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

## Theorem

Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be ordered sets.

Let $\preccurlyeq$ be the lexicographic order on $S_1 \times S_2$**:**

- $\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preceq_2 y_2}$

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 $\struct {S_1 \times S_2, \preccurlyeq}$ is a ordered set.

### Necessary Condition

Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be well-ordered sets.

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

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

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

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

Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in T_1 \times T_2$.

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

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

Consider the element $\tuple {m_1, m_2} \in T_1 \times T_2$.

Let $\tuple {x_1, x_2} \in T_1 \times T_2$.

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 $S_1 \times S_2$:

- $\tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$

$(2): \quad m_1 = x_1$

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 $S_1 \times S_2$:

- $\tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$

Thus it follows that $T_1 \times T_2$ has a smallest element $\tuple {m_1, m_2}$.

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 $\struct {S_1 \times S_2, \preccurlyeq}$ is a well-ordered set.

$\Box$

### Sufficient Condition

Let $\struct {S_1 \times S_2, \preccurlyeq}$ be a well-ordered set.

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

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

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

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

Let $\tuple {m_1, m_2} \in T_1 \times T_2$ be the smallest element of $T_1 \times T_2$.

Thus:

- $\forall \tuple {x_1, x_2} \in T_1 \times T_2: \tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$

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 $\tuple {x_1, x_2} \in T_1 \times T_2$ 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 $\struct {S_2, \preceq_2}$.

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 $\tuple {x_1, x_2} \in T_1 \times T_2$ 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 $\struct {S_1, \preceq_1}$.

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

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $14.19 \ \text{(b)}$