Are Turing-recognizable languages closed under intersection?

Problem Detail: 

Let $A$ and $B$ be Turing-recognizable languages. Must language $C = A \cap B$ be Turing-recognizable too?

I have a hunch that it should be because we can construct an enumerator for $C$ by enumerating all the languages in $A$ and then all the languages in $B$.

However, I also know that Turing-recognizable languages are not closed under complement, and $\overline{\bar{A} \cup \bar{B}} = A \cap B = C$, which seems to suggest that Turing-recognizable languages are not closed under intersection.

There is clearly a contradiction somewhere in my reasoning. Where is it?

Asked By : David Faux

Answered By : sdcvvc

Indeed, the recursively enumerable languages are closed under union and intersection, but not complement. This is not a contradiction: the formula $A \cap B = \overline{\overline{A} \cup \overline{B}}$ tells that a class closed under union and complement is closed under intersection, but does not tell anything otherwise.

A similar example: the set of finite subsets of $\mathbb N$ is clearly closed under union and intersection, but not complement.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/7466