Lexicographic Order on Pair of Well-Ordered Sets is Well-Ordering

Theorem
Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be ordered sets.

Let $\preccurlyeq$ be the lexicographic order on $S_1 \times S_2$''':
 * $\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right) \iff \left({x_1 \prec_1 y_1}\right) \lor \left({x_1 = y_1 \land x_2 \preceq_2 y_2}\right)$

Then:
 * $\preccurlyeq$ is a well-ordering on $S_1 \times S_2$


 * both $\preceq_1$ and $\preceq_2$ are well-orderings.

Proof
From Lexicographic Order on Totally Ordered Sets is Total Ordering we have that $\left({S_1 \times S_2, \preccurlyeq}\right)$ is a ordered set.

Necessary Condition
Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be well-ordered sets.

By definition of well-ordering we have that $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ are totally ordered sets.

From Lexicographic Order on Totally Ordered Sets is Total Ordering it follows that $\left({S_1 \times S_2, \preccurlyeq}\right)$ is a totally ordered set.

It remains to be shown that the Cartesian product of arbitrary non-empty subsets of $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ has a smallest element under $\preccurlyeq$.

Let $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$.

Let $\left({x_1, x_2}\right), \left({y_1, y_2}\right) \in T_1 \times T_2$.

As both $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ are well-ordered sets it follows that both $\left({T_1, \preceq_1}\right)$ and $\left({T_2, \preceq_2}\right)$ have smallest elements.

Let those smallest elements be $m_1$ and $m_2$ respectively.

Consider the element $\left({m_1, m_2}\right) \in T_1 \times T_2$.

Let $\left({x_1, x_2}\right) \in T_1 \times T_2$.

We have that:
 * $m_1 \preceq_1 x_1$

and:
 * $m_2 \preceq_2 x_2$

The following cases are to be examined:

$(1): \quad m_1 \ne x_1$

By defiinition of $m_1$ as the smallest element of $T_1$:
 * $m_1 \preceq_1 x_1$

Thus by definition of the lexicographic order on $S_1 \times S_2$:
 * $\left({m_1, m_2}\right) \preccurlyeq \left({x_1, x_2}\right)$

$(2): \quad m_1 = x_1$

By definition of $m_2$ as the smallest element of $T_2$:
 * $m_2 \preceq_2 x_2$

Thus by definition of the lexicographic order on $S_1 \times S_2$:
 * $\left({m_1, m_2}\right) \preccurlyeq \left({x_1, x_2}\right)$

Thus it follows that $T_1 \times T_2$ has a smallest element $\left({m_1, m_2}\right)$.

As $T_1 \times T_2$ is the Cartesian product of arbitrary non-empty subsets of of $S_1$ and $S_2$, it follows that $\left({S_1 \times S_2, \preccurlyeq}\right)$ is a well-ordered set.

Sufficient Condition
Let $\left({S_1 \times S_2, \preccurlyeq}\right)$ be a well-ordered set.

By definition of well-ordering we have that $\left({S_1 \times S_2, \preccurlyeq}\right)$ is a totally ordered set.

From Lexicographic Order on Totally Ordered Sets is Total Ordering $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ are totally ordered sets.

It remains to be shown that arbitrary non-empty subsets of $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ have a smallest elements.

Let $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$.

Let $\left({m_1, m_2}\right) \in T_1 \times T_2$ be the smallest element of $T_1 \times T_2$.

Thus:
 * $\forall \left({x_1, x_2}\right) \in T_1 \times T_2: \left({m_1, m_2}\right) \preccurlyeq \left({x_1, x_2}\right)$

There are two cases to address:

$(1): \quad$ Let $m_1 = x_1$.

Thus by definition of the lexicographic order on $S_1 \times S_2$:
 * $m_2 \preceq_2 x_2$

This is the case for all $\left({x_1, x_2}\right) \in T_1 \times T_2$ such that $m_1 = x_1$.

Therefore:
 * $\forall x_2 \in T_2: m_2 \preceq_2 x_2$

Thus by definition, $m_2$ is the smallest element of $T_2$.

We have that $T_2$ is an arbitrary non-empty subset $\left({S_2, \preceq_2}\right)$.

Therefore by definition $T_2$ is a well-ordered set.

$(2): \quad$ Let $m_1 \ne x_1$.

Thus by definition of the lexicographic order on $S_1 \times S_2$:
 * $m_1 \prec_1 x_1$

This is the case for all $\left({x_1, x_2}\right) \in T_1 \times T_2$ such that $m_1 \ne x_1$.

Therefore:
 * $\forall x_1 \in T_1: m_1 \preceq_1 x_1$

Thus by definition, $m_1$ is the smallest element of $T_1$.

We have that $T_1$ is an arbitrary non-empty subset $\left({S_1, \preceq_1}\right)$.

Therefore by definition $T_1$ is a well-ordered set.