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:

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

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $9$