Continuous implies Increasing in Scott Topological Lattices

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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

$x \preceq_1 y$

Aiming for a contradiction, suppose 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 $\map f y \in \paren {\map f y}^{\preceq_2}$

$\blacksquare$


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