Upper Section with no Minimal Element

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $U \subseteq S$.

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


 * $U = \bigcup \left\{{ {\uparrow}u: u \in U }\right\}$

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

Then $\bigcup \left\{{ {\uparrow}u: u \in U }\right\} \subseteq U$ by the definition of upper set.

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 $x \in {\uparrow}u$, so $x \in \bigcup \left\{{ {\uparrow}u: u \in U }\right\}$.

Since this holds for all $x \in U$, $U \subseteq \bigcup \left\{{ {\uparrow}u: u \in U }\right\}$.

Thus the theorem holds by Equality of Sets

Reverse implication
Suppose that $U = \bigcup \left\{{ {\uparrow}u: u \in U }\right\}$.

Then for each $u \in U$: ${\uparrow}u \subseteq U$, so $U$ is an upper set.

Furthermore, for each $x \in U$, there is a $u \in U$ such that $x \in {\uparrow}u$.

But then $u \prec x$, so $x$ is not minimal.

Since this holds for all all $x \in U$, $U$ has no minimal element.