<p>In this theorem:
<a href="https://i.sstatic.net/w4K7XCY8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/w4K7XCY8.png" alt="Theorem" /></a></p>
<p>When proving, in the direction from enumerators to decidable languages. The proof I know is as follows:
<a href="https://i.sstatic.net/76IXBZeK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/76IXBZeK.png" alt="proof" /></a></p>
<p>But how can one knows if langague <strong>L</strong> 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?</p>

CC BY-SA 4.0
2024-10-26T17:29:09.503Z
One cannot. The theorem does not say "given an enumerator, we can *compute* a TM that decides $L$".
Emil Jeřábek
CC BY-SA 4.0
2024-10-26T19:51:22.200Z
Ah yes, I see it now. Thank you!
Mohamed Elsheikh
<automata-theory><computability><decidability><enumeration>
Can we know from an enumator if a language is finite?
Mohamed Elsheikh
# Can we know from an enumator if a language is finite?

(Asked by: username_0 on 2024-10-26)

In this theorem:
[![Theorem](https://i.sstatic.net/w4K7XCY8.png)](https://i.sstatic.net/w4K7XCY8.png)

When proving, in the direction from enumerators to decidable languages. The proof I know is as follows:
[![proof](https://i.sstatic.net/76IXBZeK.png)](https://i.sstatic.net/76IXBZeK.png)

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?

**Tags:** automata-theory, computability, decidability, enumeration
> **Comments:**
> **username_1**: One cannot. The theorem does not say "given an enumerator, we can *compute* a TM that decides $L$".  
> **username_0**: Ah yes, I see it now. Thank you!  

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## 0 answers

