# Order Topology equals Dual Order Topology

Jump to navigation
Jump to search

## 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:

- $\tau' = \tau$

$\blacksquare$