A Property That Does Not Always Pass to Subspaces
A space is separable if it has a countable dense subset. It is tempting to expect that a subspace of a separable space will also be separable: one might try to intersect the subspace with a countable dense set in the larger space. That intersection, however, need not meet every nonempty open set of the subspace. The Sorgenfrey plane gives a concrete example: it is separable, but it contains an uncountable discrete subspace, which cannot be separable.
Let \(\mathbb{R}_\ell\) denote the real line with the lower limit topology, whose basis consists of the intervals \([a,b)\) with \(a<b\). The Sorgenfrey plane is \(\mathbb{R}_\ell\times\mathbb{R}_\ell\) with the product topology. Its basis consists of rectangles \([a,b)\times[c,d)\), where \(a<b\) and \(c<d\). A subset is dense if it intersects every nonempty open set. A space is separable if it has a countable dense subset. A subspace \(Y\subseteq X\) has the subspace topology: its open sets are the sets \(Y\cap U\), where \(U\) is open in \(X\).
The Sorgenfrey Plane Is Separable
Consider \(D_0=\mathbb{Q}\times\mathbb{Q}\). It is countable because \(\mathbb{Q}\) is countable and the Cartesian product of two countable sets is countable. To show that \(D_0\) is dense, it is enough to check that it meets every nonempty basic rectangle: every nonempty open set contains such a rectangle.
Proof. Let \([a,b)\times[c,d)\) be a nonempty basic rectangle, so \(a<b\) and \(c<d\). The density of \(\mathbb{Q}\) in the usual real line gives \(r\in\mathbb{Q}\) with \(a<r<b\), and \(s\in\mathbb{Q}\) with \(c<s<d\). Thus \((r,s)\in(\mathbb{Q}\times\mathbb{Q})\cap([a,b)\times[c,d))\). Every nonempty open set contains a nonempty basic rectangle, so it too meets \(\mathbb{Q}\times\mathbb{Q}\). Therefore \(\mathbb{Q}\times\mathbb{Q}\) is a countable dense subset, and the plane is separable. \(\square\)
Worked Example: Finding a Rational Point in a Basic Rectangle
Consider the basic rectangle \([1,2)\times[-3,-2)\). The rational point \((3/2,-5/2)\) belongs to it: \(1\leq 3/2<2\), and \(-3\leq-5/2<-2\). Its coordinates are both rational, so it also belongs to \(\mathbb{Q}\times\mathbb{Q}\). This verifies directly, for this particular rectangle, the intersection used in the density argument.
Check: the half-open endpoint convention matters. The point \((2,-5/2)\) does not belong to the rectangle because \(2\notin[1,2)\), whereas \((1,-3)\) does belong to it. The point \((1,-3)\) is itself rational in both coordinates.
The Anti-Diagonal Is Discrete
Now consider the anti-diagonal \(A=\{(x,-x):x\in\mathbb{R}\}\). We prove discreteness by finding, for each point of \(A\), an open set in the plane whose intersection with \(A\) is exactly that point. A subspace is discrete when each of its singletons is open in the subspace topology.
Proof. Fix any \(x\in\mathbb{R}\), and consider the basic open rectangle \(U_x=[x,x+1)\times[-x,-x+1)\). It contains \((x,-x)\). Suppose \((t,-t)\in A\cap U_x\). The first coordinate condition gives \(x\leq t<x+1\). The second gives \(-x\leq -t<-x+1\); multiplying these inequalities by \(-1\), and reversing their directions, gives \(x-1<t\leq x\). Together the two conditions imply \(t=x\). Hence \(A\cap U_x=\{(x,-x)\}\). By the subspace topology, \(A\cap U_x\) is open in \(A\). This holds for every \(x\), so every singleton in \(A\) is open in \(A\), and \(A\) is discrete. \(\square\)
Worked Example: Isolating a Point on the Anti-Diagonal
Take the point \((2,-2)\in A\). The rectangle \([2,3)\times[-2,-1)\) is open in the Sorgenfrey plane. If \((t,-t)\) belongs to this rectangle, then \(2\leq t<3\) and \(-2\leq-t<-1\). The latter inequalities give \(1<t\leq2\). Together they force \(t=2\). Thus the rectangle meets \(A\) in exactly \(\{(2,-2)\}\).
Check: \(t=2\) satisfies both coordinate conditions, so \((2,-2)\) is in the intersection. If \(t=5/2\), the first coordinate is allowed, but \(-t=-5/2\notin[-2,-1)\). If \(t=3/2\), then \(-t=-3/2\) is allowed, but \(t\notin[2,3)\). These tests confirm that neither nearby parameter gives another point of the intersection.
Why This Discrete Subspace Is Not Separable
The essential feature of a discrete space is that every singleton is open. A dense subset must meet every nonempty open set, so in a discrete space it must meet every singleton. That forces it to contain every point.
