Mapping Assigning to Element Its Compact Closure is Order Isomorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $C = \struct {\map K L, \preceq'}$ be an ordered subset of $L$

where $\map K L$ denotes the compact subset of $L$.

Let $I = \struct {\map {\mathit {Ids} } C, \precsim}$ be an inclusion ordered set

where $\map {\mathit {Ids} } C$ denotes the set of all ideals in $C$.

Let $f: S \to \map {\mathit {Ids} } C$ be a mapping such that

$\forall x \in S: \map f x = x^{\mathrm {compact} }$

where $x^{\mathrm {compact} }$ denotes the compact closure of $x$.


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


Proof

We will prove that

$f$ is an order embedding.

Let $x, y \in S$.


Sufficient condition

Assume that

$x \preceq y$

By Compact Closure is Increasing:

$x^{\mathrm {compact} } \subseteq y^{\mathrm {compact} }$

By definition of $f$:

$\map f x \subseteq \map f y$

Thus by definition of inclusion ordered set:

$\map f x \precsim \map f y$

$\Box$


Necessary condition

Assume that:

$\map f x \precsim \map f y$

By definition of inclusion ordered set:

$\map f x \subseteq \map f y$

By definition of $f$:

$x^{\mathrm {compact} } \subseteq y^{\mathrm{compact} }$

By Supremum of Subset:

$\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {y^{\mathrm {compact} } }$

By definition of algebraic:

$L$ satisfies the axiom of $K$-approximation.

Thus by the axiom of $K$-approximation:

$x \preceq y$

$\Box$


We will prove that

$f$ is a surjection.

Let $y \in \map {\mathit {Ids} } C$

Define $x = \sup_L y$

Thus $x \in S$.

We will prove that:

$x^{\mathrm {compact} } \subseteq y$

Let $d \in x^{\mathrm {compact} }$

By definition of compact closure:

$d$ is a compact element and $d \preceq x$

By definition of compact subset:

$d \in \map K L$

By definition of compact element:

$d \ll d$

where $\ll$ denotes the way below relation.

By definition of way below relation:

$\exists z \in y: d \preceq z$

By definition of ordered subset:

$d \preceq' z$

Thus by definition of lower section:

$d \in y$

$\Box$


We will prove that

$y \subseteq x^{\mathrm {compact} }$

Let $d \in y$.

By definition of subset:

$d \in \map K L$

By definition of compact subset:

$d$ is a compact element.

By definition of supremum:

$x$ is upper bound for $y$.

By definition of upper bound:

$d \preceq x$

Thus by definition of compact closure:

$d \in x^{\mathrm {compact} }$

$\Box$


By definition of set equality:

$y = x^{\mathrm {compact} }$

Hence:

$y = \map f x$

$\Box$


Hence $f$ is order isomorphism.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Mapping_Assigning_to_Element_Its_Compact_Closure_is_Order_Isomorphism&oldid=576636"