Frank_R comments on The Learning-Theoretic Agenda: Status 2023

Frank_R 17 May 2023 12:03 UTC
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You are right and now it is clear, why your original statement is correct, too. Let $U$ be an arbitrary computable utility function. As above, let $u = U (o_{0} a_{0} o_{1} a_{1} \dots)$ and $(u - ϵ, u + ϵ) \subseteq (a, b)$ with $ϵ > 0$ and $ϵ \in Q$ . Choose $P$ as in your definition of “computable”. Since $P (s, ϵ)$ terminates, its output depends only on finitely many $o_{i_{1}}, \dots, o_{i_{k}}, a_{j_{1}}, \dots, a_{j_{l}}$ . Now
${s^{'} = o_{0}^{'} a_{0}^{'} o_{1}^{'} a_{1}^{'} \dots | o_{i_{1}}^{'} = o_{i_{1}}, \dots, a_{j_{l}}^{'} = a_{j_{l}}}$
is open and a subset of $U^{- 1} (u - ϵ, u + ϵ)$ , since $| P (s^{'}, ϵ) - u | < ϵ$ .