Show that every order topology is Hausdorff.
Solution
Let \((X,<)\) be a simply ordered set with the order topology, and let \(x,y\in X\) with \(x<y\). The open rays \(\{z\in X:z<a\}\) and \(\{z\in X:z>a\}\), for \(a\in X\), are open in the order topology.
1. If some \(c\in X\) satisfies \(x<c<y\), then \(x\) and \(y\) have disjoint open neighborhoods.
Take \(U=\{z:z<c\}\) and \(V=\{z:z>c\}\). Then \(x\in U\), \(y\in V\), and no \(z\) satisfies both \(z<c\) and \(z>c\), so \(U\cap V=\varnothing\).
2. If no \(c\in X\) satisfies \(x<c<y\), then \(x\) and \(y\) have disjoint open neighborhoods.
Take \(U=\{z:z<y\}\) and \(V=\{z:z>x\}\). Then \(x\in U\) and \(y\in V\), and a point of \(U\cap V\) would satisfy \(x<z<y\), so \(U\cap V=\varnothing\).
3. Q.E.D.
Steps 1 and 2 cover both cases, so any two distinct points of \(X\) have disjoint open neighborhoods.