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$S$ is a subset of the class of all recursively enumerable languages over some finite symbols then $S$ is recursively enumerable iff

  1. If $L$ is in $S$ and $L'$ is a language such that $L ⊆ L'$ and $L'$ is recursively enumerable, then $L'$ is in $S$
  2. If $L$ is an infinite language in $S$, then there exists at least one finite subset of $L$ that is in $S$
  3. The set of all finite languages in $S$ is enumerable, i.e. a Turing machine can list all the finite languages in $S$

Source of the statement: https://cs.stackexchange.com/q/2322 and some online notes

Isn't the 3rd one contradictory to the 1st one?

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  • $\begingroup$ No, because you are misreading the third condition. It doesn't say that $S$ contains all the finite r.e. languages. It says that those finite r.e. languages that happen to be in $S$ are r.e. $\endgroup$ – Andrej Bauer Nov 5 '18 at 13:36
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No, it is not contradictory. Both conditions are positive in the sense that they "put things in $S$", so they cannot be at odds with each other.

There is an equivalent formulation of $S$: there is an r.e. set $B$ of finite languages such that $$S = \{L \mid \text{$L$ is r.e. and $\exists L_0 \in B \,.\, L_0 \subseteq L$}\}.$$ In words: $S$ is generated by an r.e. set of finite language $B$ in the following way: $S$ contains precisely all r.e. languages which contain an element of $B$.

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  • $\begingroup$ So, whats the 3rd statement mean exactly? I still didn't get. $\endgroup$ – Saravanan Nov 4 '18 at 16:31
  • $\begingroup$ It says that there is a machine $M$ which enumerates $\{L \in S \mid \text{$L$ is finite}\}$. Is that clear? $\endgroup$ – Andrej Bauer Nov 4 '18 at 16:50
  • $\begingroup$ Then it means {$L\in S∣L$ is finite} is r.e, but by 1st statement r.e set should contains some infinite sets also, thats what I want to clarify $\endgroup$ – Saravanan Nov 5 '18 at 4:27
  • $\begingroup$ The first condition does not say that. For instance $S$ could be empty. The first condition says that if $L \in S$ and $L \subseteq L'$ then $L' \in S$. So yes, if $S$ contains a finite language, then it also contains a lot of infinite languages. $\endgroup$ – Andrej Bauer Nov 5 '18 at 7:41
  • $\begingroup$ Still, there is no contradiction. Are you using the word "contradiction" to mean "this looks strange to me"? Because that's not how it's used. $\endgroup$ – Andrej Bauer Nov 5 '18 at 7:42

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