Continuous implies Increasing in Scott Topological Lattices

Theorem
Let $T_1 = \struct {S_1, \preceq_1, \tau_1}$ and $T_2 = \struct {S_2, \preceq_2, \tau_2}$ be up-complete topological lattices with Scott topologies.

Let $f: S_1 \to S_2$ be a continuous mapping.

Then $f$ is an increasing mapping.

Proof
Let $x, y \in S_1$ such that
 * $x \preceq_1 y$

that
 * $\map f x \npreceq_2 \map f y$

By definition of lower closure of element:
 * $\map f x \notin \paren {\map f y}^\preceq$

By definition of relative complement:
 * $\map f x \in \relcomp {S_2} {\paren {\map f y}^\preceq}$

By definition of reflexivity:
 * $\map f y \preceq_2 \map f y$

By definition of lower closure of element:
 * $\map f y \in \paren {\map f y}^{\preceq_2}$

By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
 * $\set {\map f y}^- = \paren {\map f y}^{\preceq_2}$

By definition of closure:
 * $\paren {\map f y}^{\preceq_2}$ is a closed set.

By definition of closed set:
 * $\relcomp {S_2} {\paren {\map f y}^{\preceq_2} }$ is a open set.

By definition of continuous:
 * $f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ is an open set.

By definition of Definition:Scott Topology:
 * $f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ is an upper set.

By definition of preimage of set:
 * $x \in f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$

By definition of upper set:
 * $y \in f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$

By definition of preimage of set:
 * $\map f y \in \relcomp {S_2} {\paren {\map f y}^{\preceq_2} }$

Thus this contradicts $\map f y \in \paren {\map f y}^{\preceq_2}$