$A$ is an alphabet. An automaton over $A$ can be defined as a set $A_0 = (Q, E, I, T),$ where $Q$ is the set of states, $E \subseteq Q \times A \times Q$ is the set of edges or transition, $I, T \subseteq Q$ of initial and terminal states. A path in the automaton $A_0$ is a sequence $(p_0, a_1, p_1),$ $(p_1, a_2, p_2),$ $\ldots,$ $(p_{n-1}, a_n, p_n)$ of consecutive edges. A state $p$ is accessible if there is a path starting in an initial state and ending in $p.$ It is coaccessible if there is a path starting in $p$ and ending in a terminal state. An automaton is trim if every state is accessible and coacessible.
The information above is from M. Lothaire's Algebraic Combinatorics on Words. My question is whether $(p, \epsilon, p)$ is a path or not. Since $\epsilon$ is not in $A,$ my answer is no. If so, is an inital state accessible? According to the definition above, we may say it will be if there is a path starting from an initial state and ending in the initial state.Then if an automaton is trim,the initial states should be accessible.But in the following statement about the trim automaton,I don't think the $trim$ ensures the initial state(s) is(are) necessary accessible.
So actually I want to make sure whether the $(p,\epsilon,p)$ is a path. Second, if the initial states are only coaccessible(i.e there may be no path starting from an initial state and ending in the initial states), is the automaton still trim?
These two questions make me too confused to continue to read the following content in the book. Thanks in advance.
If I applied $\epsilon$ into the definition.Then the example of the first returns to state 1 below shouldn't be $X=${$b,ab$} but $X=${$\epsilon,b,ab$} which contradicts with the former $X=${$b,ab$} given in the book.The set of $first$ $returns$ to a state $q$ is the set of labels(or words) of paths from $q$ to $q$ which do not pass another time through $q$.

On the other hand,without the $\epsilon$ path,I don't know how to ensure the automaton in the following proof is trim,since the initial state doesn't seem to be accessible without the $\epsilon$ path.


All the information above is shown in http://x.co/iEVa, from Page 12 to Page 14.
Thank you for reading this long post.