The Moore Plane Is Regular but Not Normal

The Moore Plane and Its Neighborhoods

The Moore plane, also called the Niemytzki plane, has the same underlying points as the closed upper half-plane, but its topology treats boundary points differently from interior points. At an interior point, neighborhoods are Euclidean. At a boundary point, a basic neighborhood consists of that point together with an open disk tangent to the boundary there. This change makes the space regular, yet it still does not allow every pair of disjoint closed sets to be separated by disjoint open sets.

The disk in \(N(a,r)\) has center \((a,r)\) and radius \(r\); its Euclidean closure meets the boundary line only at \((a,0)\). The basis description is consistent: intersections that contain an interior point contain a sufficiently small Euclidean open disk around that point, while two basic neighborhoods containing the same boundary point contain a smaller tangent-disk neighborhood there. Thus these sets satisfy the basis condition. In particular, the topology restricted to the open half-plane \(\{y>0\}\) is its usual Euclidean topology.

Closures and Separation

Proof. At every point with \(y>0\), the Moore topology agrees locally with the Euclidean topology. Therefore an interior point lies in the closure of \(N(a,r)\) exactly when it lies in the Euclidean closed disk centered at \((a,r)\) with radius \(r\). The point \((a,0)\) belongs to \(N(a,r)\), so it belongs to its closure. Now take a boundary point \((b,0)\) with \(b\ne a\). Put \(d=|b-a|>0\), and choose \(t>0\) with \(t<d/4\). Every interior point \((x,y)\) of \(N(b,t)\) satisfies \(|x-b|<2t\): indeed, \((x-b)^2+y^2<2ty\) implies \((x-b)^2<2ty-y^2\leq t^2\), since \(2ty-y^2=t^2-(y-t)^2\leq t^2\). Thus \(|x-b|<t\). Points of \(N(b,t)\) have horizontal coordinate within \(t\) of \(b\), whereas every point of the closed disk centered at \((a,r)\) has horizontal coordinate within \(r\) of \(a\). To show this neighborhood can miss the disk, choose \(t\) still smaller if needed so that \(t<d/2\), and note that the disk itself has positive distance from \((b,0)\): its closed disk meets \(y=0\) only at \((a,0)\). Consequently some sufficiently small tangent neighborhood at \((b,0)\) misses the closed disk. Hence \((b,0)\) is not in the closure. These cases give the stated formula. \(\square\)

The Moore plane is Hausdorff. Two distinct interior points can be separated by disjoint sufficiently small Euclidean open disks lying above the boundary. For an interior point and a boundary point, choose a Euclidean disk around the interior point small enough to stay above the boundary; a sufficiently small tangent neighborhood of the boundary point misses that disk. For distinct boundary points \((a,0)\) and \((b,0)\), let \(d=|a-b|>0\), and choose \(r=s=d/8\). The centers of their tangent disks are distance \(d\) apart, while the sum of the disk radii is \(r+s=d/4<d\); the disks are disjoint, and their boundary points differ. Thus the tangent neighborhoods are disjoint. A Hausdorff space is \(T_1\): for distinct \(x,y\), a neighborhood of \(x\) missing \(y\) shows \(X\setminus\{y\}\) is a neighborhood of \(x\), so each singleton is closed.

Every subset of the boundary \(L=\mathbb{R}\times\{0\}\) is closed in \(M\). To check this directly, let \(A\subseteq L\). Every point \((x,y)\notin A\) with \(y>0\) has a Euclidean open disk around it lying above \(y=0\), and that disk misses \(A\). If \((b,0)\notin A\), then either \(A\) is empty or each of its points has boundary coordinate different from \(b\); a tangent neighborhood \(N(b,r)\) meets the boundary only at \((b,0)\), so it misses \(A\). Thus \(M\setminus A\) is open.

Regularity of the Moore Plane

Proof. Since \(M\setminus F\) is open and contains \(p\), first suppose \(p=(a,0)\) is on the boundary. The basis definition gives \(r>0\) with \(N(a,r)\subseteq M\setminus F\). Choose \(s\) with \(0<s<r\). The preceding calculation proves \(\overline{N(a,s)}\subseteq N(a,r)\subseteq M\setminus F\). Therefore \(N(a,s)\) and \(M\setminus\overline{N(a,s)}\) are disjoint open sets containing \(p\) and \(F\), respectively. Here \(M\setminus\overline{N(a,s)}\) is open because the closure is closed, and it contains \(F\) because the closure misses \(F\).

