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Thinking about surreal numbers, I’ve now got doubts that they are actually well-defined in ZFC. Here’s my reasoning:

The first thing to notice is that the surreal numbers (assuming they *are* well defined, of course) form a proper class. Now, quoting from Wikipedia:

The surreal numbers are constructed in stages, along with an ordering

≤ such that for any two surreal numbers $a$ and $b$ either $a \le b$ or $b \le a$.

(Both may hold, in which case $a$ and $b$ are equivalent and denote the

same number.) Numbers are formed by pairing subsets of numbers already

constructed: given subsets $L$ and $R$ of numbers such that all the

members of $L$ are strictly less than all the members of $R$, then the

pair { $L$ \mid $R$ } represents a number intermediate in value between all

the members of $L$ and all the members of $R$.

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Different subsets may end up defining the same number: $\{ L \mid R \}$ and

$\{ L′ \mid R′ \}$ may define the same number even if $L \ne L′$ and $R \ne R′$. (A

similar phenomenon occurs when rational numbers are defined as

quotients of integers: $1/2$ and $2/4$ are different representations of

the same rational number.) So strictly speaking, the surreal numbers

are equivalence classes of representations of form $\{ L \mid R \}$ that

designate the same number.

OK, so surreal numbers are equivalence classes of pairs of sets of surreal numbers. So far, so good. However, how large are those equivalence?

Quoting Wikipedia again:

Two numeric forms $x$ and $y$ are forms of the same number (lie in the same equivalence class) if and only if both $x \le y$ and $y \le x$.

And

Given numeric forms $x = \{ X_L \mid X_R \}$ and $y = \{ Y_L \mid Y_R \}$, $x \le y$ if and

only if:

- there is no $x_L \in X_L$ such that $y \le x_L$ (every element in the left part of $x$ is smaller than $y$), and
- there is no $y_R \in Y_R$ such that $y_R \le x$ (every element in the right part of $y$ is bigger than $x$).

OK, let’s consider the special case of $0$. The “canonical” form of $0$ is $\{\mid\}$, that is $L=R=\emptyset$. Now consider an arbitrary surreal number $x\ne 0$. Since the surreal numbers are totally ordered, either $x>0$ or $x<0$ is true. In the first case, $\{\mid x\}$ is equivalent to $\{\mid\}$, and in the second case $\{x\mid\}$ is eqivalent to $\{\mid\}$. That is, the equivalence class $0$ contains at least as many elements as there are surreal numbers.

However since the surreal numbers form a proper class, this means that each equivalence class forms a proper class. But proper classes cannot be members of sets, including the left and right set of forms that make up surreal numbers.

Also, take addition of surreal numbers. It is a function which maps a pair of surreal numbers to a surreal number. But since a surreal number is a proper class, and a pair is ultimately a set, surreal numbers cannot be members of a pair.

On the other hand, I cannot imagine that all those who are studying surreal numbers would not have considered whether they are actually well-defined. Therefore I guess there’s some error in my reasoning. If so, where is it?

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Surreal numbers do form a class, but the surreal numbers with “birthday” equal to x are all sets. The birthday concept is basically that 0 = {|} is “born” on day 0, then 1 = {0|} and -1 = {|0} are born on day 1, and so on to w = {0, 1, 2,…|} and 1/w = {0|1, 1/2, 1/4, …} which are born on day w, and on to infinity. In general the birthday of x = {X_l|X_r} is the (ordinal) supremum of the birthdays of all the numbers in X_l and X_r.

Restricting yourself to all the surreal numbers with a birthday x or less, where x is some ordinal, you produce a set. On this set you can produce equivalence classes, addition, and multiplication. It can then be proved that these definitions on the set of surreals with birthday y or less, where y is greater than x, are extensions of the definition on surreals with birthday x or less. Therefore, the definitions can be extended to as high an ordinal as you want, so intuitively they can be applied to the class produced from the union of each birthday set for every ordinal.

To expand a bit on my comment:

In set theory one often runs into troubles when defining an equivalence relation whose universe is a class. It is quite possible to have an equivalence class (or all) that have a proper-class many members in it (e.g., all singletons).

The common trick around it is known as **Scott’s trick**, invented by Dana Scott to allow an internal definition of cardinality in models of ZF:

Suppose that $\varphi(x,y)$ describes the equivalence relation (that is, $x$ is equivalent to $y$ if and only if $\varphi(x,y)$ is true), we then define the equivalence class of $x$ as:

$$[x]_\varphi = \{y\mid \varphi(x,y)\land\mathrm{rank}(y)\text{ is minimal}\}$$

We make a heavy use of the axiom of regularity, from which follows that every set has a von Neumann rank. We take the collection of those with minimal rank.

Note that $[x]_\varphi$ is not empty. $\varphi(x,x)$ is always true so there is some $y\in[x]_\varphi$ whose rank is at most the rank of $x$, and therefore by the replacement/subset schema $[x]_\varphi$ is indeed a set.

A minor remark about the existence of such $\varphi$ is that in ZF [proper] classes are described by a formula. If $R$ is a proper class which is also an equivalence relation then there is $\varphi$ such that $\varphi(x,y)$ is true if and only if $\langle x,y \rangle\in R$.

See the Appendix to Part 0 in Conway’s *On Numbers and Games*.

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