Correct complement of a regular language when the union of the languages do not lead to entire set of strings over the given alphabet?

Question

I have a question that says that the complement of a regular language given as: $L_1=\{a^nb^m|(n+m)<5\}$ is $L_2=\{a^nb^m|(n+m)\geq5\}$ over $\Sigma=\{a,b\}$, and therefore, we can simply construct finite automaton (FA) of $L_2$ by simply interchanging the accepting and non-accepting states of $L_1$ .

However, as per my understanding if I define the complement as: $$\bar{L}=\Sigma^*-L$$ where $\sum^*$ is every possible string, then $L_2$ can't be a complement of $L_1$ as there must also be strings that do not have order $b$ after $a$.

The given approach may work for languages like $L_3=\{w|w~\text{has an odd number of a's}\}$ and $L_4=\{w|w~\text{has an even number of a's}\}$, as disjoint sets lead to $\Sigma^*=L_3+L_4$. However, as per my understanding, this logic will not work to derive FA for $L_2=\{a^nb^m|(n+m)\geq5\}$ from FA of $L_1=\{a^nb^m|(n+m)<5\}$.

Am I correct? Or the definition of complement is different from my understanding.

Answer 1

You are correct. The definition of the complement is exactly what you wrote, and indeed it is not true to say that $L_2$ is the complement of $L_1$. However, changing the accepting and non-accepting states is in fact a correct way to generate a finite automaton for the complement language, so I think whoever wrote $L_2$ just miss-typed. It won't make a difference for the solution, so just substitute the correct complement instead of $L_2$ wherever you need to.