Supremum is Increasing relative to Product Ordering

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Theorem

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

Let $I$ be a set.

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

Let $f \left[{I}\right]$ denote the image of $I$ under $f$.


Let:

$\forall i \in I: f \left({i}\right) \preceq g \left({i}\right)$

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


Let $f \left[{I}\right]$ and $g \left[{I}\right]$ admit suprema.


Then:

$\sup f \left[{I}\right] \preceq \sup g \left[{I}\right]$


Proof

Let $x \in f \left[{I}\right]$.

Then:

$\exists j \in I: f \left({j}\right) = x$

Then:

$f \left({j}\right) \prec g \left({j}\right)$

By the definition of supremum:

$\sup g \left[{I}\right]$ is an upper bound of $g \left[{I}\right]$

Thus:

$g \left({j}\right) \preceq \sup g \left[{I}\right]$

Since $\preceq$ is transitive:

$x = f \left({j}\right) \preceq \sup g \left[{I}\right]$

Since this holds for all $x \in f \left[{I}\right]$, $\sup g \left[{I}\right]$ is an upper bound of $f \left[{I}\right]$.

Thus by the definition of supremum:

$\sup f \left[{I}\right] \preceq \sup g \left[{I}\right]$

$\blacksquare$