Let \(A\) be the set of all points of \(\mathbb{R}_\ell^2\) of the form \(x \times (-x)\), for \(x\) rational; let \(B\) be the set of all points of this form for \(x\) irrational. If \(V\) is an open set of \(\mathbb{R}_\ell^2\) containing \(B\), show there exists no open set \(U\) containing \(A\) that is disjoint from \(V\), as follows:
Let \(K_n\) consist of all irrational numbers \(x\) in \([0, 1]\) such that \([x, x + 1/n) \times [-x, -x + 1/n)\) is contained in \(V\). Show that \([0, 1]\) is the union of the sets \(K_n\) and countably many one-point sets.
Use Exercise 5 of §27 to show that some set \(\overline{K}_n\) contains an open interval \((a, b)\) of \(\mathbb{R}\).
Show that \(V\) contains the open parallelogram consisting of all points of the form \(x \times (-x + \epsilon)\) for which \(a < x < b\) and \(0 < \epsilon < 1/n\).
Conclude that if \(q\) is a rational number with \(a < q < b\), then the point \(q \times (-q)\) of \(\mathbb{R}_\ell^2\) is a limit point of \(V\).
Solution
For each irrational \(x\in[0,1]\), the point \((x,-x)\) lies in the open set \(V\). Hence there is \(\varepsilon_x>0\) such that \[ [x,x+\varepsilon_x)\times[-x,-x+\varepsilon_x)\subset V. \] Choose \(n\) with \(1/n<\varepsilon_x\). Then \(x\in K_n\). Thus every irrational point of \([0,1]\) lies in some \(K_n\), while the rational points form a countable union of one-point sets. Therefore \[ [0,1]=\bigcup_{n\ge1}K_n\ \cup\ \bigcup_{q\in\mathbb Q\cap[0,1]}\{q\}. \]
If every \(\overline{K_n}\) had empty interior, then the compact Hausdorff space \([0,1]\) would be a countable union of closed sets with empty interior, together with countably many singleton closed sets, also of empty interior. This contradicts the Baire-category exercise of §27. Hence some \(\overline{K_n}\) contains a nonempty open interval \((a,b)\).
Since \((a,b)\subset\overline{K_n}\), the set \(K_n\) is dense in \((a,b)\). Fix \[ a<x<b,\qquad 0<\epsilon<1/n. \] Choose \[ t\in K_n\cap(\max\{a,x-\epsilon\},x). \] Then \(t<x<t+1/n\), and \(t>x-\epsilon\) implies \[ -t<-x+\epsilon. \] Also \(t<x<x-\epsilon+1/n\), so \[ -x+\epsilon<-t+1/n. \] Therefore \[ (x,-x+\epsilon)\in[t,t+1/n)\times[-t,-t+1/n)\subset V. \] Thus \(V\) contains the stated open parallelogram.
Let \(q\in(a,b)\cap\mathbb Q\). Every basic neighborhood of \((q,-q)\) in the Sorgenfrey plane contains a point \[ (q,-q+\epsilon) \] with \(0<\epsilon<1/n\) sufficiently small. By (c) this point lies in \(V\), and it differs from \((q,-q)\). Hence \((q,-q)\) is a limit point of \(V\).
Now if \(U\) were an open set containing \(A\) and disjoint from \(V\), then \(U\) would contain \((q,-q)\), but every neighborhood of this point meets \(V\), contradiction. Therefore no open neighborhood of \(A\) is disjoint from \(V\).