Kleene star closure of a context free grammar

I have this question about closure of a context free grammar, and if someone can check my answer and see if it makes sense, and if not, what is missing, I would be very grateful.

Give an counter-example to show that the following contruction fails in to proof that class of languages free of context is closed under the operation star. By A a language free of context that it is generated by the GLC G = {V,,R,S}. Add the new rule S -> SS e call the resulting grammar G'. That grammar is expected to generate a*"

I have only a superficial comprehension, I believe it is necessary that Rg', the set of production rules of G' must be equal to Rg united with {Sg'->SgSg'|e} or in another words Rg': Rg U {Sg'->SgSg'|e}. Even if that it is correct I don't understand why {Sg'->SgSg'|e} is necessary

There is a silly problem that you might need to add a production of the form $S\to\epsilon$, since otherwise you don't necessarily generate $\epsilon$. But the real problem is that $S$ can be used in the right-hand side of productions, and when you add the rule $S\to SS$, things could break. The "correct" way to turn a grammar of $A$ into a grammar of $A^*$ is to add a new start symbol $S'$ and the productions $S' \to S'S|\epsilon$.
• How to be sure that $$S\to\epsilon$$ wasn't part of the grammar before? And how could $$S\to SS$$ break things up? I also don't understand why it is $$S' \to S'S|\epsilon$$ necessary. Commented Jan 25, 2014 at 23:50