Exercise 10, § 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 $E$ and $F$ be two lattices and let $f$ be a mapping of $E$ into $F$.

First suppose that $f$ is increasing. Let $x, y \in E$. Since $E$ is a lattice (Definition 8, tag 03QM), the infimum $\inf(x, y)$ exists and satisfies $\inf(x, y) \leqslant x$ and $\inf(x, y) \leqslant y$. Because $f$ is increasing (Definition 1, tag 03JC), we obtain $f(\inf(x, y)) \leqslant f(x)$ and $f(\inf(x, y)) \leqslant f(y)$. Thus $f(\inf(x, y))$ is a lower bound of the set $\{f(x), f(y)\}$. As $\inf(f(x), f(y))$ is the greatest lower bound of this set in the lattice $F$, we have $f(\inf(x, y)) \leqslant \inf(f(x), f(y))$.

Conversely, assume that $f(\inf(x, y)) \leqslant \inf(f(x), f(y))$ for all $x, y \in E$. Let $x, y \in E$ with $x \leqslant y$. Then $\inf(x, y) = x$, since $x$ is a lower bound of $\{x, y\}$ and every lower bound is $\leqslant x$. Hence $f(x) = f(\inf(x, y)) \leqslant \inf(f(x), f(y))$. By definition of the infimum, $\inf(f(x), f(y)) \leqslant f(y)$. Therefore $f(x) \leqslant f(y)$, which proves that $f$ is increasing.

For the example, consider the ordered set $\mathbf{N}$ of natural numbers with its usual order. The product ordered set $\mathbf{N} \times \mathbf{N}$ is ordered by $(a, b) \leqslant (c, d)$ iff $a \leqslant c$ and $b \leqslant d$. Define $f : \mathbf{N} \times \mathbf{N} \to \mathbf{N}$ by $f(a, b) = a + b$. If $(a, b) \leqslant (c, d)$, then $a \leqslant c$ and $b \leqslant d$, so $a + b \leqslant c + d$; thus $f(a, b) \leqslant f(c, d)$ and $f$ is increasing. Now take $x = (1, 0)$ and $y = (0, 1)$. Their infimum in $\mathbf{N} \times \mathbf{N}$ is $(\min(1, 0), \min(0, 1)) = (0, 0)$. Then $f(\inf(x, y)) = 0 + 0 = 0$. On the other hand, $f(x) = 1$ and $f(y) = 1$, so $\inf(f(x), f(y)) = \min(1, 1) = 1$. Hence $f(\inf(x, y)) = 0 \neq 1 = \inf(f(x), f(y))$, and the equality fails for this pair.