Why is $\left\{a\right\}^*\cap L(A)=\emptyset$? where $\delta(q,a)=q$

Question

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 $\emptyset$?

Answer 1

Let $\mathcal{A}$ be a DFA with initial state $q$. Since $q \cdot a = q$, one has $q \cdot a^n = q$ for all $n \geqslant 0$. Therefore if $q$ is a final state, then $a^* \subseteq L(\mathcal{A})$ and if $q$ is not a final state, then $a^* \cap L(\mathcal{A}) = \emptyset$.