Order Topology equals Dual Order Topology

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Theorem

Let $\left({S, \preceq}\right)$ 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

Let $U$ be an open ray in $\left({S, \preceq}\right)$.

By Open Ray is Dual to Open Ray, $U$ is an open ray in $\left({S, \preceq}\right)$.

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


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

Thus by definition of set equality:

$\tau' = \tau$

$\blacksquare$