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

if and only if:

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.

$\Box$

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.

This article, or a section of it, needs explaining. In particular: 1955: John L. Kelley: General Topology, from which this has been taken, is unclear here exactly what is being demonstrated by the above. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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.

This needs considerable tedious hard slog to complete it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Orderings: Theorem $10$