# Why does a DFA either contain all or no words $a^k$ if it loops for $a$ in all states?

I am trying to solve this particular problem from Automata Theory by Ullman, Hopcroft, it is as shown below:

Let $$A$$ be a $$DFA$$ and $$a$$ be a particular input symbol of $$A$$, such that for all states $$q$$ of $$A$$ we have $$\delta(q,a)=q$$.

Show that either $$\left\{a\right\}^*\subseteq L(A)$$ or $$\left\{a\right\}^*\cap L(A)=\emptyset$$.

As far as my understanding of the first part of the problem, the language accepted by $$A$$ is $$L(A)=\left\{a, aa, aaa, aaaa, aaaa...\right\}$$.

Since, $$\left\{a\right\}^*$$ represents strings that belong to $$L(A)$$ including $$\varepsilon$$, therefore $$\left\{a\right\}^*\subseteq L(A)$$ is true.

How to show that $$\left\{a\right\}^*\cap L(A)=\emptyset$$? Doesn't $$L(A)$$ represent strings generated by $$\left\{a\right\}^*$$? How can the intersection be $$\emptyset$$?

Hint: Suppose that $a^k \in L(A)$ for some integer $k$. Consider the initial state $s$ of $A$. From the premises it follows that $\delta(s,a^k) = s$, and since $a^k \in L(A)$, $s$ must be an accepting state. For every integer $\ell$ we have $\delta(s,a^\ell) = s$ and so $a^\ell \in L(A)$.
• Your answer is for the first part of the problem right? How and why is the intersection of $\left\{a\right\}^*\cap L(A)=\phi$? – Siddharth Thevaril Sep 27 '14 at 20:23
• Is it because $a^*$ produces $\varepsilon$ which is not a member of $L(A)$? And hence $\left\{\varepsilon\right\}\cap L(A)=\phi$ – Siddharth Thevaril Sep 27 '14 at 20:27
• My argument shows that if $a^k \in L(A)$ for some $k$ then $a^k \in L(A)$ for all $k$. This implies your problem. – Yuval Filmus Sep 27 '14 at 20:28