Bourbaki exercise about "ramified" ordered sets This is Exercise III.2.8 of Bourbaki's Theory of Sets.
An ordered set $E$ is said to be ramified if, for each pair of elements $x,y$ of $E$ such that $x<y$, there exists $z>x$ such that $y$ and $z$ are not comparable. $E$ is said to be completely ramified if it is ramified and has no maximal elements.
(a) Let $E$ be an ordered set and let $a$ be an element of $E$. Let $\mathfrak{R}_a$ denote the set of ramified subsets of $E$ which have $a$ as least element. Show that $\mathfrak{R}_a$, ordered by inclusion, has a maximal element.
(b) If $E$ is branched, show that every maximal element of $\mathfrak{R}_a$ is completely ramified. 
(An ordered set $E$ is said to be branched if for each $x\in E$ there exist $y,z$ in $E$ such that $x\leq y$, $x\leq z$ and the intervals $\left[y,\rightarrow\right[$ and $\left[z,\rightarrow\right[$ do not intersect.)
The proof of (a) is a straightforward application of Zorn's Lemma. 
For (b), I would like to argue in the following way. Suppose $F$ is a maximal element of $\mathfrak{R}_a$ and $x$ is a maximal element of $F$. Then there is $y,z$ in $E$ such that $x\leq y$, $x\leq z$ and $\left[y,\rightarrow\right[\cap\left[z,\rightarrow\right[=\emptyset$. Then $F\cup\{y,z\}$ is ramified.
But I haven't been able to show that $F\cup\{y,z\}$ is ramified, and I'm not sure that it's true, since if $b\in F$ and $b<y$, I don't see a way to conclude that $b\leq x$ or $b<z$. On the other hand, there is no natural way of adding further elements to $F\cup\{y,z\}$ such that it becomes a ramified set.


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*Any hints at the solution or general thoughts about the problem are greatly appreciated.

*Bourbaki seems to be the only mathematician to use the words "branched" and "ramified" for those properties. Are there other more common words, or are these properties without interest?   
 A: I'm not sure if this counts as a rigourous proof, but what if you think of $F$ as a directed graph.  Then $F' = F \cup \{y,z\}$ is formed from $F$ by adding $y$ and $z$ as vertices, and just two edges, one from $x$ to $y$ and one from $x$ to $z$. Now two elements satisfy $u<v$ if and only if there is a directed path from $u$ to $v$. Then it's clear that if $b<y$ then also $b\leqslant x$ since you can just take a directed path from $b$ to $y$ and remove the last arrow, and you get a directed path from $b$ to $x$; the same logic shows that $b<z$, and since $y$ and $z$ are by construction incomparible, this shows that $F'$ is a ramified set, contradicting the maximality of $F$.
A: Here's a counterexample. Let $\{0,1\}^*$ denote the set of finite binary strings. Define E to be the set
$$E=\{a\} \cup \{bw \mid w\in\{0,1\}^*\} \cup \{cw \mid w\in\{0,1\}^*\}$$
with $a\leq bw\leq cw$ for all $w$, and $cw\leq cw'$ if $w$ is a prefix of $w'$.
Let $E'=\{a\}\cup \{bw \mid w\in\{0,1\}^*\}$. I claim that $E$ is branched, and $E'$ is a maximal element of $\mathfrak{R}_a$, but $E'$ is not completely ramified.
Every element $x\in E$ satisfies $x\leq cw$ for some $w$, so $x\leq cw0,cw1$, but $\left[cw0,\rightarrow\right[$ and $\left[cw1,\rightarrow\right[$ do not intersect. Hence $E$ is branched. (This is pretty much 1, Exercise 24 (b).)
Let $E''$ be a proper superset of $E'$. There is a prefix-minimal $w$ such that $cw\in E''$. There is no $z>bw$ incomparable with $cw$, but $bw,cw\in E''$. So $E''$ is not ramified. Hence $E'$ is a maximal element of $\mathfrak{R}_a$.
Finally, $b01010$ (and every other element of the form $bw$) is maximal in $E'$. So $E'$ is not completely ramified.
