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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?

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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.

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  • $\begingroup$ Thanks! Does my enumerator work as well? $\endgroup$
    – David Faux
    Dec 17, 2012 at 18:02
  • $\begingroup$ @David: Um, enumerators does not output languages as you wrote, but words but in a language. An enumerator for $A \cap B$ can be constructed in this way: run in parallel enumerators for $A$ and $B$. Whenever at some point a word has been enumerated by both enumerators, output it in the joint enumerator. $\endgroup$
    – sdcvvc
    Dec 17, 2012 at 21:05

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