The Lower Limit Topology
The usual real line has a countable dense set, the rational numbers, and it also has a countable basis of open intervals with rational endpoints. The lower limit topology separates those two properties: the rational numbers remain dense, but no countable basis can describe all open neighborhoods. The distinction comes from the shape of its basic neighborhoods, which include their left endpoint but exclude their right endpoint.
These intervals do form a basis. Every \(x\in\mathbb{R}\) belongs to \([x,x+1)\), so the intervals cover \(\mathbb{R}\). If \(x\in[a,b)\cap[c,d)\), then \(x\geq a,c\) and \(x<b,d\). Hence \(\max(a,c)\leq x<\min(b,d)\), so \([\max(a,c),\min(b,d))\) is a basis interval containing \(x\) and contained in the intersection. These are the two basis conditions.
Worked Examples: Neighborhoods and Density
Worked Example: A Rational Point in a Basic Interval
Consider the basic open set \([1/2,3/4)\). The rational number \(5/8\) belongs to it because \(1/2=4/8\leq5/8<6/8=3/4\). Thus this particular nonempty basic open set meets \(\mathbb{Q}\).
The same idea works for every basic interval, though the rational point must be chosen in the interval being considered. Let \(a<b\). Choose \(n\in\mathbb{N}\) such that \(n(b-a)>1\), which is possible by the Archimedean property of \(\mathbb{R}\). Let \(m=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer not exceeding \(na\). Then \(na<m\leq na+1<nb\). Dividing by the positive integer \(n\) gives \(a<m/n<b\), so \(m/n\in\mathbb{Q}\cap[a,b)\). This verifies the claim for every choice of \(a<b\).
Check: for \(a=1/2\) and \(b=3/4\), \(n=5\) satisfies \(n(b-a)=5(1/4)=5/4>1\). Here \(m=\lfloor5/2\rfloor+1=3\), and the construction gives \(m/n=3/5\), which satisfies \(1/2<3/5<3/4\).
Worked Example: Refining a Neighborhood
Let \(U=[-2,0)\cup[3,5)\) in \(\mathbb{R}_\ell\), and take \(x=7/2\). This point belongs to \(U\), since \(3\leq7/2<5\). The basis interval \([7/2,4)\) contains \(x\), and it lies in \([3,5)\), since every \(t\in[7/2,4)\) satisfies \(3\leq t<5\). Thus it is a basis neighborhood of \(x\) contained in \(U\).
This refinement property is part of what it means for \(\mathcal{B}\) to be a basis: whenever an open set contains a point, there is a basis element containing that point and lying inside the open set. The right endpoint \(4\) is excluded, so the interval does not extend as far as \(5\).
The calculation in the first example proves that every nonempty basis interval meets \(\mathbb{Q}\). To pass from basis intervals to arbitrary open sets, recall that every open set is a union of basis elements. If \(U\) is nonempty and open, choose \(x\in U\). The basis property gives a basis interval \([a,b)\) with \(x\in[a,b)\subseteq U\); the rational-point calculation gives a rational in \([a,b)\), hence in \(U\).
Proof. The rational numbers are countable: each rational has a representation \(p/q\) with \(p\in\mathbb{Z}\) and \(q\in\mathbb{N}\), and \(\mathbb{Z}\times\mathbb{N}\) is countable. By the argument above, every nonempty open set in \(\mathbb{R}_\ell\) contains a basis interval and every such interval contains a rational. Therefore every nonempty open set meets \(\mathbb{Q}\), which is the open-set characterization of density. Thus \(\mathbb{Q}\) is a countable dense subset, as required. \(\square\)
Why No Countable Basis Can Work
Second countability requires a single countable family of open sets that can refine neighborhoods at every point. In \(\mathbb{R}_\ell\), the left endpoints of the basic intervals force any basis to contain a distinct element for each real number.
