Exercise 1, Appendix 0

This solution was written by nemotron-3-ultra-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

Let $A$ be a word in $L_0(S)$ and let $B$, $C$ be two significant segments of $A$. By Proposition 2 a word is significant if and only if it is balanced; hence $B$ and $C$ are balanced words. Recall that a word $X$ is balanced when

1. $l(X)=n(X)+1$ (in particular $X$ is not empty), 2. for every proper initial segment $Y$ of $X$ one has $l(Y)\le n(Y)$.

The weight function $n$ is additive: $n(UV)=n(U)+n(V)$ for any words $U,V$, and the length is likewise additive.

We must prove that either $B$ is a segment of $C$, or $C$ is a segment of $B$, or $B$ and $C$ are disjoint. Suppose, for a contradiction, that $B$ and $C$ are not disjoint and neither is a segment of the other. Then their occurrences in $A$ overlap properly: one starts strictly before the other, the second starts before the first ends, and the second ends after the first ends. Let the starting positions of $B$ and $C$ in $A$ be $b$ and $c$ (with $1\le b,c\le l(A)$). Without loss of generality assume $b<c$. Because the segments overlap we have $c\le b+l(B)-1$; because $C$ is not contained in $B$ we have $c+l(C)-1>b+l(B)-1$. Thus

Define three words:

• $F$ = the segment of $B$ from its first position up to the $(c-b)$-th position (the part of $B$ that precedes $C$);
• $D$ = the segment of $B$ from the $(c-b+1)$-st position to the end of $B$ (the overlap of $B$ and $C$);
• $E$ = the segment of $C$ that follows $D$ (the part of $C$ that follows $B$).

By the strict inequalities above, $F$ and $E$ are non‑empty, and we have the factorisations

Since $B$ is balanced and $F$ is a proper initial segment of $B$ (because $D\neq\emptyset$), condition 2 of the definition gives $l(F)\le n(F)$. Since $C$ is balanced and $D$ is a proper initial segment of $C$ (because $E\neq\emptyset$), we likewise have $l(D)\le n(D)$. Using additivity of $l$ and $n$ we obtain

But $B$ is balanced, so $l(B)=n(B)+1$ by condition 1. This is a contradiction.

Therefore our assumption was false: if $B$ and $C$ are not disjoint, one must be a segment of the other. By the definition given in the text, two segments of $A$ that do not overlap are disjoint. Hence in all cases either $B$ is a segment of $C$, or $C$ is a segment of $B$, or $B$ and $C$ are disjoint.