Why The Sorgenfrey Line Is Separable But Not Second Countable

Two Different Countability Properties

The Sorgenfrey line is the real line with a topology generated by half-open intervals. It provides a useful example in which having a countable dense subset does not mean that the topology has a countable base. The distinction comes from how neighborhoods begin: near each real number \(x\), the topology allows intervals that start exactly at \(x\).

The intervals \([a,b)\) are the basic open sets in \(\mathbb{R}_\ell\). In particular, if \(U\) is open and \(x\in U\), there is some \(b>x\) for which \([x,b)\subseteq U\). The left endpoint matters: a neighborhood of \(x\) can include \(x\) and points immediately to its right without including any points to its left.

The Rationals Are Dense

Proof. The rationals are countable: each rational can be written as \(p/q\) for integers \(p\) and positive integers \(q\), and the pairs \((p,q)\) can be listed in order of increasing \(|p|+q\), listing the finitely many pairs at each fixed value in any fixed order. To prove density, let \(U\) be a nonempty open set and choose \(x\in U\). By the basis property, there are real numbers \(a<b\) such that \(x\in[a,b)\subseteq U\). Since \(b-a>0\), the Archimedean property gives \(n\in\mathbb{N}\) with \(1/n<b-a\). Let \(m=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer not exceeding \(na\). Then \(na<m\le na+1<nb\), so \(a<m/n<b\). Thus \(m/n\in\mathbb{Q}\cap[a,b)\subseteq\mathbb{Q}\cap U\). Every nonempty open set therefore meets \(\mathbb{Q}\), proving density. \(\square\)

This proof finds a rational strictly between the endpoints of any basic interval. That is more than is needed for density: the left endpoint is included, but the argument works even if that endpoint itself is irrational.

Why No Countable Base Can Work

To rule out second countability, it is not enough to observe that the particular basis of all intervals \([a,b)\) is uncountable. A topology might have a different, countable base. We show that any base, whatever its members look like, must contain enough distinct open sets to account for every real number.

Proof. Suppose, for contradiction, that \(\mathbb{R}_\ell\) has a countable base \(\mathcal{B}\). List its members as \(B_1,B_2,\ldots\); if the base is finite, repeat its members to make such a list. For each \(x\in\mathbb{R}\), the set \([x,x+1)\) is open and contains \(x\). The base property therefore gives at least one index \(n\) such that \(x\in B_n\subseteq[x,x+1)\). Let \(n(x)\) be the least such index.

We claim that \(n(x)\ne n(y)\) whenever \(x<y\). Since \(B_{n(x)}\) is open and contains \(x\), the basis for \(\mathbb{R}_\ell\) gives some \(\delta>0\) with \([x,x+\delta)\subseteq B_{n(x)}\). Choose \(t=\min(\delta,y-x)/2\). Then \(t>0\), \(t<\delta\), and \(t<y-x\). Thus \(z=x+t\) belongs to \(B_{n(x)}\) and satisfies \(z<y\). But \(B_{n(y)}\subseteq[y,y+1)\), so \(z\notin B_{n(y)}\). Hence \(B_{n(x)}\ne B_{n(y)}\), and therefore \(n(x)\ne n(y)\).

It follows that \(x\mapsto n(x)\) assigns distinct positive integers to distinct real numbers. This would make \(\mathbb{R}\) countable, which is false. For completeness, the set of infinite sequences with entries \(0\) or \(1\) is uncountable by the diagonal argument: from any proposed list, form a sequence whose \(k\)-th entry differs from the \(k\)-th entry of the \(k\)-th listed sequence. These sequences inject into \([0,1]\) by the map \((\varepsilon_n)\mapsto\sum_{n=1}^{\infty}2\varepsilon_n/3^n\). Indeed, if two sequences first differ at index \(k\), the difference contributed there has size \(2/3^k\), while the sum of all possible later opposing contributions is at most \(\sum_{n>k}2/3^n=1/3^k\); their images are distinct. Thus \([0,1]\), and hence \(\mathbb{R}\), is uncountable. The contradiction proves that no countable base exists. \(\square\)

The proof pinpoints the local obstruction. Each point \(x\) needs a base element inside \([x,x+1)\), and openness forces that element to contain a short interval beginning at \(x\). If \(x<y\), such a neighborhood of \(x\) contains points below \(y\), whereas a neighborhood chosen inside \([y,y+1)\) cannot. The requirements at different points cannot all be met by a countable collection of sets.

A Common Mistake: Rational Endpoints Do Not Give a Base

A tempting guess is that the intervals \([r,s)\) with rational \(r<s\) form a countable base. They are countable, and they are open, but they do not satisfy the base condition. For example, take the open neighborhood \(U=[\sqrt{2},\sqrt{2}+1)\) of \(\sqrt{2}\). If a rational-endpoint interval \([r,s)\) were to satisfy \(\sqrt{2}\in[r,s)\subseteq U\), then \(\sqrt{2}\in[r,s)\) would imply \(r\le\sqrt{2}\). The containment in \(U\) would imply \(r\ge\sqrt{2}\), since any \(r<\sqrt{2}\) would put points below \(\sqrt{2}\) into \([r,s)\). Thus \(r=\sqrt{2}\), contradicting that \(r\) is rational.

The incorrect calculation is to check only that rational endpoints can be chosen with \(r<\sqrt{2}<s\). That makes \([r,s)\) contain \(\sqrt{2}\), but it also makes the interval contain points less than \(\sqrt{2}\), so it cannot be contained in \(U\). For the lower limit topology, the left endpoint of a sufficiently small basic neighborhood at \(x\) must be \(x\) itself. This explains how \(\mathbb{Q}\) can be dense without the rational-endpoint intervals forming a base.

The contrast is about two different demands. Density asks whether every nonempty open set contains at least one point of a chosen subset. A base must provide a suitable open set inside every open neighborhood of every point. Rational points satisfy the first demand; the point-specific left endpoints prevent any countable family from satisfying the second.