Order Topology equals Dual Order Topology

Theorem

Let $\struct {S, \preceq}$ be a totally ordered set.

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

Let $\tau'$ be the $\succeq$-order topology on $S$, where $\succeq$ is the dual ordering of $\preceq$.

Then $\tau' = \tau$.

Proof

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Let $U$ be an open ray in $\struct {S, \preceq}$.

By Open Ray is Dual to Open Ray, $U$ is an open ray in $\struct {S, \preceq}$.

Since the open rays in a totally ordered set form a sub-basis for the topology on that set, $\tau'$ is finer than $\tau$.

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By the same argument, $\tau$ is finer than $\tau'$.

Thus by definition of set equality:

$\tau' = \tau$

$\blacksquare$