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In this theorem: Theorem

When proving, in the direction from enumerators to decidable languages. The proof I know is as follows: proof

But how can one knows if langague L is finite or not, as we only have an enumerator for this language, and I don't think an enumator halts (or should halt) if the language is finite?

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    $\begingroup$ One cannot. The theorem does not say "given an enumerator, we can compute a TM that decides $L$". $\endgroup$
    – Emil Jeřábek
    Commented Oct 26 at 17:29
  • $\begingroup$ Ah yes, I see it now. Thank you! $\endgroup$
    – Mohamed Elsheikh
    Commented Oct 26 at 19:51

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