This is from book, Introduction-To-The-Theory-Of-Computation-Michael-Sipser, Third Edition, P56.
Now we need to consider the ε arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along ε arrows, including the members of R themselves. Formally, for R ⊆ Q let
E(R) = {q| q can be reached from R by traveling along 0 or more ε arrows}.
Then we modify the transition function of M to place additional fingers on all states that can be reached by going along ε arrows after every step. Replacing δ(r, a) by E(δ(r, a)) achieves this effect. Thus δ′(R, a) = {q ∈ Q| q ∈ E(δ(r, a)) for some r ∈ R}. Additionally, we need to modify the start state of M to move the fingers initially to all possible states that can be reached from the start state of N along the ε arrows. Changing q0′ to be E({q0}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the computation ofM on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete.
How can we say "q can be reached from R by traveling along 0 or more ε arrows" differently
Let's have a NFA, What would E(R) be?
