Order Topology is Hausdorff


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Theorem

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.


Then $\struct {X, \tau}$ 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 $z^\prec$ and $z^\succ$ separate $x$ and $y$.


Otherwise, by Upper Closure is Strict Upper Closure of Immediate Predecessor, $y^\prec$ and $x^\succ$ separate $x$ and $y$.

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

$\blacksquare$

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