# Thread:

1. ## 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?

2. ## Re: Simply ordered set

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 $X$ is a space where you haven given it the order topology for some total ordering then the resulting topological space is Hausdorff.

3. ## Re: Simply ordered set

Originally Posted by Drexel28
You are misinterpreting the question. It's saying that if $X$ 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.

4. ## Re: Simply ordered set

Originally Posted by dwsmith
Ok. However, I am not sure how to do that one either.
Let $x,y\in X$ be distinct. Since $<$ is a total ordering on $X$ we may assume without loss of generality that $x<y$. We have two choices, either there exists $z$ with $x<z<y$ in which case take $U=(-\infty,z)$ and $V=(z,\infty)$, else $(x,y)$ is empty and take $U=(-\infty,y)$ and $V=(x,\infty)$. Regardless, $U$ and $V$ are disjoint neighborhoods of $x,y$ respectively. The conclusion follows.
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