The Sorgenfrey Plane Is Not Normal

The Product and Its Anti-Diagonal

The Sorgenfrey line is a useful reminder that separation properties need not survive products. Each factor has a topology with particularly small one-sided neighborhoods, but in the product those neighborhoods combine to create a closed discrete set with many points. We will split that set into two closed pieces and show that no disjoint open neighborhoods can separate them.

Write \(\mathbb{R}_\ell\) for the real line with the topology whose basis consists of the half-open intervals \([a,b)\), where \(a<b\). The Sorgenfrey plane is \(\mathbb{R}_\ell\times\mathbb{R}_\ell\) with the product topology. Its basis rectangles are \([a,b)\times[c,d)\), where \(a<b\) and \(c<d\). A space is normal if every pair of disjoint closed subsets has disjoint open sets containing them, respectively.

The special role of the anti-diagonal comes from the opposing directions in its coordinates. A rectangle based at \((x,-x)\) extends to the right in both coordinates. To meet another anti-diagonal point \((y,-y)\), its first coordinate requires \(y\ge x\), while its second coordinate requires \(y\le x\). Thus it can meet the anti-diagonal only at its base point.

The Anti-Diagonal Is Closed and Discrete

Proof. To prove that \(D\) is closed, take \((u,v)\notin D\) and put \(s=u+v\), so \(s\ne0\). If \(s>0\), choose any \(\delta>0\). Every \((x,y)\in[u,u+\delta)\times[v,v+\delta)\) satisfies \(x+y\ge s>0\), so this rectangle misses \(D\). If \(s<0\), choose \(\delta=-s/4>0\). In that rectangle, \(x+y<s+2\delta=s/2<0\), so it again misses \(D\). Each point outside \(D\) therefore has an open neighborhood disjoint from \(D\), proving that \(D\) is closed.

For each \((x,-x)\in D\), the rectangle \([x,x+1)\times[-x,-x+1)\) meets \(D\) only at \((x,-x)\): the two coordinate conditions for \((t,-t)\) imply \(t\ge x\) and \(t\le x\). Hence \(D\) is discrete. Every subset \(E\subseteq D\) is closed in the subspace \(D\), since every subset of a discrete space is closed. A closed subset of a closed subspace is closed in the whole space: indeed, if \(E=D\cap F\) for some closed set \(F\) in the plane, then \(E\) is an intersection of closed sets. Thus every \(E\subseteq D\) is closed in the plane. \(\square\)

What Disjoint Rectangles Would Require

Suppose, for contradiction, that the Sorgenfrey plane is normal. Then there are disjoint open sets \(U\) and \(V\) with \(D_{\mathbb{Q}}\subseteq U\) and \(D_{\mathrm{irr}}\subseteq V\). At each \((x,-x)\) in either set, openness gives a basic rectangle around that point contained in its open neighborhood. By reducing the two side lengths to their minimum, choose a number \(a_x>0\) such that \([x,x+a_x)\times[-x,-x+a_x)\) lies in that neighborhood.

Here is the constraint imposed by disjointness. Suppose \(x\) is rational, \(y\) is irrational, and \(|x-y|<\min(a_x,a_y)\). The two rectangles based at \((x,-x)\) and \((y,-y)\) intersect. For example, if \(x<y\), the point \((y,-x)\) belongs to both: \(y<x+a_x\) and \(-x<-y+a_y\), while the other endpoint conditions hold as well. If \(y<x\), the same check with the roles reversed gives the intersection point \((x,-y)\). Either case contradicts \(U\cap V=\varnothing\). Therefore \[ |x-y|\ge\min(a_x,a_y) \] for every rational \(x\in[0,1]\) and irrational \(y\in[0,1]\).

A Covering Lemma for the Interval

The widths \(a_x\) need not have one common positive lower bound. Instead, group points according to a lower bound \(1/n\). This produces countably many closed sets, and the following interval lemma ensures that one of them must contain an interval.

