So I already posted this in Computer Science Stack Exchange but haven't received any real answers and my exam is tomorrow, so I'll post it here too hoping for some clarification:
I'm having problem understanding the first part of this proof. I don't understand why it needs to hang at $(p,\epsilon,\epsilon)$ why can't the automaton just keep going, just reading the rest of the characters without touching the stack? For example what if I have a DPDA that always keeps its stack empty? like, it only reads characters, but doesn't alter the stack. taking $(q_0,a,\epsilon)$ and returning $(q_0,\epsilon)$ every time. Is it because it necessarily needs to do something when given $\epsilon$ as input? In that case can't I just say that if I read $\epsilon$ I return the empty set? Here's the definition I was given of DPDA.
Here's the definition given in Sipser's book
I don't see from these definitions why I shouldn't be able to define a DPDA that has a single state $q_0$ and just reads the whole input without ever touching the stack, like this: $\delta(q_0,a,\epsilon)=(q_0,\epsilon)$ $\forall a\in\Sigma$
