Three Different Ways Separation Can Fail
Regularity and normality are separation properties, but they ask for different separations. A space can separate every point from every closed set and still fail to separate some pair of disjoint closed sets. Moreover, normality passes from a normal space to its closed subspaces, but not to arbitrary subspaces; even the product of two normal spaces need not be normal. The deleted plank, the Moore plane, and the Sorgenfrey plane make these distinctions explicit.
For a regular space, an equivalent useful test is: if \(x\in U\) and \(U\) is open, there is an open set \(V\) such that \(x\in V\) and \(\overline V\subseteq U\). Indeed, separating \(x\) from the closed set \(X\setminus U\) gives such a \(V\); conversely, \(V\) and \(X\setminus\overline V\) separate \(x\) from any closed set contained in \(X\setminus U\). We will use this shrinking test for the Moore plane.
The closed-subspace inheritance result says that if \(X\) is normal and \(Y\subseteq X\) is closed, then \(Y\), with its subspace topology, is normal. It does not say the same for an arbitrary subspace. The first example pinpoints that distinction. The other two show that a product can fail even when each factor is normal.
The Deleted Tychonoff Plank
Give each ordinal interval \([0,\omega_1]\) and \([0,\omega]\) its order topology, and give their product the product topology. Here \(\omega\) is the first infinite ordinal, and \(\omega_1\) is the first uncountable ordinal. Let \(P=[0,\omega_1]\times[0,\omega]\), and delete its top-right point to obtain \(Y=P\setminus\{(\omega_1,\omega)\}\). The ordinal intervals are compact Hausdorff, so their finite product is compact Hausdorff and hence normal. The deleted point is closed, but \(Y\) is not a closed subspace of \(P\).
Worked Example: The Deleted Plank Is Not Normal
In \(Y\), consider \(F=\{(\omega_1,n):n<\omega\}\) and \(G=\{(\alpha,\omega):\alpha<\omega_1\}\). These sets are disjoint. The vertical edge \(\{\omega_1\}\times[0,\omega]\) and horizontal edge \([0,\omega_1]\times\{\omega\}\) are closed in \(P\). Intersecting each with \(Y\) gives \(F\) and \(G\), so both are closed in \(Y\).
Suppose there were disjoint open sets \(U,V\) in \(Y\) with \(F\subseteq U\) and \(G\subseteq V\). For each finite ordinal \(n\), openness at \((\omega_1,n)\) gives an ordinal \(\alpha_n<\omega_1\) such that \((\alpha,n)\in U\) whenever \(\alpha_n<\alpha\leq\omega_1\). This follows because \(n\) is isolated in \([0,\omega]\), while a neighborhood of \(\omega_1\) contains a final segment.
The supremum \(\alpha_*=\sup_{n<\omega}\alpha_n\) is still less than \(\omega_1\): a countable set of countable ordinals has countable supremum. Choose \(\alpha\) with \(\alpha_*<\alpha<\omega_1\). Then \((\alpha,n)\in U\) for every \(n<\omega\). On the other hand, \(V\) is a neighborhood of \((\alpha,\omega)\). A neighborhood of \(\omega\) in \([0,\omega]\) contains all sufficiently large finite ordinals, so \((\alpha,n)\in V\) for all sufficiently large \(n\). Thus \(U\cap V\ne\varnothing\), a contradiction. Hence \(Y\) is not normal.
Check why the inheritance result does not apply: every neighborhood of \((\omega_1,\omega)\) in \(P\) contains points other than \((\omega_1,\omega)\), including points \((\omega_1,n)\) for sufficiently large finite \(n\). Thus \((\omega_1,\omega)\in\overline{Y}\), so \(Y\) is not closed in \(P\). The example is consistent with normality passing to closed subspaces.
The Moore Plane: Regular but Not Normal
Let \(H=\{(x,y)\in\mathbb R^2:y>0\}\), and let \(M=H\cup(\mathbb R\times\{0\})\). At points of \(H\), use ordinary Euclidean open disks contained in \(H\) as basic neighborhoods. At a boundary point \((a,0)\), use the sets
as basic neighborhoods. These sets, together with the interior disks, define the Moore plane topology. In particular, the boundary point is explicitly included in each of its basic neighborhoods.
The basis condition can be checked directly. At a fixed boundary point, the tangent disks are nested as their radii increase, so two basic neighborhoods there contain a smaller one. Two different boundary points have no boundary point in common; any intersection of their neighborhoods lies in \(H\), where ordinary Euclidean disks provide smaller basis neighborhoods. At an interior point, intersections are handled by Euclidean disks. This also explains why the added singleton matters.
