Assume $\{f_i:\mathbb N^k\to\mathbb N\}^\infty$ is a countable set of computable functions. Prove that a computable universal function exists for this class.
This more general question stems from the special case of the polynomials. I have reduced the less general one to this one, but I don't know how to proceed.
EDIT: By a universal function for a class $\{f_i:\mathbb N^k\to\mathbb N\}_{i\in I}$ I mean a function $\Phi:\mathbb N^{k+1}\to\mathbb N$ such that $\forall i\in I\quad\exists j\in\mathbb N\quad \Phi(j,x)=f_i(x)$ and the converse $\forall j\in \mathbb N\quad\exists i\in I\quad \Phi(j,x)=f_i(x)$. Therefore, a universal function exists for a class if and only if the class is of countable or finite cardinality.
I believe that if the all the functions of a countable set are computable, a computable universal function exists. Isn't that so?
Let $A \subseteq \omega$ be a set that is not recursively enumerable. For each $i \in A$ let $f_i$ be the constant function $\lambda x.i$; each $f_i$ is computable. Then $\{f_i : i \in A\}$ has no computable universal function. Because, if $\Phi$ was a computable universal function, then $A = \{ \Phi_j(0) : j \in \omega\}$ would be recursively enumerable.
There is terminology for a collection $C = \{g_j : j \in \omega\}$ of partial computable functions that does have a universal function: we would say $C$ is uniformly computably enumerable. More generally, if we have a sequence of functions $\langle g_j : j \in\omega\rangle $ we would say $g_j$ is uniformly computable if there is a computable function $\Phi$ such that $g_j(i) = \lambda i. \Phi(j,i)$ for all $i,j \in \omega$.