Is there a surjective map between the (class of) ordinal numbers On and the set No (Conway's surreal numbers) and is it constructable, In Conway's system we have for example:

$\omega_0 = < 0,1,2,3,... | > $


$\epsilon = < 0 | 1, 1/2, 1/4, 1/8, ... > $

(where $\epsilon$ is not the first uncountable ordinal, but the "reciprocal" of $\omega_0$). My question is thus: can you device this for every ordinal, or does Conway's system On eventually run "out of space").

  • $\begingroup$ All ordinals are surreal numbers. If they were not then (since the ordinals are well-ordered) take the smallest ordinal $\beta$ which is not a surreal number and consider the surreal number $\{\alpha \in On:\alpha < \beta \mid \}$, which by definition is $\beta$ $\endgroup$
    – Henry
    Feb 26 at 22:00

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.


I''m not entirely sure what your question is. In Conway's notation On denotes the ordinal numbers (and No denotes the set of all surreal Numbers). Basically the elements of On are just von Neumann ordinals. In von Neumann's construction an ordinal $\alpha$ is just the set $\{\beta:\beta<\alpha\}$ of all preceding ordinals and this corresponds to the Conway ordinal $\langle\beta:\beta<\alpha\mid\ \rangle$. So the von Neumann and Conway ordinals naturally correspond.

But maybe your question is whether there is a one-to-one correspondence between the ordinals and the whole class of surreal numbers. These are both proper classes with respect to ZFC. There are subtleties associated to defining bijections between proper classes, which I don't really understand, but given a strong version of AC I believe that any two proper classes can be paired off. (Maybe an experienced set theorist could say more?)

  • $\begingroup$ Just edited the question (On = Ordinal and No = Surreal numbers). I am inded looking for a 1-1 correspondence between On and No, or (if it does not exist), a counter example (i.e. which ordinals cannot be represented by surreals). $\endgroup$
    – Willem Noorduin
    Nov 5 '10 at 8:16
  • 1
    $\begingroup$ I agree with Robin -- whether any two proper classes can be put in bijection (or, perhaps, "bijection") depends on your particular brand of set theory or (or, worse, class theory). So from the perspective of mainstream mathematics it is a sort of unattractive question. Note that here you have more structure on both sides -- namely orderings -- and you could ask for a "bijection" which preserves the orderings. But of course it is easily seen that no such thing exists (e.g. exactly one of the classes has a least element). $\endgroup$ Nov 5 '10 at 8:21

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