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I'd like to consider the principle asserting that the surreal number line is universal for all class linear orders, or in other words, that every linear order (including proper-class-sized) linear orders, order-embeds into the surreal number line.

This principle is the class-analogue of the corresponding result for countable linear orders, proved by Cantor, namely, that every countable linear order embeds into the rational line $\newcommand\Q{\mathbb{Q}}\Q$. Using the "forth" part of Cantor's back-and-forth method, one enumerates a given countable linear order and then maps each new point to a rational number filling the corresponding cut in the images of the previous points.

A similar transfinite extension of this argument works with the surreal numbers. Namely, if $\langle A,\leq_A\rangle$ is any class-sized linear order and $A$ is $\newcommand\Ord{\text{Ord}}\Ord$-enumerable, then we may map each new point to the first-born surreal number filling the corresponding cut in the images of the previous elements of $A$, and thereby construct an embedding of $A$ to the surreal numbers.

What this argument shows is that, provided the elements of the linear order $A$ are $\Ord$-enumerable, then we may recursively build up an embedding of the linear order into the surreal number line. In other words, if the principle of global choice holds, which is equivalent to the assertion that every class is $\Ord$-enumerable, then every class-sized linear order embeds into the surreal line. So global AC implies that the surreals are universal for class linear orders.

Similar uses of global choice underlie many other assertions of universality for the surreal numbers, with respect to class-sized structures, as ordered fields and so on. It is fundamental in all the arguments that I know for such universality that one should be able to enumerate the elements of the domain structure in order type $\Ord$, in order to carry out the transfinite embedding procedure.

Since this use of the $\Ord$-enumerations has always struck me as essential, I have long believed that the global axiom of choice was required in order to prove that the surreal number line is universal for class linear orders.

But upon recent reflection, I realized that I don't actually have a proof that global AC is required. I simply don't have a proof that global AC is required for universality, and at the same time, I don't know how to omit it.

Hence these questions:

  • Is the global axiom of choice required to prove that the surreal number line is universal for all class-sized linear orders?

  • Can one prove the universality of the surreal number line merely in GB+AC, without the global axiom of choice?

  • Or is the universality of $\text{No}$ an intermediate principle between AC and global AC?

To answer these questions, it would seem to be useful to settle the issue of the surreal-numbers-universality principle (which I propose we denote by $\text{UNo}$), in some of the known models where global AC fails. For example, in my answer to the question Does ZFC prove the universe is linearly orderable?, I showed that there are models of ZFC with no definable linear order of the universe. If we could settle the universality of the surreals in that model, it would answer at least one of the questions above. For example, if the model had surreal-universality UNo, then we would know that universality is not equivalent to global AC; if it is didn't have it, then we would know that UNo isn't provable in GB+AC.

(This topic is related to some ideas I'll be speaking about at my talk on The hypnagogic digraph, which I am giving this Friday at the JMM in Seattle.)

  • Hypnagogic digraph! Quite an intriguing title. Will there be forthcoming discussions of the hypnopompic digraph as well? :-) – Todd Trimble Jan 07 '16 at 06:14
  • I'll post my slides on Friday, and you can read all about it! – Joel David Hamkins Jan 07 '16 at 06:18
  • When you say global choice, does it come with parameters? Specifically, if you add a Cohen real to $L$, you have global choice with parameters, but not global choice itself. It stands to reason that in such model you might have what you were looking for. – Asaf Karagila Jan 07 '16 at 06:20
  • I think of global-AC as an assertion in second-order set theory, as in Goedel-Bernays set theory, so one doesn't need to worry about definability. That is, one simply asserts that there is a proper class choice function on all nonempty sets, or equivalently, that there is an Ord-enumeration of V. In particular, global AC definitely holds after adding a Cohen real. – Joel David Hamkins Jan 07 '16 at 06:22
  • @AsafKaragila But let me say that I'm very glad you're looking at this question. I'd appreciate any insight you might have. – Joel David Hamkins Jan 07 '16 at 06:25
  • I just opened my eyes ten minutes ago, it's my pleasure to have choice related questions to think about when I wake up. :-) – Asaf Karagila Jan 07 '16 at 06:27
  • If you think about $L[G]$ in the model in my answer to your question about linear orders of $V$, then the question is whether UNo holds there. We know there is no linear order of all of $L[G]$ there, so perhaps there is a paucity of definable linear orders there (always allowing parameters). If so, then we might expect UNo to hold. But I can't quite prove it. – Joel David Hamkins Jan 07 '16 at 06:29
  • Joel, what I'd try to do is to show that every linear ordering comes from a bounded part. – Asaf Karagila Jan 07 '16 at 06:32
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    Also, when you appeal to UNo, do you have to shout UNo when you have one line left in your proof, or else you have to take four more lines to finish? – Asaf Karagila Jan 07 '16 at 06:34
  • Keep in mind that the surreal numbers No themselves are definable in L[G], and as elements these define every object in L[G]. – Joel David Hamkins Jan 07 '16 at 07:41
  • The question intrigues me although I have extremely little knowledge of such matters (but I'm pondering on variations of surreal numbers for constructive math). If I understand correctly, No itself is Ord-enumerable iff Global AC holds? (I don't know the first thing about this, but read it in another post). If that is correct, then intuitively I'd say: global AC is not necessary for UNo, because one can embed No in No without global AC... This may seem a strange argument, but it is meant to illustrate that one can have weaker conditions which already give embeddability. – Franka Waaldijk Jan 08 '17 at 13:34
  • Or to put it the other way round: the fact that a class-sized linear order A is embeddable in No apparently does not give a way to Ord-enumerate A. If one wants to somehow use the property `all class-sized linear orders are embeddable in No', then again, we trivially know that all class-sized linear orders isomorphic to a subclass of No are embeddable...so you would have to come up with a class-sized linear order B defined outside of No, and from B's embeddability in No deduce that No is Ord-enumerable. [Well, I hope I'm making some sense here, feel free to disregard my question/comments]. – Franka Waaldijk Jan 08 '17 at 14:33

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