Comparing Two Topologies on the Real Line
The same underlying set can carry different topologies, so a set that is open in one topology need not be open in another. A useful way to compare topologies is to compare their basis elements: if basis elements for one topology can be refined by basis elements for the other, then every open set in the first topology is open in the second. We apply this method to the usual topology on \(\mathbb{R}\) and the lower limit topology.
Write \(\mathcal{T}_{\mathrm{us}}\) for the usual topology on \(\mathbb{R}\), whose basis consists of the open intervals \((a,b)\), where \(a<b\). Write \(\mathcal{T}_{\ell}\) for the lower limit topology, whose basis consists of the half-open intervals \([a,b)\), where \(a<b\). Thus a set is open in either topology exactly when it is a union of basis elements for that topology.
The direction of the inclusion matters. To prove that the lower limit topology is finer, we must show that every usual open set is lower-limit open. It is not enough to note that both topologies have interval-shaped basis elements: the relevant question is whether the basis elements of the usual topology can be refined by lower-limit basis elements.
The Basis-Refinement Test
Proof. Suppose first that \(\mathcal{T}_1\subseteq\mathcal{T}_2\). Each \(B\in\mathcal{B}_1\) is \(\mathcal{T}_1\)-open, and hence is \(\mathcal{T}_2\)-open. For any \(x\in B\), the basis property of \(\mathcal{B}_2\) gives a set \(C\in\mathcal{B}_2\) with \(x\in C\subseteq B\).
Conversely, suppose the stated refinement condition holds. Let \(U\in\mathcal{T}_1\). For each \(x\in U\), the basis property of \(\mathcal{B}_1\) gives \(B\in\mathcal{B}_1\) with \(x\in B\subseteq U\). By the assumed condition, there is \(C\in\mathcal{B}_2\) with \(x\in C\subseteq B\subseteq U\). Therefore every point \(x\in U\) lies in a \(\mathcal{T}_2\)-basis element contained in \(U\). The union of all such elements is \(U\), so \(U\) is \(\mathcal{T}_2\)-open. This proves \(\mathcal{T}_1\subseteq\mathcal{T}_2\). \(\square\)
For the comparison at hand, a usual basis element is \((a,b)\). Given \(x\in(a,b)\), the lower limit basis element \([x,b)\) contains \(x\) and is contained in \((a,b)\). Indeed, \(a<x\), and every \(y\in[x,b)\) satisfies \(a<y<b\). This refinement at each point gives the inclusion of topologies.
Proof. Let \(U\) be open in the usual topology, and let \(x\in U\). By the basis property for the usual topology, there are \(a,b\in\mathbb{R}\) with \(a<x<b\) and \(x\in(a,b)\subseteq U\). The lower limit basis element \([x,b)\) satisfies \(x\in[x,b)\subseteq(a,b)\subseteq U\). Thus every \(x\in U\) lies in a lower limit basis element contained in \(U\). Consequently \(U\) is a union of lower limit basis elements and is open in \(\mathcal{T}_{\ell}\). Since this holds for every usual open \(U\), the claimed inclusion follows. \(\square\)
Worked Example: Expressing a Usual Interval as a Union
Show directly that the usual open interval \((2,5)\) is open in the lower limit topology by writing it as a union of lower limit basis elements.
Use the lower limit basis intervals \([x,5)\) for \(x\in(2,5)\). Their union is contained in \((2,5)\): if \(y\in[x,5)\) for such an \(x\), then \(2<x\leq y<5\), so \(2<y<5\). Conversely, for any \(y\in(2,5)\), choose \(x=y\). Then \(x\in(2,5)\) and \(y\in[y,5)\). Therefore every point of \((2,5)\) belongs to the union, and
Each interval in this union is a lower limit basis element, so the union is lower-limit open. The equality also checks both endpoint conditions: \(2\) is not included because every \(x\) in the union's indexing set is greater than \(2\), and \(5\) is excluded from every interval \([x,5)\).
Strictness: An Open Set That Separates the Topologies
The inclusion alone does not tell us whether the two topologies are equal. To prove strictness, it is enough to find one lower-limit open set that is not open in the usual topology. The basis element \([0,1)\) provides such a witness: it is open by the definition of the lower limit topology, but it contains its left endpoint without containing any points immediately to the left.