Proof. Suppose \(\mathcal{C}\) is any basis for \(\mathbb{R}_\ell\). For each \(x\in\mathbb{R}\), the interval \([x,x+1)\) is an open neighborhood of \(x\). By the basis property, choose \(C_x\in\mathcal{C}\) such that \(x\in C_x\subseteq[x,x+1)\). We show that different real numbers give different basis elements. If \(x<y\) and \(C_x=C_y\), then \(x\in C_x=C_y\subseteq[y,y+1)\). But every point of \([y,y+1)\) is at least \(y\), whereas \(x<y\). This is a contradiction. Consequently \(x\mapsto C_x\) is injective. Since \(\mathbb{R}\) is uncountable, \(\mathcal{C}\) is uncountable. This applies to every basis, so no countable basis exists. \(\square\)
Worked Example: The Obstruction on an Uncountable Set of Points
Apply the proof just to the points \(x\in[0,1]\). For each such \(x\), choose \(C_x\) with \(x\in C_x\subseteq[x,x+1)\). If \(x=1/4\) and \(y=1/2\), then \(C_{1/4}\) cannot equal \(C_{1/2}\): the former contains \(1/4\), while the latter is contained in \([1/2,3/2)\), which does not contain \(1/4\). The same argument works for every pair \(x<y\) in \([0,1]\): \(C_y\subseteq[y,y+1)\) excludes \(x\), whereas \(C_x\) contains \(x\).
Thus the assignment \(x\mapsto C_x\) gives distinct basis elements for all \(x\in[0,1]\). Since \([0,1]\) is uncountable, even this restricted collection of points forces uncountably many elements in the basis. Restricting the argument to \([0,1]\) is enough; the full proof uses all of \(\mathbb{R}\).
A Common Pitfall: Rational Endpoints Do Not Give a Basis
In the usual topology, intervals with rational endpoints form a countable basis. It is tempting to assume they also form a basis for \(\mathbb{R}_\ell\), because rational numbers are dense. Density alone does not guarantee that a proposed family can fit inside every open neighborhood.
Common pitfall. Consider the open neighborhood \([\sqrt{2},\sqrt{2}+1)\) of \(\sqrt{2}\). A proposed basis interval \([r,s)\) with rational endpoints that contains \(\sqrt{2}\) must satisfy \(r\leq\sqrt{2}<s\). Since \(\sqrt{2}\) is irrational, \(r<\sqrt{2}\). Therefore \(r\in[r,s)\), but \(r\notin[\sqrt{2},\sqrt{2}+1)\). So this candidate interval is not contained in the neighborhood. No interval with rational endpoints containing \(\sqrt{2}\) can refine this particular neighborhood. The error is to use density to find points without checking the required containment.
The comparison with the usual topology also makes the difference concrete. Every usual open set is open in \(\mathbb{R}_\ell\): if a usual open set \(V\) contains \(x\), some usual interval \((x-\varepsilon,x+\varepsilon)\) lies in \(V\), and \([x,x+\varepsilon)\subseteq V\). But \([0,1)\) is open in \(\mathbb{R}_\ell\) and is not open in the usual topology, since no usual interval around \(0\) is contained in \([0,1)\). The two topologies are therefore not the same.
| Property | Conclusion for \(\mathbb{R}_\ell\) | Reason |
|---|---|---|
| Separable | Yes | Every nonempty basic interval contains a rational. |
| Second countable | No | Every basis must contain distinct elements indexed by all real points. |
The example shows that separability does not by itself imply second countability: a countable dense set can meet every open set even when no countable family of open sets can refine all neighborhoods. The two definitions ask for different things. Separability asks for a countable collection of points; second countability asks for a countable collection of open sets with a local refinement property at every point.
Check Your Understanding
Use the half-open basic intervals and the basis property in your answers.
- Find a rational number in \([2/3,5/6)\), and verify both endpoint inequalities.
- Explain why every nonempty open set in \(\mathbb{R}_\ell\) contains a rational.
- For \(x<y\), why can basis elements \(C_x\subseteq[x,x+1)\) and \(C_y\subseteq[y,y+1)\), each containing its indexed point, not be equal?
- Why does the countability of \(\mathbb{Q}\) not imply that \(\mathbb{R}_\ell\) is second countable?
- Show that \([1,2)\) is open in \(\mathbb{R}_\ell\) but not in the usual topology.