The Sorgenfrey Line Is Normal

The Lower-Limit Topology

The Sorgenfrey line has the same underlying set as the real line, but a different topology. Its basic neighborhoods extend to the right from their left endpoint. The resulting space is regular, and it also has a countable-subcover property. Together those facts imply that any two disjoint closed sets can be separated by disjoint open sets.

Here \([a,b)\) includes \(a\) and excludes \(b\). In particular, the topology is not the standard topology on \(\mathbb{R}\). We will first verify the separation properties that follow directly from this basis.

Clopen Basis Intervals and Regularity

Every basis interval \([a,b)\) is open by definition. It is also closed: its complement is \((-\infty,a)\cup[b,\infty)\). For \(z<a\), the basis neighborhood \([z,a)\) lies in \((-\infty,a)\); for \(z\ge b\), the neighborhood \([z,z+1)\) lies in \([b,\infty)\). These neighborhoods show that the complement is open, so \([a,b)\) is closed as well.

The Sorgenfrey line is \(T_1\). To check this from the definition, fix \(y\in\mathbb{R}\). For each \(x\ne y\), if \(x<y\), then \([x,y)\) is a neighborhood of \(x\) that avoids \(y\). If \(x>y\), then \([x,x+1)\) is a neighborhood of \(x\) that avoids \(y\). Thus every point of \(\mathbb{R}\setminus\{y\}\) has an open neighborhood contained in that complement, proving that \(\mathbb{R}\setminus\{y\}\) is open and \(\{y\}\) is closed.

Proof. Since \(F\) is closed, \(\mathbb{R}\setminus F\) is open and contains \(x\). By the basis definition, there are \(a<b\) with \(x\in[a,b)\subseteq\mathbb{R}\setminus F\). The set \([a,b)\) is open and closed, as shown above. Its complement is therefore open, contains \(F\), and is disjoint from \([a,b)\). This separates \(x\) and \(F\). Together with the \(T_1\) property, this proves regularity. \(\square\)

A Countable-Subcover Property

We now prove that \(\mathbb{R}_{\ell}\) is Lindelöf. The proof turns a cover by lower-limit neighborhoods into a family of ordinary open intervals, while carefully accounting for the points those intervals might miss.

Proof. Let \(\mathcal{U}\) be an open cover of \(\mathbb{R}_{\ell}\). For each \(x\in\mathbb{R}\), choose \(U_x\in\mathcal{U}\) containing \(x\). Since \(U_x\) is open in \(\mathbb{R}_{\ell}\), there is \(b_x>x\) such that \([x,b_x)\subseteq U_x\). Set \(J_x=(x,b_x)\), an open interval in the standard topology on \(\mathbb{R}\), and let \(O=\bigcup_{x\in\mathbb{R}}J_x\).

There is a countable subfamily of the intervals \(J_x\) covering \(O\). To see this, use the countable basis of rational open intervals for the standard topology on \(\mathbb{R}\). For each rational open interval \(Q\) that is contained in at least one \(J_x\), choose one such \(J_x\). The chosen intervals are countably many. If \(y\in O\), then \(y\in J_x\) for some \(x\); a rational open interval \(Q\) satisfies \(y\in Q\subseteq J_x\). The interval chosen for \(Q\) contains \(Q\), and hence contains \(y\). Thus the chosen intervals cover \(O\). The corresponding sets \(U_x\) cover \(O\) as well, because \(J_x\subseteq[x,b_x)\subseteq U_x\).

It remains to account for \(C=\mathbb{R}\setminus O\). If \(c\in C\), then \((c,b_c)\subseteq O\), so this interval contains no point of \(C\). If \(c,d\in C\) and \(c<d\), then \(d\notin(c,b_c)\), which forces \(d\ge b_c\). Consequently the intervals \((c,b_c)\), for \(c\in C\), are pairwise disjoint. Each is a nonempty real open interval and therefore contains a rational number. Choosing one rational in each interval gives distinct rationals for distinct \(c\), so \(C\) is countable.

