Let \(D\) be the closed unit disk in the complex plane. Let \(\sim\) be the equivalence relation on \(D\) defined by \(z_1\sim z_2\) if and only if \(z_1=z_2\) or \(|z_1|=|z_2|<1\). Is the quotient topological space Hausdorff? (Prove your assertion.)
Solution
Let \[ q:D\longrightarrow D/{\sim} \] be the quotient map. The quotient is not Hausdorff.
1. Distinct points of the boundary circle determine distinct points of the quotient.
Choose distinct points \[ u,v\in S^1=\{z\in\mathbb C:|z|=1\}. \] By definition of the relation, two distinct points can be equivalent only when their common modulus is strictly less than \(1\). Since \[ |u|=|v|=1 \quad\text{and}\quad u\ne v, \] we have \(u\not\sim v\). Thus \[ q(u)\ne q(v). \]
2. Every open neighborhood of \(q(u)\) has preimage containing an entire circle of some radius arbitrarily close to \(1\).
Let \(O\subseteq D/{\sim}\) be open and suppose \(q(u)\in O\). Then \[ U=q^{-1}(O) \] is an open subset of \(D\) containing \(u\). It is also saturated: if \(x\in U\) and \(x\sim y\), then \[ q(y)=q(x)\in O, \] so \(y\in U\).
Consider the radial path \[ \gamma_u:[0,1]\longrightarrow D, \qquad \gamma_u(t)=(1-t)u. \] It is continuous and \(\gamma_u(0)=u\in U\). Hence, for some \(\varepsilon>0\), \[ (1-t)u\in U \qquad(0<t<\varepsilon). \] For each such \(t\), the radius \(1-t\) is strictly less than \(1\). Therefore every point \(z\in D\) satisfying \[ |z|=1-t \] is equivalent to \((1-t)u\) and, by saturation, also belongs to \(U\).
3. The two quotient points \(q(u)\) and \(q(v)\) cannot have disjoint open neighborhoods.
Suppose, for contradiction, that there are disjoint open sets \(O_u,O_v\subseteq D/{\sim}\) such that \[ q(u)\in O_u, \qquad q(v)\in O_v. \] Set \[ U=q^{-1}(O_u), \qquad V=q^{-1}(O_v). \] Then \(U\) and \(V\) are disjoint open saturated subsets of \(D\), with \(u\in U\) and \(v\in V\).
By 2 applied to \(U\), there is \(\varepsilon_u>0\) such that \[ (1-t)u\in U \qquad(0<t<\varepsilon_u). \] The same argument applied to \(V\) gives \(\varepsilon_v>0\) such that \[ (1-t)v\in V \qquad(0<t<\varepsilon_v). \] Choose \[ 0<t<\min\{\varepsilon_u,\varepsilon_v,1\}. \] Then \[ |(1-t)u|=|(1-t)v|=1-t<1, \] so \[ (1-t)u\sim(1-t)v. \] Since \(U\) is saturated and \((1-t)u\in U\), it follows that \((1-t)v\in U\). But \((1-t)v\in V\) as well. Thus \[ U\cap V\ne\varnothing, \] contradicting the disjointness of \(O_u\) and \(O_v\).
4. Therefore \(D/{\sim}\) is not Hausdorff.
By 1, \(q(u)\) and \(q(v)\) are distinct points of the quotient. By 3, they cannot be separated by disjoint open neighborhoods. This violates the Hausdorff separation axiom. Hence \[ \boxed{D/{\sim}\text{ is not Hausdorff}.} \]