Upper Section with no Minimal Element

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $U \subseteq S$.

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

{{iff}:
 * $\displaystyle U = \bigcup \left\{{u^\succ: u \in U}\right\}$

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:
 * $\displaystyle \bigcup \left\{{u^\succ: u \in U}\right\} \subseteq U$

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:
 * $\displaystyle x \in \bigcup \left\{{u^\succ: u \in U }\right\}$

Since this holds for all $x \in U$:
 * $\displaystyle U \subseteq \bigcup \left\{{u^\succ: u \in U}\right\}$

Thus the theorem holds by definition of set equality.

Reverse implication
Let:
 * $\displaystyle U = \bigcup \left\{{u^\succ: u \in U }\right\}$

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