Characteristic of Increasing Mapping from Toset to Order Complete Toset

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets.
Let $T$ be order complete.
Let $H \subseteq S$ be a subset of $S$.
Let $f: H \to T$ be an increasing mapping from $H$ to $T$.
Then:
:$f$ has an extension to $S$ which is increasing
:
:for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$
where $f \sqbrk A$ denotes the image set of $A$ under $f$.
== Proof ==
=== Necessary Condition ===
Suppose $f$ is such that it is not the case that:
:for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$.
Proof by Counterexample:
Let $S = \R_{>0}$ be the set of all (strictly) positive real numbers.
Let $H \subseteq S$ be the open real interval $H = \openint 0 1$.
Let $T = H$.
By the axiomatic definition of real numbers, $S$ and $T$ are totally ordered.
From the Continuum Property, $T$ is order complete.
Let $f: H \to T$ be the identity mapping.
Note that while $H$ has an upper bound in $S$, for example $1$
Let $y \in T$ be an upper bound of $f \sqbrk H$.
By construction:
:$y < 1$
and so:
:$\exists \epsilon \in \R_{>0}: y + \epsilon = 1$
Thus:
:$y < y + \dfrac \epsilon 2 < 1$
and so there exists $y' = y + \dfrac \epsilon 2 \in T$ such that:
:$y < y'$
and such that:
:$y' = \map f {y'} \in f \sqbrk H$
So $y$ is not an upper bound of $f \sqbrk H$.
Thus $f \sqbrk H$ has no upper bound in $T$.
Let $g: S \to T$ be an extension of $f$ to $S$.
Consider $\map g 1$.
We have that $1$ is an upper bound of $H$.
Let $\map g 1 = y$.
As $y \in \openint 0 1$ it follows that:
:$y > 1$
and so:
:$\exists \epsilon \in \R_{>0}: y + \epsilon = 1$
But then:
:$y < y + \dfrac \epsilon 2 < 1$
and so:
:$\map g 1 < \map g {y + \dfrac \epsilon 2}$
and so $g$ is not increasing.
Thus it has been demonstrated that if it is not the case that:
:for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$
where:
:$\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets
and:
:$T$ is order complete
and:
:$f: H \to T$ is an increasing mapping from $H$ to $T$.
and:
:$H \subseteq S$ is a subset of $S$
then it is not necessarily the case that $f$ always has an extension to $S$ which is increasing.
=== Sufficient Condition ===
Let $f$ be such that:
:for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$.
An increasing mapping $g: S \to T$ is to be constructed such that $g$ is an extension of $f$.
So, let $A \subseteq H$ such that $A$ is bounded in $S$.
Let $x \in A$.
Consider the set:
:$A' = \set {y \in A: y \preceq x}$
$A'$ is bounded in $S$:
:bounded below by a lower bound of $A$
:bounded above by $y$.
Thus by hypothesis $f \sqbrk {A'}$ is bounded in $T$.
A lower bound of $f \sqbrk {A'}$ is also a lower bound of $f \sqbrk A$.
Similarly for upper bounds.
For each $x \in S$ let $L_x$ be defined as:
:$L_x = \set {y \in H: y \le x}$
Note that $L_x$ is bounded above by $x$ by definition of bounded above.
Suppose $L_x = \O$.
Then $x$ is a lower bound for $H$.
Thus $f \sqbrk H$ has an infimum $v$, and $\map g x$ can be defined as:
:$\map g x = v$
Suppose $L_x \ne O$.
We have that $x$ is an upper bound of $L_x$.
Hence by hypothesis $f \sqbrk {L_x}$ has an upper bound.
As $T$ is order complete, $f \sqbrk {L_x}$ admits a supremum.
Thus $\map g x$ can be defined as:
:$\map g x = \sup \set {f \sqbrk {L_x} }$
It remains to be proved that $g$ is an increasing mapping.
== Sources ==
* : Chapter $0$: Orderings: Theorem $10$
Category:Bounded Sets
Category:Total Orderings
Category:Order Complete Sets