# Condition for Uniqueness of Increasing Mappings between Tosets

## Theorem

Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets.

Let $f: S \to T$ and $g: S \to T$ be increasing mappings from $S$ to $T$.

Let $H \subseteq S$ be a subset of $S$.

Let $f$ and $g$ agree on $H$.

Let $K = f \sqbrk H$ be the image set of $H$ under $f$.

Let the intersection of $K$ with every set of the form:

- $\set {y \in T: u < y < v: u, v \in T, u < v}$

be non-empty.

Then $f = g$.

## Proof

By hypothesis, let the intersection of $K$ with every set of the form:

- $\set {y \in T: u \prec y \prec v: u, v \in T, u \prec v}$

be non-empty.

Aiming for a contradiction, suppose $f \ne g$.

Then:

- $\exists x \in S: \map f x \ne \map g x$

Without loss of generality, suppose that $\map f x < \map g x$.

Let:

- $a \in H$ such that $a \preceq x$
- $b \in H$ such that $b \succeq x$

We have that $f$ and $g$ are increasing.

Thus:

- $\map f a \preccurlyeq \map f x$
- $\map g b \succcurlyeq \map g x$

Thus no element of $H$ maps to the set:

- $\set {y \in T: \map f x \prec y \prec \map g x}$

That is, the intersection of $K$ with this set, which is of the form:

- $\set {y \in T: u \prec y \prec v: u, v \in T, u \prec v}$

is empty.

This contradicts our hypothesis.

Thus, by Proof by Contradiction:

- $f = g$

$\blacksquare$

## Sources

- 1955: John L. Kelley:
*General Topology*... (previous) ... (next): Chapter $0$: Orderings: Theorem $11$