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?