Operand is Upper Bound of Way Below Closure

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

Let $x \in S$.

Then
 * $x$ is upper bound for $x^\ll$

where $x^\ll$ denotes the way below closure of $x$.

Proof
Let $y \in x^\ll$

By definition of way below closure:
 * $y \ll x$

where $\ll$ denotes the way below relation.

Thus by Way Below implies Preceding:
 * $y \preceq x$

Thus by definition:
 * $x$ is upper bound for $x^\ll$