Infimum of Upper Closure of Set

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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 $T$ if and only if it is the infimum of $U$.


Proof

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

$\blacksquare$


Sources