Supremum is Increasing relative to Product Ordering

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Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $I$ be a set.

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

Let $f \sqbrk I$ denote the image of $I$ under $f$.

Let:

$\forall i \in I: \map f i \preceq \map g i$

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

Let $f \sqbrk I$ and $g \sqbrk I$ admit suprema.

Then:

$\sup f \sqbrk I \preceq \sup g \sqbrk I$

Proof

Let $x \in f \sqbrk I$.

Then:

$\exists j \in I: \map f j = x$

Then:

$\map f j \prec \map g j$

By the definition of supremum:

$\sup g \sqbrk I$ is an upper bound of $g \sqbrk I$

Thus:

$\map g j \preceq \sup g \sqbrk I$

Since $\preceq$ is transitive:

$x = \map f j \preceq \sup g \sqbrk I$

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

Thus by the definition of supremum:

$\sup f \sqbrk I \preceq \sup g \sqbrk I$

$\blacksquare$