Thread: Simply ordered set

Every simply ordered set is a Hausdorff space in the order topology.

Since it is simply ordered, the relation is reflexive, anti-symmetric, and transitive. How can this coupled with the Hausdorff condition help show that it is the order topology?

Originally Posted by dwsmith

Every simply ordered set is a Hausdorff space in the order topology.

Since it is simply ordered, the relation is reflexive, anti-symmetric, and transitive. How can this coupled with the Hausdorff condition help show that it is the order topology?

You are misinterpreting the question. It's saying that if is a space where you haven given it the order topology for some total ordering then the resulting topological space is Hausdorff.

Originally Posted by Drexel28

You are misinterpreting the question. It's saying that if is a space where you haven given it the order topology for some total ordering then the resulting topological space is Hausdorff.

Ok. However, I am not sure how to do that one either.

Originally Posted by dwsmith

Ok. However, I am not sure how to do that one either.

Let be distinct. Since is a total ordering on we may assume without loss of generality that . We have two choices, either there exists with in which case take and , else is empty and take and . Regardless, and are disjoint neighborhoods of respectively. The conclusion follows.