Is there a chess position with a finite number of pieces on the infinite chess board $\mathbb{Z}^2$ such that White to move has a forced win, but Black can stave off mate for at least $n$ moves for every $n$?
This question is motivated by a question posed here a few months ago by Richard Stanley. He asked whether chess with finitely many pieces on $\mathbb{Z}^2$ is decidable.
A compactness observation is that if Black has only short-range pieces (no bishops, rooks or queens), then the statement "White can force mate" is equivalent to "There is some $n$ such that White can force mate in at most $n$ moves".
This probably won't lead to an answer to Stanley's question, because even if there are only short-range pieces, there is no general reason the game should be decidable. It is well-known that a finite automaton with a finite number of "counters" can emulate a Turing machine, and there seems to be no obvious reason why such an automaton could not be emulated by a chess problem, even if we allow only knights and the two kings.
But it might still be of interest to have an explicit counterexample to the idea that being able to force a win means being able to do so in some specified number of moves. Such an example must involve a long-range piece for the losing side, and one idea is that Black has to move a rook (or bishop) out of the way to make room for his king, after which White forces Black's king towards the rook with a series of checks, finally mating thanks to the rook blocking a square for the king.
If there are such examples, we can go on and define "mate in $\alpha$" for an arbitrary ordinal $\alpha$. To say that White has a forced mate in $\alpha$ means that White has a move such that after any response by Black, White has a forced mate in $\beta$ for some $\beta<\alpha$.
For instance, mate in $\omega$ means that after Black's first move, White is able to force mate in $n$ for some finite $n$, while mate in $2\omega + 3$ means that after Black's fourth move, White will be able to specify how many more moves it will take until he can specify how long it will take to mate.
With this definition, we can ask exactly how long-winded the solution to a chess problem can be:
What is the smallest ordinal $\gamma$ such that having a forced mate implies having a forced mate in $\alpha$ for some $\alpha<\gamma$?
Obviously $\gamma$ is infinite, and since there are only countably many positions, $\gamma$ must be countable. Can anyone give better bounds?
