The Order Topology on Q and Real Intervals

Two Ways to Give the Rationals a Topology

The rational numbers inherit an order from the real line, so they can be given the order topology. They also inherit a topology by being viewed as a subset of the real line. These are different constructions: one starts with the order on \(\mathbb{Q}\), while the other takes intersections with open subsets of \(\mathbb{R}\). In this case the resulting topologies agree. The same comparison clarifies why neighborhoods at included endpoints of real intervals have a one-sided form, while excluded endpoints have no neighborhoods in the interval at all.

For an ordered set, the order topology is generated by open order intervals and open order rays. In \(\mathbb{Q}\), for example, the bounded basic intervals are \((r,s)_{\mathbb{Q}}=\{q\in\mathbb{Q}:r<q<s\}\), where \(r,s\in\mathbb{Q}\) and \(r<s\). The subspace topology on \(\mathbb{Q}\subseteq\mathbb{R}\) consists of sets \(O\cap\mathbb{Q}\), where \(O\) is open in \(\mathbb{R}\).

Proof. We show that each topology is contained in the other. First, every basic order interval \((r,s)_{\mathbb{Q}}\) is \((r,s)\cap\mathbb{Q}\), where \((r,s)\) is open in \(\mathbb{R}\). Each basic order ray, such as \(\{q\in\mathbb{Q}:q<s\}\), is \((-\infty,s)\cap\mathbb{Q}\), also the intersection of \(\mathbb{Q}\) with a real open set. Thus the order-topology basis sets are open in the subspace topology, so every order-open set is subspace-open.

For the reverse inclusion, let \(V=O\cap\mathbb{Q}\) be subspace-open, with \(O\) open in \(\mathbb{R}\), and take \(q\in V\). Since \(O\) is open, there is \(\varepsilon>0\) such that \((q-\varepsilon,q+\varepsilon)\subseteq O\). The rationals are dense in the reals: there are rationals \(r,s\) such that \(q-\varepsilon<r<q<s<q+\varepsilon\). Consequently \(q\in(r,s)_{\mathbb{Q}}\subseteq V\). Each point of \(V\) therefore lies in an order-open basic interval contained in \(V\), so \(V\) is order-open. The two topologies have exactly the same open sets. \(\square\)

The density fact used in the proof follows from the Archimedean property of \(\mathbb{R}\). Given real numbers \(x<y\), choose \(n\in\mathbb{N}\) with \(n(y-x)>1\), and choose an integer \(m\) with \(nx<m\le nx+1\). Then \(nx<m<ny\), so \(x<m/n<y\). Applying this once on each side of \(q\) gives the rationals \(r\) and \(s\) used above.

Intervals as Ordered Sets and as Subspaces

Now fix real numbers \(a<b\), and let \(J\) be one of \((a,b)\), \([a,b]\), \([a,b)\), or \((a,b]\). We can give \(J\) its order topology using the usual order inherited from \(\mathbb{R}\), or its subspace topology as a subset of \(\mathbb{R}\). In each case these topologies agree. To state their basic neighborhoods precisely, write \((u,v)_J=(u,v)\cap J\). At an included left endpoint the corresponding relative interval is \([a,v)\), and at an included right endpoint it is \((u,b]\).

Proof. The order topology on \(J\) has a basis consisting of intersections of \(J\) with open intervals and open rays in the inherited order. These sets are also intersections of \(J\) with open intervals or rays in \(\mathbb{R}\), so they are open in the subspace topology. This proves that every order-open subset of \(J\) is subspace-open.

Conversely, take a subspace-open set \(W=O\cap J\), where \(O\) is open in \(\mathbb{R}\), and a point \(x\in W\). Choose \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq O\). The intersection \((x-\varepsilon,x+\varepsilon)\cap J\) is an order-open neighborhood of \(x\) contained in \(W\). If \(x\) is an interior point of \(J\), this is an interval neighborhood. If \(x=a\) is included, the intersection is of the form \([a,v)\) after taking \(v=x+\varepsilon\) (or a smaller right endpoint). If \(x=b\) is included, it is of the form \((u,b]\). These are exactly the listed endpoint forms. An excluded endpoint is not a point of \(J\), so no neighborhood of that point in \(J\) is being asserted. Thus every subspace-open set is order-open, and the topologies agree. \(\square\)

Reading Endpoint Neighborhoods Correctly

The interval notation in these neighborhood descriptions is relative to the space \(J\). For instance, \([a,v)\) is open as a subset of \(J=[a,b]\), even though it is not open as a subset of \(\mathbb{R}\). Indeed, \([a,v)=J\cap(a-\varepsilon,v)\) for any \(\varepsilon>0\). The intersection formula is the subspace-topology test: a set in \(J\) is open precisely when it is \(J\cap O\) for some real open set \(O\).

A useful summary is that interior points always have two-sided interval neighborhoods. An included left endpoint has one-sided neighborhoods extending to the right; an included right endpoint has one-sided neighborhoods extending to the left. An excluded endpoint is not a point of the space and has no neighborhood there.

Common pitfall. For \(J=(0,1)\), it is incorrect to call \((0,1/2)_J\) a neighborhood of \(0\). Although the displayed interval has a relative-interval form, \(0\notin J\), and the intersection \((0,1/2)\cap J=(0,1/2)\) does not contain \(0\). A neighborhood must contain the point in question. The correction is to describe neighborhoods of points \(x\in(0,1)\) using \((u,v)_J\) with \(u<x<v\); there is no endpoint-neighborhood claim at \(0\) or \(1\) for this space. By contrast, in \(J=[0,1]\), the set \([0,1/2)\) is a neighborhood of \(0\), because it contains \(0\) and equals \(J\cap(-1,1/2)\), with \((-1,1/2)\) open in \(\mathbb{R}\).

These comparisons are useful whenever a set carries both an order and a topology inherited from a larger space. The rationals are a particularly important case: density supplies enough rational endpoints to refine every real neighborhood, which is why the two topologies coincide. For intervals, the intersection description makes the endpoint behavior explicit without confusing a relative-open set with an open subset of the whole real line.