I'm trying to understand analogies and disanalogies between ${\Bbb R}$, the reals numbers, and ${\bf No}$, the surreal numbers.

${\Bbb R}$ admits countable dense sets such as the rationals. This leads me to wonder whether ${\bf No}$ might contain a dense subclass somehow smaller than the whole.

On one hand, any sub-set of ${\bf No}$ must be closed in the interval "topology" since every element of ${\bf No}$ lies in a nest of shrinking intervals ordered like ${\bf On}$, the class of ordinals. So that makes any dense subclass a proper class.

On the other hand, ${\bf ZFC}$ doesn't entail a well-ordering of the whole universe (right?) so that means one might (consistently) regard the proper class ${\bf On}$ as small compared with the class of all sets.

${\bf Question}$: Can we also regard ${\bf No}$ as a larger proper class than ${\bf On}$, in the sense that there does not exist any surjection from ${\bf On}$ to ${\bf No}$?

In other words, might ${\bf ZFC}$ not guarantee any well-ordering of all of ${\bf No}$. (This despite a well-ordered class of birthdays and well-orderings for all elements with any set of birthdays).

It nevertheless seems reasonable to wonder whether ${\bf ZFC}$, or even ${\bf ZF}$, entails a dense subclass of ${\bf No}$ that admits a well-ordering hence my main

${\bf Question}$: Does there exists a (definable) map from all the ordinals to the surreal numbers that has a dense image?

As with my last question, I confess uneasiness about working with proper classes, so please feel free to correct any misconception that underlies the formulation of my question. In particular, please tell me if I need to change set theories to get a better question.

  • $\begingroup$ I think this might be relevant to your first question: mathoverflow.net/questions/44303/cardinality-of-classes $\endgroup$ – Michael Greinecker Apr 8 '12 at 1:44
  • $\begingroup$ Thanks Michael Greinecker, but I think the answer there just says that global choice moots my question, as I suggested already. Gonshor's book defines a notation of surreal integer and proves every element of No equals the quotient of two integers. But his class of integers doesn't admit any obvious well-ordering. It seems not entirely unreasonable that some narrower concept of integer might lead to a definably well-ordered class and merely a dense set of quotients. So I think I have a real question about the surreals and not just about classes. $\endgroup$ – David Feldman Apr 8 '12 at 3:53
  • $\begingroup$ While not directly related to this question, this m.SE thread might be helpful in understanding a bit how the surreals work in a small scale: math.stackexchange.com/q/77992/622 $\endgroup$ – Asaf Karagila Apr 8 '12 at 12:13

The answer is that the existence of a definable class embedding like that is independent of ZFC. In fact, it is equivalent to the axiom V=HOD.

Theorem. The following are equivalent.

  1. There is a definable bijection from Ord to No.

  2. There is a definable surjection from Ord to No.

  3. There is a definable map from Ord to No with dense image.

  4. V=HOD.

Proof. If V=HOD holds, then there is definable bijection between the class of ordinals and the entire universe $V$, and from such an embedding one can construct a definable bijection between Ord and No. So 4 implies 1, which implies 2, which implies 3.

Convesely, suppose that V=HOD fails. Since in ZFC every set is coded by a set of ordinals, it follows that there must be some $\alpha$-length binary sequence $s$ that is not in HOD, which means that $s$ is not definable from ordinals, and neither therefore is any longer sequence than $s$. But the binary sequence $s$ determines a certain interval in No, by following the digits of $s$ left-and-right through the tree representation of the surreal numbers. If there were a definable map from Ord to No with dense image, then there would be OD element of No in this interval. Suppose that the $\beta^{\rm th}$ element was in that interval. It follows that $s$ would be definable from $\beta$ and $\alpha$ as the $\alpha$ length sequence describing the interval at that level of the tree inside of which the $\beta$-th surreal was to be found. So $s$ would be ordinal definable, a contradiction. So 3 implies 4. QED

The standard way to force $V\neq \text{HOD}$ is to add a Cohen real $V[c]$. This forcing is almost homogeneous, which implies that every ordinal definable object in the extension $V[c]$ is already in the ground model. In particular, in the extension $c$ itself is not ordinal definable, but it still determines an interval in the surreals, which no ordinal definable element can fill.

A generalization of the argument will handle the case that one might allow parameters in the definition. This is because by class forcing one can ensure that $V\neq HOD(A)$ for any fixed set $A$. In this case, there wouldn't even be a map from Ord to No with dense image, which was definable from parameters.

Meanwhile, I paste below the answer I had posted on math.SE to a similar question, concerning surjections from Ord to No.

