# 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 $b^\prec = a^\preccurlyeq$.

$\blacksquare$