Infima Preserving Mapping on Filters is Increasing

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Theorem

Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.

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


For every filter $F$ in $\left({S, \preceq}\right)$, 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:

$\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ admit infima in $\left({S, \preceq}\right)$

By Infimum of Upper Closure of Set:

$\left\{ {x}\right\}^\succeq$ and $\left\{ {y}\right\}^\succeq$ admit infima in $\left({S, \preceq}\right)$

where $\left\{ {x}\right\}^\succeq$ denotes the upper closure of $\left\{ {x}\right\}$.

By Upper Closure of Singleton

$x^\succeq$ and $y^\succeq$ admit infima in $\left({S, \preceq}\right)$

By Upper Closure of Element is Filter:

$x^\succeq$ and $y^\succeq$ are filter in $\left({S, \preceq}\right)$

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

$f^\to \left({x^\succeq}\right)$ and $f^\to \left({y^\succeq}\right)$ admit infima in $\left({T, \precsim}\right)$

and

$\inf \left({f^\to \left({x^\succeq}\right)}\right) = f \left({\inf\left({x^\succeq}\right)}\right)$ and $\inf \left({f^\to \left({y^\succeq}\right)}\right) = f \left({\inf \left({y^\succeq}\right)}\right)$

By Infimum of Upper Closure of Element:

$\inf \left({x^\succeq}\right) = x$ and $\inf \left({y^\succeq}\right) = y$

By Upper Closure is Decreasing:

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

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

$f^\to \left({y^\succeq}\right) \subseteq f^\to \left({x^\succeq}\right)$

Thus by Infimum of Subset:

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

Thus by definition:

$f$ is increasing.

$\blacksquare$


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