An example showing that a quotient space of an hausdorff space is not hausdorff.

Question

Let be $\mathscr P$ the partition of $[0,1]^2$ defined through the position $$ \mathscr P:=\big\{\{x\}\times[0,1]:x\in\Bbb Q\big\}\cup\big\{\{(x,y)\}:(x,y)\in \mathbb R \setminus \mathbb Q\times[0,1]\big\} $$ So I ask to prove that the quotient topology of $\mathscr P$ is not hausdorff: unfortunately I have not many ideas about although I worked sometimes. So could someone help me, please?

Answer 1

Your partition consists of all line segments $L_x = \{x\} \times I$ with $x \in \mathbb Q \cap I$ and all singletons outside $(\mathbb Q \cap I) \times I$. Let $p : I^2 \to \mathscr P$ denote the function assigning to each $(x,y)$ the member of $\mathscr P$ containing it. Give $\mathscr P$ the quotient topology induced by $p$.

Take any irrational $\xi \in I$ and define $\xi_i = p(\xi,i) = \{(\xi,i)\} \in \mathscr P$ for $i = 0,1$. Clearly $\xi_0 \ne \xi_1$. Assume that there exist disjoint open neighborhoods $U_i$ of $\xi_i$. Then the sets $p^{-1}(U_i)$ are disjoint open subsets of $I^2$. Since $(\xi,i) \in p^{-1}(U_i)$, there exist open neigborhoods $V_i$ of $\xi$ and $W_i$ of $i$ in $I$ such that $V_i \times W_i \subset p^{-1}(U_i)$. Pick any rational $\eta \in V_1 \cap V_2$. Since $(\eta,i) \in p^{-1}(U_i)$, we have $p(\eta,i) \in U_i$, thus $L_\eta = p^{-1}(p(\eta,i)) \subset p^{-1}(U_i)$. Therefore the $p^{-1}(U_i)$ are not disjoint, a contradiction.