Give an example of a quotient map in which the domain is Hausdorff, but the quotient is not.
Solution
Define an equivalence relation on \(\RR\) by \[ x\sim y \quad\Longleftrightarrow\quad x-y\in\QQ, \] and let \[ q:\RR\longrightarrow\RR/{\sim} \] be the quotient map.
1. The domain \(\RR\) is Hausdorff.
The usual topology on \(\RR\) is induced by the metric \[ d(x,y)=|x-y|, \] and every metric space is Hausdorff.
2. Every nonempty open subset of \(\RR/{\sim}\) is the whole quotient.
Let \[ V\subseteq\RR/{\sim} \] be nonempty and open. By definition of the quotient topology, \[ U=q^{-1}(V) \] is a nonempty open subset of \(\RR\). Choose \(x\in U\) and \(\varepsilon>0\) such that \[ (x-\varepsilon,x+\varepsilon)\subseteq U. \]
Fix arbitrary \(y\in\RR\). The interval \[ (x-y-\varepsilon,x-y+\varepsilon) \] contains some rational number \(r\), since \(\QQ\) is dense in \(\RR\). Therefore \[ y+r\in(x-\varepsilon,x+\varepsilon)\subseteq U. \] But \[ (y+r)-y=r\in\QQ, \] so \[ q(y+r)=q(y). \] Since \(y+r\in U=q^{-1}(V)\), one has \(q(y+r)\in V\), hence \(q(y)\in V\).
As \(y\) was arbitrary and \(q\) is surjective, \[ V=\RR/{\sim}. \] Thus the quotient topology is indiscrete.
3. The quotient \(\RR/{\sim}\) has at least two points.
The classes of \(0\) and \(\sqrt2\) are distinct because \[ \sqrt2-0\notin\QQ. \]
4. The quotient \(\RR/{\sim}\) is not Hausdorff.
By 2, the only nonempty open subset is the whole space. By 3 there are distinct points in the quotient. Any neighborhoods of two such points are therefore both the whole quotient and cannot be disjoint.
Hence \[ \boxed{q:\RR\to\RR/{\sim}} \] is a quotient map from a Hausdorff space onto a non-Hausdorff space.