Consider the finite alphabet $\sum=\{0,1,\cdots k-1\}$, the set $\sum^*$ of all words over it, and a mapping $T:\sum\to \sum^*$ such that each $i\in\sum$ is the initial symbol of $T(i)$. Now the natural order on $\sum$ gives a lexicographic order on $\sum^*$. We may extend $T$ to a map $T:\sum^*\to\sum^*$ by defining $T(v_1\cdots v_n)=T(v_1)\cdots T(v_n)$. Now we define the $D0L$ infinite word as the limit word of $0,T(0),T^2(0),T^3(0)\cdots$.
I want to establish that this is the least non-empty word $\sigma$ in lexicographic order such that $T(\sigma)=\sigma$. How can I establish that? By explicitly writing out the first few entries it is clear that this is a fixed point, but I am looking for a more formal proof.
Thanks.