Supremum of Subset
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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.
$\blacksquare$