Proof: Every lattice has a maximal filter iff AC

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I'm working through a proof of Herrlich's book Axiom of Choice, p.58 (Google books):

Equivalent are

  1. Every lattice has a maximal filter.
  2. Axiom of Choice.

In this book, a lattice is defined as

  • a poset
  • in which each finite subset has an infimum and a supremum and
  • there exists a smallest element $0$ and a largest element $1$
  • such that $1 \neq 0$.

A filter $F$ is a subset of a lattice containing $1$ but not $0$, and $x,y \in F$ iff $x \land y \in F$ (I suppose $x \land y := \inf(\{x,y\})$ as the book doesn't say so explicitly).


In the proof of (1)$\Rightarrow$(2), the author starts off with an indexed Family of non-empty sets $(X_i)_{i \in I}$ and then considers the set $S$ of all pairs $(J,x)$ with $J \subseteq I$ and $x \in \prod_{i \in J} X_i$. The set $S$ can be made a poset by the following relation:

$$ (J,x) \leq (K,y) \iff (\text{$J \subseteq K$ and $y$ restricted to $J$ equals $x$}). $$

Referring to 1. the author adds a maximal element $1$ to $S$, saying that it is now a lattice.

I can't seem to verify all axioms of a lattice. $S$ is a poset but I'm struggling with the infimum and supremum.

One way to define $(J,x) \land (K,y)$ could be as $(D(x \cap y), x\cap y)$, interpreting $x, y$ as relations (i.e. sets of ordered pairs) with domain $D(x)=J$ and $D(y)=K$.

Is this correct? How would you define the supremum, i.e. $x \lor y$?