Why Every Order Topology Is Hausdorff

Separating Points in an Ordered Space

The order topology turns the order on a set into a supply of open neighborhoods. Its basic open sets are intervals and rays, so the relative position of two points suggests how to separate them. The only apparent difficulty is that a pair of points may be adjacent: there may be no point between them at which to split the order. Rays handle that case as well.

Let \(X\) be a set with a linear order \(<\). A linear order compares every pair of elements, and it is transitive and antisymmetric. For \(a,b\in X\) with \(a<b\), write \((a,b)=\{t\in X:a<t<b\}\). The initial and final rays are \((-\infty,b)=\{t\in X:t<b\}\) and \((a,\infty)=\{t\in X:a<t\}\). These are relative to \(X\); for example, they need not be unbounded in any numerical sense.

In particular, the rays \((-\infty,b)\) and \((a,\infty)\) are open in the order topology. We say that \(x<y\) are adjacent if there is no \(z\in X\) with \(x<z<y\). The proof of the Hausdorff property uses one choice of rays when such a \(z\) exists and another when it does not.

The Separation Theorem

Proof. If \(X\) has fewer than two points, there are no distinct points to separate, so the Hausdorff condition holds vacuously. Otherwise take any two distinct points \(x,y\in X\). Linearity lets us name them in increasing order, so assume \(x<y\). If there is a point \(z\in X\) with \(x<z<y\), let \(U=(-\infty,z)\) and \(V=(z,\infty)\). These are open rays, \(x\in U\), and \(y\in V\). No point can belong to both: membership in both would require that point to be strictly less than and strictly greater than \(z\). Thus \(U\cap V=\varnothing\). If there is no such \(z\), then \(x,y\) are adjacent. Let \(U=(-\infty,y)\) and \(V=(x,\infty)\). These are open rays, \(x\in U\), and \(y\in V\). If \(t\in U\cap V\), then \(x<t<y\), contradicting adjacency. Hence \(U\cap V=\varnothing\) in this case too. We have found disjoint open neighborhoods for every pair of distinct points, so \(X\) is Hausdorff. \(\square\)

The distinction between the cases matters. When there is an element between \(x\) and \(y\), cutting at that element makes disjoint rays. When the points are adjacent, the rays ending at \(y\) and beginning at \(x\) are disjoint precisely because the interval between them is empty. The proof does not require that \(X\) have a first or last element, or that between any two points there be another point.

A Consequence: Limits Are Unique

Hausdorffness is useful because it prevents a sequence from converging to two different points. Here convergence has its neighborhood meaning: a sequence \((x_n)\) converges to \(x\) if every open neighborhood of \(x\) contains all terms \(x_n\) for sufficiently large \(n\).

Proof. Suppose \((x_n)\) converges to both \(x\) and \(y\), and assume \(x\ne y\). Since the space is Hausdorff, there are disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Convergence to \(x\) gives an index \(N_1\) such that \(x_n\in U\) whenever \(n\ge N_1\). Convergence to \(y\) gives an index \(N_2\) such that \(x_n\in V\) whenever \(n\ge N_2\). For \(n\ge\max(N_1,N_2)\), the term \(x_n\) would belong to \(U\cap V=\varnothing\), which is impossible. Thus \(x=y\), and the limit is unique whenever it exists. The theorem proves that every order topology is Hausdorff, so the conclusion applies there. \(\square\)

This consequence does not assert that every sequence in an ordered space converges. It says only that two different limits cannot occur. The distinction between existence and uniqueness is important: the separation argument supplies disjoint neighborhoods, not a limit for any particular sequence.

Common Pitfall: The Open Interval May Not Separate

A tempting but incorrect approach is to use the interval \((x,y)\) itself as a neighborhood of either \(x\) or \(y\). By definition, its elements satisfy \(x<t<y\), so neither endpoint belongs to it. For an adjacent pair it is empty as well. An empty set is open, but it is not a neighborhood of either endpoint because it contains neither one.

For a concrete calculation, take \(X=\{0,1\}\) with \(0<1\). The interval \((0,1)=\{t\in X:0<t<1\}\) is \(\varnothing\). It cannot be used as a neighborhood of \(0\) or \(1\). The correction is to use the endpoint rays \(U=(-\infty,1)=\{0\}\) and \(V=(0,\infty)=\{1\}\). These are open in the order topology, contain \(0\) and \(1\), respectively, and satisfy \(\{0\}\cap\{1\}=\varnothing\).

A related error is to begin by choosing an element \(z\) with \(x<z<y\) without checking that one exists. In the two-point example, no such \(z\) exists, so that step cannot be carried out. The proof avoids this gap by splitting into the two exhaustive possibilities: an intermediate element exists, or the pair is adjacent. In both cases, it specifies open sets, verifies which points they contain, and checks that their intersection is empty.