The Sorgenfrey Plane: Separable But Not Lindelof

The Plane Built from Half-Open Intervals

The usual Euclidean plane has a countable dense set, and it also satisfies the Lindelof property. The Sorgenfrey plane is a useful contrast: it still has a countable dense set, but it is not Lindelof. The difference is detected by a particular uncountable subset, the anti-diagonal, whose points can each be isolated from all the others by open rectangles.

On \(\mathbb{R}\), let \(\mathbb{R}_{\ell}\) denote the lower-limit 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. A basis for this product topology consists of rectangles \([a,b)\times[c,d)\), where \(a<b\) and \(c<d\). In particular, a set is open if every point in it lies in one of these basis rectangles that is contained in the set.

A Countable Dense Subset

Consider \(Q=\mathbb{Q}\times\mathbb{Q}\), the set of points in the plane with both coordinates rational. The rationals are countable, and a product of two countable sets is countable: for example, enumerate pairs of natural numbers by increasing sum of their coordinates, and use this to enumerate pairs of rational-number indices. Thus \(Q\) is countable.

Proof. Let \(G\) be a nonempty open set in the Sorgenfrey plane. Choose a point of \(G\). Since the product rectangles form a basis, there are real numbers \(a<b\) and \(c<d\) such that \([a,b)\times[c,d)\subseteq G\). Each interval \([a,b)\) contains a rational: choose \(n\in\mathbb{N}\) with \(n(b-a)>1\), and let \(m\) be the least integer greater than or equal to \(na\). Then \(m\ge na\), and \(m<na+1<nb\), so \(a\le m/n<b\). Applying the same argument to \([c,d)\) gives a rational in that interval as well. The pair of rationals belongs to \([a,b)\times[c,d)\), so \(G\) meets \(\mathbb{Q}\times\mathbb{Q}\). Hence this countable set is dense. \(\square\)

The Anti-Diagonal Is Closed and Discrete

Now define the anti-diagonal \[ D=\{(x,-x):x\in\mathbb{R}\}. \] It is uncountable because different real numbers \(x\) give different points \((x,-x)\). To use it to test the Lindelof property, we need both its closedness and its discreteness in the Sorgenfrey plane. These facts follow directly from the half-open rectangle basis.

Proof. First let \((a,b)\notin D\), so \(s=a+b\ne0\). If \(s>0\), the open rectangle \([a,a+1)\times[b,b+1)\) contains \((a,b)\). Every point \((u,v)\) in this rectangle satisfies \(u\ge a\) and \(v\ge b\), so \(u+v\ge s>0\). No point of the rectangle lies on \(D\), since the coordinates of a point on \(D\) sum to zero.

If \(s<0\), set \(\delta=-s/4>0\). The rectangle \([a,a+\delta)\times[b,b+\delta)\) contains \((a,b)\). For every \((u,v)\) in it, \(u<a+\delta\) and \(v<b+\delta\), so \(u+v<s+2\delta=s/2<0\). Again the rectangle misses \(D\). Thus every point outside \(D\) has an open neighborhood disjoint from \(D\), which proves that the complement of \(D\) is open and \(D\) is closed.

To prove discreteness, fix \(x\in\mathbb{R}\) and consider the open rectangle \([x,x+1)\times[-x,-x+1)\), which contains \((x,-x)\). If a point \((t,-t)\in D\) lies in this rectangle, then its first coordinate gives \(t\ge x\), while its second coordinate gives \(-t\ge -x\), or \(t\le x\). Therefore \(t=x\). The rectangle intersects \(D\) in exactly \(\{(x,-x)\}\). This works for every \(x\), so \(D\) is discrete as a subspace. \(\square\)

An Open Cover with No Countable Subcover

For each \(x\in\mathbb{R}\), define the open rectangle \[ U_x=[x,x+1)\times[-x,-x+1). \] The isolation calculation above shows that \(U_x\cap D=\{(x,-x)\}\). The complement of \(D\) is open because \(D\) is closed. These sets together form an open cover: \[ \mathcal{U}=\{\,\mathbb{R}^{2}\setminus D\,\}\cup\{\,U_x:x\in\mathbb{R}\,\}. \] Every point outside \(D\) belongs to \(\mathbb{R}^{2}\setminus D\), and every point of \(D\) belongs to its corresponding \(U_x\).

Proof. Suppose a countable subfamily of \(\mathcal{U}\) is chosen. If it does not contain \(\mathbb{R}^{2}\setminus D\), it covers no point outside \(D\), so it cannot cover the plane. If it does contain \(\mathbb{R}^{2}\setminus D\), that set covers no point of \(D\). Each remaining selected rectangle \(U_x\) covers exactly one point of \(D\). A countable collection of such rectangles therefore covers at most countably many points of \(D\). Since \(D\) is uncountable, some point of \(D\) is not covered. In either case, the chosen countable subfamily is not a cover. Thus \(\mathcal{U}\) has no countable subcover, and the plane is not Lindelof. \(\square\)

What the Example Shows

The Sorgenfrey plane is separable because a countable set meets every nonempty open set. It is not Lindelof because a particular open cover requires uncountably many of its listed rectangles to cover the anti-diagonal. These conclusions concern different requirements: density asks whether a single countable subset meets every open set, whereas the Lindelof property asks whether every open cover can be reduced to a countable subcover.

Common pitfall. One might argue, “The plane has a countable dense subset, so every open cover has a countable subcover.” That implication is false: the proof of density only shows that each nonempty open set contains at least one point of \(\mathbb{Q}\times\mathbb{Q}\). It does not select a countable collection of members from an arbitrary open cover that covers the entire space. In the cover \(\mathcal{U}\), the single set \(\mathbb{R}^{2}\setminus D\) misses every point of \(D\), and each \(U_x\) covers just one of those uncountably many points. The countable dense subset does not change this count of the cover’s members.