Upper Closure is Strict Upper Closure of Immediate Predecessor

Theorem
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

Let $b$ be the immediate successor element of $a$:

Then:
 * $a^\succ = b^\succcurlyeq$

where:
 * $a^\succ$ is the strict upper closure of $a$
 * $b^\succcurlyeq$ is the upper closure of $b$.

Proof
Let:
 * $x \in a^\succ$

By the definition of strict upper closure:
 * $a \prec x$

By the definition of total ordering:
 * $x \prec b$ or $x \succcurlyeq b$

If $x \prec b$ then $a \prec x \prec b$, contradicting the premise.

Thus:
 * $x \succcurlyeq b$

and so:
 * $x \in b^\succcurlyeq$

By definition of subset:
 * $a^\succ \subseteq b^\succcurlyeq$

Let:
 * $x \in b^\succcurlyeq$

By the definition of upper closure:
 * $b \preccurlyeq x$

Since $a \prec b$, Extended Transitivity shows that $a \prec x$.

Thus:
 * $x \in a^\succ$

By definition of subset:
 * $b^\succcurlyeq \subseteq a^\succ$

Therefore by definition of set equality:
 * $a^\succ = b^\succcurlyeq$