computably enumerable set

the definitions computably_enumerable_set.html and computably_enumerable_set.html below define an enumerable set to be the domain of some recursive function, but roger's defines it to be the range of some recursive function.

a set $\text{[math]}$ is computably enumerable (c.e.) if $\text{[math]}$ is the domain of some partial computable function.
let the $\text{[math]}$ th c.e. set be denoted by $\text{[math]}$

[cite:;taken from @computability_soare_2016 definition 1.6.2]

note that any computable set is c.e. since if $\text{[math]}$ is computable, then $\text{[math]}$ , where $\text{[math]}$ if the characteristic function $\text{[math]}$ and $\text{[math]}$ otherwise.
[cite:;taken from @computability_soare_2016 chapter 1 defining computability]

the set $\text{[math]}$ is called recursively enumerable if there is a partially computable function $\text{[math]}$ such that
$\text{[math]}$ [cite:;taken from @computability_davis_1994 chapter 4.4 recursively enumerable sets]

$\text{[math]}$ is recursively enumerable if either $\text{[math]}$ or there exists a recursive $\text{[math]}$ such that $\text{[math]}$ .
[cite:;taken from @computability_rogers_1987 chapter 5 recursive and recursively enumerable sets]

let $\text{[math]}$ be a nonempty set. then there is a primitive recursive function $\text{[math]}$ such that $\text{[math]}$ . that is, $\text{[math]}$ is the range of $\text{[math]}$ .
[cite:;taken from @computability_davis_1994 chapter 4.4 a universal program; theorem 4.9]

let $\text{[math]}$ be a partially computable function and let $\text{[math]}$ . (that is, $\text{[math]}$ is the range of $\text{[math]}$ .) then $\text{[math]}$ is r.e.
[cite:;taken from @computability_davis_1994 chapter 4.4 a universal program; theorem 4.10]

suppose that $\text{[math]}$ . then the following statements are all equivalent

$\text{[math]}$ is r.e.;
$\text{[math]}$ is the range of a primitive recursive function;
$\text{[math]}$ is the range of a recursive function;
$\text{[math]}$ is the range of a broken link: blk:def-recursive-function-soare.

[cite:;taken from @computability_davis_1994 chapter 4.4 a universal program; theorem 4.11]

$\text{[math]}$ is c.e..
[cite:;taken from @computability_soare_2016 theorem 1.6.4]

$\text{[math]}$ is the domain of the following p.c. function:
$\text{[math]}$

$\text{[math]}$ is not computable.
[cite:;taken from @computability_soare_2016 theorem 1.6.5]

let $\text{[math]}$ .
[cite:;taken from @computability_soare_2016 definition 1.6.3]

by quantifier contraction theorem, the projection of a c.e. relation is c.e.
[cite:;taken from @computability_soare_2016 corollary 2.1.6]

the following sets and relations are c.e.
$\text{[math]}$ [cite:;taken from @computability_soare_2016 chapter 2 computably enumerable sets]

the c.e. sets are closed under union and intersection uniformly, i.e., there are computable functions $\text{[math]}$ and $\text{[math]}$ such that $\text{[math]}$ and $\text{[math]}$ .
[cite:;taken from @computability_soare_2016 theorem 2.1.11]

a set $\text{[math]}$ is computable iff both $\text{[math]}$ and $\text{[math]}$ are c.e., i.e. iff $\text{[math]}$ .
[cite:;taken from @computability_soare_2016 theorem 2.1.14], [cite:;also in @computability_davis_1994 chapter 4.4 recursively enumerable sets; theorem 4.4]

a partial function $\text{[math]}$ is partial computable iff its graph is computably enumerable.
[cite:;taken from @computability_soare_2016 theorem 2.1.9]

let $\text{[math]}$ be a nonempty r.e. set. then there is a primitive recursive function $\text{[math]}$ such that $\text{[math]}$ . that is, $\text{[math]}$ is the range of $\text{[math]}$ .
[cite:;taken from @computability_davis_1994 theorem 4.9]

if $\text{[math]}$ is a recursive set, then $\text{[math]}$ is r.e.
[cite:;taken from @computability_davis_1994 chapter 4.4 a universal program; theorem 4.3]

is the set of all recursively enumerable sets $\text{[math]}$ computably enumerable itself?

no, this cant be true, otherwise RE in RE, which isnt correct in usual set theory, but what if the set theory foundation allowed such sets?. if we drop that axiom from set theory, then we could have $\text{[math]}$ , we can write a program that receives an element of RE and iterates through all numbers, and checks if the number is an encoding of a function that converges on all inputs, this is demonstrated in computability_practice_problems.html. using this technique, we would diverge if there is no function that converges on all inputs of the elements, and converge otherwise. perhaps this result wouldnt be important however.

what is an example of a recursively enumerable set that is neither decidable nor m-complete?