In a previous post I asked about the definability of a function that can be "approximated" by a uniformly definable family of functions. Nevertheless, the notion of approximation I gave was too weak and a counterexample was given. Here is another try. Let $T$ be a first order theory, $M\prec N$ models of $T$ equipped with a topology with a uniform definable basis (i.e. every basic open is definable with parameters with the same formula). Suppose furthermore that $N$ is saturated enough. Let $F:M\rightarrow M$ be a partial function such that $dom(F)$ is $M$-definable and open (you can also suppose $rng(F)$ is $M$-definable). Let $(f_a)_{a\in N^k}$ be a uniformly definable family of functions (the only parameters in the formula $f_a$ are $a\in N^k$) such that for every positive integer $n$ and $b_1,...,b_n\in dom(F)$ there is $a\in M^k$ and a basic open $U \subsetneq dom(F)$ containing $b_1,...,b_n$ such that $F\upharpoonright U=f_a\upharpoonright U$. In addition, suppose there is $d\in N^k$ such that $F=f_d\upharpoonright M$. Can we conclude that $F$ is $M$-definable? The question is still interesting dropping the assumption $F=f_d\upharpoonright M$, but in case this question has a negative answer, I would still like to have an answer for the former question.

In a previous post I asked about the definability of a function that can be "approximated" by a uniformly definable family of functions. Nevertheless, the notion of approximation I gave was too weak and a counterexample was given. Here is another try. Let $T$ be a first order theory, $M\prec N$ models of $T$ equipped with a topology with a uniform definable basis (i.e. every basic open is definable with parameters with the same formula). Suppose furthermore that $N$ is saturated enough. Let $F:M\rightarrow M$ be a partial function such that $dom(F)$ is $M$-definable and open (you can also suppose $rng(F)$ is $M$-definable). Let $(f_a)_{a\in N^k}$ be a uniformly definable family of functions (the only parameters in the formula $f_a$ are $a\in N^k$) such that for every positive integer $n$ and $b_1,...,b_n\in dom(F)$ there is $a\in M^k$ and a basic open $U \subsetneq dom(F)$ containing $b_1,...,b_n$ such that $F\upharpoonright U=f_a\upharpoonright U$. In addition, suppose there is $d\in N^k$ such that $F=f_d\upharpoonright M$. Can we conclude that $F$ is $M$-definable? The question is still interesting dropping the assumption $F=f_d\upharpoonright M$, but in case this question has a negative answer, I would still like to have an answer for the former question.