# Does every set in ZFC have a value when interpreted as a misère game?

A misère game is one where a player with no moves wins. The object of the game is to be presented with a board position in which you cannot do anything.

If we imagine a two-player game where the players take alternate turns, then we can think of a game as being isomorphic to a set of games that we can choose to present to our opponent. This means we can represent any game fitting our criteria as a set.

Suppose the value to the player is $$+1$$ if they win and $$-1$$ if the opponent wins.

The game $$\{\}$$ has value $$+1$$ because the player cannot make any moves and therefore wins.

The game $$\{\{\}\}$$ has value $$-1$$ because the player must pick the game $$\{\}$$ and give it to the opponent, who then wins.

The game $$\{\{\}, \{\{\}\}\}$$ has value $$+1$$ because the player picks the game $$\{\{\}\}$$ (which has value -1), to give to the opponent.

Does every set in ZFC have a value as a misère game? I think the axiom of regularity, guarantees that there must be some kind of "trace" through any set which has a value in some sense, but I don't know whether it's possible to extend this to a value for the entire set.

This is mainly about set theory; the misère game business is a red herring. Note that the outcome of a misère game can be calculated recursively: It's $$\mathcal N$$, which you called "value $$+1$$", if there are no moves or there's a move to a $$\mathcal P$$-position. It's $$\mathcal P$$, which you called "value $$-1$$", otherwise. Therefore, all that matters is whether play in the set ends in finite time, so that the induction can get to the base case.
• Is the [combinatorial-game-theory] tag appropriate? I assumed "no" based on your answer and removed it. Jan 23, 2021 at 2:54