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?
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2$\begingroup$ I'm not sufficiently conversant with hyperreals to know whether $\sqrt{\infty}$ is one. $\endgroup$– Gerry MyersonApr 22, 2018 at 1:06
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6$\begingroup$ @MJD $+_2$ and $*$ are games, not surreal numbers. Every number is a game but not vice versa. $\endgroup$– Mike EarnestApr 22, 2018 at 1:15
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2$\begingroup$ 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). $\endgroup$– Eric WofseyApr 22, 2018 at 1:17
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$\begingroup$ @MikeEarnest I don't see why $+_2$ isn't a number? As far as I can tell, $0 < $ {$0 | 2$} is true. $\endgroup$– RothXApr 22, 2018 at 2:27
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3$\begingroup$ 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. $\endgroup$– RothXApr 22, 2018 at 15:15
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3 Answers
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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.
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$\begingroup$ What does it mean when you say "canonically"? $\endgroup$– RothXApr 22, 2018 at 20:52
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$\begingroup$ 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. $\endgroup$– nombreApr 22, 2018 at 23:11
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$\begingroup$ 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 $\endgroup$ May 1, 2020 at 11:28
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1$\begingroup$ @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. $\endgroup$– nombreMay 1, 2020 at 13:46
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1$\begingroup$ @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. $\endgroup$– nombreMay 1, 2020 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.
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3$\begingroup$ 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 $\omega^\omega$. $\endgroup$– Mark S.May 5, 2020 at 18:07
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$\begingroup$ 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. $\endgroup$– RothXMay 5, 2020 at 18:20
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1$\begingroup$ @MarkS. Do you mean $\{1, \sqrt{2}, \sqrt{3}, \ldots\ |\ \}$? If so, that's $= \omega$ by the simplicity theorem $\endgroup$ May 5, 2020 at 18:30
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1$\begingroup$ @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. $\endgroup$– Mark S.May 5, 2020 at 19:17
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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.
answered May 5, 2020 at 18:04
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
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2$\begingroup$ 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 $\mathbf{No}$. $\endgroup$– nombreMay 5, 2020 at 18:17
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1$\begingroup$ @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 $\omega_1$? $\endgroup$– RothXMay 5, 2020 at 18:36
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1$\begingroup$ 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 $\omega_{\omega_1}$. $\endgroup$– nombreMay 5, 2020 at 18:47
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1$\begingroup$ @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. $\endgroup$– nombreMay 7, 2020 at 19:01
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1$\begingroup$ @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. $\endgroup$– nombreMay 8, 2020 at 17:47