Simple Order Product of Pair of Ordered Sets is Lattice iff Factors are Lattices

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

Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the simple (order) product of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.

Then $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is a lattice both $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ are lattices.

Proof
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.

First we note that from Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {T, \preccurlyeq_s}$ is an ordered set.

Sufficient Condition
Let $\struct {T, \preccurlyeq_s}$ be a lattice.

Let $x_1$ and $y_1$ be arbitrary elements of $S_1$.

Let $x_2$ and $y_2$ be arbitrary elements of $S_2$.

Then we have that $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ are elements of $T$.

By definition of lattice, $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ admits both a supremum and an infimum.

Let $\tuple {c_1, c_2} = \sup \set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ be the supremum of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$.

Then by definition:
 * $\tuple {c_1, c_2}$ is an upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ in $T$


 * $\tuple {c_1, c_2} \preccurlyeq_s \tuple {d_1, d_2}$ for all upper bounds $\tuple {d_1, d_2}$ of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ in $T$.

Let $\tuple {d_1, d_2}$ be an arbitrary upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ in $T$

Similarly:

Thus:
 * $d_1$ is an upper bound of $\set {x_1, y_1}$

and:
 * $d_2$ is an upper bound of $\set {x_2, y_2}$.

As $\tuple {c_1, c_2}$ is also an upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$, it similarly follows that:
 * $c_1$ is an upper bound of $\set {x_1, y_1}$

and:
 * $c_2$ is an upper bound of $\set {x_2, y_2}$.

Now we have:

Thus:
 * $c_1$ is an upper bound of $\set {x_1, y_1}$

and:
 * if $d_1$ is an upper bound of $\set {x_1, y_1}$, then $c_1 \preccurlyeq_1 d_1$

so $\set {x_1, y_1}$ admits a supremum $c_1$ in $\struct {S_1, \preccurlyeq_1}$.

Also:
 * $c_2$ is an upper bound of $\set {x_2, y_2}$

and:
 * if $d_2$ is an upper bound of $\set {x_2, y_2}$, then $c_2 \preccurlyeq_1 d_2$

so $\set {x_2, y_2}$ admits a supremum $c_2$ in $\struct {S_2, \preccurlyeq_2}$.

Now let $\tuple {c_1, c_2} = \inf \set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ be the infimum of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$.

Mutatis mutandis, we use a similar argument to the above to show that:


 * $\set {x_1, y_1}$ admits an infimum $c_1$ in $\struct {S_1, \preccurlyeq_1}$

and:


 * $\set {x_2, y_2}$ admits an infimum $c_2$ in $\struct {S_2, \preccurlyeq_2}$.

As $x_1$, $x_2$, $y_1$ and $y_2$ are arbitrary, it follows that both $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ are lattices.

Necessary Condition
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be lattices.

Let $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ be arbitrary elements of $T$.

Then:


 * $x_1$ and $y_1$ are elements of $S_1$


 * $x_2$ and $y_2$ are elements of $S_2$.

By definition of lattice, both $\set {x_1, y_1}$ and $\set {x_2, y_2}$ admit both a supremum and an infimum.

Let $c_1 = \sup \set {x_1, y_1}$ and $c_2 = \sup \set {x_2, y_2}$ be the suprema of $\set {x_1, y_1}$ and $\set {x_2, y_2}$ respectively.

Then by definition:
 * $c_1$ is an upper bound of $\set {x_1, y_1}$ in $S_1$


 * $c_1 \preccurlyeq_1 d_1$ for all upper bounds $d_1$ of $\set {x_1, y_1}$ in $S_1$

and:


 * $c_2$ is an upper bound of $\set {x_2, y_2}$ in $S_2$


 * $c_2 \preccurlyeq_2 d_2$ for all upper bounds $d_2$ of $\set {x_2, y_2}$ in $S_2$

Let:
 * $d_1$ be an arbitrary upper bound of $\set {x_1, y_1}$ in $S_1$


 * $d_2$ be an arbitrary upper bound of $\set {x_2, y_2}$ in $S_2$.

Similarly:

Thus:
 * $\tuple {d_1, d_2}$ is an upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$

As $c_1$ and $c_2$ are also upper bounds of of $\set {x_1, y_1}$ and $\set {x_2, y_2}$ respectively, it similarly follows that:
 * $\tuple {c_1, c_2}$ is an upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$.

Now we have:

Thus:
 * $\tuple {c_1, c_2}$ is an upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$

and:
 * if $\tuple {d_1, d_2}$ is an upper bound of $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$, then $\tuple {c_1, c_2} \preccurlyeq_s \tuple {d_1, d_2}$

so $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ admits a supremum $\tuple {c_1, c_2}$ in $\struct {T, \preccurlyeq_s}$.

Now let:
 * $c_1 = \inf \set {x_1, y_1}$ be the infimum of $\set {x_1, y_1}$

and:
 * $c_2 = \inf \set {x_2, y_2}$ be the infimum of $\set {x_2, y_2}$.

Mutatis mutandis, we use a similar argument to the above to show that:


 * $\set {\tuple {x_1, x_2}, \tuple {y_1, y_2} }$ admits an infimum $\tuple {c_1, c_2}$ in $\struct {T, \preccurlyeq_s}$.

As $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ are arbitrary, it follows that $\struct {T, \preccurlyeq_s}$ is a lattice.