3 added 42 characters in body; added 72 characters in body

Theorem. The following are equivalent.

• There is a definable class embedding bijection from Ord to No.

• There is a dense subclass of definable surjection from Ord to Noif and only if .

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

• No. So this is a case where we have such a definable dense class asyou requested4 implies 1, which implies 2, which implies 3.

Conversely

Convesely, suppose that V=HOD fails. Since in ZFC every set isSo $s$ would be ordinal definable, a contradiction. So 3 implies 4. QED

2 added 443 characters in body

One

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 generalize fill.

A generalization of the argument to will handle the case that one might allow parameters , since within the definition. This is because by class forcing one can also ensure that $V\neq HOD(A)$ for any fixed

1

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. There is a definable class embedding from Ord to a dense subclass of No if and only if 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 this is a case where we have such a definable dense class as you requested.

Conversely, 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. QED

One can generalize the argument to allow parameters, since with class forcing one can also 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.