Order Isomorphism on Well-Ordered Set preserves Well-Ordering

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Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Let $\phi: \struct {S_1, \preccurlyeq_1} \to \struct {S_2, \preccurlyeq_2}$ be an order isomorphism.

Then $\struct {S_1, \preccurlyeq_1}$ is a well-ordered set if and only if $\struct {S_2, \preccurlyeq_2}$ is also a well-ordered set.


Let $\struct {S_1, \preccurlyeq_1}$ be a well-ordered set.

Then by definition $\preccurlyeq_1$ is a well-ordering.

By Well-Ordering is Total Ordering, $\preccurlyeq_1$ is a total ordering.

From Order Isomorphism on Totally Ordered Set preserves Total Ordering, it follows that $\struct {S_2, \preccurlyeq_2}$ is a totally ordered set.

All that remains is to show that $\preccurlyeq$ is a well-founded relation.

Let $T_1 \subseteq S_1$.

As $\struct {S_1, \preccurlyeq_1}$ is a well-ordered set:

$\exists m_1 \in T_1: \forall t \in T_1: t = m_1 \lor \neg \paren {t \preccurlyeq_1 m_1}$

Now consider the image $m_2$ of $m_1$ in $S_2$:

$m_2 = \map \phi {m_1}$

Aiming for a contradiction, suppose $\struct {S_2, \preccurlyeq_2}$ is not a well-ordered set.

Let there exist $T_2 \subseteq S_2$ such that:

$\exists x \in T_2: x \ne m_2, x \preccurlyeq_2 m_2$

By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$.


$\map {\phi^{-1} } x \ne \map {\phi^{-1} } {m_2}, \map {\phi^{-1} } x \preccurlyeq_2 \map {\phi^{-1} } {m_2}$

But this contradicts the fact that $\struct {S_1, \preccurlyeq_1}$ is well-ordered.

Hence by Proof by Contradiction $\struct {S_2, \preccurlyeq_2}$ must be a well-ordered set.

The same technique is used to show that if $\struct {S_2, \preccurlyeq_2}$ is a well-ordered set then so is $\struct {S_1, \preccurlyeq_1}$.