Upper Section with no Minimal Element
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $U \subseteq S$.
Then:
- $U$ is an upper section in $S$ with no minimal element
- $\ds U = \bigcup \set {u^\succ: u \in U}$
where $u^\succ$ is the strict upper closure of $u$.
Proof
Forward implication
Let $U$ be an upper section in $S$ with no minimal element.
Then by the definition of upper section:
- $\ds \bigcup \set {u^\succ: u \in U} \subseteq U$
Let $x \in U$.
Since $U$ has no minimal element, $x$ is not minimal.
Thus there is a $u \in U$ such that $u \prec x$.
Then $x \in u^\succ$, so:
- $\ds x \in \bigcup \set {u^\succ: u \in U }$
Since this holds for all $x \in U$:
- $\ds U \subseteq \bigcup \set {u^\succ: u \in U}$
Thus the theorem holds by definition of set equality.
$\Box$
Reverse implication
Let:
- $\ds U = \bigcup \set {u^\succ: u \in U}$
Then:
- $\forall u \in U: u^\succ \subseteq U$
so $U$ is an upper section.
Furthermore:
- $\forall x \in U: \exists u \in U: x \in u^\succ$
But then:
- $u \prec x$
so $x$ is not minimal.
Since this holds for all $x \in U$, $U$ has no minimal element.
$\blacksquare$