0

$S$ 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?

Saravanan
  • 151
  • 8
  • 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. – Andrej Bauer Nov 05 '18 at 13:36

1 Answers1

2

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 $B$.

Andrej Bauer
  • 30,396
  • 1
  • 70
  • 117
  • So, whats the 3rd statement mean exactly? I still didn't get. – Saravanan Nov 04 '18 at 16:31
  • It says that there is a machine $M$ which enumerates ${L \in S \mid \text{$L$ is finite}}$. Is that clear? – Andrej Bauer Nov 04 '18 at 16:50
  • Then it means {$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 – Saravanan Nov 05 '18 at 04:27
  • 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. – Andrej Bauer Nov 05 '18 at 07:41
  • 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. – Andrej Bauer Nov 05 '18 at 07:42
  • So, every r.e $S$ which is nonempty contains an infinite language. But by 3. "The set of all finite languages in $S$ is enumerable", consider the set of all finite languages in $S$($S$ is r.e) which itself is a r.e set, then it doesn't contain an infinite set. – Saravanan Nov 05 '18 at 08:16
  • Precisely, if $S$ is an r.e. set of r.e. langauges, then ${L \in S \mid \text{$L$ is finite}}$ is not an r.e. set of r.e. languages, according to the definition (unless $S = \emptyset$). What is the problem? Note that $S$ itself is not a language, it is a set of languages. There are deeper reasons why the definition is given in the way it is given. – Andrej Bauer Nov 05 '18 at 08:42
  • "Note that S itself is not a language, it is a set of languages" ? Its a language, contains encoding of TMs as per the definition I know https://chat.stackexchange.com/rooms/85322/discussion-between-saravanan-and-andrej-bauer – Saravanan Nov 05 '18 at 08:57
  • No, $S$ is not a language. It is a set of languages because nowhere did we say that the languages are encoded using TMs. This is why we have a special definition what it means for $S$ to be an "r.e. set of languages". Note that "$S$ is an r.e. set of r.e. languages" is different from stating that the set ${n \in \mathbb{N} \mid \exists L \in S ,., \text{$n$ encodes $L$}}$ is r.e. – Andrej Bauer Nov 05 '18 at 13:33
  • Ok, you incorrectly transcribed the definition from https://cs.stackexchange.com/q/2322 by filiing to use encodings of languages. I give up. – Andrej Bauer Nov 05 '18 at 13:35
  • @Andrej Bauer I think that linked answer itself has flaws which lead to this misunderstanding, let alone the "language/machine" notation which adds up to the confusion. See my answer to the same question here https://math.stackexchange.com/a/3063866/300521. I also commented on the linked answer. – Beleg Jan 06 '19 at 16:01
  • I gave up a while ago. – Andrej Bauer Jan 06 '19 at 16:19
  • @Andrej Bauer btw, is the mentioned alternative form in your answer exactly equivalent of the original Rice-Shapiro formulation? I've seen it in the Rogers' book, but its converse case also applies, i.e. all sets of the mentioned form are c.e., which is not correct for the theorem stated in the compactness+upward-closure form, so I guess. – Beleg Jan 06 '19 at 16:24