# Supremum of Cartesian Product

## Theorem

Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be an ordered set.

Let $\left({S_1 \times S_2, \precsim}\right)$ be the Cartesian product of $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$.

Let $X_1$ be a non-empty subset of $S_1$, $X_2$ be a non-empty subset of $S_2$ such that

$X_1$ and $X_2$ admit suprema.

Then

$X_1 \times X_2$ admits a supremum

and

$\sup \left({X_1 \times X_2}\right) = \left({\sup X_1, \sup X_2}\right)$

## Proof

We will prove that

$\left({\sup X_1, \sup X_2}\right)$ is upper bound for $X_1 \times X_2$

Let $\left({a, b}\right) \in X_1 \times X_2$.

By definition of Cartesian product:

$a \in X_1$ and $b \in X_2$

By definitions of supremum and upper bound:

$a \preceq_1 \sup X_1$ and $b \preceq_2 \sup X_2$

Thus by definition Cartesian product of ordered sets:

$\left({a, b}\right) \precsim \left({\sup X_1, \sup X_2}\right)$

$\Box$

We will prove that

$\forall \left({a, b}\right) \in S1 \times S_2: \left({a, b}\right)$ is upper bound for $X_1 \times X_2 \implies \left({\sup X_1, \sup X_2}\right) \precsim \left({a, b}\right)$

Let $\left({a, b}\right) \in S1 \times S_2$ such that

$\left({a, b}\right)$ is upper bound for $X_1 \times X_2$

We will prove as sublemma that

$a$ is upper bound for $X_1$

Let $c \in X_1$.

By definition of non-empty set:

$\exists d: d \in X_2$

By definition of Cartesian product:

$\left({c, d}\right) in X_1 \times X_2$

By definition of upper bound:

$\left({c, d}\right) \precsim \left({a, b}\right)$

Thus be definition of Cartesian product of ordered sets:

$c \preceq_1 a$

This ends the proof of sublemma.

Analogically we have that

$b$ is upper bound for $X_2$

By definition of supremum;

$\sup X_1 \preceq_1 a$ and $\sup X_2 \preceq_2 b$

Thus by definition of Cartesian product of ordered sets:

$\left({\sup X_1, \sup X_2}\right) \precsim \left({a, b}\right)$

$\Box$

Thus by definition

$X_1 \times X_2$ admits a supremum

and

$\sup \left({X_1 \times X_2}\right) = \left({\sup X_1, \sup X_2}\right)$

$\blacksquare$