Antilexicographic Order is Ordering

Theorem

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

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

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {y_1, y_2} \iff \paren {x_2 \prec_2 y_2} \lor \paren {x_2 = y_2 \land x_1 \preccurlyeq_1 y_1}$

Then $\preccurlyeq_a$ is an ordering on $S_1 \times S_2$.

Proof

In the following, $\tuple {x_1, x_2}, \tuple {y_1, y_2}, \tuple {z_1, z_2} \in S_1 \times S_2$.

Checking in turn each of the criteria for an ordering:

Reflexivity

From Equality of Ordered Pairs:

$x_1 = x_2 \land y_1 = y_2 \iff \tuple {x_1, x_2} = \tuple {y_1, y_2}$

Thus:

$\tuple {x_1, x_2} = \tuple {x_1, x_2}$

and so:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {x_1, x_2}$

by definition of lexicographic order on $S_1 \times S_2$.

So $\preccurlyeq_a$ has been shown to be reflexive.

$\Box$

Transitivity

Let:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {y_1, y_2}$

and:

$\tuple {y_1, y_2} \preccurlyeq_a \tuple {z_1, z_2}$

$(1): \quad$ Let $x_2 = y_2 = z_2$.

Then by definition of lexicographic order on $S_1 \times S_2$:

$x_1 \preccurlyeq_1 y_1$

and:

$y_1 \preccurlyeq_1 z_1$

As $\preccurlyeq_1$ is an ordering, it is transitive.

Thus:

$x_1 \preccurlyeq_1 z_1$

and it follows by definition of lexicographic order on $S_1 \times S_2$ that:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {z_1, z_2}$

$(2): \quad$ Let $x_2 = y_2$ and $y_2 \ne z_2$.

Then by definition of lexicographic order on $S_1 \times S_2$:

$y_2 \prec z_2$

But as $x_2 = y_2$:

$x_2 \prec z_2$

and so by definition of lexicographic order on $S_1 \times S_2$:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {z_1, z_2}$

$(3): \quad$ Let $x_2 \ne y_2$ and $y_2 = z_2$.

Then by definition of lexicographic order on $S_1 \times S_2$:

$x_2 \prec y_2$

But as $y_2 = z_2$:

$x_2 \prec z_2$

and so by definition of lexicographic order on $S_1 \times S_2$:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {z_1, z_2}$

$(4): \quad$ Let $x_2 \ne y_2$ and $y_2 \ne z_2$.

Then by definition of lexicographic order on $S_1 \times S_2$:

$x_2 \prec y_2$

and:

$y_2 \prec z_2$

As $\preccurlyeq_2$ is an ordering, it is transitive.

$x_2 \prec z_2$

So by definition of lexicographic order on $S_1 \times S_2$:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {z_1, z_2}$

Thus in all cases it can be seen that:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {z_1, z_2}$

So $\preccurlyeq_a$ has been shown to be transitive.

$\Box$

Antisymmetry

Suppose that:

$\tuple {x_1, x_2} \preccurlyeq_a \tuple {y_1, y_2}$

and:

$\tuple {y_1, y_2} \preccurlyeq_a \tuple {x_1, x_2}$

Suppose $x_2 \ne y_2$.

Then by definition of lexicographic order on $S_1 \times S_2$:

$x_2 \preccurlyeq_2 y_2$

and:

$y_2 \preccurlyeq_2 x_2$

But $\preccurlyeq_2$ is an ordering and so $x_2 = y_2$.

From that contradiction it follows that $x_2 = y_2$.

Then by definition of lexicographic order on $S_1 \times S_2$:

$x_1 \preccurlyeq_1 y_1$

and:

$y_1 \preccurlyeq_1 x_1$

As $\preccurlyeq_1$ is an ordering, it is antisymmetric.

Therefore:

$x_1 = y_1$

and so:

$\tuple {x_1, x_2} = \tuple {y_1, y_2}$

So $\preccurlyeq_a$ has been shown to be antisymmetric.

$\Box$

$\preccurlyeq_a$ has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.

$\blacksquare$