Suprema Preserving Mapping on Ideals is Increasing

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Theorem

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

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


For every ideal $I$ in $\struct {S, \preceq}$, let $f$ preserve the supremum on $I$.


Then $f$ is increasing.


Proof

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

$x \preceq y$

By Supremum of Singleton:

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

By Supremum of Lower Closure of Set:

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

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

By Lower Closure of Singleton:

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

By Lower Closure of Element is Ideal:

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

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

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

and:

$\map \sup {\map {f^\to} {x^\preceq} } = \map f {\map \sup {x^\preceq} }$

and:

$\map \sup {\map {f^\to} {y^\preceq} } = \map f {\map \sup {y^\preceq} }$

By Supremum of Lower Closure of Element:

$\map \sup {x^\preceq} = x$ and $\map \sup {y^\preceq} = y$

By Lower Closure is Increasing:

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

By Image of Subset under Mapping is Subset of Image:

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

Thus by Supremum of Subset:

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

Thus by definition:

$f$ is increasing.

$\blacksquare$


Sources