Infima Preserving Mapping on Filters is Increasing

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Theorem

Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be ordered sets.

Let $f: S \to T$ be a mapping.


For every filter $F$ in $\struct {S, \preceq}$, let $f$ preserve the infimum on $F$.


Then $f$ is increasing.


Proof

Let $x, y \in S$ such that:

$x \preceq y$

By Infimum of Singleton:

$\set x$ and $\set y$ admit infima in $\struct {S, \preceq}$

By Infimum of Upper Closure of Set:

$\set x^\succeq$ and $\set y^\succeq$ admit infima in $\struct {S, \preceq}$

where $\set x^\succeq$ denotes the upper closure of $\set x$.

By Upper Closure of Singleton

$x^\succeq$ and $y^\succeq$ admit infima in $\struct {S, \preceq}$

By Upper Closure of Element is Filter:

$x^\succeq$ and $y^\succeq$ are filter in $\struct {S, \preceq}$

By assumption and definition of mapping preserves the infimum on subset:

$\map {f^\to} {x^\succeq}$ and $\map {f^\to} {y^\succeq}$ admit infima in $\struct {T, \precsim}$

and

$\inf \set {\map {f^\to} {x^\succeq} } = \map f {\inf \set {x^\succeq} }$ and $\inf \set {\map {f^\to} {y^\succeq} } = \map f {\inf \set {y^\succeq} }$

By Infimum of Upper Closure of Element:

$\inf \set {x^\succeq} = x$ and $\inf \set {y^\succeq} = y$

By Upper Closure is Decreasing:

$y^\succeq \subseteq x^\succeq$

By Image of Subset under Relation is Subset of Image/Corollary 2:

$\map {f^\to} {y^\succeq} \subseteq \map {f^\to} {x^\succeq}$

Thus by Infimum of Subset:

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

Thus by definition:

$f$ is increasing.

$\blacksquare$


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