Upper Section with no Minimal Element

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

if and only if:

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

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.

$\blacksquare$

Also see

• Lower Section with no Maximal Element