properties of wellordered sets
The purpose of this entry is to collect properties of wellordered sets. We denote all orderings uniformly by $\le $. If you are interested in history, have a look at [C].
The following properties are easy to see:

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If $A$ is a totally ordered set^{} such that for every subset $B\subseteq A$ the set of all elements of $A$ strictly greater than the elements of $B$,
$${B}_{>}:=\{a\in A\setminus B\mid b\le a\text{for all}b\in B\},$$ is either empty or has a least element, then $A$ is wellordered.

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Every subset of a wellordered set is wellordered.

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If $A$ is wellordered and $B$ is a poset such that there is a bijective^{} order morphism $\phi :A\to B$, then $B$ is wellordered.
Now we define an important ingredient for understanding the structure^{} of wellordered sets.
Definition 1 (section).
Let $A$ be wellordered. Then for every $a\in A$ we define the section^{} of $a$:
$$\widehat{a}:=\{b\in A\mid b\le a\}.$$ 
A section is also known as an initial segment. We denote the set of all sections of $A$ by $\widehat{A}$. This set is ordered by inclusion.
Theorem 1.
Let $A$ be a wellordered set. Then the mapping $\widehat{\mathrm{\cdot}}\mathrm{:}A\mathrm{\to}\widehat{A}$ defined by $a\mathrm{\mapsto}\widehat{a}$ is a bijective order morphism. In particular, $\widehat{A}$ is wellordered.
Proof.
Let $a,b\in A$ with $a\le b$. Then $\widehat{a}\subseteq \widehat{b}$, so $\widehat{\cdot}$ is an order morphism. Now assume that $\widehat{a}=\widehat{b}$. If $a$ didn’t equal $b$, we would have $b\notin \widehat{a}$, leading to a contradiction^{}. Therefore $\widehat{\cdot}$ is injective^{}. Now let $C\in \widehat{A}$, then there exists a $c\in A$ such that $\widehat{c}=C$, so $\widehat{\cdot}$ is surjective. ∎
Theorem 2.
Let $A$ and $B$ be wellordered sets and $\phi \mathrm{:}A\mathrm{\to}B$ a bijective order morphism. Then there exists a bijective order morphism $\widehat{\phi}\mathrm{:}\widehat{A}\mathrm{\to}\widehat{B}$ such that for all $a\mathrm{\in}A$
$$\widehat{\phi}(\widehat{a})=\widehat{\phi (a)}.$$ 
Proof.
Setting $\widehat{\phi}(\widehat{a}):=\widehat{\phi (a)}$ is welldefined by Theorem^{} 1. The rest of the theorem follows since $\widehat{\cdot}$ and $\phi $ are bijective order morphisms. ∎
Theorem 3.
Let $A$ be a wellordered set and $a\mathrm{\in}A$ such that there is an injective order morphism $\phi \mathrm{:}A\mathrm{\to}\widehat{a}$. Then $A\mathrm{=}\widehat{a}$.
Proof.
The image of a section of $A$ under $\phi $ 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:
${B}_{0}$  $:=\phi (\widehat{a}),$  ${A}_{0}$  $:=\text{section defined by maximal element of}{B}_{0},$  
${B}_{n}$  $:=\phi ({A}_{n1}),$  ${A}_{n}$  $:=\text{section defined by maximal element of}{B}_{n}.$ 
Now $\widehat{A}$ is wellordered, 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 ${\phi}^{1}$ $N+1$ times to the latter equation yields $\widehat{a}={A}_{0}$, that is $a$ is the maximal element of $B$, and thus of $A$. ∎
Theorem 4.
Let $A$ and $B$ be wellordered sets. Then there exists at most one bijective order morphism $\phi \mathrm{:}A\mathrm{\to}B$.
Proof.
