Element equals to Supremum of Infima of Open Sets that Element Belongs implies Topological Lattice is Continuous

Theorem

Let $L = \left({S, \preceq, \tau}\right)$ be a complete Scott topological lattice.

Let

$\forall x \in S: x = \sup \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$

Then $L$ is continuous.

$\forall x \in S:x^\ll$ is directed.

Thus by definition pf complete lattice:

Let $x \in S$.

Define $W := \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$

By definition of way below closure:

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

By definition of supremum:

$\sup \left({x^\ll}\right) \preceq x$

We will prove that

$W \subseteq x^\ll$

Let $d \in W$.

By definition of $W$:

$\exists V: d = \inf V \land x \in V \land V \in \sigma\left({L}\right)$

$\tau = \sigma\left({L}\right)$

By definition:

$V$ is open.

$d \ll x$

Thus by definition of way below closure:

$d \in x^\ll$

$\Box$

$\sup W \preceq \sup\left({x^\ll}\right)$

By assumption:

$x \preceq \sup\left({x^\ll}\right)$

Hence by definition of antisymmetry:

$x = \sup \left({x^\ll}\right)$

$\blacksquare$