Exercise 9, § 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 prove that $\sup\nolimits_j\,(\inf\nolimits_i\,x_{ij})\leqslant\inf\nolimits_i\,(\sup\nolimits_j\,x_{ij})$ in a lattice $E$ for every finite double family $(x_{ij})$, we proceed as follows. Since $E$ is a lattice, all infima and suprema of finite subsets exist. Let $m_j = \inf\nolimits_i\,x_{ij}$ for each $j$, and $n_i = \sup\nolimits_j\,x_{ij}$ for each $i$. We aim to show that $\sup_j\,m_j \leqslant \inf_i\,n_i$.

For any $i$ and $j$, we have $m_j \leqslant x_{ij}$ by the definition of infimum. Moreover, $x_{ij} \leqslant n_i$ since $n_i$ is an upper bound for the $x_{ij}$ for fixed $i$. Combining these inequalities gives $m_j \leqslant n_i$ for all $i$, $j$.

Fixing $i$, this shows that $n_i$ is an upper bound of the set $\{m_j\}$. Therefore, the supremum $\sup_j\,m_j$ is less than or equal to each $n_i$, so $\sup_j\,m_j \leqslant n_i$ for every $i$. Hence, $\sup_j\,m_j$ is a lower bound of the set $\{n_i\}$. The infimum $\inf_i\,n_i$ is the greatest lower bound, so $\sup_j\,m_j \leqslant \inf_i\,n_i$, as required.

\boxed{\sup\nolimits_j\,\bigl(\inf\nolimits_i\,x_{ij}\bigr)\leqslant\inf\nolimits_i\,\bigl(\sup\nolimits_j\,x_{ij}\bigr)}

The section does not provide specific result tags necessary for this proof, as it directly applies the definitions of suprema, infima, and order relations in a lattice. The key steps rely on standard properties of lattices and ordering, which are foundational to the material presented here.

However, no result tags listed in the provided exercises or propositions were explicitly required for the solution.

USES: 03QM