1. ## Show not Hausdorff

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

My solution:
Let $\pi:X\rightarrow X/\equiv$ be the projection map. The equivalence class $\pi((s,0))$ of $(s,0)$, where $s\in[0.1]$, is $(s,0)$ since $t=0$.
If $(s_1,t_1)\equiv(s_2,t_2)$ then $t_1=t_2$. Then the equivalence class $\pi((s,t))$ of $(s,t)$, where $t\in(0,1]$, is $\{(s_i,t)|s_i\in[0,1]\}$.
Thus the set $X/\equiv$ is the union of $\{(s,0)|s\in[0,1]\}$ and $\{(s_i,t)|s_i\in[0,1]\}$ for all $t$.

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

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

My solution:
Let $\pi:X\rightarrow X/\equiv$ be the projection map. The equivalence class $\pi((s,0))$ of $(s,0)$, where $s\in[0.1]$, is $(s,0)$ since $t=0$.
If $(s_1,t_1)\equiv(s_2,t_2)$ then $t_1=t_2$. Then the equivalence class $\pi((s,t))$ of $(s,t)$, where $t\in(0,1]$, is $\{(s_i,t)|s_i\in[0,1]\}$.
Thus the set $X/\equiv$ is the union of $\{(s,0)|s\in[0,1]\}$ and $\{(s_i,t)|s_i\in[0,1]\}$ for all $t$.

Can someone help me show that this space is not a Hausdorff space?
If a space is Hausdorff then a convergent sequence has a unique limit point. You could show that this space is non-Hausdorff by showing that the sequence $(x_n)$ given by $x_n = (0,1/n)$ converges to all the points $(s,0)\;\,(s\in[0,1])$.

3. Originally Posted by Opalg
If a space is Hausdorff then a convergent sequence has a unique limit point. You could show that this space is non-Hausdorff by showing that the sequence $(x_n)$ given by $x_n = (0,1/n)$ converges to all the points $(s,0)\;\,(s\in[0,1])$.

Thanks, but why does this sequence converges to (s,0)? For every open neigborhood $(s_i,0)\times(s_{i+1},0), s_i\leq s\leq s_{i+1}$ there is no $N$ such that $x_i\in(s_i,0)\times(s_{i+1},0)$ for all $i>N$..?

4. Originally Posted by bram kierkels
Thanks, but why does this sequence converges to (s,0)? For every open neigborhood $(s_i,0)\times(s_{i+1},0), s_i\leq s\leq s_{i+1}$ there is no $N$ such that $x_i\in(s_i,0)\times(s_{i+1},0)$ for all $i>N$..?
I don't understand what you mean by the "open neigborhood $(s_i,0)\times(s_{i+1},0), s_i\leq s\leq s_{i+1}$".

The way I think of it is that the projection map $\pi$ is continuous. So if $x_n\to x$ in $X$ then $\pi(x_n)\to\pi(x)$. Since $(s,1/n)\to(s,0)$ in $X$ (for each $s\in[0,1]$), it follows that $\pi(s,1/n)\to\pi((s,0)$. But $\pi(s,1/n) = \pi(0,1/n) = \pi(x_n)$. It follows that $\pi(x_n)\to(s,0)$ in the quotient space.

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

My solution:
Let $\pi:X\rightarrow X/\equiv$ be the projection map. The equivalence class $\pi((s,0))$ of $(s,0)$, where $s\in[0.1]$, is $(s,0)$ since $t=0$.
If $(s_1,t_1)\equiv(s_2,t_2)$ then $t_1=t_2$. Then the equivalence class $\pi((s,t))$ of $(s,t)$, where $t\in(0,1]$, is $\{(s_i,t)|s_i\in[0,1]\}$.
Thus the set $X/\equiv$ is the union of $\{(s,0)|s\in[0,1]\}$ and $\{(s_i,t)|s_i\in[0,1]\}$ for all $t$.

Can someone help me show that this space is not a Hausdorff space? Thanks
Another possible way is to notice the following theorem:
Theorem: Let $X$ be a compact Hausdorff space and $q:X\to Y$ a quotient map. Then, $Y$ is Hausdorff if and only if $q$ is closed.