Upper Section with no Minimal Element

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

Let $U \subseteq S$.

Then:
 * $U$ is an upper set in $S$ with no minimal element


 * $\ds U = \bigcup \set {u^\succ: u \in U}$
 * $\ds U = \bigcup \set {u^\succ: u \in U}$

where $u^\succ$ is the strict upper closure of $u$.

Forward implication
Let $U$ be an upper set in $S$ with no minimal element.

Then by the definition of upper set:
 * $\ds \bigcup \set {u^\succ: u \in U} \subseteq U$

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.

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 set.

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.

Also see

 * Lower Set with no Maximal Element