Strict Upper Closure is Upper Section

Theorem

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

Let $p \in S$.

Let $p^\succ$ denote the strict upper closure of $p$.

Then $p^\succ$ is an upper section.

Proof

Let $u \in p^\succ$.

Let $s \in S$ with $u \preceq s$.

Then by the definition of strict upper closure:

$p \prec u$

Thus by Extended Transitivity:

$p \prec s$

So by the definition of strict upper closure:

$s \in p^\succ$

Since this holds for all such $u$ and $s$, $p^\succ$ is an upper section.

$\blacksquare$

Also see

• Upper Closure is Upper Section
• Strict Lower Closure is Lower Section