Correctness of Definition of Increasing Mappings Satisfying Inclusion in Lower Closure

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Theorem

Let $R = \left({S, \preceq}\right)$ be an ordered set.

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

Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{ {\it Ids}\left({R}\right) \times {\it Ids}\left({R}\right)}$

Let

$M = \left({F, \preccurlyeq}\right)$

where

$F = \left\{ {f: S \to {\it Ids}\left({R}\right): f}\right.$ is increasing mapping $\left.{\land \forall x \in S: f\left({x}\right) \subseteq x^\preceq}\right\}$

and

$\preccurlyeq$ is ordering on mappings generated by $\precsim$

where $x^\preceq$ denotes the lower closure of $x$.


Then

$M$ is an ordered set.

Proof

Reflexivity

Let $f \in F$.

By definition of reflexivity:

$\forall x \in S: f\left({x}\right) \precsim f\left({x}\right)$

Thus by definition of ordering on mappings:

$f \preccurlyeq f$

$\Box$

Transitivity

Let $f, g, h \in F$ such that

$f \preccurlyeq g \preccurlyeq h$

By definition of ordering on mappings:

$\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \precsim h\left({x}\right)$

By definition of transitivity:

$\forall x \in S: f\left({x}\right) \precsim h\left({x}\right)$

Thus by definition of ordering on mappings:

$f \preccurlyeq h$

$\Box$

Antisymmetry

Let $f, g \in F$ such that

$f \preccurlyeq g$ and $g \preccurlyeq f$

By definition of ordering on mappings:

$\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \land g\left({x}\right) \precsim f\left({x}\right)$

By definition of antisymmetry:

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

Thus by equality of mappings:

$f = g$

$\Box$

Thus by definition:

$M$ is an ordered set.

$\blacksquare$


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