Mind the Gap

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Theorem

Let $\left({S, \preccurlyeq}\right)$ be a totally ordered set.

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

Suppose that:

$\left\{ {x \in S: a \prec x \prec b}\right\} = \varnothing$


Then:

$a^\succ = b^\succcurlyeq$

and

$b^\prec = a^\preccurlyeq$

where:

$a^\succ$ is the strict upper closure of $a$
$b^\succcurlyeq$ is the weak upper closure of $b$
$b^\prec$ is the strict lower closure of $b$
$a^\preccurlyeq$ is the weak lower closure of $a$.


Proof

Let:

$p \in a^\succ$

By the definition of strict upper closure:

$a \prec p$

By the definition of total ordering, $p \prec b$ or $p \succcurlyeq b$.

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

Thus $p \succcurlyeq b$, so $p \in b^\succcurlyeq$.

By definition of subset:

$a^\succ \subseteq b^\succcurlyeq$


Let:

$p \in b^\succcurlyeq$

By the definition of weak upper closure:

$b \preccurlyeq p$

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

Thus:

$p \in a^\succ$


By definition of subset:

$b^\succcurlyeq \subseteq a^\succ$


Therefore by definition of set equality:

$a^\succ = b^\succcurlyeq$

$\Box$


A similar argument shows that $b^\prec = a^\preccurlyeq$.

$\blacksquare$