# Characterization of Strictly Increasing Mapping on Woset

## Lemma

Let $J$ and $E$ be well-ordered sets.

Let $h: J \to E$ be a mapping.

Let $S_\alpha$ denote an initial segment determined by $\alpha$.

The following are equivalent:

$(1):\quad$ $h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
$(2):\quad$ $\forall \alpha \in J: h\left({\alpha}\right) = \min \left({E\setminus h\left[{S_\alpha}\right]}\right)$, and $h[S_\alpha] = S_{h(\alpha)}$

where:

$h\left[{S_\alpha}\right]$ denotes the image of $S_\alpha$ under $h$
$\min$ denotes the smallest element of the set.

## Proof

### $(1)$ implies $(2)$

Suppose $h$ satisfies:

$h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$

Then for any $x,y \in J$:

 $\ds x$ $\prec$ $\ds y$ $\ds \implies \ \$ $\ds h(x)$ $\prec$ $\ds h(y)$ Definition of strictly increasing $\ds h[S_y]$ $=$ $\ds \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(y) } \right\}$ $\ds$ $=$ $\ds S_{h(y)}$ $\ds \min\left({E \setminus h\left[{S_y}\right] }\right)$ $=$ $\ds \min\left({E \setminus S_{h(y)} }\right)$ $\ds$ $=$ $\ds h(y)$ Definition of smallest and of initial segment

$\Box$

### $(2)$ implies $(1)$

Suppose $h$ satisfies:

$h(\alpha) = \min\left({E \setminus h\left[{S_\alpha}\right] }\right)$

By the Principle of Recursive Definition for Well-Ordered Sets, $h$ is thus uniquely determined.

Then:

 $\ds h(y)$ $=$ $\ds \min\left({E \setminus h\left[{S_y}\right] }\right)$ $\ds h[S_y]$ $=$ $\ds \left\{ { h(x) \in E: \exists x \in J: h(x) = \min\left({E \setminus h\left[{S_y }\right] }\right)} \right\}$ $\ds$ $=$ $\ds \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(y)} \right\}$ $\ds$ $=$ $\ds S_{h(y)}$ Definition of initial segment

Thus for every $x \in S_y$, we have that $h(x) \in S_{h(y)}$.

Therefore $h$ is an strictly increasing mapping.

Furthermore, the image set of $h$ is the union of initial segments in $E$.

By Union of Initial Segments is Initial Segment or All of Woset, $h[J]$ is an initial segment of $E$ or all of $E$.

$\blacksquare$