User:IssaRice/Computability and logic/Characterization of recursively enumerable sets

Let $S$ be a set of natural numbers. The following are all equivalent.

Property	Emphasis
The elements of $S$ can be enumerated in a computable manner as $a_0,a_1,a_2,\dots$	Recursively enumerable
$S$ is the range of a computable partial function.	Recursively enumerable
$S$ is empty or the range of a computable total function.	Recursively enumerable
$S$ is empty or the range of a primitive recursive function.	Recursively enumerable
$S$ is the domain of a computable partial function.	Semi-ness
A recursive semicharacteristic function for $S$ exists.	Semi-ness
There exists a two-place recursive relation $R$ such that $x\in S⟺\mathrm{\exists }tR(x,t)$	Semi-ness
The relation $x\in S$ is ${\mathrm{\Sigma }}_1$, i.e., there exists a $(k+1)$-place recursive relation $R$ such that $x\in S⟺\mathrm{\exists }{t}_1\cdots \mathrm{\exists }{t}_{k}R(x,t_1,\dots ,t_k)$	Semi-ness
The relation $x\in S$ has a computable partial verifying function, i.e., there exists a computable partial function $f$ such that $x\in S⟺f(x)\simeq 1$.	Semi-ness

The rows labeled "Recursively enumerable" all emphasize the fact that $S$ is "recursively enumerable", i.e., that the elements of $S$ can be listed in a recursive (computable) way.

The rows labeled "Semi-ness" all emphasize the fact that $S$ is semidecidable/recognizable/verifiable/semirecursive, i.e., that if something is in $S$, then in a finite amount of time we can verify this computably, but that if something isn't in $S$, then we will run into an infinite loop.

The fundamental theorem of recursively enumerable sets says that recursively enumerable equals semi-ness.