# Lattice is Complete iff it Admits All Suprema

## Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Then $\struct {S, \preceq}$ is a complete lattice if and only if

$\forall X \subseteq S: X$ admits a supremum.

## Proof

### Sufficient Condition

Let $\struct {S, \preceq}$ be a complete lattice.

Thus by definition of complete lattice:

$\forall X \subseteq S: X$ admits a supremum.

$\Box$

### Necessary Condition

Let $\struct {S, \preceq}$ be such that

$\forall X \subseteq S: X$ admits a supremum.

We will prove that

$\forall X \subseteq S: X$ admits both a supremum and an infimum.

Let $X \subseteq S$.

Thus by assumption:

$X$ admits a supremum.

Define:

$Y := \leftset {s \in S: s}$ is a lower bound for $\rightset X$

By assumption:

$Y$ admits a supremum

We will prove that

$\sup Y$ is a lower bound for $X$

Let $x \in X$.

By definition of lower bound:

$x$ is an upper bound for $Y$

By definition of supremum:

$\sup Y \preceq x$

Thus by definition:

$\sup Y$ is a lower bound for $X$

We will prove that

$\forall s \in S: s$ is a lower bound for $X \implies s \preceq \sup Y$

Let $s \in S$ such that:

$s$ is a lower bound for $X$.

By definition of $Y$:

$s \in Y$

By definition of supremum:

$\sup Y$ is an upper bound for $Y$.

Thus by definition of upper bound:

$s \preceq \sup Y$

Thus by definition

$X$ admits an infimum

By definition

$\struct {S, \preceq}$ is a lattice.

Thus the result follows by definition of complete lattice.

$\blacksquare$