# How to prove that the decidable languages are closed against iteration only by enumerators?

We have the $L\in R$, how can we prove that $L^*\in R$ only by enumerators?

I try to use induction, but as I understand I wrong...
I'd like to get any help!

• What have you tried? Where did you get stuck? We do not want to just do your (home-)work for you; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for a relevant discussion. If you are uncertain how to improve your question, why not ask around in Computer Science Chat? You may also want to check out our reference questions. – Raphael Mar 27 '16 at 21:59
• Also, the question sounds very familiar – I believe it has been asked before. – Yuval Filmus Mar 27 '16 at 22:02

Let's take as an example $L = \{0,1\}$. Then $L^* = \{\epsilon, 0, 1, 00, 01, 10, 11, 000, \ldots\}$. You probably know how to enumerate $L^*$. Same thing works if $L = \{w_0,w_1\}$ for two arbitrary words.
The next step is to handle a language with infinitely many words. Following the earlier convention, I'll just indicate how to solve it for the special in which $L = \mathbb{N}$, and instead of $L^*$ I will use comma-separated sequences. Here is my enumeration: $$\epsilon; 1; 2; 1,1; 3; 1,2; 2,1; 1,1,1; 4; \ldots$$ and so on. See if you can figure out a pattern (or if you don't like my pattern, come up with a different one).