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"Let A be a finite alphabet, and let L1 and L2 be two Turing-recognisable languages over A such that L1 is a proper subset of L2, i.e. L1 ⊂ L2 but L1 ≠ L2. Let a language L over the alphabet A satisfy that L1 ⊂ L ⊂ L2. Does L have to be Turing-recognisable as well? Justify your answer." I am unsure about the answer to this question. My initial answer was yes, L has to be Turing-recognisable as well, because I couldn't find an example where it is not given that both L1 and L2 are, and L1 ⊂ L ⊂ L2.

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Hint: consider the empty set and the set of all strings.

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  • $\begingroup$ That kind of confuses me even more. I thought the empty set (ε) was not Turing-recognizable. Does A* (the set of all strings) contain ε as well? $\endgroup$
    – Mark
    Commented Apr 15 at 16:33
  • $\begingroup$ @Theo You are misunderstanding a few concepts here. $\varepsilon$ isn't the empty set, $\emptyset$ is. And $\emptyset$ is recognisable by a TM that simply rejects everything. $\varepsilon$ is the empty string, and is contained in $A^*$. $\endgroup$
    – Knogger
    Commented Apr 15 at 16:42

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