Extended Transition Function for NFA with $(\epsilon )$ - Transitions
For single input symbols extended transition function behaves same as normal transition function in case of DFA and NFA without $(\epsilon )$ - Transitions.
But not in case of NFA with $(\epsilon )$ - Transition.
Remember this,
$\epsilon $-closure({q1,q2,...qn})=$\epsilon $(q1)$\bigcup$$\epsilon $(q2)$\bigcup$....$\bigcup$$\epsilon $(qn) |
Now applying the above rule in the example given in fig:-
$\epsilon $-closure({q0,q1,q2,q3})=$\epsilon $(q1)$\bigcup$$\epsilon $(q2)$\bigcup$$\epsilon $(q0)
={{q0,q1,q2,q3}$\bigcup${q2}$\bigcup${q0,q3}.
After having taken the union of the above sets we finally get,
={q0,q1,q2,q3}