A well-order on a set $W$ is a total order such that every non-empty subset of $W$ has a least element.
A strict well-order on $W$ is a strict total order such that every non-empty subset of $W$ has a least element.
$\leq$ is a well-order if and only if $<$ is a strict well-order.
If $W$ is a well-ordered set and $f : W \to W$ is strictly increasing, then $x \leq f(x)$ for every $x \in W$.
Proof. Assume $A = \{x \in W : f(x) < x\}$ were non-empty. Let $z = \min A$, then
\[f(f(z)) < f(z) < z,\]a contradiction.
The only automorphism on a well-ordered set is the identity $\id_W$.
Proof. Let $f$ be an automorphism on a well-ordered set $W$. By ODR-MOR > Proposition 1, both $f$ and $f^{-1}$ are strictly increasing. By Proposition 2, $x \leq f(x)$ and $x \leq f^{-1}(x)$ for every $x \in W$, which follows that $x \leq f(x)$ and $f(x) \leq x$ for every $x \in W$.
If two well-ordered sets $W$ and $W’$ are isomorphic, then the isomorphism is unique.
Proof. If $f$ and $g$ are isomorphisms from $W$ onto $W’$, $f\circ g^{-1}$ is an automorphism on $W$. By Proposition 3, $f\circ g^{-1} = \id_W$.
Let $W$ be a well-ordered set and $u \in W$. The initial segment of $W$ given by $u$ is
\[\init_u W = \{x \in W : x < u\}.\]$S \subseteq W$ is an initial segment of $W$, if $S = \init_u W$ for some $u$.
No well-ordered set is isomorphic to an initial segment of itself.
Proof. Let $W$ be a well-ordered set. For any $u \in W$, if
\[f : W \to \init_u W,\]then $f(u) < u$. By Proposition 1, $f$ can not be an isomorphism.
For any two well-ordered sets $W$ and $W’$, exactly one of the following cases holds:
- $W$ is isomorphic to $W’$.
- $W$ is isomorphic to an initial segment of $W’$.
- $W’$ is isomorphic to an initial segment of $W$.
The following theorem is equivalent to $\AC$:
Proposition 7. Well-Ordering Theorem.
Every set can be well-ordered, i.e. for any set $X$, there exists a well-order on $X$.