# Every Point except Endpoint in Connected Linearly Ordered Space is Cut Point

## Theorem

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.

Let $A \subseteq S$ be a connected space.

Let $p \in A$ be a point of $A$ which is not an endpoint of $A$.

Then $p$ is a cut point of $A$.

## Proof

We have that $A \setminus \set p$ is separated by $\set {x \in A: x \prec p}$ and $\set {x \in A: p \prec x}$.

If $p$ is an endpoint of $A$, then either:

$\set {x \in A: x \prec p} = \O$

or:

$\set {x \in A: p \prec x} = \O$

$\blacksquare$