I've got a question on Quotient Space but I just can't visualise the space.
This is the definition:
X = $\mathbb{R}$ x $\mathbb{Z}$ (x,z) and (x',z') are said to be equivalent iff
x = x' and z = z' or
x = x' and x is not 0.
Y = X/~
Now I should prove that Y is not Hausdorff and Y is not a pseudo metric but I'm not sure how to start.
Can anyone point me in the right direction?
Thanks in advance!
I'll help you try to imagine the space. First $X$ itself is $\mathbb{R} \times \mathbb{Z}$. We can think of this as just $\mathbb{Z}$ many copies of the line, stacked horizontally on top of eachother, one at each integer.
Now the equivalence relation is saying that at the origin, i.e. points of the form $(0,z)$, you do nothing, you leave the origin of each copy alone. However, away from the origin, you identify any points of the form $(x,z)$ and $(x,z')$. That means you paste together every copy of $\mathbb{R} \setminus 0$. So what you get is something that away from the origin looks like just a line, but at the origin has $\mathbb{Z}$ many copies of the origin.
Here's a picture of the same construction but when you take only $2$ copies instead of $\mathbb{Z}$ copies to begin with. This looks like the line everywhere except with two origins. