(Jan ’08 # B9) Let \(X=\mathbb R^2\) and define an equivalence relation on \(X\) by \((x_1,x_2)\sim(y_1,y_2)\) if and only if they are equal or \(x_1=y_1=0\). Set \(Y=X/\sim\). Show that \(Y\) is Hausdorff but does not have a countable basis for its topology.
Solution
Let \(q:\mathbb R^2\to Y\) be the quotient map and let \(p=q(\{0\}\times\mathbb R)\).
First \(Y\) is Hausdorff. If two quotient points are represented by points \(x,y\notin\{0\}\times\mathbb R\), choose disjoint Euclidean balls about \(x\) and \(y\) small enough to avoid the \(y\)-axis; these balls are saturated, so their images under \(q\) are disjoint open neighborhoods. If one point is \(p\) and the other is represented by \(x=(x_1,x_2)\) with \(x_1\ne0\), choose \(0<\delta<|x_1|/3\). The saturated strip \[ S=\{(u,v):|u|<\delta\} \] contains the collapsed axis, while a sufficiently small ball about \(x\) is disjoint from \(S\) and from the axis. Their quotient images separate the two points. Thus \(Y\) is Hausdorff.
Now suppose \(p\) had a countable neighborhood basis \(U_1,U_2,\dots\). Put \(V_n=q^{-1}(U_n)\). Each \(V_n\) is an open set containing the entire \(y\)-axis. Hence for the point \((0,n)\) there exists \(r_n>0\) with \[ B((0,n),r_n)\subset V_n. \] Choose \(0<t_n<\min(r_n,1/2)\) and set \(z_n=(t_n,n)\). Then \(z_n\in V_n\). The set \(F=\{z_n:n\ge1\}\) is closed in \(\mathbb R^2\): its second coordinates are distinct integers and it has no finite accumulation point. Therefore \[ W=\mathbb R^2\setminus F \] is open, saturated, and contains the whole \(y\)-axis. Thus \(q(W)\) is an open neighborhood of \(p\). But \(U_n\nsubseteq q(W)\), since \(q(z_n)\in U_n\setminus q(W)\). This contradicts that \((U_n)\) is a neighborhood basis at \(p\).
Hence \(Y\) is not first countable at \(p\), and therefore cannot have a countable basis.