I should prove or give a counterexample to the above statement.

In my opinion, this statement is false but I don't manage to find the right counterexample.

My idea was to define $C = Σ^*$ because $Σ^*$ is decidable and contains all the undecidable languages but I fail to find an undecidable language that contains a decidable language.

  • $\begingroup$ A simple hint: have you considered finite languages? And, maybe, the smallest of them? $\endgroup$
    – Tonita
    Apr 23 at 9:49
  • $\begingroup$ @Tonita for B? aren't all finite languages decidable? $\endgroup$
    – RedYoel
    Apr 23 at 9:53
  • 1
    $\begingroup$ So how about taking a finite language as a decidable?) $\endgroup$
    – Tonita
    Apr 23 at 9:57

1 Answer 1

  1. There exists undecidable languages.
  2. $A = \emptyset$ and $C = \Sigma^*$ are decidable.
  3. ????
  4. Profit.

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