Characteristic of Increasing Mapping from Toset to Order Complete Toset

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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

if and only if:

for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \left[{A}\right]$ is bounded in $T$

where $f \left[{A}\right]$ 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 $f \left[{A}\right]$ 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 $H = \left({0 \,.\,.\, 1}\right)$.

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 $f \left[{H}\right]$.

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:

$y' = f \left({y'}\right) \in f \left[{H}\right]$

So $y$ is not an upper bound of $f \left[{H}\right]$.

Thus $f \left[{H}\right]$ has no upper bound in $T$.


Let $g: S \to T$ be an extension of $f$ to $S$.

Consider $g \left({1}\right)$.

We have that $1$ is an upper bound of $H$.

Let $g \left({1}\right) = y$.

As $y \in \left({0 \,.\,.\, 1}\right)$ 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:

$g \left({1}\right) < g \left({y + \dfrac \epsilon 2}\right)$

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 $f \left[{A}\right]$ is bounded in $T$

where:

$\left({S, \preceq}\right)$ and $\left({T, \preccurlyeq}\right)$ 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.

$\Box$


Sufficient Condition

Let $f$ be such that:

for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \left[{A}\right]$ 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:

$A' = \left\{ {y \in A: y \preceq x}\right\}$

$A'$ is bounded in $S$:

bounded below by a lower bound of $A$
bounded above by $y$.

Thus by hypohesis $f \left[{A'}\right]$ is bounded in $T$.

A lower bound of $f \left[{A'}\right]$ is also a lower bound of $f \left[{A}\right]$.

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