Supremum is Increasing relative to Product Ordering

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $I$ be a Definition:set.

Let $f, g: I \to S$.

Suppose that for each $i \in I$, $f(i) \preceq g_(i)$.

That is, suppose that $f \preceq g$ in the product ordering.

Suppose also that $f(I)$ and $g(I)$ admit suprema.

Then $\sup f(I) \preceq \sup g(I)$.

Proof
Let $x \in f(I)$.

Then for some $j \in I$, $f(j) = x$.

Then $f(j) \prec g(j)$.

By the definition of supremum:
 * $\sup g(I)$ is an upper bound of $g(I)$.

Thus $g(j) \preceq \sup g(I)$.

Since $\preceq$ is transitive:
 * $x = f(j) \preceq \sup g(I)$.

Since this holds for all $x \in f(I)$, $\sup g(I)$ is an upper bound of $f(I)$.

Thus by the definition of supremum, $\sup f(I) \preceq \sup g(I)$.