Question about deterministic finite automaton (DFA)

Question

For $n \in \mathbb N$, an "$n-$DFA" is an automaton with exactly $n$ accepting states.

Let $\Sigma=\{0,1\}$. Prove that the language $\mathcal L=\{0,00,0000\}$ cannot be accepted by any $2-$DFA.

Answer 1

Let $q_0$ be the initial state of a DFA accepting $\mathcal L$. Suppose that the DFA has exactly two final states. Let $q_1 = q_0 \cdot 0$, $q_2 = q_0 \cdot 00$ and $q_3 = q_0 \cdot 0000$. Since $0$, $00$ and $0000$ are in $\mathcal L$, the states $q_1$, $q_2$ and $q_3$ are final. Therefore two of them have to be equal, since there are exactly two final states. Suppose for instance that $q_1 = q_2$. Then $q_1 = q_2 = q_0 \cdot 00 = q_1 \cdot 0$. It follows that for any $n > 0$, $q_0 \cdot 0^n = q_1$ and thus every word of the form $0^n$ with $n > 0$ belongs to $\mathcal L$, a contradiction. The two other cases ($q_1 = q_3$ and $q_2 = q_3$) can be treated by a similar argument.