Supremum in Ordered Subset

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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 $\sup_L X \in T$ if and only if

$X$ 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.

$\blacksquare$


Sources