What is the transition function of the given NFA?

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: $$N = (Q=\\>, \Sigma, \delta, q_ \text< is the start state>, \ < q_\> \text < is the final/accept state>)$$ Any tips or help is appreciated.

1 Answer 1

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 $\delta : Q \times \Sigma \to \mathcal(Q)$ , where $q' \in \delta(q,a)$ means that $q \overset <\longrightarrow>q'$ 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 $\ \times \emptyset = \emptyset$ .