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**bram kierkels** Define an equivalence relation in $\displaystyle X=[0,1]\times[0,1]$, by declaring $\displaystyle (s_0,t_0)\equiv(s_1,t_1)$ if and only if $\displaystyle t_0=t_1>0$. Describe the quotient space $\displaystyle X/\equiv$ and show that it is not a Hausdorff space.

**My solution:**

Let $\displaystyle \pi:X\rightarrow X/\equiv$ be the projection map. The equivalence class $\displaystyle \pi((s,0))$ of $\displaystyle (s,0)$, where $\displaystyle s\in[0.1]$, is $\displaystyle (s,0)$ since $\displaystyle t=0$.

If $\displaystyle (s_1,t_1)\equiv(s_2,t_2)$ then $\displaystyle t_1=t_2$. Then the equivalence class $\displaystyle \pi((s,t))$ of $\displaystyle (s,t)$, where $\displaystyle t\in(0,1]$, is $\displaystyle \{(s_i,t)|s_i\in[0,1]\}$.

Thus the set $\displaystyle X/\equiv$ is the union of $\displaystyle \{(s,0)|s\in[0,1]\}$ and $\displaystyle \{(s_i,t)|s_i\in[0,1]\}$ for all $\displaystyle t$.

Can someone help me show that this space is not a Hausdorff space?