Supremum of Lower Closure of Element

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Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $s$ be an element of $S$.


Then:

$\map \sup {s^\preceq} = s$

where $s^\preceq$ denotes the lower closure of $s$.


Proof

\(\ds \map \sup {s^\preceq}\) \(=\) \(\ds \map \sup {\set s^\preceq}\) Lower Closure of Singleton
\(\ds \) \(=\) \(\ds \map \sup {\set s}\) Supremum of Lower Closure of Set
\(\ds \) \(=\) \(\ds s\) Supremum of Singleton

$\blacksquare$


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