I am starting to learn about quotient spaces, but I am facing some troubles.
One of the problems that I'm stuck at is this:
Let $X = \mathbb{R} \times \{ 1, -1 \}$ and let's define the following equivalence relation: $(x,t) \sim (y,s) \iff (x,t) = (y,s) $ or $x=y \neq 0$. Determine whether the quotient space $X / \sim$ is Fréchet and if it is Hausdorff.
The solution only states that it is Fréchet, but not Hausdorff. I have absolutely no idea how to prove it though. The only way of showing that it is Fréchet or Hausdorff I know is by looking at it's neighborhoods, but I'm completely lost, because I'm not accustomed to quotient spaces. I would appreciate a solution and also how can I look at a quotient space (in this example or in general) and intuitively decide if it is Fréchet or Hausdorff.
Best Answer
If $V$ is a neighborhood of $(0,1)$ and $W$ is a neighborhood of $(0,-1)$ in $X/\sim$, then $V\cap W\ne\emptyset$, since $V$ contains a set of the form $(-\varepsilon_1,\varepsilon_1)\times\{1\}$ and $W$ contains a set of the form $(-\varepsilon_1,\varepsilon_1)\times\{-1\}$. But, if $0<\varepsilon<\min\{\varepsilon_1,\varepsilon_2\}$, $(\varepsilon,1)=(\varepsilon,-1)\in V\cap W$. Therefore, $X/\sim$ is not Hausdorff.
Can you now see why it is a Fréchet space?