Is the class of ordinal numbers bounded by a fixed ordinal number really a set when defined through surreal numbers?

Question

At his book "On Numbers and Games", Conway defines ordinal numbers as games which doesn't have right options and whose left options contain only ordinal numbers. Then, fixed an ordinal number $\alpha$, he claims that $V_\alpha=\{\beta : \beta < \alpha\} $ is a set (i.e., not a proper class). His proof (which I think is wrong) is based at showing that $$ X = {\alpha^L} \bigcup (\bigcup_{\gamma \in \alpha^L} \{\beta : \beta < \gamma\} ) $$ is equal to $V_\alpha$. Indeed, as the induction hyphotesis easily implies that $X$ is a set, it suffices to show that $V_\alpha = X$. The fact that every member of $X$ is in $V_\alpha$ is obvious. To show the converse, Conway first observes that if $\beta$ is an ordinal number, then $\beta < \alpha \Rightarrow (\exists \alpha_L \in \alpha^L | \beta \le \alpha_L)$, which is easy to check using the definition of <. So, any member $\beta$ of $V_\alpha$ satisfies either $\beta < \alpha_L$ or $\beta = \alpha_L$. The first case clearly implies $\beta \in X$. At the second case, because $\alpha_L \in \alpha^L$, he claims that $\beta \in X$. But it is not enough, because here the symbol "=" is not an equality in the sense of set theory, but a certain equivalence relation between games. So, being equal to a member of a set doesn't necessarily mean belonging to it. I might be missing something, but I believe that Conway's proof is incorrect and needs to be patched.

Generally, the class of all possible representations of a given game may not be a set. For example, simplicity's rule implies that $1 = \{0|\alpha\}$ for any ordinal number $\alpha>1$. This fact makes me wonder if something similar could happen for ordinal numbers. My intuition says no because ordinal numbers are a very special kind of games, but still I wanted to be sure before trying to patch the proof. Is the class $V_\alpha$ defined above really a set even at the context of surreal numbers, in which every number has infinite games equals to it?

Thanks in advance.

Anderson Brasil

Answer 1

Indeed this does not work as such, for instance for $\alpha=\{\{1 \ | \ \varnothing\} \ |\ \varnothing\} \equiv 3$, one gets $\{0,1 \ | \ \varnothing\}\notin X$ (for $1$ there is only one possible bracket).

So one probably needs to prove that the class $\{\beta: \beta = \alpha\}$ is a set. I think at some point you have to use the powerset axiom, and maybe Conway wanted to avoid this.

In the class $\mathbf{No}$ of surreal numbers, the class $V_{\alpha}$ is only a set if we take Conway's of ordinals as games, which forbids representations such as $0=\{ \varnothing\ | \ \{0 \ | \ \varnothing\}\}$.

Answer 2

If I understand you correctly, your issue is that the class of representations of a given surreal number is proper, therefore if you put all representations of smaller ordinal numbers in the left set of your new ordinal number, you do not get a left set, but rather a left proper class.

Well, the solution is to not put all representations in the left set, but only one representative of each of them.

Note that this is essentially the same problem one faces in set theory when defining cardinality: The class of all sets of the same cardinality as a given one is proper, and therefore cannot serve as representation of that cardinality. So what you do (in ZFC) is to choose one representative of each class and identify the cardinality with that representative (this is possible in ZFC because of the well-ordering theorem, which implies that each set has the cardinality of an ordinal, so you can just use the initial ordinal for this).

So the trick is to replace the proper class by a single, definable representative of that class. And in the surreal numbers, this is indeed possible, too.

One way to do it is to take the sign representation of the surreal number (the sign representation is unique), and then construct a canonical left and right set by the simple rule that the surreal number represented by any initial string that precedes a plus sign goes into the left set, and the surreal number represented by any initial string that precedes a minus sign goes into the right set.

This representation has the advantage that for ordinals, you get exactly the representation Conway talked about, where the left set contains all smaller ordinals, and the right set is empty. This is because in the sign representation, each ordinal is represented by the constant function that maps that ordinal to $+$.

Finally, going after the question of the title: Also in the surreal number construction, the class of all ordinal numbers is not a set. Indeed, the construction given by Conway matches the standard construction of the von-Neumann ordinals in set theory. Basically, the left set is the von-Neumann ordinal with all elements replaced by their surreal counterpart, and the right set is empty. So if that construction would imply that all ordinals form a set, it would so also for the class of von-Neumann ordinals in ZFC. Which we know is not the case.

