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

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

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

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

Let $\left({D, \preceq'}\right)$ be a directed ordered subset of $L$.

Let $f \restriction D: D \to T$, the restriction of mapping, be a Moore-Smith sequence in $T$.

Then $\liminf \left({f \restriction D}\right) = \sup \left({f\left[{D}\right]}\right)$

Proof
We will prove that
 * (lemma): $\forall j \in D: \inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right) = f\left({j}\right)$

Let $j \in D$.

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

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 Relation is Subset of Image/Corollary 2:
 * $\left({f \restriction D}\right)\left[{j^{\succeq'} }\right] \subseteq \left({f \restriction D}\right)\left[{j^{\succeq_1} }\right]$

By Infimum of Subset and Infimum of Image of Upper Closure of Element under Increasing Mapping:
 * $f\left({j}\right) \preceq_2 \inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right)$

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

By definition of image of element:
 * $j \in \preceq'\left({j}\right)$

By definition of image of set:
 * $f\left({j}\right) \in \left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]$

By definitions of infimum and lower bound:
 * $\inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right) \preceq_2 f\left({j}\right)$

Thus by definition of antisymmetry:
 * $\inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right) = f\left({j}\right)$

Thus