Suppose I play the following game against the Opponent. My moves are rational numbers $p_i$ and the Opponent's moves are real numbers $\epsilon_i>0$.

On turn $n+1$ the past move sequence is $p_1,\epsilon_1,\ldots, p_{n}, \epsilon_{n}$. I select a point $p_{n+1}\in \mathbb Q \cap (p_{n}-\epsilon_{n},p_{n}+\epsilon_{n})$ and the Opponent selects some $\epsilon_{n+1} >0$.

I win provided the sequence $p_i$ tends to an irrational number.

Does anyone know if I have a winning strategy? Certainly I don't if the rational and irrational numbers are swapped, as the Opponent can just enumerate the rationals and select the intervals small enough to exclude each rational in turn.

I imagine I can always win. But that's only because this looks like the Banach-Mazur game, which I can win if the target set is the irrationals.

The motivation behind this is I want to recursively build a set of homeomorphisms $F_1,F_2,\ldots: \mathbb R \to \mathbb R$ such that $F(x) =\displaystyle \lim_{n \to \infty} F_n \circ F_{n-1} \circ \ldots \circ F_1 (x)$ is a well-defined homeomorphism and such that some fixed $p \in \mathbb Q$ is sent to an irrational number.

I have conditions under which the limit exists. Namely I have to ensure the next $\max d(x,F_n(x)) < \epsilon_n$. Unfortunately $\epsilon_n$ are not known in advance. Each $\epsilon_n$ is determined by $F_1,\ldots, F_{n-1}$. This leads to the above game where $p_n = F_n \circ F_{n-1} \circ \ldots F_1(p)$.