In this script of Lou van den Dries I encountered (total) functions $\mathbb N^n\to\mathbb N$ and relations $R\subseteq\mathbb N^n$ that are marked as computable.
They are introduced in section 5.1. on page 82.
Addition, multiplication, coördinate functions and relation $\leq$ are computable by definition and further the collection is closed under composition and a minimalisation restricted in such a way that it does not produce undefined values.
My question is the following:
If $R_n\subseteq\mathbb N$ is computable for every $n\in\mathbb N$ and $R\subseteq\mathbb N$ is defined as: $R(a)\iff\exists n<a[R_n(a)]$ then is $R$ computable?
If I look at it only on intuition then I would expect the answer to my question to by "yes". For a fixed $a$ we just have to check for a finite ($n<a$) number of algorithms whether they confirm or deny $R_n(a)$.
But intuition has deceived me many times and am eager to find a real proof (or counterexample).
Thank you very much for taking notice of this question.
You need some "uniformness" of computability of $R_n$. Otherwise, take any set $X$ and define $$R_n = \begin{cases}{\varnothing, \ n + 1 \notin X\\ \{n + 1\},\ n + 1 \in X} \end{cases}$$ Then every $R_n$ is computable, and (assuming $0 \notin X$), we get $R = X$. Taking $X$ to be non-computable, we get non-computable set that can be represented as you wrote.
If you require uniformness - ie function $f(n, m) = \chi_{R_n}(m)$ is computable, then your argument works.
Note that with your definition we necessary have $0 \notin R$, to fix that we can, for example, change right part to $\exists n \leq a (a \in R_n)$.