Supremum in Ordered Subset

Theorem
Let $L = \left({S, \preceq}\right)$ be an ordered set.

Let $R = \left({T, \preceq'}\right)$ be an ordered subset of $L$.

Let $X \subseteq T$ such that
 * $X$ admits an supremum in $L$.

Then $\sup_L X \in T$
 * $X$ admits an supremum in $R$ and $\sup_R X = \sup_L X$

Proof
This follows by mutatis mutandis of the proof of Infimum in Ordered Subset.