Why is decidable included in Turing-recognizable? Recognizable means there is a Turing-machine that accepts all and only instances of that language So that does not mean that if the input is not of that language, the machine rejects, because the machine could also go into some infinite loop if the input is otherwise
Simple concrete example of a language that is Turing recognizable but . . . The set of string encodings of instances of the Post Correspondence Problem that have matches is a language that is recognizable but not decidable, as discussed in Sipser's Introduction to the Theory of Computation, Third Edition, chapter 5, pages 227-233 The discussion includes a proof of undecidability by reduction from the language of encodings <M,w> where M is a Turing machine that
Non-Turing Recognisable Languages - Mathematics Stack Exchange I'll answer your questions in order First, what is a "simple" example of a language which is not Turing recognizable: If $\mathcal {H}$ is the halting problem, then I claim $\mathcal {H}^c$ (that is, the complement of $\mathcal {H}$) is not recognizable Why might this be true?
Can a problem be undecidable but yet Turing recognizable? Can a promblem be undecidable but yet be fully Turing recognizable? For example - if language is Turing recognizable OR it's complement is Turing recognizable - so it still undecidable?
Why is showing a language is Turing recognizable trickier than showing . . . Later, I have written a proof to show that Turing Recognizable languages are closed under union I am supposed to identify why closing a Turing Recognizable language under some operation is trickier to prove than when dealing with Turing Decidable languages