Supremum of Ideals is Upper Adjoint implies Lattice is Continuous

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below up-complete lattice.

Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.

Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$

Let $f: \map {\mathit {Ids} } L \to S$ be a mapping such that
 * $\forall I \in \map {\mathit {Ids} } L: f \sqbrk I = \sup I$

Let $f$ be an upper adjoint of a Galois connection.

Then $L$ is continuous.

Proof
We will prove that
 * $\forall x \in S: \exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$

Let $x \in S$.

Define $I := \map \inf {f^{-1} \sqbrk {x^\succeq} }$.

By definition of $P$:
 * $I \in \map {\mathit {Ids} } L$

We will prove that
 * $\forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$

Let $J \in \map {\mathit {Ids} } L$ such that
 * $x \preceq \sup J$

By definition of $f$:
 * $x \preceq f \sqbrk J$

By definition of upper closure of element:
 * $f \sqbrk J \in x^\succeq$

By definition of image of set:
 * $J \in f^{-1} \sqbrk {x^\succeq}$

By definition of infimum:
 * $I \precsim J$

Hence by definition of $\precsim$:
 * $I \subseteq J$

By definition of upper adjoint of a Galois connection:
 * there exists a mapping $d: S \to \map {\mathit {Ids} } L$: $\struct {f, d}$ is a Galois connection.

By Galois Connection is Expressed by Minimum
 * $\map d x = \map \min {f^{-1} \sqbrk {x^\succeq} }$

By definition of smallest element:
 * $I \in f^{-1} \sqbrk {x^\succeq}$

By definition of image of set:
 * $f \sqbrk I \in x^\succeq$

By definition of upper closure of element:
 * $x \preceq f \sqbrk I$

Thus by definition of $f$:
 * $x \preceq \sup I$

Hence
 * $\exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$

Hence by Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:
 * $L$ is continuous.