Order Topology is Hausdorff

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Theorem

Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space.


Then $\left({X, \tau}\right)$ is a Hausdorff space.


Proof

Let $x, y \in X$ with $x \ne y$.

Since $\le$ is a total ordering, either $x \prec y$ or $y \prec x$.

Without loss of generality, assume that $x \prec y$.

If there is a $z \in X$ such that $x \prec z \prec y$, then ${\downarrow}z$ and ${\uparrow}z$ separate $x$ and $y$.



Otherwise, by Mind the Gap, ${\downarrow}y$ and $\uparrow x$ separate $x$ and $y$.

Since any two distinct points can be separated by neighborhoods, $\left({X, \tau}\right)$ is a Hausdorff space.

$\blacksquare$