Upper Closure is Strict Upper Closure of Immediate Predecessor

Theorem
Let $(S, \le)$ be a totally ordered set.

Let $a, b \in S$ with $a < b$.

Suppose that $\{ x \in S: a < x < b\} = \varnothing$.

Then ${\uparrow}a = {\bar\uparrow}b$,

where ${\uparrow}a$ is the strict upper closure of $a$ and ${\bar\uparrow}b$ is the weak upper closure of $b$.

Proof
We will show that

${\uparrow}a \subseteq {\bar\uparrow}b$ and ${\bar\uparrow}b \subseteq {\uparrow}a$.

Suppose that $p \in {\uparrow}a$

By the definition of strict upper closure, $a < p$.

By the definition of total ordering, $p < b$ or $p \ge b$.

But if $p < b$ then $a < p < b$, contradicting the premise.

Thus $p \ge b$, so $p \in {\bar\uparrow}b$.

Now suppose that $p \in {\bar\uparrow}b$.

By the definition of weak upper closure:
 * $b \le p$

Since $a < b$, Extended Transitivity shows that $a < p$.

Thus $p \in {\uparrow}a$.