Mapping Assigning to Element Its Lower Closure is Isomorphism

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

Let $I = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ 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 $P = \left({K\left({I}\right), \precsim'}\right)$ be an ordered subset of $I$

where
 * $K\left({I}\right)$ denotes the compact subset of $I$.

Let $f:S \to K\left({I}\right)$ be a mapping such that
 * $\forall x \in S: f\left({x}\right) = x^\preceq$

Then $f$ is an order isomorphism between $L$ and $P$.

Proof
By definition:
 * $\forall x \in S: x^\preceq$ is a principal ideal.

By Compact Element iff Principal Ideal:
 * $\forall x \in S: x^\preceq$ is a compact element in $I$.

By definition of compact subset:
 * $\forall x \in S: x^\preceq \in K\left({I}\right)$

Then $f$ is well-defined.

We will prove that
 * $f$ is an order embedding

That means
 * $\forall x, y \in S: x \preceq y \iff f\left({x}\right) \precsim' f\left({y}\right)$

Let $x, y \in S$.

We will prove as lemma that
 * $x^\preceq \subseteq y^\preceq \implies x \preceq y$

Assume that
 * $x^\preceq \subseteq y^\preceq$

By definition of reflexivity:
 * $x \preceq x$

By definition of lower closure of element:
 * $x \in x^\preceq$

By definition of subset:
 * $x \in y^\preceq$

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


 * $x \preceq y$


 * $x^\preceq \subseteq y^\preceq$ by lemma and Lower Closure is Increasing


 * $f\left({x}\right) \subseteq f\left({y}\right)$ by definition of $f$


 * $f\left({x}\right) \precsim f\left({y}\right)$ by definition of $\precsim$


 * $f\left({x}\right) \precsim' f\left({y}\right)$ by definition of ordered subset.

We will prove that
 * $\forall a \in K\left({I}\right): \exists y \in S:a = f\left({y}\right)$

Let $a \in K\left({I}\right)$

By definition of compact subset:
 * $a$ is a compact element in $I$.

by Compact Element iff Principal Ideal:
 * $a$ is a principal ideal.

By definition of principal ideal:
 * $\exists y \in S: a = y^\preceq$

Thus by definition of $f$:
 * $a = f\left({y}\right)$

Then $f$ is a surjection.

Hence $f$ is an order isomorphism.