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 $T$
 * 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 $f$.

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.