User:Dfeuer/Upper Closure of Element of Complete Lattice is Complete Lattice

Theorem

Let $(L, \preceq)$ be a complete lattice.

Let $p \in L$.

Then $({\uparrow}p, \preceq)$ is a complete lattice.

Let $U = {\uparrow} p$

Let $S \subseteq U$.

If $S$ is empty, then $p$ is its supremum in $U$.

Otherwise, let $s$ be an arbitrary element of $S$.

Then $p \preceq s \preceq \bigvee S$ by the definition of supremum, so $\bigvee S \in U$ by the definition of upper closure.

Since $p \preceq x$ for each $x \in S$, $p \preceq \bigwedge S$ by the definition of infimum.

Thus $\bigwedge S \in U$.

Since every subset of $U$ has a supremum and infimum in $U$, $U$ is a complete lattice.

$\blacksquare$