Lattice is Complete iff it Admits All Suprema

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Then $\left({S, \preceq}\right)$ is a complete lattice
 * $\forall X \subseteq S: X$ admits a supremum.

Sufficient Condition
Assume that $\left({S, \preceq}\right)$ is a complete lattice.

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

Necessary Condition
Assume 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
 * $Y := \left\{ {s \in S: s}\right.$ id lower bound for $\left.{X}\right\}$

By assumption
 * $Y$ admits a supremum

We will prove that
 * $\sup Y$ is lower bound for $X$

Let $x \in X$.

By definition of lower bound:
 * $x$ is upper bound for $Y$

By definition of supremum:
 * $\sup Y \preceq x$

thus by definition
 * $\sup Y$ is lower bound for $X$

We will prove that
 * $\forall s \in S: s$ is lower bound for $X \implies s \preceq \sup Y$

Let $s \in S$ such that
 * $s$ is lower bound for $X$.

By definition of $Y$:
 * $s \in Y$

By definition of supremum:
 * $\sup Y$ is upper bound for $Y$.

Thus by definition of upper bound:
 * $s \preceq \sup Y$

Thus by definition
 * $X$ admits an infimum

By definition
 * $\left({S, \preceq}\right)$ is a lattice.

Thus the result follows by definition of complete lattice.