# Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice

## Theorem

Let $\struct {S, \preccurlyeq}$ be an ordered set such that:

$\struct {S, \preccurlyeq}$ has a greatest element $u$
every non-empty subset of $S$ admits an infimum.

Then $\struct {S, \preccurlyeq}$ is a complete lattice.

## Proof

For $\struct {S, \preccurlyeq}$ to be a complete lattice, it has to be such that:

every non-empty subset of $S$ admits both a supremum and an infimum.

Let $T \subseteq S$ be an arbitrary subset of $S$.

We already have by hypothesis that $T$ admits an infimum.

We are to show that $T$ admits a supremum.

By definition of greatest element:

$\forall x \in T: x \preccurlyeq u$

Thus by definition $u$ is an upper bound of $T$.

Let $H \subseteq S$ be the set of all upper bounds of $T$.

Because $u$ is an upper bound of $T$, $u \in H$.

Hence $H$ is not empty.

We have by hypothesis that $H$ has an infimum.

That is:

$\exists v \in H: \forall x \in H: v \preccurlyeq x$

Hence $v$ is by definition a supremum of $T$.

Thus $T$ admits a supremum.

Hence a priori $T$ admits both a supremum and an infimum.

Hence the result.

$\blacksquare$