Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below
Theorem
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Let $\left({T, \vee_2, \wedge_2, \precsim}\right)$ be a continuous complete lattice.
Let $f: S \to T$ be a mapping such that
- $\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
Let $x \in S, y \in T$.
Then
- $y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$
Proof
By Mapping at Element is Supremum implies Mapping is Increasing:
- $f$ is an increasing mapping.
Sufficient Condition
Let $y \ll f\left({x}\right)$
By Way Below Closure is Directed in Bounded Below Join Semilattice:
- $x^\ll$ is directed
where $x^\ll$ denotes the way below closure of $x$.
By Image of Directed Subset under Increasing Mapping is Directed:
- $f\left[{x^\ll}\right]$ is directed
where $f\left[{x^\ll}\right]$ denotes the image of $x^\ll$ under $f$.
By assumption:
- $f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
By definitions of image of set and way below closure:
- $f\left[{x^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
- $\exists z \in f\left[{x^\ll}\right]: y \ll z$
By definition of image of set:
- $\exists w \in x^\ll: f\left({w}\right) = z$
Thus by definition of way below closure:
- $w \ll x$
Thus $y \ll f\left({w}\right)$
$\Box$
Necessary Condition
Let
- $\exists w \in S: w \ll x \land y \ll f\left({w}\right)$
By Way Below implies Preceding:
- $w \preceq x$
By definition of increasing mapping:
- $f\left({w}\right) \precsim f\left({x}\right)$
Thus by Preceding and Way Below implies Way Below:
- $y \ll f\left({x}\right)$
$\blacksquare$
Sources
- Mizar article WAYBEL17:14