# Winning strategies for games on the natural numbers

Define $$F=\{(l_n, k_n)_{n=1}^t: t,l_n, k_n \in \mathbb{N}, l_1<\ldots Suppose that I have a collection $$G\subset F$$, which is a set of ''good'' sequences.

Define $$A=\{(l_n)_{n=1}^\infty: (\exists (k_n)_{n=1}^\infty)(\forall t\in \mathbb{N})((l_n, k_n)_{n=1}^t\in G)\}.$$

I want to define two games between two players and ask whether the existence of a winning strategy for Player I in the first game is equivalent to a winning strategy for Player I in the second game. Roughly speaking, one game requires Player I to choose their integers with limited information during the game, while the other allows Player I to choose their integers with full information. My naive hope that existence of winning strategies for the two distinct games would be equivalent is based on the fact that the "target set" for the game is defined using $$G$$ which, in some sense, only depends on the past choices.

Game 1 Player I chooses an infinite subset $$L_1$$ of the natural numbers. Player II chooses $$l_1\in L_1$$ and an infinite subset $$M_1$$ of $$L_1$$. Player I chooses $$k_1\in \mathbb{N}$$ and an infinite subset $$L_2$$ of $$M_1$$. Player II chooses $$l_2\in L_2$$ with $$l_1 and an infinite subset $$M_2$$ of $$L_2$$. Play continues in this way. Player I wins if $$(l_n, k_n)_{n=1}^t\in G$$ for all $$t\in \mathbb{N}$$. We note that Player I is allowed to choose any $$k_n\in \mathbb{N}$$, and is not required to choose $$k_n$$ from any particular subset of $$\mathbb{N}$$.

Game 2 Player I chooses an infinite subset $$L_1$$ of the natural numbers, Player II chooses $$l_1\in L_1$$ and an infinite subset $$M_1$$ of $$L_1$$. Player I chooses an infinite subset $$L_2$$ of $$M_1$$. Player II chooses $$l_2\in L_2$$ with $$l_1 and an infinite subset $$M_2$$ of $$L_2$$. Play continues in this way. Player I wins if $$(l_n)_{n=1}^\infty\in A$$, where $$A$$ is defined above. This means that Player I wins if there exists $$(k_n)_{n=1}^\infty$$ such that $$(l_n, k_n)_{n=1}^t\in G$$ for all $$t\in \mathbb{N}$$. Here, Player I has the luxury of choosing each $$k_n$$ after all $$l_n$$ choices have been made.

Is it true that Player I has a winning strategy in the first game if and only if Player II has a winning strategy in the second game?

In the particular case that I care about, $$G$$ has some additional nice properties, so please feel free to use them if it helps:

$$1$$: If $$(l_n, k_n)_{n=1}^t\in G$$ and if $$1\leqslant r_1<\ldots , then $$(l_{r_n}, k_{r_n})_{n=1}^s\in G$$.

$$2$$: If $$(l_n, k_n)_{n=1}^t\in G$$ and if $$(m_n)_{n=1}^t$$ is any increasing sequence of integers such that $$k_n\leqslant m_n$$ for each $$1\leqslant n\leqslant t$$, then $$(l_n, m_n)_{n=1}^t\in G$$.

• mathoverflow.net/questions/338777/… – Asaf Karagila Aug 21 at 8:53
• Do not crosspost like this. – Asaf Karagila Aug 21 at 8:53
• The absence of any justification for why I should not crosspost like this, the rudeness of your order, and the downvote of my question (making it less likely to get any useful response), all make me inclined to simply ignore you. – user697250 Aug 21 at 12:13
• Your lack of effort in reading about cross-posting is the reason I do not bother adding details. The overlap between the two communities is large, and asking on both without even bothering to tell people you've asked on both is a duplication of effort, which is quite frankly a way to spit in the face of people who volunteer their free time to help you. Also, I didn't want to abuse my moderator power, but if you think that it's easier to ignore me, I can just manually migrate this to MO where it will be closed promptly as a duplicate. Will you consider that a better solution, perhaps? – Asaf Karagila Aug 21 at 12:54