Lower Closure is Strict Lower Closure of Immediate Successor

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

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

Then:
 * $a^\preccurlyeq = b^\prec$

where:
 * $a^\preccurlyeq$ is the lower closure of $a$
 * $b^\prec$ is the strict lower closure of $b$.

Proof
Let:
 * $x \in b^\prec$

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

By the definition of total ordering:
 * $a \prec x$ or $x \preccurlyeq a$

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

Thus:
 * $x \preccurlyeq a$

and so:
 * $x \in a^\preccurlyeq$

By definition of subset:
 * $b^\prec \subseteq a^\preccurlyeq$

Let:
 * $x \in a^\preccurlyeq$

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

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

Thus:
 * $x \in b^\prec$

By definition of subset:
 * $a^\preccurlyeq \subseteq b^\prec$

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