Answer 3

I've checked and if we exchange the $\{\beta : \beta < \alpha\}$ for any class $V_{\alpha}$ of ordinal numbers lesser than $\alpha$ containing at least one representative of all classes (i.e., $\forall \beta < \alpha, \exists \gamma \in V_{\alpha}|\gamma=\beta$), Conway's argument can be easily adapted to prove that $V_{\alpha}$ is a set. And all of his forthcoming proofs (at least until the end of the chapter) can be adapted to use those $V_\alpha$ instead of $\{\beta : \beta < \alpha\}$. Some of these proofs required some additional work written like this, but they are still essentially the same. Conway was a genius, we should have expected his proofs to be basically correct even if he leave us only with a draft, requiring several details to be filled in.

But, still, I wanted an argument showing that $\{\beta : \beta < \alpha\}$ was indeed a set. I don't feel very secure and confortable working with classes (only recently I felt the need of using anything beyond sets) but I think that I've got a solution.

Lemma: If $\alpha$ and $\beta$ are ordinal numbers such that $\beta < \alpha$ then $\exists \alpha_{L}\in\alpha^{L}|\beta\le\alpha_{L}$.
Proof: $\beta < \alpha \Rightarrow \{\beta^L|\}+\{|-\alpha^L\}<0 \Rightarrow \{\beta^L-\alpha|\beta-\alpha^L\}<0$. Then the definition of $<$ implies that $\exists \alpha_L\in\alpha^L|\beta-\alpha^L \le 0$. So, $\beta \le \alpha_L$, as we wanted.

Notation: Let $\alpha$ an ordinal number. We will denote the class of all ordinal numbers $\le \alpha$ by $S_\alpha $ . And the symbol $I_\alpha$ will denote the class of all ordinal numbers equals to $\alpha$.

Proposition: Let $\alpha$ an ordinal number. Then the class $S_\alpha$ is a set.
Proof: Because of the induction hypothesis, $S_\gamma$ is a set for every $\gamma \in \alpha^L$. Therefore, $L = \bigcup_{\gamma \in \alpha^L}S_\gamma$ is a set.
Now we claim that if $l$ is a set of ordinal numbers such that $\alpha=\{l|\}$, then $l \in \wp(L)$ (i.e, $l$ is a subset of $L$). In fact, if $\beta \in l$ then $\beta < \alpha$. The lemma implies that $\exists \alpha_{L}\in\alpha^{L}|\beta\le\alpha_{L}$. So $\beta \in S_{\alpha_{L}}$ with $\alpha_{L}\in\alpha^{L}$. Then $l\in L$ because of the definition of $L$.
If we understand a game as an ordered pair between two sets (the left and the right options), then the claim above means that every element of $I_{\alpha}$ belongs to $\wp (L) \times \{\emptyset\}$, which is a set. So, $I_{\alpha}$ is a set and we can define the set $X=I_{\alpha} \cup L$. To finish our proof, it is enough to show that $S_\alpha = X$.
To see that $X \subset S_\alpha$, first suppose $\beta\in I_\alpha$. Then $\beta=\alpha$ and therefore $\beta \in S_{\alpha}$, as required. At the second case, we suppose $\beta\in L$. Then $\exists \gamma \in \alpha^L|\beta \in S_\gamma$. But $\beta \in S_\gamma \Rightarrow \beta < \gamma$. And $\gamma \in \alpha^L \Rightarrow \gamma < \alpha$. So, $\beta \le \alpha$. Therefore, $\beta \in S_\alpha$.
To see that $S_\alpha \subset X$, let $\beta \in S_\alpha$, so $\beta \le \alpha$. If $\beta = \alpha$ then $\beta \in I_\alpha$ and therefore $\beta \in X$. And if $\beta < \alpha$, the lemma implies that $\exists \alpha_{L}\in\alpha^{L}|\beta\le\alpha_{L}$. Then $\beta \in S_{\alpha_{L}}$. As $\alpha_{L}\in\alpha^{L}$, then $\beta \in L$, so $\beta \in X$.
QED

It obviously imply at:

Corolary: Given an ordinal number $\alpha$, the class of all ordinal numbers $\beta$ such that $\beta < \alpha$ is a set.

As we wanted.