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 $R$.
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?
