Compact Element iff Principal Ideal

Theorem
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.

Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an inclusion ordered set

where
 * $\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$,
 * $\mathord\precsim = \mathord\subseteq \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Let $x \in \mathit{Ids}\left({L}\right)$

Then $x$ is compact element $x$ is principal ideal in $L$

Proof
By Ideals are Continuous Lattice Subframe of Power Set:
 * $P$ is continuous lattice subframe of $\left({\mathcal P\left({S}\right), \subseteq'}\right)$

where
 * $\mathcal P\left({S}\right)$ denotes the power set of $S$,
 * $\mathord\subseteq' = \mathord\subseteq \cap \left({\mathcal P\left({S}\right) \times \mathcal P\left({S}\right)}\right)$

Sufficient Condition
Assume that
 * $x$ is compact element.

By Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset:
 * $\exists F \in \mathit{Fin}\left({S}\right): x = \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\} \land F \subseteq x$

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

We will prove that
 * $\exists y \in x: y$ is upper bound for $x$.

Define $y = \sup_L F$.

By Directed in Join Semilattice with Finite Suprema:
 * $F \ne \varnothing \implies y \in x$

By Supremum of Empty Set is Smallest Element:
 * $F = \varnothing \implies y = \bot_L$

where $\bot_L$ is the smallest element in $L$.

By Bottom in Ideal:
 * $F = \varnothing \implies y \in x$

Thus $y \in x$

Let $z \in x$.

We will prove that
 * $F \subseteq y^\preceq$

Let $u \in F$.

By definitions of supremum and upper bound:
 * $u \preceq y$

Thus by definition of lower closure of element:
 * $u \in y^\preceq$

By Lower Closure of Element is Ideal:
 * $y^\preceq$ is ideal in $L$.

Then
 * $y^\preceq \in \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$

By definition of intersection:
 * $z \in y^\preceq$

Thus by definition of lower closure of element:
 * $z \preceq y$

Hence $x$ is principal ideal.

Necessary Condition
Assume that $x$ is principal ideal.

By definition of principal ideal:
 * $\exists y \in x: y$ is upper bound for $x$.

We will prove that
 * $\exists F \in \mathit{Fin}\left({S}\right): F \subseteq x \land x = \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$

Define $F = \left\{ {y}\right\}$

By Singleton is Finite:
 * $F$ is finite.

Thus by definition of $\mathit{Fin}$:
 * $F \in \mathit{Fin}\left({S}\right)$

Thus by definitions of subset and singleton:
 * $F \subseteq x$

We will prove that
 * $\forall z: z \in x \iff z \in \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$

Thus by definition of set equality:
 * $x = \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$

Thus by Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset:
 * $x$ is compact element.