Strictly Increasing Mapping Between Wosets Implies Order Isomorphism
 == Lemma == Let $J$ and $E$ be well-ordered sets. Let there exist a mapping $k: J \to E$ which is strictly increasing. Then $J$ is order isomorphic to $E$ or an initial segment of $E$. == Proof == If the sets considered are empty or singletons, the theorem holds vacuously or trivially. Suppose $J,E$ both have at least two elements. Let $e_0 = \min E$, the smallest element of $E$. Define the mapping: :$h: J \to E$: :$h(\alpha) =  \begin{cases} \min \left({E \setminus h\left[ {S_\alpha} \right]}\right) & \text{ if } h\left[ {S_\alpha} \right] \ne E \\ e_0 & \text{ if } h\left[ {S_\alpha} \right] = E \end{cases}$ where $S_\alpha$ is the initial segment determined by $\alpha$ and $h\left[ {S_\alpha} \right]$ is the image of $S_\alpha$ under $h$. By the Principle of Recursive Definition for Well-Ordered Sets, this construction is well-defined and uniquely determined. Observe that: {{begin-eqn}} {{eqn | l = h[S_\alpha]       | r = \left\{ { h(x) \in E: \exists x \in J: h(x) =              \min\left({E \setminus h\left[{S_\alpha}\right] }\right)} \right\}       | c = {{Defof|Image of Subset under Mapping|image of a subset}} }} {{eqn | r =  \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(\alpha)} \right\} }} {{eqn | r = S_{h(\alpha)}       | c = {{Defof|Initial Segment|initial segment}} }} {{end-eqn}} This equality also holds if $h(\alpha) = e_0$, by Initial Segment Determined by Smallest Element is Empty. We claim that $h(\alpha) \preceq k(\alpha)$ for all $\alpha \in J$.  there is some $a \in J$ such that $h(a) \not \preceq k(a)$. Then $k(a) \prec h(a)$ by the trichotomy law. Because $h(a)$ has an element preceding it, $h(a) \ne e_0$. Thus $k(a) \prec \min \left({E \setminus h\left[ {S_a} \right]}\right)$ by the construction of $k$. Then $k(a) \in h\left[ {S_a} \right]$, because it precedes the smallest element that isn't in $h\left[ {S_a} \right]$. Recall that $h[S_a] = S_{h(a)}$. Then $k(a) \in S_{h(a)}$. This implies that $h(a) \prec k(a)$, contradicting the assumption that $h(a) \not \preceq k(a)$ From this contradiction we can conclude: :$h(\alpha) \preceq k(\alpha)$ for all $\alpha \in J$.  there is some $\alpha \in J$ such that $h[S_\alpha] = E$. Recall that $h[S_\alpha] = S_{h(\alpha)}$. Thus, were such an $\alpha$ to exist, then it would succeed all elements in $E$. It particular, it would also succeed $k(\alpha)$. But we showed above that $h(\alpha) \preceq k(\alpha)$. From this contradiction we see that there cannot be any $\alpha \in J$ with $h[S_\alpha] = E$. Thus the definition of $h$ can be simplified: :$h: J \to E$: :$h(\alpha) = \min \left({E \setminus h\left[ {S_\alpha} \right]}\right)$ Then the hypotheses of Characterization of Strictly Increasing Mapping on Woset are satisfied. Thus $h$ is a strictly increasing mapping and its image is $E$ or an initial segment of $E$. From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $h$ is also injective. From Injection to Image is Bijection, $h$ is also bijective to its image. We conclude that there is an order isomorphism from $J$ to $E$, or from $J$ to an initial segment of $E$.  == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition}}: Supplementary Exercises $1.3$