# Interval of Totally Ordered Set is Convex

It has been suggested that this page or section be merged into Interval of Ordered Set is Convex.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Mergeto}}` from the code. |

## 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.

$\Box$

### 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.

$\Box$

### 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.

$\Box$

### 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.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $1$