Up-Complete Lower Bounded Join Semilattice is Complete

Theorem
Let $\struct {S, \preceq}$ be an up-complete lower bounded join semillattice.

Then $\struct {S, \preceq}$ is complete.

Proof
Let $X$ be a subset of $S$.

In the case when $X = \O$:

by definition of lower bounded:
 * $\exists L \in S: L$ is lower bound for $S$.

By definition of empty set:
 * $L$ is upper bound for $X$.

By definition of lower bound:
 * $\forall x \in S: x$ is upper bound for $X \implies L \preceq x$

Thus by definition
 * $L$ is a supremum of $X$.

Thus:
 * $X$ admint a supremum.

In the case when $X \ne \O$:

Define
 * $Y := \set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$

where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $X$.

By Existence of Non-Empty Finite Suprema in Join Semilattice
 * all suprema in $Y$ exist,

By definition of non-empty set:
 * $Y$ is a non-empty set.

We will prove that
 * $Y$ is directed

Let $x, y \in Y$.

By definition of $Y$:
 * $\exists A \in \map {\operatorname {Fin} } X \setminus \set \O: x = \sup A$

and
 * $\exists B \in \map {\operatorname {Fin} } X \setminus \set \O: y = \sup B$

By Finite Union of Finite Sets is Finite:
 * $A \cup B$ is finite
 * $A \cup B \ne \O$

By Union is Smallest Superset:
 * $A \cup B \subseteq X$

By definition of $Y$:
 * $\map \sup {A \cup B} \in Y$

By Set is Subset of Union:
 * $A \subseteq A \cup B$ and $B \subseteq A \cup B$

Thus by Supremum of Subset:
 * $x \preceq \map \sup {A \cup B}$ and $y \preceq \map \sup {A \cup B}$

Thus by definition:
 * $Y$ is directed.

By definition up-complete:
 * $Y$ admits a supremum

By definition of supremum
 * $\sup Y$ is upper bound for $Y$

We will prove that
 * $X \subseteq Y$

Let $x \in X$.

By definitions of subset and singleton:
 * $\set x \subseteq X$
 * $\set x$ is finite
 * $\set x \ne \O$

By definition of $Y$:
 * $\sup {\set x} \in Y$

Thus by Supremum of Singleton:
 * $x \in Y$

By Upper Bound is Upper Bound for Subset:
 * $\sup Y$ is upper bound for $X$

We will prove that
 * $\forall x \in S: x$ is upper bound for $X \implies \sup Y \preceq x$

Let $x \in S$ such that
 * $x$ is upper bound for $X$

We will prove as sublemma that
 * $x$ is upper bound for $Y$

Let $y \in Y$.

By definition of $Y$:
 * $\exists A \in \map {\operatorname {Fin} } X \setminus \set \O: y = \sup A$

By definition of $\operatorname {Fin}$:
 * $A \subseteq X$

By Upper Bound is Upper Bound for Subset
 * $x$ is upper bound for $A$

Thus by definition of supremum:
 * $y \preceq x$

Thus by definition
 * $x$ is upper bound for $Y$

This ends the proof of sublemma.

Thus by definition of supremum:
 * $\sup Y \preceq x$

This ends the proof of lemma.

By definition
 * $\sup Y$ is a supremum of $X$

and thus:
 * $X$ admits a supremum.

Thus result follows by Lattice is Complete iff it Admits All Suprema.