Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Supremum of Image is Mapping at Supremum of Directed Subset
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Theorem
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be up-complete lattices.
Let $f: S \to T$ be a mapping such that
- for all directed set $\struct {D, \precsim}$ and net $N: D \to S$ in $S: \map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Let $D$ be a directed subset of $S$.
Then:
- $\map \sup {f \sqbrk D} = \map f {\sup D}$
where $f \sqbrk D$ denotes the image of $D$ under $f$.
Proof
- $f$ is an increasing mapping.
By Image of Directed Subset under Increasing Mapping is Directed:
- $f \sqbrk D$ is directed.
By definition of up-complete:
- $D$ and $f \sqbrk D$ admit suprema.
- $\map \sup {f \sqbrk D} \preceq_2 \map f {\sup D}$
By Limit Inferior of Inclusion Net is Supremum of Directed Subset:
- $\sup D = \liminf i_D$
By assumption:
- $\map f {\sup D} \preceq_2 \map \liminf {f \circ i_D}$
By Composition of Mapping and Inclusion is Restriction of Mapping:
- $f \circ i_D = f {\restriction} D$
By Limit Inferior of Restriction Net is Supremum of Image of Directed Subset:
- $\map \sup {f \sqbrk D} = \map \liminf {f \circ i_D}$
Then
- $\map f {\sup D} \preceq_2 \map \sup {f \sqbrk D}$
Thus by definition of antisymmetry:
- $\map \sup {f \sqbrk D} = \map f {\sup D}$
$\blacksquare$
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 WAYBEL17:19