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

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

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

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

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.

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.