Linearly Ordered Space is Completely Normal

Theorem

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

Then $T$ is a completely normal space.

Proof

By Linearly Ordered Space is $T_1$, $T$ is a $T_1$ (Fréchet) space.

By Linearly Ordered Space is $T_5$, $T$ is a $T_5$ space.

Hence the result, by definition of completely normal space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $6$