1. ## Show not Hausdorff

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? Thanks

2. Originally Posted by 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?
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 $\displaystyle (x_n)$ given by $\displaystyle x_n = (0,1/n)$ converges to all the points $\displaystyle (s,0)\;\,(s\in[0,1])$.

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 $\displaystyle (x_n)$ given by $\displaystyle x_n = (0,1/n)$ converges to all the points $\displaystyle (s,0)\;\,(s\in[0,1])$.

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

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

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

5. Originally Posted by 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? Thanks
Another possible way is to notice the following theorem:
Theorem: Let $\displaystyle X$ be a compact Hausdorff space and $\displaystyle q:X\to Y$ a quotient map. Then, $\displaystyle Y$ is Hausdorff if and only if $\displaystyle q$ is closed.