I am studying GTM 139 and troubling about the proof of Bourbaki's fixed point theorem. To quote from that book:
Let $X$ be a poset such that every well ordered subset has an lub in $X$. If $f: X \rightarrow X$ is such that $f(x) \geq x$ for all $x \in X$, then $f$ has a fixed point.
Pick an element $x_{0} \in X .$ Let $\mathbf{S}$ be the collection of subsets $Y \subset X$ such that:
(1) $Y$ is well ordered with least element $x_{0}$ and successor function $\left.f\right|_{Y-\{\text { lub } Y\}}$.
(2) $x_{0} \neq y \in Y \Rightarrow \operatorname{lub}_{X}\left(\operatorname{IS}_{Y}(y)\right) \in Y$.
For example, $\left\{x_{0}\right\} \in \mathbf{S},\left\{x_{0}, f\left(x_{0}\right)\right\} \in \mathbf{S}$, etc. We need the following sublemmas (A) and (B):
(A) If $Y \in \mathbf{S}$ and $Y^{\prime} \in \mathbf{S}$, then $Y$ is an initial segment of $Y^{\prime}$ or vice versa.
To prove (A) let $V=\left\{x \in Y \cap Y^{\prime} \mid \mathrm{WIS}_{Y}(x)=\mathrm{WIS}_{Y^{\prime}}(x)\right\} .$ Suppose first that $V$ has a last element $v .$ If $v$ is not the last element of $Y$ then $\operatorname{succ}_{Y}(v)=f(v)$. If $v$ is not the last element of $Y^{\prime}$ then $\operatorname{succ}_{Y^{\prime}}(v)=f(v) .$ Hence if neither of $Y, Y^{\prime}$ is an initial segment of the other then $f(v) \in V$, whence $f(v)=v$ and we are done.
If, on the contrary, $V$ has no last element, let $z=\operatorname{lub}_{X}(V) .$ If $Y \neq V \neq Y^{\prime}$ then it follows from $(2)$ that $z \in Y \cap Y^{\prime}$ (because if $y=\inf (Y-V)$ then $V=\operatorname{IS}_{Y}(y)$ and therefore $z=\operatorname{lub}_{X}\left(\operatorname{IS}_{Y}(y)\right) \in Y$ by $\left.(2)\right) .$ Therefore, $z \in V$, a contradiction, proving (A).
(B) The set $Y_{0}=\bigcup\{Y \mid Y \in \mathbf{S}\}$ is in $\mathbf{S}$.
To prove (B) note that if $y_{0} \in Y \in \mathbf{S}$ then it follows from (A) that $\left\{y \in Y_{0} \mid y<y_{0}\right\}=\operatorname{IS}_{Y}\left(y_{0}\right)$ and so this subset is well ordered with successor function $f$. This implies immediately that $Y_{0}$ is well ordered and satisfies (1). Also lub $_{X}\left(\operatorname{IS}\left(y_{0}\right)\right) \in Y \subset Y_{0}$ which gives condition (2) for $Y_{0}$. Thus (B) is proved.
Now we complete the proof. Let $y_{0}=l u b_{X}\left(Y_{0}\right)$. If $y_{0} \notin Y_{0}$ then $Y_{0} \cup\left\{y_{0}\right\} \in \mathbf{S}$ and so $y_{0} \in Y_{0}$ after all. If $f\left(y_{0}\right)>y_{0}$ then $Y_{0} \cup\left\{f\left(y_{0}\right)\right\} \in \mathbf{S}$ contrary to the definition of $Y_{0}$. Thus $f\left(y_{0}\right)=y_{0}$ as desired.
First, I had no idea about what is happening. Could someone give me any general ideas about how this proof works?
Second, I've tried using some concrete examples to help me understand, but they induced even more questions. For example, let $X=\{1,2,3,4,5\}$ (with normal partial order) and $x_0=1$. Consider $Y=\{1,2,3\}$ and $Y'=\{1,2,4,5\}$, then $V=\{1,2\}$ and has the last element $v=2$. However, $\operatorname{succ}_{Y'}(2)=4\neq 3=f(2)$. It contradicts the proof and I don't know where is the problem.
A straightforward approach to this works:
Pick $e$ in $X,$ and use transfinite induction to define:
$$\begin{cases} e_0 &= &e, \\ e_{\alpha+1}&=&f(e_\alpha), \\ e_\lambda &= &\mathsf{lub}\{e_\alpha \mid \alpha\lt\lambda\}\text{, for }\lambda\text{ a limit ordinal.} \end{cases} $$
If some $e_\eta = e_{\eta+1},$ we're done, because $e_\eta$ is then a fixed point of $f.$
If there is no such $\eta,$ then we have $e_\alpha\lt e_\beta$ whenever $\alpha\lt\beta.$ But that gives us an injection from the class of all ordinals into the set $X,$ which contradicts Hartogs' Theorem.
If you prefer to avoid class-sized transfinite induction, first use Hartogs' Theorem to find an ordinal $\delta$ that cannot be mapped 1-1 into $X.$ Then define $e_\alpha$ as above for $\alpha\lt\delta.$ If $f$ has no fixed point, then $\langle e_\alpha\mid\alpha\lt\delta\rangle$ maps $\delta$ 1-1 into $X,$ contradicting the choice of $\delta.$