I'm trying to understand the following lemma in Bourbaki's set theory (chapter III, §2,no. 3,Lemma 3):
Lemma 3: Let $E$ be a set, let $S$ be a subset of $P(E)$, and let $p$ be a mapping of $S$ into $E$ such that $p(X)\notin X$ for all $X\in S$. Then there exists a subset $M$ of $E$ and a well-ordering $\Gamma$ on $M$ such that, if $x\leq y$ denotes $y\in \Gamma(x)$ and $S_x$ denotes the segment $]\leftarrow,x[$,
- for all $x\in M$ we have $S_x\in S$ and $p(S_x)=x$;
- $M\notin S$
(note that "segment" means "initial segment")
The following figure, shows a very simplistic scenario:
The conditions of the lemma are satisfied in the mapping shown ($p(X)\notin X$ for all $X\in S$), however, any ordered $M\subset E$ that you take, will have segments that are not in $S$ (e.g. $\emptyset$). So what's the conclusion 1 of the lemma all about?
If $\emptyset\notin S$ (as in your example), one can choose $M=\emptyset$ and condition 1 becomes trivial. Otherwise, $M$ cannot be empty, it has a least element $a$, which is $p(\emptyset)$ by condition 1. If $A= \{a\}$ is not in $S$, one can choose $M=A$, otherwise $M$ has a second element, which is $p(A)$, etc. More generally, if $x\in M$, and $B$ is the set of elements $\le x$, either $B\in S$, case where $p(B)$ is the successor of $x$ or $B\notin S$, case where $M=B$.
The proof of the lemma is a clever use of the uniqueness property given by condition 1.