Continuous implies Increasing in Scott Topological Lattices

Theorem
Let $T_1 = \left({S_1, \preceq_1, \tau_1}\right)$ and $T_2 = \left({S_2, \preceq_2, \tau_2}\right)$ 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
 * $x \preceq_1 y$

Aiming for a contradiction suppose that
 * $f\left({x}\right) \npreceq_2 f\left({y}\right)$

By definition of lower closure of element:
 * $f\left({x}\right) \notin \left({f\left({y}\right)}\right)^\preceq$

By definition of relative complement:
 * $f\left({x}\right) \in \complement_{S_2}\left({\left({f\left({y}\right)}\right)^\preceq}\right)$

By definition of reflexivity:
 * $f\left({y}\right) \preceq_2 f\left({y}\right)$

By definition of lower closure of element:
 * $f\left({y}\right) \in \left({f\left({y}\right)}\right)^{\preceq_2}$

By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
 * $\left\{ {f\left({y}\right)}\right\}^- = \left({f\left({y}\right)}\right)^{\preceq_2}$

By definition of closure:
 * $\left({f\left({y}\right)}\right)^{\preceq_2}$ is a closed set.

By definition of closed set:
 * $\complement_{S_2}\left({\left({f\left({y}\right)}\right)^{\preceq_2} }\right)$ is a open set.

By definition of continuous:
 * $f^{-1}\left[{\complement_{S_2}\left({\left({f\left({y}\right)}\right)^{\preceq_2} }\right)}\right]$ is an open set.

By definition of Definition:Scott Topology:
 * $f^{-1}\left[{\complement_{S_2}\left({\left({f\left({y}\right)}\right)^{\preceq_2} }\right)}\right]$ is an upper set.

By definition of preimage of set:
 * $x \in f^{-1}\left[{\complement_{S_2}\left({\left({f\left({y}\right)}\right)^{\preceq_2} }\right)}\right]$

By definition of upper set:
 * $y \in f^{-1}\left[{\complement_{S_2}\left({\left({f\left({y}\right)}\right)^{\preceq_2} }\right)}\right]$

By definition of preimage of set:
 * $f\left({y}\right) \in \complement_{S_2}\left({\left({f\left({y}\right)}\right)^{\preceq_2} }\right)$

Thus this contradicts $f\left({y}\right) \in \left({f\left({y}\right)}\right)^{\preceq_2}$