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the regular languages are closed under the operation:
init(L) = the set of the strings w such that for some x, wx is in L.
EDIT :
x can be any string, a character or empty string
How can I prove that ?
OK, misread the quesion on the first time, now I get it. It still trivial. Looking at the automate what you searching is a partion of the automate into two state sets S1 and S2, so that just one transition is between them (and if its from S1->S2 S1 contains of course the start node, and S2 the end node). Such exist always (exception empty language), in case there is no such node you can add one, so w is just a set containing the empty word, which is of course also regular (as well as the empty language case).