I understand that this question shouldn't be hard but still.
Let $U: \mathbb N \times \mathbb N \rightarrow \mathbb N$ be a function such that for any fixed $c$ function $D_c(y) = U(c, y)$ is computable. Does it imply that $U$ is computable?
I believe that answer is no, but I can't find an example to demonstrate it.
Thanks!
No. Let $C$ be the set of computable functions from $\Bbb{N}$. Then $C$ is (countably) infinite, and there are uncountably many functions $D : \Bbb{N} \to C$. Each such function $D$ corresponds to a unique function $U : \Bbb{N} \times \Bbb{N} \to \Bbb{N}$ such that $U(c, y) = D_c(y)$ for all $c$ and $y$ with $D_c$ computable. But there are only countably many computable functions from $\Bbb{N}\times \Bbb{N}$ to $\Bbb{N}$, so uncountably many of those functions $U$ are not computable.
For a specific counter-example: let $H$ be the set of codes of partial recursive functions (or Turing machines) $x$, such that $\{x\}(x)$ is defined (or terminates). Define $$ U(c, y) = \left\{ \begin{array}{cl} 0 &\mbox{if $c \in H$} \\ 1 & \mbox{if $c \not\in H$} \end{array}\right. $$
Then $D_c(y) = U(c, y)$ is a computable function of $y$ for every $c$, but $U$ is not computable because $H$ is not recursive.