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Let $T$ be a first order theory, $M$ a model of $T$ equipped with a topology with a definable basis (i.e. every basic open is definable with parameters). Let $F: M\rightarrow M$ be a partial function and $(f_a)_{a\in M^k}$ be a uniformly definable family of functions such that for any open set $U\subsetneq dom(F)$ there is $a\in M^k$ such that $F\upharpoonright U=f_a\upharpoonright U$. Suppose in addition both $dom(F)$ and $rg(F)$ are $M$-definable. Can we say something about the definability of $F$ in $M$? What about if $F=f_d\upharpoonright M$ for some $d$ in an elementary extension of $M$, i.e., $F$ is an externally definable function? If the exchange property is assumed I know in this later case that $F$ is indeed $M$-definable. Is it true without the exchange property? (NIP can be assumed if needed)

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Where you wrote "any open set $U\subsetneq dom(F)$", did you literally mean that, and not, for example, that $U$ should belong to the basis assumed in the first sentence? Under the literal reading, if the topology is $T_1$, then $F$ would have to be definable on a set $U$ consisting of all but one point of its domain. – Andreas Blass Apr 18 '13 at 13:01
Thanks, you are totally right, the set $U$ should be a basic open from the definable basis. We can also assume that $\dom(F)$ is open, maybe the notion of ``approximation'' makes more sense. Can you explain to me the argument with the $T_1$ topology assumption? – Cubikova Apr 18 '13 at 14:13
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With your clarification (in a comment) that $U$ should be from the definable basis, the answer to your question seems to be negative. Notice first that the discrete topology on any model has a definable (with parameters, as you wrote in the question) basis, consisting of the singletons. Now you can uniformly define a family of functions $(f_a)_{a\in M}$ to be the family of constant functions, i.e., $f_a$ is constant with value $a$. Then any $F:M\to M$ whatsoever will agree locally with this family, i.e., $F$ is constant on each singleton. So you can't conclude anything about definability of $F$ (unless your model $M$ is such that all functions on it are definable, which can only happen when either $M$ is finite or its language is bigger than the model itself).

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Thank you very much for you answer. I will try to make another version of the question that excludes this case, making the notion of approximation a little bit finer. In your example, every function is locally constant so constant functions indeed approximate any function. This seems to show that the notion of approximation should be defined differently. What I need to add is probably something like "for any finite number of points in $dom(F)$ there is basic open $U$ containing them and a parameter $a\in M$ such that $f_a=F$ on U$", but I will think about it. – Cubikova Apr 22 '13 at 7:48
Btw, I still do not know what is the argument you were having in mind with the $T_1$ topology, am I missing something trivial? – Cubikova Apr 22 '13 at 7:49
For a topology to be $T_1$ means that singletons are closed and therefore the complements of singletons are open. Use such complements as the $U$ in your question (or $U\cap dom(F)$, when $dom(F)$ is an open set but not the whole space). – Andreas Blass Apr 22 '13 at 11:30

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