Limit Inferior of Restriction Net is Supremum of Image of Directed Subset

Theorem
Let $L = \struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be up-complete lattices.

Let $f: S \to T$ be an increasing mapping.

Let $D \subseteq S$ be a directed subset of $S$.

Let $\struct {D, \preceq'}$ be a directed ordered subset of $L$.

Let $f \restriction D: D \to T$, the restriction of mapping, be a net in $T$.

Then $\map \liminf {f \restriction D} = \map \sup {f \sqbrk D}$

Proof
We will prove that
 * (lemma): $\forall j \in D: \map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } = \map f j$

Let $j \in D$.

By definitions of image of element and upper closure of element:
 * $\map {\preceq'} j = j^{\succeq'}$

By Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure:
 * $j^{\succeq'} = D \cap j^{\succeq_1}$

By Intersection is Subset:
 * $j^{\succeq'} \subseteq j^{\succeq_1}$

By Image of Subset under Mapping is Subset of Image:
 * $\paren {f \restriction D} \sqbrk {j^{\succeq'} } \subseteq \paren {f \restriction D} \sqbrk {j^{\succeq_1} }$

By Infimum of Subset and Infimum of Image of Upper Closure of Element under Increasing Mapping:
 * $\map f j \preceq_2 \map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} }$

By definition of reflexivity:
 * $j \preceq' j$

By definition of image of element:
 * $j \in \map {\preceq'} j$

By definition of image of set:
 * $\map f j \in \paren {f \restriction D} \sqbrk {\map {\preceq'} j}$

By definitions of infimum and lower bound:
 * $\map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } \preceq_2 \map f j$

Thus by definition of antisymmetry:
 * $\map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } = \map f j$

Thus