Prove that the class of regular languages is closed under the Kleene + operation. That is, show that if L is regular, then so is $L^{+}$

Question

This is my attempt at a proof:

Let $E$ be a $REGEX$ accepting $L$. We claim the $REGEX$ $E^{'} = E^{+}$ accepts L. i.e. $L(E^{+}) = (L(E))^{+}$
$L^{+}$ is regular since there is a $REGEX$ $E^{+}$ accepting $L$

I am very unsure if my proof makes sense or is correct. Can someone please advise?

Answer 1

In this type of poblems , we usually construct a DFA that recognises the the language under some operation (in your example L⁺) using the DFA recognising L

So , bearing in mind the construction we use to build a DFA of L* , can you build a DFA that accepts L⁺ (please try before viewing the answer )

It is the same construction without adding a new start state

Let M be a DFA recognising L , we add sigma-moves from all of M accept states to the start state

Formally , if δ is the transition function for M , add the following rules to δ :

δ(q_a,ε) = q₀ ,where q₀ is the start state of M , and q_a ∈ F is an accept state of M

M now accepts a string in L , then returns to start state , and repeats again till end of string , allowing for concatenations of strings in L

Hopefully you can see that the only difference between this construction and that of L* is that now we don't just accept ε , we need to have at least one string in L before we can accept