Infimum of Upper Closure of Set

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

Let $T \subseteq S$.

Let $U = T^\succeq$ be the upper closure of $T$ in $S$.

Let $s \in S$.

Then $s$ is the infimum of $T$ it is the infimum of $U$.

Proof
This follows by mutatis mutandis of the proof of Supremum of Lower Closure of Set.