One thing I understand is that the complement of every Turing recognizable(but not decidable) language is non Turing recognizable.
What about the complement of non Turing recognizable language? Is the complement of every non r.e a recognizable language.
I don't know how to approach this problem.
But one idea I had was this:
If the complement of every non r.e were a r.e language then there would be a bijection from r.e to non r.e. This bijection maps a r.e to a non r.e . The set of r.e languages is countable(even though infinite). If a bijection exists then non r.e would also be countable. But the union of r.e and non r.e is uncountable. This means one of them must be uncountable. So the original supposition must be wrong.
I am new to coutability and such things. So it may be possible that the idea is compeletely wrong.
Anyhow can someone provide an approach to this problem.