# Can a game be an option of itself?

My question is, can a game contain itself as an option? and can it be a surreal number? For example $A=\{A|\}$ or $B=\{C|B\}$ where $C$ is a surreal number. from the point of view of games, it is equivalent to having the player play by doing nothing an effectively passing his turn to the second player. But I am not sure if this is okay from the point of view as surreal numbers. But as a number it is a problem, since for example to proof that $A=\{A|\}$ is a number, we  will need to prove that all of its options are number, which means that we need to prove that $A$ is a number! So is it even possible to have a game with itself as an option? or will I have a "set which contains itself" paradox or something like that. Thanks!

• Short answer: no. – Wojowu Aug 9 '17 at 16:55
• Slightly longer answer: games that allow this are called "loopy" and are a different theory. Unless expressedly qualified as "loopy", games in the Conway sense are well-founded ("enders" or whatever Conway calls this) so they cannot be options of themselves (they satisfy the equivalent of the Foundation axiom in set theory). – Gro-Tsen Aug 9 '17 at 17:36
• Game $\{\;|\;\}$ is not the empty set. So $\{\;|\;\} \ne\{ \{\;|\;\}\;|\;\}$ – Gerald Edgar Aug 9 '17 at 18:03
• See Hypergame ... – Nik Weaver Aug 9 '17 at 19:33

A surreal number cannot have itself as an option, because numbers are defined inductively: if $L$ and $R$ are sets of numbers, then $\{L|R\}$ is a number. Notice that $L$ and $R$ have to already contain things that are numbers before we can say that $\{L|R\}$ is a number.