Add the sets \(U_c\) for \(c\in C\) to the countable family already obtained. Each \(c\in C\) belongs to \(U_c\); the other selected sets cover \(O\). This is a countable subcover of \(\mathcal{U}\), proving the theorem. \(\square\)

Why Regularity and Lindelöfness Give Normality

The countable-subcover property can be combined with regularity in any space. We give the argument, since controlling the open neighborhoods one at a time is not enough: a countable union of neighborhoods whose closures avoid a closed set need not itself have closure disjoint from that set.

Proof. Let \(X\) be regular and Lindelöf, and let \(A,B\subseteq X\) be disjoint closed sets. If either set is empty, the open sets \(\varnothing\) and \(X\) separate them. Suppose both are nonempty.

A closed subspace of a Lindelöf space is Lindelöf. Indeed, given an open cover of \(A\) in its subspace topology, write its members as intersections with open sets of \(X\), and add \(X\setminus A\). This gives an open cover of \(X\); a countable subcover restricts to a countable subcover of \(A\). For each \(a\in A\), regularity gives an open set \(U_a\) containing \(a\) whose closure misses \(B\): separate \(a\) from \(B\) by disjoint open sets \(U_a,W_a\), so \(\overline{U_a}\subseteq X\setminus W_a\) and \(B\subseteq W_a\). By Lindelöfness of \(A\), countably many such sets cover \(A\). Write them as \(U_1,U_2,\ldots\), with \(\overline{U_n}\cap B=\varnothing\) for every \(n\). Applying the same argument to \(B\), obtain open sets \(V_1,V_2,\ldots\) covering \(B\) with \(\overline{V_n}\cap A=\varnothing\) for every \(n\). Finite covers can be repeated to give sequences.

Define \(G_n=U_n\setminus\bigcup_{i=1}^{n}\overline{V_i}\) and \(H_n=V_n\setminus\bigcup_{i=1}^{n}\overline{U_i}\). Each is open, since it is an open set with a finite union of closed sets removed. The open sets \(G=\bigcup_{n\ge1}G_n\) and \(H=\bigcup_{n\ge1}H_n\) contain \(A\) and \(B\), respectively: every point of \(A\) belongs to some \(U_n\) and to none of the \(\overline{V_i}\), and the corresponding statement holds for \(B\).

To check disjointness, suppose a point belonged to \(G_n\cap H_m\). If \(n\le m\), then membership in \(H_m\) excludes \(\overline{U_n}\), but membership in \(G_n\) implies membership in \(U_n\subseteq\overline{U_n}\), a contradiction. If \(m\le n\), membership in \(G_n\) excludes \(\overline{V_m}\), but membership in \(H_m\) implies membership in \(V_m\subseteq\overline{V_m}\), also a contradiction. The two cases cover all pairs \(m,n\). Thus \(G\cap H=\varnothing\), and \(G,H\) separate the closed sets. \(\square\)

The theorem applies to \(\mathbb{R}_{\ell}\): we proved it is \(T_1\) and regular, and the countable-subcover theorem proves it is Lindelöf. Therefore the Sorgenfrey line is normal.

A Common Pitfall

A tempting shortcut in the Lindelöf proof is to assert that the ordinary interiors \((x,b_x)\) cover the whole line. They need not. In the integer-interval example, the lower-limit open cover \(\{[n,n+1):n\in\mathbb{Z}\}\) covers every real number, but the corresponding ordinary intervals \((n,n+1)\) omit every integer. The correction is not to ignore those endpoints: prove that the leftover set \(C\) is countable, then add one member of the original cover for each point of \(C\).

This distinction matters because openness depends on the topology being used. A half-open interval \([a,b)\) is open in \(\mathbb{R}_{\ell}\), even though it is not open in the standard topology on \(\mathbb{R}\). The proof succeeds by using ordinary open intervals only as an intermediate tool, and then explicitly covering every point they leave out.