Fundamental Theorem of Well-Ordering

== Theorem ==
Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Then either:
:$\struct {A, \preccurlyeq_A}$ is order isomorphic to a lower section of $\struct {B, \preccurlyeq_B}$, or perhaps all of $\struct {B, \preccurlyeq_B}$
or:
:$\struct {B, \preccurlyeq_B}$ is order isomorphic to a lower section of $\struct {A, \preccurlyeq_A}$, or perhaps all of $\struct {A, \preccurlyeq_A}$.
== Proof ==
Let $N$ be the class of all order isomorphisms from lower sections of $\struct {A, \preccurlyeq_A}$ which are sets to lower sections of $\struct {B, \preccurlyeq_B}$ which are sets.
Each such order isomorphism is itself a set.
By Isomorphisms between Lower Sections of Well-Ordered Classes are Nested, $N$ is a nest.
$N$ may or may not contain a maximal element the subset relation.
However, note that no order isomorphism from a proper lower section of $\struct {A, \preccurlyeq_A}$ onto a proper lower section of $\struct {B, \preccurlyeq_B}$ can be a maximal element of $N$.
Indeed, suppose $\psi$ is an order isomorphism from a proper lower section $a^\prec$ of $\struct {A, \preccurlyeq_A}$ onto a proper lower section $b^\prec$ of $\struct {B, \preccurlyeq_B}$.
Then $\psi \cup \tuple {a, b}$ is an element of $N$ which is a proper extension of $\psi$.
Let $I = \bigcup N$ be the union of $N$.
Because $N$ is a nest, $I$ must be an order isomorphism from a lower section of $\struct {A, \preccurlyeq_A}$ (not necessarily proper) onto a lower section of $\struct {B, \preccurlyeq_B}$ (not necessarily proper).
It remains to be shown that either:
:the domain of $I$ is the whole of $A$
or:
:the image of $I$ is the whole of $B$.
Equivalently, we can instead show that:
:the domain of $I$ cannot be a proper lower section of $\struct {A, \preccurlyeq_A}$
and:
:the image of $I$ cannot be a proper lower section of $\struct {B, \preccurlyeq_B}$.
, the above were the case.
Because $I = \bigcup N$, $I$ would then be a maximal element of $N$.
But we have shown that no order isomorphism from a proper lower section of $\struct {A, \preccurlyeq_A}$ onto a proper lower section of $\struct {B, \preccurlyeq_B}$ can be a maximal element of $N$.
Therefore $I$ cannot be an order isomorphism from a proper lower section of $\struct {A, \preccurlyeq_A}$ onto a proper lower section of $\struct {B, \preccurlyeq_B}$.
The result follows.
== Also presented as ==
This can also be presented as:
Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Then either:
:$\struct {A, \preccurlyeq_A}$ is order isomorphic to a lower section of $\struct {B, \preccurlyeq_B}$
or:
:$\struct {B, \preccurlyeq_B}$ is order isomorphic to a lower section of $\struct {A, \preccurlyeq_A}$
or:
:$\struct {A, \preccurlyeq_A}$ is order isomorphic to $\struct {B, \preccurlyeq_B}$.
== Also known as ==
The Fundamental Theorem of Well-Ordering is also known as the comparability theorem for well-orderings, or simply the comparability theorem.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 2$ Isomorphisms of well orderings: Theorem $2.9$
Category:Fundamental Theorem of Well-Ordering
Category:Well-Orderings
Category:Fundamental Theorems
Category:Order Isomorphisms