From my answer to Willem Norduin's question on math.SE

The existence of a bijection between the class of ordinals $On$ and the class of surreal numbers $No$ is independent of the axioms of set theory. There are several interesting possibilities:

  • If ZFC is consistent, then there is a model of ZFC in which there is a definable such bijection. This is true in Goedel's constructible universe $L$, for example, for in $L$ there is a definable well-ordering of the universe, and we can use this well-ordering to well-order the surreals, which provides the desired bijection.

  • More generally, there is a first-order definable bijection between $On$ and $No$ if and only if the axiom known as $V=HOD$ holds. For the one direction, if $V=HOD$ holds, then there is a definable well-ordering of the universe and hence in particular a definable well-ordering of the surreals. Conversely, under ZFC if there is a definable bijection between $On$ and $No$, then there is a definable well-ordering of $No$. This allows us to construct a definable well-ordering of the class of sets of ordinals, since any set of ordinals determines a transfinite binary sequence of some ordinal length, and we can interpret this sequence as a $\pm 1$ sequence, which determines a unique surreal number by climbing through the tree of left-right cuts. Thus, we can well-order the class of sets of ordinals. But in ZFC every set is coded by a set of ordinals, and so we can construct a well-ordering of the entire universe, by looking for the least ordinal mapping to a surreal whose $\pm 1$ representation codes that set. So in this case, V=HOD holds.

  • Another way to summarize this argument is to say that if you can well-order $No$---and this is what your bijection to $On$ amounts to---then you can well-order every class.

  • If you drop the requirement that the bijection be definable, then we should move to the Goedel-Bernays context, in order to treat classes. The assertion that there is a bijection between $On$ and $No$ is equivalent over ZFC+GB to the axiom of Global Choice, which asserts that there is a well-ordering of the universe. This is by the same argument as above. (Note, we need AC for sets in order to make the last step of the argument; the class bijection in effect allows us to sew the set sized well-orderings together into a class well-ordering.) Thus, the theory ZFC+GB+(your bijection) is equivalent to GBC.

  • Because of this, if ZFC is consistent, then there are models of ZFC that have no bijection between $On$ and $No$, either definable or definable-from-parameters or otherwise. This is because it is known that ZFC does not imply global choice. One can construct such models by performing a class forcing iteration, adding a Cohen subset to every regular cardinal.

  • Meanwhile, every model of ZFC has a class forcing extension in which there is a class well-ordering of the universe, simply by forcing to add a global well-ordering, and this forcing extension adds no new sets, only classes. In this sense, it is compatible with every model of ZFC set theory to have the desired bijection as a class, without adding any new sets.

  • Further, every model of ZFC has a class forcing extension in which there is a definable bijection between $On$ and $No$, since we can force $V=HOD$. (This forcing, however, does add new sets.)

Lastly, upon reading your question again, I see that you asked for a surjection from $On$ onto $No$, rather than a bijection. But these are equivalent, since if there is a surjection, then we can remove the redundant ordinals from the domain by only using the least ordinal that maps to a given surreal, and this gives a bijection from a proper class of ordinals to $No$. But every proper class of ordinals is bijection with $On$ simply by collapsing to the order type of the predecessors.

  • $\begingroup$ I think your argument may settle a related question I posted recently, but I'm not sure. See mathoverflow.net/questions/188430/… $\endgroup$ – Jesse Elliott Dec 1 '14 at 0:24
  • $\begingroup$ @JesseElliott My argument here is really only about the equinumerosity relation on those classes, and not about the algebraic structure. So I don't think that this answer actually has much bearing on your other question. $\endgroup$ – Joel David Hamkins Dec 1 '14 at 1:17
  • $\begingroup$ One point is confusing me about `coding' sets as sets of ordinals. Is there a definable bijection $No \longrightarrow V$ in ZFC, or is that, too, equivalent to $V = HOD$? $\endgroup$ – Jesse Elliott Dec 8 '14 at 2:06
  • $\begingroup$ @JesseElliott Good Question! That statement isn't provable in ZFC alone, since this would imply that there is a definable linear order of $V$, but it is consistent with ZFC that there is no definable linear order of the universe. But I don't see without further thought whether it implies V=HOD, and I'm inclined to think it doesn't, but I'd need to think more about it. Why not post a question? $\endgroup$ – Joel David Hamkins Dec 8 '14 at 2:17
  • $\begingroup$ OK. mathoverflow.net/questions/189165/… $\endgroup$ – Jesse Elliott Dec 8 '14 at 3:34

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