# 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:

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.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(a)}$