# Why do we need Kleene Star when there is concatenation?

For an alphabet $$A = \{ a_1, a_2..., a_n \}$$, the set of regular langages $$L_r$$ on $$A$$ are recursively defined by closed union, concatenation, and Kleene star's operator. I understood that languages ($$A^*$$) and regular languages (a subset of $$A^*$$) are different. Why do we need Kleene star, isn't concatenation enough for this definition?

Very simple "proof" that should be obviously wrong:

If $$X \in L_r$$ a regular language on $$A$$, and $$E \in X^*$$ (if i'm right also $$X^* \in L_r$$) a word then we could write $$E$$ as $$e_1e_2\dots e_n = e_1 . e_2 \ldots e_{n-1} . e_n$$, with each $$e_i \in X$$. Then $$E$$ is explicitly constructible by concatenation.

I forgot $$\epsilon$$ but so just add a simple rule that allow $$\epsilon$$. My intuition says that it has something to do with infinity, that Kleene Star allows infinite-lengh chains whereas concatenation doesn't. Is it that?

• For the same reason we need while-loops in programming languages. If we can't iterate all programs will stop within a fixed number of steps. Feb 3 '21 at 17:50
• @HendrikJan Something something recursion :) Which actually has an analogue in this context as well - CFGs allow recursion, regular expressions do not.
– orlp
Feb 3 '21 at 19:13
• @orlp Agreed. I was comparing regular expressions to "sequence, selection, iteration" programs. Feb 4 '21 at 13:27

Regular expressions without Kleene star define finite languages. You can prove this by induction on the structure of the regular expression. In contrast, $$a^*$$ is a regular expression which defines an infinite language.
We could try to define $$a^*$$ using concatenation and union: $$a^* = \epsilon + a + a^2 + a^3 + \cdots$$ Unfortunately, the required regular expression is infinite, which we do not allow (infinite expressions do make sense in some contexts, for example in infinitary logic, but not here).
• +1. And not only don't we allow infinite unions in the definition of regular expressions, we actually can't do so -- or at least, that definition wouldn't be equivalent to the standard one -- because that would also allow languages such as $a^n b^n = \epsilon + ab + aabb + aaabbb + aaaabbbb + \cdots$ that are known to be non-regular according to the standard definition. Feb 4 '21 at 4:53