Show that $X\subseteq \mathbb{N}$ is computably enumerable iff it is the image of a computable function

Question

A computably enumerable relation is a $R\subseteq \mathbb{N}^k$ such that there is some computable $S\subseteq \mathbb{N}^{k+1}$ and $R(\vec{x})\Leftrightarrow \exists y(S(\vec{x},y))$. A relation $S$ is computable if its characteristic function is, and a function is computable if it has an expression in terms of projection, constants, plus, times, the characteristic function of less-than, composition, and minimization.

I want to prove that $X\subseteq \mathbb{N}$ is computably enumerable iff there is a computable function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $X = \{f(x):x\in\mathbb{N}\}$. Besides the basic definition of a computable function I can also use induction, definition by cases, and the coding functions. We found a way of injectively mapping each tuple of natural numbers of arbitrary length to a single natural number, and the function $\ell(x)$ maps any natural number to the length of the tuple represented by $x$. The function $th(x,y)$ maps to the $x$th element of tuple $y$. A hint given for the $\Rightarrow$ direction (I haven't even though about the $\Leftarrow$ direction because supposedly it's easy) is to use projection on coordinates 0 and 1, and to define it by cases, using the assumption that $X\ne \emptyset$.

My instincts tell me that I want to define $f(0)$ in some way as to make this map to the least value of $R(x)$, and $f(1)$ to the next largest, and so on. I can't define $f(0) = \mu x R(x)$, the minimization of $x$ in $R$, since $R$ is not necessarily computable and therefore $f$ would not necessarily be computable (a c.e. function being not necessarily computable). My brain keeps thinking of something like

$$f(0) = \begin{cases} \mu x S(x,0) \text{ if } \exists x (S(x,0)) \\ ...otherwise \end{cases}$$

but then this looks like it's heading down the wrong tracks and I'm not sure how to basically "scan down the list" of all possible values.

Answer 1

At the start of the second paragraph, do you mean to say $f$ is a computable function like in the title? If it is, you can enumerate $X$ just by computing $f(0), f(1), f(2) \ldots $ For the other direction, if $X$ is computably enumerable just enumerate it and define $f(0)$ as the first element you get, $f(1)$ as the second, and so on. If $X$ is finite and you run out of values, you can just define $f$ of anything higher as the same as $f(0)$

Answer 2

I think I've found an answer. I don't traverse $X$ from least to greatest, but instead leverage the existence of an enumeration of all pairs. With this I define inductively a function that enumerates all pairs that satisfy $S$ and with that I define a function that maps to the pair and projects on the first coordinate.

[Edit, realized this is wrong because functions defined inductively are not necessarily computable.]