1
$\begingroup$

The union of a recursively enumerable set and its complement is $\Sigma^*$, which is definitely recursively enumerable. What if instead we consider the following operation, on an RE set $S$? $$ \{ \# w : w \in S \} \cup \{ \\\$ w : w \in \overline{S} \} $$ Is the result recursively enumerable?

$\endgroup$
1
  • 4
    $\begingroup$ The question in the body doesn't really match the title $\endgroup$
    – Ariel
    Apr 9 at 5:47
2
$\begingroup$

The operation you described in actually the join operator in the semilattice of Turing degrees, defined as $A\sqcup B=\{0x | x\in A\}\cup \{1x | x\in B\}$. It is not hard to show that $A,B\le_T A\sqcup B$, thus if $A\in RE\setminus R$ then $\overline{A}\notin RE$ and thus $A\sqcup\overline{A}\notin RE$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.