Proof. Suppose every \(E_k\) has empty interior relative to \(I\). We construct nested nondegenerate closed intervals \(J_k\subset(0,1)\) such that \(J_k\cap E_k=\varnothing\), and, for \(k>1\), \(J_k\) lies in the interior of \(J_{k-1}\). Because \(E_1\) is closed and has empty interior, its complement contains a nonempty open subinterval of \((0,1)\); choose \(J_1\) inside that subinterval. At each subsequent stage, the interior of \(J_{k-1}\) is a nonempty open interval. Since \(E_k\) has empty interior and is closed, this interval contains a nonempty open subinterval disjoint from \(E_k\). Choose a nondegenerate closed interval \(J_k\) inside it, with length at most \(1/k\). The nested intervals have a common point by completeness of \(\mathbb{R}\). That point belongs to none of the \(E_k\), contradicting that they cover \(I\). \(\square\)

Applying the Lemma to the Widths

For each \(n\in\mathbb{N}\), define \(F_n=\{x\in\mathbb{Q}\cap I:a_x\ge1/n\}\) and \(G_n=\{y\in I\setminus\mathbb{Q}:a_y\ge1/n\}\). Every rational point of \(I\) belongs to some \(F_n\), because its width is positive; every irrational point belongs to some \(G_n\) for the same reason. Write \(\overline{F_n}^{\,I}\) and \(\overline{G_n}^{\,I}\) for the closures in the usual subspace topology on \(I\). These are closed subsets of \(I\), and together they cover \(I\).

For every \(n,m\), every \(x\in F_n\) and \(y\in G_m\) satisfy \[ |x-y|\ge\min(a_x,a_y)\ge\min(1/n,1/m). \] The same positive separation holds between their closures. To justify this, if \(u\in\overline{F_n}^{\,I}\) and \(v\in\overline{G_m}^{\,I}\) had \(|u-v|<c\), where \(c=\min(1/n,1/m)\), choose \(\eta>0\) with \(|u-v|+2\eta<c\). The intervals of radius \(\eta\) around \(u\) and \(v\) meet \(F_n\) and \(G_m\), respectively, by the definition of closure. Points \(x,y\) chosen in those intersections would satisfy \(|x-y|<c\), a contradiction. In particular, \(\overline{F_n}^{\,I}\) and \(\overline{G_m}^{\,I}\) are disjoint.

Apply the lemma to the countable family \(\overline{F_1}^{\,I},\overline{G_1}^{\,I},\overline{F_2}^{\,I},\overline{G_2}^{\,I},\ldots\), which covers \(I\). Some member has nonempty relative interior. If \(\overline{F_n}^{\,I}\) contains a nonempty relative open interval, then every \(G_m\) misses that interval because \(\overline{F_n}^{\,I}\cap\overline{G_m}^{\,I}=\varnothing\). But the interval contains irrational numbers, and every irrational in \(I\) belongs to some \(G_m\). This is a contradiction. If instead \(\overline{G_n}^{\,I}\) has nonempty relative interior, every \(F_m\) misses that interval, although it contains rational numbers and every rational in \(I\) belongs to some \(F_m\). This is also a contradiction. The density of both rational and irrational numbers in every nonempty interval now contradicts the assumed normality. Therefore the Sorgenfrey plane is not normal.

Why the Product Matters

The proof depends on the one-sided basis rectangles, not just on the fact that the anti-diagonal is discrete. Disjoint open neighborhoods force a numerical separation between opposite parts of the diagonal; grouping their widths by \(1/n\) turns that separation into a contradiction on the ordinary interval. Thus normality of the Sorgenfrey line does not imply normality of its product with itself.

Common pitfall. It is not enough to say that a space is normal because its factors are normal. Normality is not preserved by arbitrary products. Nor does closed discreteness alone prove nonnormality: the contradiction here requires splitting the closed discrete set into two pieces and using the special rectangle geometry to constrain their widths.