Supremum of Ideals is Upper Adjoint

Theorem

Let $L = \struct {S, \vee, \preceq}$ be a bounded below continuous join semilattice.

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

$\forall I \in \map {\mathit {Ids} } L: \map f I = \sup I$

Then $f$ is an upper adjoint of Galois connection.

Proof

Define $d: S \to \map {\mathit {Ids} } L$

$\forall t \in S: \map d t = \map \inf {f^{-1} \sqbrk {t^\succeq} }$

where

$t^\succeq$ denotes the upper closure of $t$,

$f^{-1} \sqbrk {t^\succeq} $ denotes the image of $t^\succeq$ over $f^{-1}$.

We will prove that

$\forall t \in S: \map d t = \map \min {f^{-1} \sqbrk {t^\succeq} }$

Let $t \in S$.

By Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:

there exists an ideal $J$ in $L$ such that

$t \preceq \sup J$ and for every ideal $K$ in $L$: $t \preceq \sup K \implies J \subseteq K$

We will prove that

$\forall K \in \map {\mathit {Ids} } L: K$ is lower bound for $f^{-1} \sqbrk {t^\succeq} \implies K \precsim J$

Let $K \in \map {\mathit {Ids} } L$ such that

$K$ is a lower bound for $f^{-1} \sqbrk {t^\succeq}$

By definition of $f$:

$t \preceq \map f J$

By definition of upper closure of element:

$\map f J \in t^\succeq$

By definition of image of set:

$J \in f^{-1} \sqbrk {t^\succeq}$

Thus by definition of lower bound:

$K \precsim J$

$\Box$

We will prove that:

$J$ is a lower bound for $f^{-1} \sqbrk {t^\succeq}$

Let $K \in \map {\mathit {Ids} } L$ such that:

$K \in f^{-1} \sqbrk {t^\succeq}$

By definition of image of set:

$\map f K \in t^\succeq$

By definition of upper closure of element:

$t \preceq \map f K$

By definition of $f$:

$t \preceq \sup K$

Then

$J \subseteq K$

Thus by definition of $\precsim$:

$J \precsim K$

$\Box$

By definition of supremum:

$t \preceq \map \sup {\map \inf {f^{-1} \sqbrk {t^\succeq} } }$

By definition of $f$:

$t \preceq \map f {\map \inf {f^{-1} \sqbrk {t^\succeq} } }$

By definition of upper closure of element:

$\map f {\map \inf {f^{-1} \sqbrk {t^\succeq} } } \in t^\succeq$

By definition of image of set:

$\map \inf {f^{-1} \sqbrk {t^\succeq} } \in f^{-1} \sqbrk {t^\succeq}$

Thus by definition of smallest element:

$\map d t = \map \min {f^{-1} \sqbrk {t^\succeq} }$

$\Box$

By Supremum of Ideals is Increasing:

$f$ is an increasing mapping.

By Galois Connection is Expressed by Minimum:

$\struct {f, d}$ is a Galois connection.

Hence $f$ is an upper adjoint of Galois connection.

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL_5:3