User:Dfeuer/Order Completion is Minimal

Note: our current definition of completion is as a superset. We presumably want also a definition as an order embedding. This theorem will of course take on a slightly different form in that case.

Theorem
Let $(T,\le)$ be a complete lattice.

Let $S \subseteq T$.

Suppose that $T$ is an order completion of $S$.

Then for any $U$ such that $S \subseteq U \subsetneqq T$,


 * $(U, \le)$ is not complete.

Proof
Suppose $U$ is complete.

Clearly, the identity function on $T$ restricted to $U$ is an order-preserving mapping from $U$ to $T$.

We will show that it is not the only such.

Let $p \in T \setminus U$.

Let $f\colon U \to T$ be defined by
 * $f(x) = x \vee p$.

Suppose that $x,y \in U$ and $x \le y$.

Then $x \vee y = y$.

So $(x \vee p) \vee (y \vee p) = (x \vee y) \vee p = y \vee p$,

so $x \vee p \le y \vee p$.

Thus $f$ is an order-preserving mapping.

Similarly, the mapping $g$ defined by $g(x) = x \wedge p$ is an order-preserving mapping.

All that remains is to show that $f$ or $g$ is not the identity mapping.

Suppose for the sake of contradiction that $f$ and $g$ are the identity mapping.

Then $f(\inf U)=\inf U \vee p = \inf U$.

So $p \le \inf U$.

$g(\inf U) = \inf U \wedge p = \inf U$,

so $\inf U \le p$.

By antisymmetry, $p = \inf U$.

But since $U$ is complete, $\inf U \in U$, so $p \ne \inf U$, a contradiction.

So $f$ and $g$ are not both the identity mapping. Thus there is not a unique order-preserving mapping from $U$ to $T$, contradicting the assumption that $T$ is a completion of $S$.