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?
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?
Let $\displaystyle x,y\in X$ be distinct. Since $\displaystyle <$ is a total ordering on $\displaystyle X$ we may assume without loss of generality that $\displaystyle x<y$. We have two choices, either there exists $\displaystyle z$ with $\displaystyle x<z<y$ in which case take $\displaystyle U=(-\infty,z)$ and $\displaystyle V=(z,\infty)$, else $\displaystyle (x,y)$ is empty and take $\displaystyle U=(-\infty,y)$ and $\displaystyle V=(x,\infty)$. Regardless, $\displaystyle U$ and $\displaystyle V$ are disjoint neighborhoods of $\displaystyle x,y$ respectively. The conclusion follows.