The text of the problem is the following:
$X = [0,1] \times \{0,1\} $ with the topology induced by the euclidean topology, is an Hausdorff space. Let's define an equivalence relation: $(x_1,y_1)R(x_2,y_2)$ iff:
1)$(x_1,y_1)=(x_2,y_2);$
2)$x_1 = y_1 $ and $x_2 = 0, \ y_2 = 1$ or viceversa.
Show that $X/R$ is not Hausdorff.
Firstly it isn't to clear to me what viceversa does it mean. I interpreted it in the following way: $x_2=y_2$ and $x_1 =0, y_1 = 1$.
Now I will try to explain my thoughts. The quotient space $X/R$ can be thought as a segment of the following form: $[0,1)$. Let's define the set $S = \{(0,0),(1,0),(0,1),(1,1)\}$ and the projection:
$\pi(x,y) \to x \ if (x,y) \notin S$
$\pi(x,y) \to \{0\} \ if (x,y) \in S$
$\forall z_1,z_2 \in X/R $, without loss of generality we can take $z_1 < z_2$. Now I will define two disjoint and open sets (in the quotient topology), containing $z_1$ and $z_2$.
$A_1 = [0,(z_1+z_2)/2) \ \bigcup \ ((z_2+1)/2,1);$
$A_2= ((z_1+z_2)/2,(z_2+1)/2)$.
So the quotient seems to be Hausdorff. Where is my mistake?
Any help is appreciated. Thanks in advance.