Supremum of Subset

Theorem

Let $\left({U, \preceq}\right)$ be an ordered set.

Let $S \subseteq U$.

Let $T \subseteq S$.

Let $S$ admit a supremum (in $U$).

If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.

Proof

Let $B = \sup \left({S}\right)$.

Then $B$ is an upper bound for $S$.

As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$.

Because $x \in S \implies x \preceq B$ (as $B$ is an upper bound for $S$) it follows that $x \in T \implies x \preceq B$.

So $B$ is an upper bound for $T$.

Therefore $B$ succeeds the supremum of $T$ in $S$.

Hence the result.

$\blacksquare$