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It's quite easy to show that there are only three recursive sets up to m-equivalence, namely $\emptyset,\{1\},\mathbb N.$ Can we state something "similarly short" for recursively enumerable sets, like there are only 4 recursively enumerable sets up to m-equivalence – $\emptyset,\{1\},K,\mathbb N$ (where $K=\{i\mid\varphi_i(i)\text{ is def.}\}$)? More specifically, is every non-recursive recursively enumerable set m-equivalent to $K$?

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  • $\begingroup$ See en.wikipedia.org/wiki/…. Recursively enumerable Turing degrees are dense. $\endgroup$
    – Ariel
    Dec 6 '20 at 13:56
  • $\begingroup$ I am not sure if that answers my question, because I don't know much about Turing equivalence. Does it imply that there is a non-recursive r.e. set $A$ such that $K\nleq_m A$ ? $\endgroup$
    – byk7
    Dec 6 '20 at 14:58
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    $\begingroup$ Since Turing reducibility extends many-one reducibility, yes it does: if $A\not\ge_TB$ then a fortiori $A\not\ge_m B$. $\endgroup$ Dec 6 '20 at 20:18
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There are lots of these. Even with respect to Turing reducibility (which extends $m$-reducibility: if $A\not\ge_TB$ then $A\not\ge_mB$), the structure of c.e. sets is extremely rich.

That said, constructing an infinite non-recursive r.e. $A$ with $A<_mK$ is significantly easier than constructing one with $A<_T K$. The former was done by Post in 1944 who introduced the notions of creative and simple sets (each of which are necessarily non-recursive), proved that a simple set exists and that $K$ is creative, and proved that no creative set is $m$-reducibile to any simple set; the latter wasn't achieved until a decade later by Friedberg and Muchnik independently, and required a fundamentally new technique (the priority method).

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