Worked Example: The Boundary Neighborhood Must Include Its Point
For \(r>0\), the distance from \((a,0)\) to the center \((a,r)\) is \(\sqrt{(a-a)^2+(0-r)^2}=r\). An open disk of radius \(r\) contains points at distance strictly less than \(r\), so \(D((a,r),r)\) does not contain \((a,0)\). Using just \(D((a,r),r)\cap M\) therefore fails to give a neighborhood of the boundary point and does not cover \(M\) by the proposed basic neighborhoods.
The corrected set \(N(a,r)\) includes \((a,0)\) and the part of the tangent disk lying in \(H\). For example, when \(a=0\) and \(r=1\), the point \((0,0)\) is included by the singleton term, while \((0,1)\) belongs to the disk because its distance from \((0,1)\) is \(0<1\). The point \((0,2)\) does not belong, since its distance from \((0,1)\) is \(1\), which is not strictly less than \(1\).
The Moore plane is \(T_1\). For an interior point, a sufficiently small Euclidean disk avoids any specified different point. A boundary neighborhood \(N(a,r)\) avoids every other boundary point; for an interior point \((u,v)\), it avoids that point whenever \(2rv\leq (u-a)^2+v^2\). Choosing \(r\) small enough ensures this inequality.
Proof. We verify the shrinking test. Let \(W\) be open and contain an interior point \(x\). There is a Euclidean disk of radius \(R>0\), centered at \(x\) and contained in \(H\), that lies in \(W\). The closed Euclidean disk of radius \(R/2\) centered at \(x\) lies in that open disk. Its interior, used as a Moore-plane neighborhood \(V\), therefore has \(\overline V\subseteq W\).
Now let \(W\) contain a boundary point \((a,0)\). It contains \(N(a,r)\) for some \(r>0\). Choose \(s\) with \(0<s<r\), and set \(V=N(a,s)\). The Euclidean closed disk centered at \((a,s)\) with radius \(s\), together with its tangency point \((a,0)\), is closed in \(M\) and contains \(V\). To see closedness, interior points outside this closed disk have small Euclidean disk neighborhoods disjoint from it, and boundary points other than \((a,0)\) have sufficiently small tangent neighborhoods disjoint from it. This closed set is contained in \(N(a,r)\): the two Euclidean disks are internally tangent at \((a,0)\), and every other point of the smaller closed disk lies strictly inside the larger open disk. Consequently \(\overline V\subseteq N(a,r)\subseteq W\). The shrinking test and the \(T_1\) property prove regularity. \(\square\)
To see why regularity does not extend to normality here, let \(Q_0=\mathbb Q\times\{0\}\) and \(J_0=(\mathbb R\setminus\mathbb Q)\times\{0\}\). The boundary \(\mathbb R\times\{0\}\) is closed in \(M\), since its complement \(H\) is open; it is discrete in its subspace topology because \(N(a,r)\) meets the boundary only at \((a,0)\). Therefore every subset of the boundary is closed in \(M\), so \(Q_0\) and \(J_0\) are disjoint closed sets.
Suppose disjoint open sets separated \(Q_0\) and \(J_0\). For each rational \(q\), choose \(r_q>0\) so that \(D((q,r_q),r_q)\cap H\) lies in the open set containing \((q,0)\). For each irrational \(p\), choose \(s_p>0\) similarly for the open set containing \((p,0)\). The two Euclidean disks must be disjoint. The distance between their centers is \(\sqrt{(p-q)^2+(r_q-s_p)^2}\); disjoint open disks imply this is at least \(r_q+s_p\). Squaring and subtracting \((r_q-s_p)^2\) gives \((p-q)^2\geq4r_qs_p\).
For each \(n\in\mathbb N\), let \(G_n\) be the union, over \(q\in\mathbb Q\), of the open intervals centered at \(q\) with radius \(2\sqrt{r_q/n}\). It is open and dense because it contains \(\mathbb Q\). If \(s_p\geq1/n\), the inequality just proved gives \(|p-q|\geq2\sqrt{r_q/n}\) for every rational \(q\). Thus \(p\notin G_n\). The sets \(C_n=\mathbb R\setminus G_n\) are closed with empty interior, and every irrational \(p\) belongs to some \(C_n\), since \(s_p>0\). Hence \(\mathbb R\) would be the union of the countable set \(\mathbb Q\) and the countably many nowhere-dense sets \(C_n\). This contradicts the Baire category theorem for \(\mathbb R\), a standard result whose proof is not needed here. Therefore \(M\) is not normal.
