Order Topology is Normal

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Theorem

Let $\left({S, \preceq}\right)$ be a toset.

Let $\tau$ be the order topology on $S$.


Then $\left({S, \tau}\right)$ is normal.


Proof

From Linearly Ordered Space is Completely Normal, $\left({S, \tau}\right)$ is a completely normal space.

The result follows from Completely Normal Space is Normal Space.

$\blacksquare$