Every computably enumerable set is the domain of a partial computable function

If $A=\mathrm{\varnothing }$, take the nowhere-defined partial function from $X$ to the one-point set $\{0\}$. Its machine diverges on every input, so [L2] shows that its domain is exactly $\mathrm{\varnothing }=A$.

Suppose now that $A\mathrm{\ne }\mathrm{\varnothing }$, and let $e:\mathbb{N}\to X$ be the total computable enumerator from [L1]. Define a machine on an input code $z\in \{0,1{\}}^{\ast }$ that computes the outputs $e(0),e(1),e(2),\dots$ one after another until it first sees the codeword $z$, at which point it halts and outputs $0$. If $z=e_X(x)$ with $x\in A$, some index $n$ satisfies $e(n)=x$, so the machine eventually halts on $z$. If $x\mathrm{\notin }A$, no enumerated value has code $z$, so the machine never halts on $z$. Thus [L2] gives a partial computable function whose domain is exactly $A$.

In either case, $A$ is the domain of a partial computable function.