Finite Suprema Set and Lower Closure is Smallest Ideal

Theorem
Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.

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

Then $X \subseteq \map {\operatorname {finsups} } X^\preceq$ and
 * for every ideal $I$ in $L$: $X \subseteq I \implies \map {\operatorname {finsups} } X^\preceq \subseteq I$

where
 * $\map {\operatorname {finsups} } X$ denotes the finite suprema set of $X$,
 * $X^\preceq$ denotes the lower closure of $X$.

Proof
By Set is Subset of Finite Suprema Set:
 * $X \subseteq \map {\operatorname {finsups} } X$

By Lower Closure of Subset is Subset of Lower Closure:
 * $X^\preceq \subseteq \map {\operatorname {finsups} } X^\preceq$

By Set is Subset of Lower Closure:
 * $X \subseteq X^\preceq$

Thus by Subset Relation is Transitive:
 * $X \subseteq \map {\operatorname {finsups} } X^\preceq$

Let $I$ be a ideal in $L$ such that
 * $X \subseteq I$

Let $x \in \map {\operatorname {finsups} } X^\preceq$

By definition of lower closure of subset:
 * $\exists y \in \map {\operatorname {finsups} } X: x \preceq y$

By definition of finite suprema set:
 * $\exists A \in \map {\operatorname {Fin} } X: y = \sup A \land A$ admits a supremum

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

By Subset Relation is Transitive:
 * $A \subseteq I$

By Directed in Join Semilattice with Finite Suprema:
 * $A \ne \O \implies \sup A \in I$

By Supremum of Empty Set is Smallest Element and Bottom in Ideal:
 * $A = \O \implies \sup A = \bot \in I$

where $\bot$ denotes the smallest element of $S$.

So
 * $y \in I$

Thus by definition of lower set:
 * $x \in I$