Let $A \subseteq B$, and A is unrecognizable. I know in general that doesn't mean B is unrecognizable. However, are there some limitations we could put on A and B that would make it true? The only thing I could come up with was A=B. Plus, we could say B is unrecognizable as part of our assumptions, but that's not really helpful.

(recognizable means recursively enumerable, in case you use different terminology)

  • 3
    $\begingroup$ You can transfer the condition on $B$ to a condition on $B\setminus A$. Specifically, if $B\setminus A$ is decidable, then $B$ is unrecognizable (as otherwise you could decide $A$ as well). $\endgroup$
    – Shaull
    Feb 3, 2015 at 21:41
  • $\begingroup$ @Shaull Make an answer? $\endgroup$ Feb 4, 2015 at 5:01

1 Answer 1


The condition about $B$ can be transformed into a condition about $B\setminus A$, as follows. If $B\setminus A$ is decidable, then $B$ is unrecognizable. Indeed, if you could recognize $B$, you could determine whether a word $w$ is in $B\setminus A$. If it is, then it's not in $A$, otherwise it is. So $A$ would be recognizable as well.

Note, however, that the converse does not work - there are sets $C,D$ such that $C\cap D=\emptyset$, both $C$ and $D$ are not recognizable, and $C\cup D$ is unrecognizable as well.


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