Let $\phi ,\psi :A\to B$ be two bijective order morphisms. Let $a\in A$ and set $b:=\phi (a)$ and $c:=\psi (a)$. Then the restrictions^{} ${\phi }_{\widehat{a}}:\widehat{a}\to \widehat{b}$ and ${\psi }_{\widehat{a}}:\widehat{a}\to \widehat{c}$ are bijective order morphisms, so the restriction of $\psi {\phi}^{1}$ to $\widehat{b}$ is a bijective order morphism to $\widehat{c}$. Now either $\widehat{b}\subseteq \widehat{c}$ or $\widehat{c}\subseteq \widehat{b}$, so by Theorem 3 $\widehat{b}=\widehat{c}$, hence $b=c$ and thus $\phi =\psi $. ∎
Theorem 5.
Let $A$ and $B$ be wellordered sets such that for every section $\widehat{a}\mathrm{\in}\widehat{A}$ there is a bijective order morphism to a section $\widehat{b}\mathrm{\in}\widehat{B}$ and viceversa, then there is a bijective order morphism $\phi \mathrm{:}A\mathrm{\to}B$.
Proof.
Let $a\in A$ and let $\widehat{b}\in \widehat{B}$ be a section such that there is a bijective order morphism ${\psi}_{a}:\widehat{a}\to \widehat{b}$. By Theorem 3, $\widehat{b}$ is unique, and so is ${\psi}_{a}$ by Theorem 4. Defining $\phi :A\to B$ by setting $\phi (a)=b$ gives therefore a welldefined (by Theorem 1) and injective order morphism. But $\phi $ is also surjective, since any $b\in B$ maps uniquely to $A$ via $\widehat{b}\to \widehat{a}$, and back again by $\phi $. ∎
Theorem 6.
Let $A$ and $B$ be wellordered sets. Then there is an injective order morphism $\iota \mathrm{:}A\mathrm{\to}B$ or $\iota \mathrm{:}B\mathrm{\to}A$. If $\iota $ cannot be chosen bijective, then it can at least be chosen such that its image is a section.
Proof.
Let ${\widehat{A}}_{0}$ be the set of sections of $A$ from which there is an injective order morphism to $B$. If ${\widehat{A}}_{0}$ is the empty set^{}, then $B$ must be empty, since otherwise we could map the least element of $A$ to $B$. If ${\widehat{A}}_{0}$ is not empty, we may consider the set ${A}_{0}:=\cap {\widehat{A}}_{0}$. 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 ${\widehat{a}}_{0}$ to $B$, but there is an injective order morphism from ${\phi}_{a}:\widehat{a}\to B$ 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 $\widehat{b}\to A$. Then we can similarly construct a least element ${b}_{0}\in B$ for which there is no injective order morphism ${\widehat{b}}_{0}\to A$. Surely, ${b}_{0}$ is greater than all the elements from the images of the functions ${\phi}_{a}$, but then there is a bijective order morphism from ${\widehat{a}}_{0}$ to ${\widehat{b}}_{0}$ by Theorem 5 which is a contradiction. Therefore, all sections of $B$ and $B$ itself map injectively and orderpreserving to $A$. ∎
Theorem 7.
Let $A$ be a wellordered set and $B\mathrm{\subseteq}A$ a nonempty subset. Then there is a bijective order morphism from $B$ to one of the sets in $\widehat{A}\mathrm{\cup}\mathrm{\{}A\mathrm{\}}$.
Proof.
The set $B$ is wellordered 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 $\iota :A\to B$ whose image is a section $\widehat{b}\in \widehat{B}$. The element $b$ defines a section in $\widehat{A}$ which is identical to $A$ by Theorem 3. Thus $\iota $ is surjective which is a contradiction. ∎
References
 C G. Cantor, Beiträge zur Begründung der transfiniten Mengenlehre (Zweiter Artikel), Math. Ann. 49, 207–246 (1897).
Title  properties of wellordered sets 

Canonical name  PropertiesOfWellorderedSets 
Date of creation  20130322 15:23:42 
Last modified on  20130322 15:23:42 
Owner  GrafZahl (9234) 
Last modified by  GrafZahl (9234) 
Numerical id  10 
Author  GrafZahl (9234) 
Entry type  Theorem 
Classification  msc 06A05 
Synonym  initial segment 
Defines  section 