TITLE: Is the class of ordinal numbers bounded by a fixed ordinal number really a set when defined through surreal numbers?
QUESTION [5 upvotes]: 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

REPLY [2 votes]: 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.