Continuous Lattice and Way Below implies Preceding implies Preceding

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a continuous complete lattice.

Let $a, b \in S$.

Let
 * $\forall c \in S: c \ll a \implies c \preceq b$

Then $a \preceq b$

Proof
By definition of way below closure:
 * $\forall c \in a^\ll: c \preceq b$

By definition:
 * $b$ is an upper bound for $a^\ll$

By definition of supremum:
 * $\sup \left({a^\ll}\right) \preceq b$

By definition of continuous:
 * $L$ satisfies axiom of approximation.

Thus by axiom of approximation:
 * $a = \sup \left({a^\ll}\right) \preceq b$