Supremum of Ideals is Upper Adjoint implies Lattice is Continuous

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

Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.

Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$

Let $f: \mathit{Ids}\left({L}\right) \to S$ be a mapping such that
 * $\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$

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

Then $L$ is continuous.

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

Let $x \in S$.

Define $I := \inf \left({f^{-1}\left[{x^\succeq}\right]}\right)$.

By definition of $P$:
 * $I \in \mathit{Ids}\left({L}\right)$

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

Let $J \in \mathit{Ids}\left({L}\right)$ such that
 * $x \preceq \sup J$

By definition of $f$:
 * $x \preceq f\left({J}\right)$

By definition of upper closure of element:
 * $f\left({J}\right) \in x^\succeq$

By definition of image of set:
 * $J \in f^{-1}\left[{x^\succeq}\right]$

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 $d: S \to \mathit{Ids}\left({L}\right)$: $\left({f, d}\right)$ is a Galois connection.

By Galois Connection is Expressed by Minimum
 * $d\left({x}\right) = \min \left({f^{-1} \left[{x^\succeq}\right]}\right)$

By definition of smallest element:
 * $I \in f^{-1} \left[{x^\succeq}\right]$

By definition of image of set:
 * $f\left({I}\right) \in x^\succeq$

By definition of upper closure of element:
 * $x \preceq f\left({I}\right)$

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

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

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