Continuous iff Way Below iff There Exists Element that Way Below and Way Below
Theorem
Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be complete continuous topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous if and only if
- $\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
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Proof
Sufficient Condition
Assume that
- $f$ is continuous.
By Continuous iff Directed Suprema Preserving:
By Directed Suprema Preserving Mapping at Element is Supremum:
- $\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$
- $\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
$\Box$
Necessary Condition
Assume that
- $\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
Let $U$ be an open subset of $T$.
By definition of Scott topology:
- $U$ is upper.
By definition of interior:
- $\paren {f^{-1} \sqbrk U}^\circ \subseteq f^{-1} \sqbrk U$
We will prove that
- $f^{-1} \sqbrk U \subseteq \paren {f^{-1} \sqbrk U}^\circ$
Let $x \in f^{-1} \sqbrk U$
By definition of preimage of set:
- $p := \map f x \in U$
By Open implies There Exists Way Below Element:
- $\exists u \in T: u \ll p \land u \in U$
By assumption:
- $\exists w \in S: w \ll x \land u \ll \map f w$
We will prove that
- $f \sqbrk {w^\gg} \subseteq U$
Let $h \in f \sqbrk {w^\gg}$
By definition of image of set:
- $\exists z \in w^\gg: h = \map f z$
By definition of way above closure:
- $w \ll z$
By assumption:
- $u \ll h$
By Way Below implies Preceding:
- $u \preceq_2 h$
Thus by definition of upper section:
- $h \in U$
$\Box$
By Preimage of Subset is Subset of Preimage:
- $f^{-1} \sqbrk {f \sqbrk {w^\gg} } \subseteq f^{-1} \sqbrk U$
By Preimage of Image under Left-Total Relation is Superset:
- $w^\gg \subseteq \paren {f^{-1} \circ f} \sqbrk {w^\gg}$
By definition of composition of relations:
- $w^\gg \subseteq f^{-1} \sqbrk {f \sqbrk {w^\gg} }$
By Subset Relation is Transitive:
- $w^\gg \subseteq f^{-1} \sqbrk U$
By Interior is Union of Way Above Closures:
- $\paren {f^{-1} \sqbrk U}^\circ = \bigcup \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$
By definition of way above closure:
- $x \in w^\gg$
We have:
- $w^\gg \in \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$
Thus by definition of union:
- $x \in \paren {f^{-1} \sqbrk U}^\circ$
$\Box$
By definition of set equality:
- $\paren {f^{-1} \sqbrk U}^\circ = f^{-1} \sqbrk U$
Thus by definition of interior:
- $f^{-1} \sqbrk U$ is open.
Hence $f$ is continuous.
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL17:23