A well-order on $W$ 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.
$\preceq$ is a well-order if and only if $\prec$ is a strict well-order.
If $(W,\preceq)$ is a well-ordered set and $f:W\to W$ is strictly increasing, then $x\preceq f(x)$ for every $x\in W$.
Proof.Assume $A=\{x\in W:f(x)\prec x\}$ were non-empty. Let $z=\min A$, then
\[f(f(z))\prec f(z)\prec z,\]a contradiction.
The only automorphism on a well-ordered set is the identity function $\id_W$.
Proof.Let $f$ be an automorphism on a well-ordered set $(W,\preceq)$. By PROP-OH-MON, both $f$ and $f^{-1}$ are strictly increasing. By PROP-WO-INC, $x\preceq f(x)$ and $x\preceq f^{-1}(x)$ for every $x\in W$, which follows that $x\preceq f(x)$ and $f(x)\preceq x$ 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 PROP-WO-AUT, $f\circ g^{-1}=\id_W$.
Let $(W,\preceq)$ be a well-ordered set and $u\in W$. The initial segment of $W$ given by $u$ is
\[\{x\in W : x\prec u\}.\]$S\subseteq W$ is an initial segment of $W$, if $S$ is the initial segment of $W$ given by $u$ for some $u$.
No well-ordered set is isomorphic to an initial segment of itself.
Proof.Let $(W,\preceq)$ be a well-ordered set and $f:W\to X$. For any $u\in W$, if
\[X = \{x\in W:x\prec u\},\]then $f(u)\prec u$. By PROP-WO-INC, $f$ can not be an isomorphism.
If $W$ and $W’$ are well-ordered sets, then 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 $W$.