Show that every order topology is hausdorff?

In an order topology, the basis consists of open intervals defined by the order relation. For any two distinct points ( x ) and ( y ) in a totally ordered set, without loss of generality, assume ( x < y ). We can find open sets ( U = (x - \epsilon, y) ) and ( V = (x, y + \epsilon) ) for some small ( \epsilon > 0 ) such that ( U ) contains ( x ) and ( V ) contains ( y ). Since ( U ) and ( V ) are disjoint, this shows that every pair of distinct points can be separated by neighborhoods, confirming that the order topology is Hausdorff.