Mapping Assigning to Element Its Compact Closure is Order Isomorphism

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

Let $C = \left({K\left({L}\right), \preceq'}\right)$ be an ordered subset of $L$

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

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

where $\mathit{Ids}\left({C}\right)$ denotes the set of all ideals in $C$.

Let $f: S \to \mathit{Ids}\left({C}\right)$ br a mapping such that
 * $\forall x \in S: f\left({x}\right) = x^{\mathrm{compact} }$

Thus $f$ is order isomorphism between $L$ and $I$.