Interval of Totally Ordered Set is Convex

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

Let $I \subseteq S$ be an interval in $S$.

Then $I$ is convex.

Proof
There are a number of cases to investigate.

Open Interval
Let $I = \openint a b$ be an open interval:
 * $I = \set {x \in S: a \prec x \prec b}$

Let $s, t, x \in I$ such that $s \prec x \prec t$.

Then by definition:
 * $a \prec s \prec x$

and:
 * $x \prec t \prec b$

and so:
 * $a \prec x \prec b$

and $x \in I$.

Thus we have:
 * $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$

and $I$ is convex by definition.

Left Half-Open Interval
Let $I = \hointl a b$ be a left half-open interval:
 * $I = \set {x \in S: a \prec x \preceq b}$

Let $s, t, x \in I$ such that $s \prec x \prec t$.

Then by definition:
 * $a \prec s \prec x$

and:
 * $x \prec t \preceq b$

and so:
 * $a \prec x \preceq b$

and $x \in I$.

Thus we have:
 * $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$

and $I$ is convex by definition.

Right Half-Open Interval
Let $I = \hointr a b$ be a right half-open interval:
 * $I = \set {x \in S: a \preceq x \prec b}$

Let $s, t, x \in I$ such that $s \prec x \prec t$.

Then by definition:
 * $a \preceq s \prec x$

and:
 * $x \prec t \prec b$

and so:
 * $a \preceq x \prec b$

and $x \in I$.

Thus we have:
 * $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$

and $I$ is convex by definition.

Closed Interval
Let $I = \closedint a b$ be a closed interval:
 * $I = \set {x \in S: a \preceq x \preceq b}$

Let $s, t, x \in I$ such that $s \prec x \prec t$.

Then by definition:
 * $a \preceq s \prec x$

and:
 * $x \prec t \preceq b$

and so:
 * $a \preceq x \preceq b$

and $x \in I$.

Thus we have:
 * $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$

and $I$ is convex by definition.