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



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> 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 $\epsilon\_{n+1} >0$. 
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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 $p\_n = F\_n \circ F\_{n-1} \circ \ldots F\_1(p)$.



Can I win this variant of the Banach-Mazur Game?
Sure you can win. Let's enumerate rationals as $q\_n, n = 1, 2,\ldots$. Also we can WLOG assume that $\varepsilon\_{i+1} \le \frac{\varepsilon\_i}{20000}$. We will make it so that $d(q\_n, p\_{n+1}) \ge 10\varepsilon\_{n+1}$. Then for $m > n$ we have $d(q\_n, p\_m) \ge 10\varepsilon\_{n+1} - \varepsilon\_{n + 2} - \ldots \ge \varepsilon\_{n+1}$.


For $p\_{n+1}$ the possible values are an interval of length $\varepsilon\_n \ge 20000\varepsilon\_{n+1}$. Therefore there exists at least one admissible value of $p\_{n+1}$ such that $d(q\_n, p\_{n+1})\ge 10\varepsilon\_{n+1}$ whatever the value of $q\_n$ is.


Thus, $p\_n$ can not converge to any rational $q$ since if $q = q\_n$ for some $n$ then $d(q, p\_m) \ge \varepsilon\_{n+1} > 0$ for $m > n$. On the other hand since $\varepsilon\_{i+1} \le \frac{\varepsilon\_i}{20000}$ the sequence $p\_n$ is Cauchy and thus converge for some $q\in \mathbb{R}$. Therefore $q\in \mathbb{R}\backslash \mathbb{Q}$.


