Lower Closure is Strict Lower Closure of Immediate Successor
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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:
- $a^\succ = b^\succcurlyeq$
$\blacksquare$