Let \(U\subset\RR^n\) be an open set which is bounded in the standard Euclidean metric. Prove that the quotient space \(\RR^n/U\) is not Hausdorff.
Solution
1. As written, the statement has an exceptional case: if \(U=\varnothing\), then the quotient is Hausdorff.
Collapsing the empty subset makes no identifications, so \[ \RR^n/\varnothing\cong\RR^n, \] which is Hausdorff. Thus the intended assertion requires \(U\neq\varnothing\).
Assume henceforth that \(n\ge1\) and that \(U\) is nonempty. Let \[ q:\RR^n\longrightarrow\RR^n/U \] be the quotient map, and write \(p=q(U)\) for the point obtained by collapsing \(U\).
2. The boundary \(\partial U\) is nonempty.
Because \(U\) is bounded and \(n\ge1\), it is a proper subset of \(\RR^n\). If \(\partial U\) were empty, then, since \(U\) is open, \[ \overline U=U\cup\partial U=U, \] so \(U\) would also be closed. This would make \(U\) a nonempty proper clopen subset of the connected space \(\RR^n\), a contradiction.
3. For any \(x\in\partial U\), the two distinct quotient points \(p\) and \(q(x)\) cannot be separated by disjoint open neighborhoods.
Since \(U\) is open, no boundary point lies in \(U\), so \[ x\notin U \qquad\text{and hence}\qquad q(x)\neq p. \] Suppose there were disjoint open sets \(O_p,O_x\subseteq\RR^n/U\) with \[ p\in O_p, \qquad q(x)\in O_x. \] Their inverse images \[ V=q^{-1}(O_p), \qquad W=q^{-1}(O_x) \] are disjoint open subsets of \(\RR^n\). Because \(p=q(U)\) lies in \(O_p\), one has \[ U\subseteq V. \] Because \(x\in W\) and \(x\in\partial U\), every open neighborhood of \(x\) meets \(U\); in particular, \[ W\cap U\neq\varnothing. \] Since \(U\subseteq V\), this gives \[ W\cap V\neq\varnothing, \] contradicting the disjointness of \(O_p\) and \(O_x\).
4. Therefore, for every nonempty bounded open set \(U\subset\RR^n\) with \(n\ge1\), the quotient \(\RR^n/U\) is not Hausdorff.
By 2 choose \(x\in\partial U\). Then 3 exhibits two distinct points of the quotient that admit no disjoint open neighborhoods, which is exactly the failure of the Hausdorff property.