1
$\begingroup$

Suppose that $G( \omega, A, X)$ denotes a sequential game of perfect information in which player I and player II play an element in $A$ in each turn with a total number of $\omega$ moves. $X$, the winning set, is a subset of $A ^{\omega}$. If the outcome is in $X$, then player I wins. Otherwise, player II wins.

$G(\omega, A, X_1)$ and $G(\omega, A, X_2)$ are two games in which player I has a winning strategy. Clearly in the game $G(\omega, A, X_1\cup X_2)$, player I also has a winning strategy. Let $W_{X_1}$, $W_{X_2}$ and $W_{X_1 \cup X_2}$ be the sets of player I's winning strategies respectively. I'm interested to know their relationship given their non-emptiness in general.

In particular, is it true that $W_{X_1} \cap W_{X_2} =\varnothing$, then $W_{X_1 \cup X_2} = W_{X_1} \cup W_{X_2}$?

$\endgroup$
1
$\begingroup$

This family of games is known as Gale-Stewart games, after the name of the two authors who were the first to prove that

  • for $X$ closed or open in $A^\omega$, for the product topology with $A$ discrete, the game $G(A,X)$ is determined.
  • with the axiom of choice, there is a subset $X$ of $\omega^\omega$ for which the game $G(\omega,X)$ is not determined.

Where a game is said to be determined if one of the two player has a winning strategy.

The best possible result in ZFC about determinacy of these games is due to D. A. Martin, it says that all Borel subsets are determined. More determinacy can be obtained by assuming Large Cardinal Hypothesis, cf. D. A. Martin and J. Steel.

I am not aware of any result saying that if $X_1$ and $X_2$ are determined, then so is their union or intersection. I once asked an expert in the field, who says this might well be wrong in ZFC.

For your particular question, the answer is no. Take $A=2$, $$ X_1=\{01x\mid x\in 2^\omega\}\cup\{10x\mid x\in 2^\omega\} $$ and $$ X_2=2^\omega \setminus X_1=\{11x\mid x\in 2^\omega\}\cup \{00x\mid x\in 2^\omega\} $$ then every strategy for player I is winning in $X_1\cup X_2=2^\omega$, while there are strategies for player I which are not winning neither in $G(2,X_1)$ nor in $G(2,X_2)$, take for example the strategy which consists in playing $1$ all the time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.