The Lower Limit Topology And Half-Open Basis Intervals

Building a Topology from Half-Open Intervals

On the real line, the intervals \([a,b)\) behave differently from the familiar open intervals \((a,b)\): they include their left endpoint and exclude their right endpoint. Declaring these half-open intervals to be basic open sets gives a topology finer than the usual one. The details of the basis check matter, since a proposed collection of sets generates a topology only when it satisfies the basis conditions.

Throughout, \(\mathbb{R}\) with its usual topology is called the usual real line. Let \(\mathcal{B}_\ell\) be the collection of all intervals \([a,b)\) with \(a,b\in\mathbb{R}\) and \(a<b\). The topology generated by this basis is called the lower limit topology, and the resulting space is denoted by \(\mathbb{R}_\ell\).

For this particular collection, the second condition can be checked directly: the intersection of two basis intervals is either empty or another interval of the same form. Thus, when the intersection contains \(x\), it itself can serve as \(B_3\).

Proof. First, every \(x\in\mathbb{R}\) belongs to the basis interval \([x,x+1)\), since \(x<x+1\). Now take \([a,b),[c,d)\in\mathcal{B}_\ell\). Their intersection is empty if \(\max(a,c)\geq\min(b,d)\). If \(\max(a,c)<\min(b,d)\), then \[ [a,b)\cap[c,d)=[\max(a,c),\min(b,d)). \] This is a member of \(\mathcal{B}_\ell\). Therefore, whenever the intersection contains a point, the intersection itself is a basis element containing that point and contained in both original intervals. Both basis conditions hold, so \(\mathcal{B}_\ell\) is a basis. \(\square\)

For completeness, the basis conditions ensure that all unions of basis elements form a topology. The empty set is the empty union, and \(\mathbb{R}\) is a union of basis elements because every point belongs to one. Arbitrary unions of such unions are still unions of basis elements. For finite intersections, distribute intersection over unions; each intersection of two basis elements is either empty or a basis element, so the result is again a union of basis elements. These are exactly the topology axioms.

Comparing the Two Topologies

Write \(\mathcal{T}_{\mathrm{usual}}\) for the usual topology on \(\mathbb{R}\) and \(\mathcal{T}_\ell\) for the lower limit topology. A topology is finer than another when it contains every open set of the other topology. Here the lower limit topology is finer, and the inclusion is strict.

Proof. Let \(U\) be usual open and \(x\in U\). By the definition of the usual topology, there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). Then \([x,x+\varepsilon)\) is a basis element of \(\mathcal{T}_\ell\) containing \(x\), and \([x,x+\varepsilon)\subseteq(x-\varepsilon,x+\varepsilon)\subseteq U\). This provides a lower-limit basis neighborhood inside \(U\) for each \(x\in U\), so \(U\in\mathcal{T}_\ell\). The empty set is open in both topologies as well.

To show that the inclusion is strict, take \([0,1)\). It is a basis element, hence lower-limit open. It is not usual open: every usual open interval containing \(0\) contains some negative real numbers, and therefore cannot be contained in \([0,1)\). Thus \([0,1)\notin\mathcal{T}_{\mathrm{usual}}\), proving strict inclusion. \(\square\)

Continuity of the Identity Maps

For topological spaces \(X\) and \(Y\), a function \(f:X\to Y\) is continuous if the preimage \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). The identity function \(\operatorname{id}_{\mathbb{R}}\) sends each real number to itself, but its continuity depends on the topology of its domain and codomain. The inclusion of topologies just proved determines one direction immediately; the strictness of that inclusion rules out the other.

Proof. First give the domain of the identity map the lower limit topology and its codomain the usual topology. If \(V\) is usual open, then \(\operatorname{id}_{\mathbb{R}}^{-1}(V)=V\). The theorem on topology inclusion says \(V\) is lower-limit open. Therefore the preimage of every open set in the codomain is open in the domain, and this identity map is continuous.

Now give the domain the usual topology and the codomain the lower limit topology. The set \([0,1)\) is open in the codomain, but its preimage under the identity map is \([0,1)\), which is not open in the usual topology. This violates the defining condition for continuity, so this identity map is not continuous. \(\square\)

A Common Pitfall

Common pitfall: Do not assume that a set open in the lower limit topology is also open in the usual topology. The incorrect argument is: “\([0,1)\) is a basic open set, so it is open in the usual topology.” Being basic open establishes openness only in the topology generated by that basis. In the usual topology, every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\); that interval contains \(-\varepsilon/2<0\), which is not in \([0,1)\). So \([0,1)\) is not usual open.

A related calculation must also account for whether the endpoints overlap. For example, \([0,1)\cap[1,2)=\varnothing\), not an interval containing \(1\): the first interval requires \(y<1\), while the second requires \(y\geq1\). In the general intersection formula, if the larger left endpoint is greater than or equal to the smaller right endpoint, the intersection is empty.

The basis description is useful because it turns questions about arbitrary lower-limit open sets into local questions: each point in an open set must lie in a half-open basis interval contained in that set. Comparing topologies then gives a direct continuity test for identity maps. The direction from a finer topology to a coarser one is continuous here, while the reverse direction fails because some codomain-open sets have non-open preimages.