Exercise 7, § 1

This solution was written by laguna-s-2.1-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

To establish the necessary and sufficient condition for every mapping of an ordered set $E$ into an ordered set $F$ with at least two elements, which is both increasing and decreasing, to be constant on $E$, we proceed as follows:

Necessity. Suppose that every such mapping $f: E \to F$ is constant. Assume, for contradiction, that $E$ is not connected with respect to the reflexive and symmetric relation " $x$ and $y$ are comparable" (denoted $x \sim y$). By Exercise 10 of Chapter II, §6 (cited as ens-ii-s6-ex-10, tag 03J8), $E$ would then admit a nontrivial partition into its connected components. Let $A$ and $B$ be two such components. Define $f$ by setting $f(x) = a$ for all $x \in A$ and $f(x) = b$ for all $x \in B$, with $a \neq b$. Since elements in $A$ and $B$ are not comparable, the order conditions for $f$ impose no constraints between them. Within each component, $f$ is constant, so $f$ is both increasing and decreasing on $E$ but not globally constant, contradicting our hypothesis. Hence, $E$ must be connected with respect to comparability.

Sufficiency. Conversely, suppose $E$ is connected with respect to comparability. Let $f: E \to F$ be a mapping that is both increasing and decreasing. For any $x, y \in E$, since $E$ is connected, there exists a finite sequence $x = x_0, x_1, \dots, x_n = y$ such that each consecutive pair $x_i$ and $x_{i+1}$ are comparable. For each $x_i \sim x_{i+1}$, say $x_i \leq x_{i+1}$, the increasing property gives $f(x_i) \leq f(x_{i+1})$, and the decreasing property gives $f(x_i) \geq f(x_{i+1})$, hence $f(x_i) = f(x_{i+1})$. Iteratively, $f(x) = f(y)$. Therefore, $f$ is constant on $E$.

Connection with directed sets. If $E$ is right-directed (or left-directed), then for any $x, y \in E$, there exists $z \in E$ such that $x \leq z$ and $y \leq z$ (respectively, $z \leq x$ and $z \leq y$). The elements $x$ and $y$ are connected via $z$ in the comparability graph, forming a single connected component. Therefore, $E$ is connected with respect to comparability, implying every such mapping $f$ is constant.

This completes the proof. The referenced results are those defining connected components (Exercise 10 of Chapter II, §6) and the definition of directedness (Definition 7 of §10).