Up-Complete Lower Bounded Join Semilattice is Complete

Theorem
Let $\left({S, \preceq}\right)$ be an up-complete lower bounded join semillattice.

Then $\left({S, \preceq}\right)$ is complete.

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

Define
 * $Y := \left|[ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$

where ${\it Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$.

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

We will prove that
 * $Y$ is directed

Let $x, y \in Y$.

By definition of $Y$:
 * $\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: x = \sup A$

and
 * $\exists B \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \sup B$

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

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

By definition of $Y$:
 * $\sup\left({A \cup B}\right) \in Y$

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

Thus by Supremum of Subset:
 * $x \preceq \sup\left({A \cup B}\right)$ and $y \preceq \sup\left({A \cup B}\right)$

Thus by definition:
 * $Y$ is directed.

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