Proof. The first theorem gives \(\mathcal{T}_{\mathrm{us}}\subseteq\mathcal{T}_{\ell}\). The set \([0,1)\) is a basis element for \(\mathcal{T}_{\ell}\), so \([0,1)\in\mathcal{T}_{\ell}\). Suppose, for contradiction, that \([0,1)\) were usual-open. Since \(0\in[0,1)\), the basis property of the usual topology would give \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq[0,1)\). But \(-\varepsilon/2\in(-\varepsilon,\varepsilon)\) and \(-\varepsilon/2<0\), so \(-\varepsilon/2\notin[0,1)\). This contradicts the proposed containment. Hence \([0,1)\) is not usual-open, and the inclusion of topologies is proper. \(\square\)
Worked Example: Checking a Half-Open Basis Element
Compare \(A=[-1,2)\) in the two topologies.
In the lower limit topology, \(A\) is open because it is itself a basis element: its endpoints satisfy \(-1<2\), and the interval has the required form \([a,b)\).
It is not open in the usual topology. If it were, the point \(-1\in A\) would have a usual open interval \((-\varepsilon,\varepsilon)+(-1) =(-1-\varepsilon,-1+\varepsilon)\) contained in \(A\), for some \(\varepsilon>0\). The point \(-1-\varepsilon/2\) belongs to that interval because \(-\varepsilon<-\varepsilon/2<\varepsilon\), but it is less than \(-1\), so it does not belong to \(A\). This contradiction shows that \(A\) is not usual-open.
Check: the same endpoint test applies to \([0,1)\): it is a lower limit basis element, and any usual interval around \(0\) contains a negative point outside \([0,1)\). Thus both examples distinguish the topologies, while the first theorem ensures that neither example can be usual-open.
Worked Example: A Union of Lower Limit Basis Elements
Let \(V=[-2,-1)\cup[1,3)\). Determine whether \(V\) is open in each topology.
Each component is a lower limit basis element: the first has endpoints \(-2<-1\), and the second has endpoints \(1<3\). A union of open sets is open, so \(V\) is lower-limit open.
For the usual topology, consider the point \(-2\in V\). If \(V\) were usual-open, there would be an \(\varepsilon>0\) such that \((-2-\varepsilon,-2+\varepsilon)\subseteq V\). The point \(-2-\varepsilon/2\) is in this interval, since \(-\varepsilon<-\varepsilon/2<\varepsilon\), but it is less than \(-2\) and therefore is in neither component of \(V\). No such usual interval can be contained in \(V\), so \(V\) is not usual-open.
Check: the half-open endpoints are treated strictly as specified: \(-1\notin[-2,-1)\), \(3\notin[1,3)\), while \(-2\) and \(1\) are included. The failure of usual openness occurs at the included left endpoint \(-2\), not at either excluded right endpoint.
A Common Direction Error
Common pitfall. A mistaken argument might say, “Every lower limit basis element \([a,b)\) is open in the usual topology because it is an interval; therefore the lower limit topology is contained in the usual topology.” The calculation fails at the left endpoint: for example, \([0,1)\) is a lower limit basis element but is not usual-open, as shown above. Moreover, even if every basis element of one topology is open in another, that proves inclusion in the direction from the topology generated by those basis elements to the other topology—not the reverse.
For this comparison, the successful direction is to start with a usual basis interval and find lower limit basis elements inside it around each of its points. For strictness, one then checks a lower limit basis element that fails the usual open-neighborhood test. The two parts answer different questions: refinement proves inclusion, and a set in the larger topology but not the smaller proves that the inclusion is proper.
| Question | What to check | Conclusion here |
|---|---|---|
| Is every usual-open set lower-limit open? | Refine each usual basis interval around each point | Yes: \(\mathcal{T}_{\mathrm{us}}\subseteq\mathcal{T}_{\ell}\) |
| Are the two topologies equal? | Look for a lower-limit open set that is not usual-open | No: \([0,1)\) is such a set |
| Which topology is finer? | Compare the collections of open sets | The lower limit topology is strictly finer |
Check Your Understanding
Use basis elements and the usual open-neighborhood test to justify each answer.
- Write \((0,3)\) as a union of lower limit basis elements.
- Is \([2,4)\) open in the lower limit topology? Is it open in the usual topology?
- For a usual-open set \(U\) and \(x\in U\), describe a lower limit basis element containing \(x\) and contained in \(U\).
- Explain why showing \(\mathcal{T}_{\mathrm{us}}\subseteq\mathcal{T}_{\ell}\) does not by itself prove that the topologies are unequal.
- Find a point of \([-3,0)\) that prevents this set from being usual-open, and give the interval-neighborhood contradiction.