Proof. Let \(Y\) be a discrete space. If \(E\subseteq Y\) is dense, then for every \(y\in Y\), the open singleton \(\{y\}\) must meet \(E\). This means \(y\in E\), so \(E=Y\). Therefore, if \(Y\) is separable, it has a countable dense subset, which must equal \(Y\); hence \(Y\) is countable. Conversely, if \(Y\) is countable, then \(Y\) itself is a countable dense subset of \(Y\), since it intersects every nonempty open subset of \(Y\). This also covers \(Y=\varnothing\), which is countable and dense in itself. \(\square\)
The anti-diagonal is uncountable because the map \(x\mapsto(x,-x)\) is a bijection from \(\mathbb{R}\) to \(A\). For completeness, uncountability of \(\mathbb{R}\) follows from the diagonal argument: the set of binary sequences \(\{0,1\}^{\mathbb{N}}\) is uncountable, since from any proposed enumeration \(s^{(1)},s^{(2)},\ldots\) the sequence with \(n\)-th term \(1-s^{(n)}_n\) differs from the \(n\)-th listed sequence in its \(n\)-th term. These sequences inject into \(\mathbb{R}\) by the map \((a_n)\mapsto\sum_{n=1}^{\infty}2a_n/3^n\). Indeed, if two sequences first differ at index \(k\), the contribution \(2/3^k\) at that index exceeds the largest possible tail difference, \(\sum_{n>k}2/3^n=1/3^k\), so their sums differ. Thus \(\mathbb{R}\), and consequently \(A\), is uncountable. Since \(A\) is discrete, the theorem shows that it is not separable.
Worked Example: Why Intersecting with the Dense Set Fails
The countable dense set \(\mathbb{Q}\times\mathbb{Q}\) in the plane intersects \(A\) in \((\mathbb{Q}\times\mathbb{Q})\cap A=\{(q,-q):q\in\mathbb{Q}\}\). This intersection is countable. It is not dense in \(A\): for any irrational \(x\), the set \(A\cap U_x=\{(x,-x)\}\) is a nonempty open subset of \(A\), and it does not meet \(\{(q,-q):q\in\mathbb{Q}\}\). The irrational \(x\) is not rational, so its point is not in that intersection.
Check: the calculation uses both coordinates: \((q,-q)\) has rational coordinates whenever \(q\in\mathbb{Q}\), and if \((x,-x)\) has both coordinates rational then in particular \(x\in\mathbb{Q}\). Thus the displayed intersection contains exactly the points with rational parameter. Its failure to be dense follows from the open singleton around each point with irrational parameter.
A Contrast: The Ordinary Diagonal
The anti-diagonal is not the only natural subspace to examine. The ordinary diagonal \(\Delta=\{(x,x):x\in\mathbb{R}\}\) helps show that the product topology can induce different behavior on different subsets. In fact, its subspace topology is the lower limit topology, so it is separable.
Worked Example: The Induced Topology on the Diagonal
Intersect a basic rectangle \([a,b)\times[c,d)\) with \(\Delta\). A point \((x,x)\) lies in this intersection exactly when \(a\leq x<b\) and \(c\leq x<d\). If the intersection is nonempty, these conditions describe the interval \([\max(a,c),\min(b,d))\) in the diagonal parameter \(x\). Conversely, any lower-limit basic interval \([r,s)\), with \(r<s\), is obtained by intersecting \([r,s)\times[r,s)\) with \(\Delta\). Thus the induced topology on \(\Delta\), under the parameterization \(x\mapsto(x,x)\), is precisely the lower limit topology.
The rational-parameter points \(\{(q,q):q\in\mathbb{Q}\}\) are dense in \(\Delta\): every nonempty basic open interval \([r,s)\) in its parameter has a rational \(q\) with \(r<q<s\). This set is countable, so \(\Delta\) is separable. In contrast, the anti-diagonal is discrete and uncountable, so it is not separable.
Check: when \(r<s\), a rational strictly between \(r\) and \(s\) belongs to \([r,s)\); the right endpoint \(s\) is excluded and is not needed. The two diagonal examples therefore have different induced topologies even though each is parameterized by all real numbers.
The General Lesson
Separability is not, in general, inherited by subspaces. The specific argument is worth keeping distinct from the definition of density: a dense set in \(X\) meets every nonempty open set of \(X\), while a dense set in a subspace \(Y\) must meet every nonempty set of the form \(Y\cap U\), with \(U\) open in \(X\). Intersecting a dense subset of \(X\) with \(Y\) is not guaranteed to meet all such intersections. In this example, the ambient countable dense set misses many open singletons of the anti-diagonal.
Common pitfall. A mistaken argument says that \(\mathbb{Q}\times\mathbb{Q}\) is dense in the plane, so its intersection with every subspace must be dense there. The computation above disproves that inference: the intersection with \(A\) is \(\{(q,-q):q\in\mathbb{Q}\}\), and the nonempty open singleton \(\{(x,-x)\}\) in \(A\), for irrational \(x\), misses it. Density in the ambient space does not by itself imply density after intersection with an arbitrary subspace.
Check Your Understanding
Use the definitions of density, discreteness, and the subspace topology to answer the following.
- Why does every nonempty basic rectangle in the Sorgenfrey plane meet \(\mathbb{Q}\times\mathbb{Q}\)?
- For a general point \((x,-x)\) of the anti-diagonal, write down a basic open rectangle whose intersection with the anti-diagonal is a singleton.
- Prove directly that a countable discrete space is separable.
- Why can no proper subset of a discrete space be dense in that space?
- Which step fails in the claim that intersecting an ambient dense set with a subspace always produces a dense subset of that subspace?