The Sorgenfrey Plane: A Product of Normal Spaces
The Sorgenfrey line \(\mathbb R_\ell\) is \(\mathbb R\) with basis intervals \([a,b)\), where \(a<b\). The Sorgenfrey plane is \(\mathbb R_\ell\times\mathbb R_\ell\) with the product topology, whose basic sets are \([a,b)\times[c,d)\). The earlier result that the Sorgenfrey line is normal applies to each factor. The plane nevertheless is not normal.
Worked Example: The Sorgenfrey Plane Is Not Normal
Let \(D=\{(x,-x):x\in\mathbb R\}\), the anti-diagonal. It is closed: if \(a+b>0\), a basic rectangle based at \((a,b)\) stays in the region where the coordinate sum is positive; if \(a+b<0\), choose its two side lengths small enough that every point in the rectangle still has negative coordinate sum. In either case the rectangle misses \(D\). Also, \(D\) is discrete: the basic rectangle \([x,x+1)\times[-x,-x+1)\) meets \(D\) only at \((x,-x)\), because a diagonal point \((t,-t)\) in it would require both \(t\geq x\) and \(t\leq x\).
Every subset of \(D\) is closed: it is closed in the discrete subspace \(D\), and \(D\) itself is closed in the plane. In particular, \(A=\{(q,-q):q\in\mathbb Q\}\) and \(B=\{(p,-p):p\notin\mathbb Q\}\) are disjoint closed sets.
Assume disjoint open sets contain \(A\) and \(B\), respectively. For each \(x\in\mathbb R\), choose \(\varepsilon_x>0\) such that the rectangle \(R_x=[x,x+\varepsilon_x)\times[-x,-x+\varepsilon_x)\) lies in the open set containing \((x,-x)\). If \(x<y\), these two rectangles intersect exactly when \(y-x<\varepsilon_x\) and \(y-x<\varepsilon_y\). The first inequality makes their first-coordinate intervals overlap; the second makes their second-coordinate intervals overlap. Consequently, disjointness of the open sets requires \(|p-q|\geq\min(\varepsilon_p,\varepsilon_q)\) for every rational \(q\) and irrational \(p\).
For each \(n\in\mathbb N\), define \(G_n\) as the union over rational \(q\) of the intervals centered at \(q\) with radius \(\min(\varepsilon_q,1/n)\). Each \(G_n\) is open and dense because it contains every rational. If \(\varepsilon_p\geq1/n\), the required separation inequality implies \(|p-q|\geq\min(\varepsilon_q,1/n)\) for every rational \(q\), so \(p\notin G_n\). Thus \(p\in C_n=\mathbb R\setminus G_n\). Every irrational \(p\) satisfies \(\varepsilon_p\geq1/n\) for some \(n\), so \(\mathbb R\) is covered by \(\mathbb Q\) and the closed sets \(C_n\). Each \(C_n\) has empty interior, as \(G_n\) is dense. This again contradicts the Baire category theorem. No such disjoint open sets exist, and the plane is not normal.
Common pitfall. There are two different failed inferences to avoid. The deleted plank does not contradict closed-subspace inheritance, because it is not a closed subspace of its normal ambient product. The Sorgenfrey plane does not contradict that inheritance result either: its failure concerns products, not closed subspaces. Normality of each factor alone does not guarantee normality of their product.
| Space | What fails | Why the example matters |
|---|---|---|
| Deleted plank | Normality of a nonclosed subspace | Closed-subspace inheritance has a closedness hypothesis |
| Moore plane | Normality, despite regularity | Point-versus-closed-set separation is weaker than closed-set separation |
| Sorgenfrey plane | Normality of a product | Two normal factors need not have a normal product |
Check Your Understanding
For each question, identify the relevant separation condition and its hypotheses.
- Why does the deleted plank not violate the theorem that normality passes to closed subspaces?
- In the Moore plane, why must a basic neighborhood of \((a,0)\) include the point \((a,0)\) in addition to the tangent disk?
- Which two disjoint closed sets witness nonnormality in the Moore plane?
- Why is every subset of the anti-diagonal closed in the Sorgenfrey plane?
- What does the Sorgenfrey-plane example show about products of normal spaces?