Exercise 1, § 1

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 $E$ be an ordered set; denote its order relation by $\leqslant$. For $x, y \in E$ we write $x < y$ to mean $x \leqslant y$ and $x \neq y$. The relation $R$ is defined by: $R\{x, y\}$ holds if and only if $x \in E$, $y \in E$ and $x < y$.

We verify the three conditions of no. 1 for $R$.

First condition (transitivity). Assume $R\{x, y\}$ and $R\{y, z\}$. Then $x < y$ and $y < z$, so $x \leqslant y$, $y \leqslant z$, $x \neq y$ and $y \neq z$. By transitivity of $\leqslant$ we have $x \leqslant z$. If $x = z$, then $x \leqslant y$ and $y \leqslant x$, whence $x = y$ by antisymmetry of $\leqslant$, contradicting $x \neq y$. Thus $x \neq z$, so $x < z$, and therefore $R\{x, z\}$ holds.

Second condition. Assume $R\{x, y\}$ and $R\{y, x\}$. Then $x < y$ and $y < x$, so $x \leqslant y$ and $y \leqslant x$. Antisymmetry of $\leqslant$ gives $x = y$. But $x < y$ implies $x \neq y$, a contradiction. Hence the antecedent $R\{x, y\} \wedge R\{y, x\}$ is false for all $x, y$, and the implication $(R\{x, y\} \wedge R\{y, x\}) \Rightarrow (x = y)$ is true (vacuously).

Third condition. We show that this condition fails. By hypothesis there exists at least one pair of distinct comparable elements in $E$; thus there are $a, b \in E$ with $a \neq b$ and either $a < b$ or $b < a$. Without loss of generality suppose $a < b$. Then $R\{a, b\}$ holds. However $R\{a, a\}$ would mean $a < a$, which is false because $a \leqslant a$ is true but $a \neq a$ is false. Similarly $R\{b, b\}$ is false. Consequently $R\{a, b\}$ is true while $R\{a, a\} \wedge R\{b, b\}$ is false, so the implication $R\{a, b\} \Rightarrow (R\{a, a\} \wedge R\{b, b\})$ is false. Therefore the third condition is not satisfied.