I have a question regarding the following NFA:
When I provide the formal definition, I am stuck at the alphabet $\Sigma$ and $\delta$ parts. Since the alphabet is not given, and no transitions are present, does this mean that there is no transition table? If there is one how would it look like? I am not sure how to complete the 5-tuple without the $\Sigma$ and $\delta$.
What I have so far:
$$N = (Q=\{q_{0}\}, \Sigma, \delta, q_{0} \text{ is the start state}, \{ q_{0} \} \text{ is the final/accept state} )$$
Any tips or help is appreciated.
Best Answer
Welcome to MSE!
This NFA could actually work for any alphabet you like (including the empty alphabet $\Sigma = \emptyset$). This is basically because it ignores its input and accepts, and of course we can always do that, regardless of our alphabet.
Now, if your alphabet is $\Sigma$, what should the transition function be? Well remember that (for an NFA) we know $\delta : Q \times \Sigma \to \mathcal{P}(Q)$, where $q' \in \delta(q,a)$ means that $q \overset{a}{\longrightarrow} q'$ is an arrow in our graph.
For this NFA, there are no arrows, so $\delta(q_0,a) = \emptyset$ for each $a \in \Sigma$. Notice if $\Sigma = \emptyset$, then this means $\delta$ is the empty function, since its domain is $\{q_0\} \times \emptyset = \emptyset$.
I hope this helps ^_^