Directed Suprema Preserving Mapping at Element is Supremum
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Theorem
Let $\struct {S, \vee, \wedge, \preceq}$ and $\struct {T, \vee_2, \wedge_2, \precsim}$ be bounded below continuous lattices.
Let $f: S \to T$ be a mapping such that
Let $x \in S$.
Then:
- $\map f x = \sup \set {\map f w: w \in S \land w \ll x}$
Proof
By definition of continuous:
- $x^\ll$ is directed
and
- $\struct {S, \vee, \wedge, \preceq}$ is up-complete
and
- $\struct {S, \vee, \wedge, \preceq}$ satisfies the Axiom of Approximation.
By definition of mapping preserves directed suprema:
- $f$ preserves the supremum of $x^\ll$.
By definition of up-complete:
- $x^\ll$ admits a supremum.
Thus
\(\ds \map f x\) | \(=\) | \(\ds \map f {\map \sup {x^\ll} }\) | Axiom of Approximation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \sup {f \sqbrk {x^\ll} }\) | Definition of Mapping Preserves Supremum: Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \sup {f \sqbrk {\set {w \in S: w \ll x} } }\) | Definition of Way Below Closure | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup \set {\map f w: w \in S \land w \ll x}\) | Definition of Image of Subset under Mapping |
$\blacksquare$
Sources
- Mizar article WAYBEL17:12