Continuous iff Mapping at Element is Supremum of Compact Elements

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Theorem

Let $L = \struct {S, \preceq_1, \tau_1}$ and $R = \struct {T, \preceq_2, \tau_2}$ be complete algebraic topological lattices with Scott topologies.

Let $f: S \to T$ be a mapping.


Then $f$ is continuous if and only if:

$\forall x \in S: \map f x = \sup \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset {}$


Proof

By Algebraic iff Continuous and For Every Way Below Exists Compact Between:

$L$ and $R$ are continuous.


Sufficient Condition

Assume that:

$f$ is continuous.

By Continuous iff Mapping at Element is Supremum:

$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$

Let $x \in S$.

By definitions of image of set and compact closure:

$\leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset {} = f \sqbrk {x^{\mathrm{compact} } }$

By Compact Closure is Directed:

$D := x^{\mathrm{compact} }$ is directed.

By Continuous iff Directed Suprema Preserving:

$f$ preserves directed suprema.

By definition of mapping preserves directed suprema:

$f$ preserves the supremum of $D$.

By definition of complete lattice:

$D$ admits a supremum.

By definition of algebraic:

$L$ satisfies the axiom of $K$-approximation.

Thus

\(\ds \map f x\) \(=\) \(\ds \map f {\sup D}\) Axiom of $K$-Approximation
\(\ds \) \(=\) \(\ds \map \sup {f \sqbrk D}\) Definition of Mapping Preserves Supremum of Subset
\(\ds \) \(=\) \(\ds \sup \set {\map f w: w \in S \land w \preceq_1 x \land w \text { is compact} }\)

$\Box$


Necessary Condition

Assume that

$\forall x \in S: \map f x = \sup \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$

By Mapping at Element is Supremum of Compact Elements implies Mapping at Element is Supremum that Way Below:

$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$

Thus by Continuous iff Mapping at Element is Supremum:

$f$ is continuous.

$\blacksquare$


Sources

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