# Question on Gale-Stewart Theorem and Axiom of Choice

I am reading Kechris' descriptive set theory text book, and there is this Theorem regarding infinite games:

Gale-Stewart: Let $$T$$ be a non-empty pruned tree on $$A$$. Let $$X\subset[T]$$ be closed or open. Then $$G(T,X)$$ is determined.

There is this exercise: show that AC is equivalent to Gale-Stewart Theorem.

Assuming the exercise is true, if we consider the statement:

$$\blacksquare$$ Every infinite game is determined.

Then the statement $$\blacksquare$$ implies Gale-Stewart Theorem, which implies AC, which implies the negation of $$\blacksquare$$.

Does this mean ZF$$+\blacksquare$$ is inconsistent? Can we conclude that we can find a non-determined infinite game without using AC? If so, do you have any example in your mind?

Thanks!

• What is $A$ in this context? Commented Oct 17, 2022 at 21:06
• $A$ is a non-empty set. I meant to have $T\subset A^{<\mathbb{N}}$, a non-empty pruned tree. Commented Oct 17, 2022 at 21:37

Your reasoning is correct: $$\mathsf{ZF}$$ proves $$\lnot \blacksquare$$.
Note that $$\blacksquare$$ is not $$\mathsf{AD}$$, the Axiom of Determinacy. $$\mathsf{AD}$$ states that every Gale-Stewart game on $$\omega^\omega$$ is determined, which is much weaker than $$\blacksquare$$.
It is possible to prove in $$\mathsf{ZF}$$ alone that there is a game on $$\omega_1^\omega$$ which is undetermined ($$\omega_1$$ is the first uncountable ordinal). A reference is Theorem 10.2 in Games with perfect information by Mycielski, Chapter 3 in Handbook of Game Theory with Economic Applications, 1992, vol. 1, pp 41-70 (link).
For a more explicit example of an undetermined game in $$\mathsf{ZF}$$, see the answer here by Joel David Hamkins.