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

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

Thus by definition of complete lattice:
 * $\forall X \subseteq S: X$ admits a supremum.

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.