Definition 1 (section)   Let $A$ be well-ordered. Then for every $a\in A$ we define the section of $a$ : \begin{equation*} \h{a}:=\{b\in A\mid b\leq a\}. \end{equation*}A section is also known as an initial segment. We denote the set of all sections of $A$ by $\h{A}$ . This set is ordered by inclusion.

Theorem 1   Let $A$ be a well-ordered set. Then the mapping $\h{\cdot}\colon A\to\h{A}$ defined by $a\mapsto\h{a}$ is a bijective order morphism. In particular, $\h{A}$ is well-ordered.

Proof. Let $a,b\in A$ with $a\leq b$ . Then $\h{a}\subseteq\h{b}$ , so $\h{\cdot}$ is an order morphism. Now assume that $\h{a}=\h{b}$ . If $a$ didn't equal $b$ , we would have $b\notin\h{a}$ , leading to a contradiction. Therefore $\h{\cdot}$ is injective. Now let $C\in\h{A}$ , then there exists a $c\in A$ such that $\h{c}=C$ , so $\h{\cdot}$ is surjective.

Theorem 2   Let $A$ and $B$ be well-ordered sets and $\varphi\colon A\to B$ a bijective order morphism. Then there exists a bijective order morphism $\h{\varphi}\colon\h{A}\to\h{B}$ such that for all $a\in A$ \begin{equation*} \h{\varphi}(\h{a})=\h{\varphi(a)}. \end{equation*}

Proof. Setting $\h{\varphi}(\h{a}):=\h{\varphi(a)}$ is well-defined by Theorem 1. The rest of the theorem follows since $\h{\cdot}$ and $\varphi$ are bijective order morphisms.

Theorem 3   Let $A$ be a well-ordered set and $a\in A$ such that there is an injective order morphism $\varphi\colon A\to\h{a}$ . Then $A=\h{a}$ .

Proof. The image of a section of $A$ under $\varphi$ has a maximal element which in turn defines a smaller section of $A$ . We may therefore define the following two monotonically decreasing sequences of sets:
Now $\h{A}$ is well-ordered, so the set defined by the elements of the sequence $(A_n)$ has a minimal element, that is $A_N=A_{N+1}$ and hence $B_N=B_{N+1}$ for some sufficiently large $N$ . Applying $\varphi^{-1}$ $N+1$ times to the latter equation yields $\h{a}=A_0$ , that is $a$ is the maximal element of $B$ , and thus of $A$ .

Theorem 4   Let $A$ and $B$ be well-ordered sets. Then there exists at most one bijective order morphism $\varphi\colon A\to B$ .

Proof. Let $\varphi,\psi\colon A\to B$ be two bijective order morphisms. Let $a\in A$ and set $b:=\varphi(a)$ and $c:=\psi(a)$ . Then the restrictions $\left.\varphi\right|_{\h{a}}\colon\h{a}\to\h{b}$ and $\left.\psi\right|_{\h{a}}\colon\h{a}\to\h{c}$ are bijective order morphisms, so the restriction of $\psi\varphi^{-1}$ to $\h{b}$ is a bijective order morphism to $\h{c}$ . Now either $\h{b}\subseteq\h{c}$ or $\h{c}\subseteq\h{b}$ , so by Theorem 3 $\h{b}=\h{c}$ , hence $b=c$ and thus $\varphi=\psi$ .

Theorem 5   Let $A$ and $B$ be well-ordered sets such that for every section $\h{a}\in\h{A}$ there is a bijective order morphism to a section $\h{b}\in\h{B}$ and vice-versa, then there is a bijective order morphism $\varphi\colon A\to B$ .

Proof. Let $a\in A$ and let $\h{b}\in\h{B}$ be a section such that there is a bijective order morphism $\psi_a\colon\h{a}\to\h{b}$ . By Theorem 3, $\h{b}$ is unique, and so is $\psi_a$ by Theorem 4. Defining $\varphi\colon A\to B$ by setting $\varphi(a)=b$ gives therefore a well-defined (by Theorem 1) and injective order morphism. But $\varphi$ is also surjective, since any $b\in B$ maps uniquely to $A$ via $\h{b}\to\h{a}$ , and back again by $\varphi$ .

Theorem 6   Let $A$ and $B$ be well-ordered sets. Then there is an injective order morphism $\iota\colon A\to B$ or $\iota\colon B\to A$ . If $\iota$ cannot be chosen bijective, then it can at least be chosen such that its image is a section.

Proof. Let $\h{A}_0$ be the set of sections of $A$ from which there is an injective order morphism to $B$ . If $\h{A}_0$ is the empty set, then $B$ must be empty, since otherwise we could map the least element of $A$ to $B$ . If $\h{A}_0$ is not empty, we may consider the set $A_0:=\cap\h{A}_0$ . If $A_0=A$ , nothing remains to be shown. Otherwise the set $A\setminus A_0$ is nonempty an hence has a least element $a_0\in A$ . By construction, there is no injective order morphism from $\h{a}_0$ to $B$ , but there is an injective order morphism from $\varphi_a\colon\h{a}\to B$ for every element $a\in A$ which is strictly smaller than $a_0$ . Now assume there is an element $b\in B$ such that there is no injective order morphism from $\h{b}\to A$ . Then we can similarly construct a least element $b_0\in B$ for which there is no injective order morphism $\h{b}_0\to A$ . Surely, $b_0$ is greater than all the elements from the images of the functions $\varphi_a$ , but then there is a bijective order morphism from $\h{a}_0$ to $\h{b}_0$ by Theorem 5 which is a contradiction. Therefore, all sections of $B$ and $B$ itself map injectively and order-preserving to $A$ .

Theorem 7   Let $A$ be a well-ordered set and $B\subseteq A$ a nonempty subset. Then there is a bijective order morphism from $B$ to one of the sets in $\h{A}\cup\{A\}$ .

Proof. The set $B$ is well-ordered with respect to the order induced by $A$ . Assume a bijective order morphism as stated by the theorem does not exist. Then, by virtue of Theorem 6, there is an injective but not surjective order morphism $\iota\colon A\to B$ whose image is a section $\h{b}\in\h{B}$ . The element $b$ defines a section in $\h{A}$ which is identical to $A$ by Theorem 3. Thus $\iota$ is surjective which is a contradiction.