Now suppose \(p=(a,y)\) with \(y>0\). There is a Euclidean open disk \(B_{\mathbb{R}^2}(p,\delta)\) contained in \(M\setminus F\) and in \(\{y>0\}\), for some \(\delta>0\), because the Moore topology is Euclidean at \(p\). Choose \(0<\rho<\delta\). The Euclidean closed disk of radius \(\rho\) about \(p\) lies in \(B_{\mathbb{R}^2}(p,\delta)\), hence misses \(F\). It is also the closure in \(M\) of the Euclidean open disk of radius \(\rho\), since it lies entirely above the boundary and the topology there is Euclidean. The open disk and the complement of its closure are therefore disjoint open sets containing \(p\) and \(F\). This proves regularity in both cases. \(\square\)

Why the Moore Plane Is Not Normal

Normality requires disjoint open neighborhoods for every pair of disjoint closed sets. We use the closed sets \(A\) and \(B\) above. The key obstruction is that a tangent disk of a fixed positive radius around a rational boundary point cannot approach an irrational boundary point whose own tangent disk has positive radius.

Proof. Suppose \(C_1,C_2,\ldots\) are closed subsets of \([0,1]\) with empty relative interior. Construct nested nonempty closed intervals \(I_k\) as follows. Start with \(I_0=[0,1]\). Given \(I_{k-1}\), its interior contains a nonempty open interval. Since \(C_k\) has empty relative interior, that interval contains a point outside \(C_k\). Since \(C_k\) is closed, its complement is relatively open, so there is a nonempty closed interval \(I_k\) around that point, contained in the interior of \(I_{k-1}\), disjoint from \(C_k\), and of length at most \(1/k\). The nested closed intervals have a common point: their left endpoints increase and are bounded above, and their supremum belongs to every \(I_k\) because the intervals are closed and nested. This point is outside every \(C_k\), since \(I_k\cap C_k=\varnothing\). Thus the sets cannot cover \([0,1]\). \(\square\)

Proof. Suppose, to obtain a contradiction, that disjoint open sets \(U,V\subseteq M\) contain \(A\) and \(B\), respectively. For each rational \(q\in[0,1]\), choose \(r_q>0\) with \(N(q,r_q)\subseteq U\). For each irrational \(x\in[0,1]\), choose \(s_x>0\) with \(N(x,s_x)\subseteq V\). These choices are possible because \(U,V\) are open and contain the indicated boundary points.

If two tangent disks \(N(q,r)\) and \(N(x,s)\) are disjoint, then \((q-x)^2+(r-s)^2\geq(r+s)^2\), since otherwise the distance between their centers is less than the sum of their radii and their open Euclidean disks intersect. Subtracting \((r-s)^2\) gives \(|q-x|\geq2\sqrt{rs}\).

For each \(n\in\mathbb{N}\), define \(E_n=\{q\in\mathbb{Q}\cap[0,1]:r_q\geq1/n\}\) and \(F_n=\{x\in[0,1]\setminus\mathbb{Q}:s_x\geq1/n\}\). Every point of \([0,1]\) belongs to some \(E_n\) or \(F_n\): its chosen radius is positive, so there is an integer \(n\) with \(1/n\) no greater than that radius. Hence the sets \(\overline{E_n}^{[0,1]}\) and \(\overline{F_n}^{[0,1]}\), where these closures are taken in the usual interval, together cover \([0,1]\).

No irrational \(x\in[0,1]\) belongs to \(\overline{E_n}^{[0,1]}\). If it did, there would be \(q_j\in E_n\) with \(q_j\to x\). The neighborhoods \(N(q_j,r_{q_j})\subseteq U\) and \(N(x,s_x)\subseteq V\) are disjoint, so the distance inequality gives \(|q_j-x|\geq2\sqrt{r_{q_j}s_x}\geq2\sqrt{s_x/n}>0\) for every \(j\), contradicting \(q_j\to x\). Thus \(\overline{E_n}^{[0,1]}\) contains only rationals and has empty relative interior, since every nonempty relative interval contains an irrational. Interchanging rationals and irrationals in the same argument shows that \(\overline{F_n}^{[0,1]}\) contains only irrationals; it too has empty relative interior, since every nonempty relative interval contains a rational. Each of these closures is closed in \([0,1]\). The countable family of all of them would therefore cover \([0,1]\) by closed sets with empty relative interior, contradicting the lemma. No such \(U,V\) exist, so \(M\) is not normal. \(\square\)

The distinction between regularity and normality is important here. Regularity separates one point from a closed set, and the shrinking of tangent neighborhoods supplies exactly that separation. Normality asks for a single pair of disjoint open sets around two whole closed sets. The rational-versus-irrational boundary argument shows why pointwise choices of neighborhoods need not combine into such a pair.