Limit Inferior of Inclusion Moore-Smith Sequence is Supremum of Directed Subset

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be an up-complete lattice.

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

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

Let $i_D: D \to S$, the inclusion mapping, be a Moore-Smith sequence in $S$.

Then $\liminf i_D = \sup D$

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

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:
 * $j^{\succeq'} = D \cap j^\succeq$

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

By Infimum of Subset and Infimum of Upper Closure of Element:
 * $j \preceq \map {\inf_L} {\map {\preceq'} j}$

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

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

By definitions of infimum and lower bound:
 * $\map {\inf_L} {\map {\preceq'} j} \preceq j$

Thus by definition of antisymmetry:
 * $\map {\inf_L} {\map {\preceq'} j} = j$

Thus