If $L \cdot \{\epsilon, a, b\}$ is regular, is $L$? > Given that $L \cdot \\{\epsilon, a, b\\}$ is regular, is $L$ regular too? (Our alphabet is $\Sigma = \\{a,b,c,d\\}$ What I thought was yes, and here is why: > If it is regular, then we know there is a finite automat such that $L(M) = L \cdot \\{\epsilon, a, b\\}$. Now, since it accepts that language, we know that at most there are three accepting states. One that we get to by reading the last $\epsilon$, one by reading the last $a$, and one by reading the last $b$. They can be also one state that is acheived by reading any of thee three.. etc. Now if we remove that (those) accepting states, and put the one previous to the last reading of either $a$ or $b$ or $\epsilon$, and mark them as accepting states, then we will have a finite automat that accepts $L$. Is that correct or?

For suitable $A\subseteq 2\mathbb N$ we have that $$ L=\\{\,\alpha\in\\{a,b\\}^*:|\alpha|\
otin A\,\\}$$ is not regular (use the pumping lemma). But $L\cdot\\{\epsilon,a,b\\}=\\{a,b\\}^*$ is clearly regular.