apr 22 '18 at 23:11
  • How does "largest possible ordered field" differ from "every ordered field embeds in it"? Those seem like the same thing, and in fact every reference on surreals I've seen claims they are the same thing. It's even mentioned on the wikipedia page – BlueRaja - Danny Pflughoeft May 01 '20 at 11:28
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    @BlueRaja-DannyPflughoeft This is a matter of language. I am okay with saying "the largest" if this is properly defined (and correct). Just saying "the largest" conveys to me the sense that it is maximal in that it would have no proper extension, and that it is unique with that property. However it is not maximal, and for that matter it is not the unique universal ordered field (in the sense defined above). There are similar characterizations that do hold but I think "largest" is too vague a qualification to designate them. – nombre May 01 '20 at 13:46
  • My understanding is that there is no (ordered) field extension, by definition of being "universal" – BlueRaja - Danny Pflughoeft May 01 '20 at 15:48
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    @BlueRaja-DannyPflughoeft Just because every ordered field embeds in $\mathbf{No}$ doesn't mean that $\mathbf{No}$ has no extension. There are proper extensions, but they all embed in $\mathbf{No}$ anyways. The minimal example is the field $\mathbf{No}(X)$ of rational functions over $\mathbf{No}$. It can be ordered in many ways, each one reflecting that $X$ fills a certain gap in $\mathbf{No}$. For instance, setting $F(X)>0$ if $F(a)>0$ for sufficiently large $a \in \mathbf{No}$, you get $X$ to fill "the gap ${\mathbf{On} \ | \ }$" if I am using this denotation correctly. – nombre May 01 '20 at 16:05
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For a simple answer to the original question, per Gerry Myerson's comment, $$ \sqrt{\omega} \equiv \left\{1,2,3,\ldots \;\vert\; \omega, \omega/2, \omega/3, \ldots \right\} $$ is a surreal number that isn't part of the usual hyperreals. (Its name is justified because you can verify that $\sqrt{\omega}\cdot\sqrt{\omega}=\omega$.)

My understanding is that surreal numbers like $\omega^\omega\equiv\left\{\omega,\omega^2,\omega^3,\ldots\;\vert\;\right\}$ are also outside the usual hyperreals.

mjqxxxx
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    If $\omega$ denotes the hyperreal that is the equivalence class of $(1,2,3,\dots)$ then the equivalence class of $(1,\sqrt 2,\sqrt 3,\dots)$ would be $\sqrt\omega$. And the equivalence class of $(1,2^2,3^3,\ldots)$ would be a number deserving of a name like $\omega^\omega$. – Mark S. May 05 '20 at 18:07
  • Seems like this works. I'm wondering how this lines up with @nombre's answer. Do we run into any issues here due to the ambiguity associated with the definition of the hyperreals, and the way in which they interact with the surreals? I feel like it shouldn't, since $\omega$ has a proper definition in the surreals, if I'm not mistaken. – RothX May 05 '20 at 18:20
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    @MarkS. Do you mean ${1, \sqrt{2}, \sqrt{3}, \ldots\ |\ }$? If so, that's $= \omega$ by the simplicity theorem – BlueRaja - Danny Pflughoeft May 05 '20 at 18:30
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    @BlueRaja, your statement is correct, but no I did not mean that. I was referring to the standard ultrapower construction of the hyperreals, rather than the "ordered pair of sets of surreals" construction of the surreals. – Mark S. May 05 '20 at 19:17
  • But it is well part of Hardy fields or formal power series. – Anixx Jun 15 '24 at 20:24
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Another example I've just learned. Let $\Omega$ be the set of all hyperreals (whatever your definition for that is). Then $$\omega_1=\{\ \Omega\ |\ \}$$ defines a new number $\omega_1$ larger than every hyperreal, which will be reached on the $\omega_1$-th day.

BlueRaja - Danny Pflughoeft
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    In fact, this depends on the definition, since there is no natural embedding of a field of hyperreal numbers (which can have arbitrary cardinality) in $\mathbf{No}$. – nombre May 05 '20 at 18:17
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    @nombre Surely it always works an an answer to the question though? For any hyperreal field, and any embedding of that field into the surreals, this element would not be hyperreal.

    I guess the more interesting question, is: when generating $\omega_1$ given a hyperreal field and an embedding, is there a different hyperreal-embedding pair such that one of those hyperreals maps to $\omega_1$?

     – RothX May 05 '20 at 18:36
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    Yes but isn't it a bit too tautological that there be somehting which is not in the range of each embedding? The thing is, any surreal number can lie in the range of such embedding. Even if you take a free ultrafilter on $\mathbb{N}$ to build $^\mathbb{R}$ and decide to send the class $c$ of $(0,1,2,...)$ onto $\omega$, then anything that lies above $\mathbb{N}[c]$ in $^\mathbb{R}$ can be sent onto anything that lies above $\mathbb{R}[\omega]$ in $\mathbf{No}$, including $\omega_1$ or $\omega_{\omega_1}$. – nombre May 05 '20 at 18:47
  • @nombre So, for any surreal number, there is an embedding that sends a hyperreal to that number? That's actually fairly interesting and answers the question in a way. If you can elaborate on that more, might be worth an answer. – RothX May 07 '20 at 17:54
  • I don't know much about ultrafilters, but I think @nombre is saying that $\omega_1$ can be generated in the hyperreals in a similar way. In which case, are the two number systems isomorphic (as claimed by another answer on this site)? – BlueRaja - Danny Pflughoeft May 07 '20 at 18:26
  • @RothX Yes but this is not a feature of the surreal numbers. This is actually true of any sufficiently saturated ordered field containing said field of hyperreal numbers. This is yet another reason why it does not make sense to wonder whether this or that number exists in one of two systems. – nombre May 07 '20 at 18:58
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    @BlueRaja-DannyPflughoeft I don't think that is what I meant. Like I said there are several non-isomorphic things called "field of hyperreal numbers". One of them is isomorphic to $\mathbf{No}$, one of them is an object of models of Internal Set Theory, some of them contain ordinals in a sense, some not particularly... So you have to say which system you are talking about and which embedding into $\mathbf{No}$ is under consideration to make sense of the type of questions discussed here. – nombre May 07 '20 at 19:01
  • @nombre That fact doesn't seem relevant given that you already decided which one we're talking about, above. And you've contradicted both my answer and subsequent comment, but they can't both be wrong: either $\omega_1$ is a part of that hyperreal field, or it's not. – BlueRaja - Danny Pflughoeft May 08 '20 at 17:42
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    @BlueRaja-DannyPflughoeft Is $0$ an element of $\mathbb{R}$? Is $\varnothing$? Those things depend on some context. If you fix a field of hyperreal numbers as well as an embedding into $\mathbf{No}$, then you can ask whether $\omega_1$ seen as a surreal number lies in the range of that embedding yes. But that it does not help you decide the meaningless statement that $\omega_1$ lies in the field itself. – nombre May 08 '20 at 17:47