Limited subsets of a free monoid

0 • Ix IT == 0 (mod 2)}. 3 We leave it to the reader to show that A* = (luA) , hence A is limited. If we put B= I*\A, B* = (1 uB) 3, hence B is also limited. 0 EXAMPLE 1. Let F be a finite set containing some nonempty word. It is easy to see that F is not limited, while E = I*\F is limited. 0 2 n A* = luAuA while iff there exists an integer m? 1, such that A-A' = (1 uA) m. Sinee Thue proved that given and integer m~ 1, there exists a word x of lengh m in I*, such that for every U,v and w, x= uv 2 w inlplies that v= 1 [5,20J. Hence, xE(luA)m, while x~(luA)m-1. 0 it follows that A is limited iff the infinite union in (1) can be replaced by the finite union in (2), for some m. Equivalently, A is limited iff any concatenation of a finite number of elements of A can also be obtained by the concatenation of at most m elements of A. We say that a subset A of I* is iimde.d