Mapping Assigning to Element Its Lower Closure is Isomorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

where

$\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$,
$\mathord\precsim = \mathord\subseteq \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Let $P = \left({K\left({I}\right), \precsim'}\right)$ be an ordered subset of $I$

where

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

Let $f:S \to K\left({I}\right)$ be a mapping such that

$\forall x \in S: f\left({x}\right) = x^\preceq$


Then $f$ is an order isomorphism between $L$ and $P$.


Proof

By definition:

$\forall x \in S: x^\preceq$ is a principal ideal.

By Compact Element iff Principal Ideal:

$\forall x \in S: x^\preceq$ is a compact element in $I$.

By definition of compact subset:

$\forall x \in S: x^\preceq \in K\left({I}\right)$

Then $f$ is well-defined.

We will prove that

$f$ is an order embedding

That means

$\forall x, y \in S: x \preceq y \iff f\left({x}\right) \precsim' f\left({y}\right)$

Let $x, y \in S$.

We will prove as lemma that

$x^\preceq \subseteq y^\preceq \implies x \preceq y$

Assume that

$x^\preceq \subseteq y^\preceq$

By definition of reflexivity:

$x \preceq x$

By definition of lower closure of element:

$x \in x^\preceq$

By definition of subset:

$x \in y^\preceq$

Thus by definition of lower closure of element:

$x \preceq y$

$\Box$

$x \preceq y$

if and only if

$x^\preceq \subseteq y^\preceq$ by lemma and Lower Closure is Increasing

if and only if

$f\left({x}\right) \subseteq f\left({y}\right)$ by definition of $f$

if and only if

$f\left({x}\right) \precsim f\left({y}\right)$ by definition of $\precsim$

if and only if

$f\left({x}\right) \precsim' f\left({y}\right)$ by definition of ordered subset.

$\Box$

We will prove that

$\forall a \in K\left({I}\right): \exists y \in S:a = f\left({y}\right)$

Let $a \in K\left({I}\right)$

By definition of compact subset:

$a$ is a compact element in $I$.

by Compact Element iff Principal Ideal:

$a$ is a principal ideal.

By definition of principal ideal:

$\exists y \in S: a = y^\preceq$

Thus by definition of $f$:

$a = f\left({y}\right)$

$\Box$

Then $f$ is a surjection.

Hence $f$ is an order isomorphism.

$\blacksquare$


Sources