Upper Closure is Strict Upper Closure of Immediate Predecessor
Jump to navigation
Jump to search
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:
- $a^\succ = b^\succcurlyeq$
$\blacksquare$