Languages

Problem:

Mixing it up: Let $ F $ denote some finite language, $ R $ denote some regular language, $ C $ denote some context-free language, $ D $ denote some decidable language, $ E $ denote some
recognizable language, and $ N $ denote some non-recognizable language. For each one of the
following statements, prove whether it always holds, sometimes holds, or never holds:


$ N - D $ is recognizable

Short answer: never

Support I could enumerate $ N - D $, I can also enumerate $ D $ because $ D $ is decidable. Now I can dovetail to enumerate $ N = (N - D) \cup D $, therefore I should not be able to enumerate $ N - D $.

$ RE $ is context free 

Short answer: sometimes

A finite language is recursive enumerable and also context free.
A recursively enumerable language include $ a^n b^n c^n $, which is not context free.

$ N \cap F $ is decidable

Short answer: always

$ N \cap F $ is finite, and a finite language is always decidable.

$ E \cup F $ is recognizable

Short answer: sometimes

E is a recognizable language, it could be finite. F is finite, so E union F could be finite, and a finite language is regular.

E could be $ a^n b^n $, and $ F $ could be the empty language, so $ E \cup F $ is $ a^n b^n $ and is not regular.

$ N^* $ is non recognizable

Short answer: always

If $ N* $ is recognizable, we will have a Turing machine that halt on $ N* $. We can modify that machine such that whenever it reach a final state, any input will go to a state that just consume the input and do nothing and never halt, that would become a recognizer for $ N $.
Therefore $ N^* $ is not recognizable.

$ N \cup C $ is finite

Short answer: never

If $ N \cup C $ is finite, then $ N $ is finite, but finite language is always recognizable.