TOC Recursive languages

Question

How does complement of a Recursively enumerable but not recursive is not recursively enumerable?

Answer 1

If recursively enumerable languages were closed under complementation, then they would also be decidable, and this would make the Halting problem decidable. 

Let M the TM accepting a recursively enumerable language L. Now assume by contradiction that L', the complement of L, is also recursively enumerable.
This means that there exists a TM M' that enumerates all the elements in L'.
Now we can run M and M' in parallel (see this to understand what I mean by in parallel: Dovetailing (computer science)) and output YES if M halts, and NO if M' halts. Notice that since both L and L' are recursively enumerable this machine is guarantee to halt. But this means that L is decidable which contradicts the fact that the halting problem is undecidable.

Hope it will help!

Source:https://www.quora.com/Why-are-recursively-enumerable-languages-not-closed-under-complementation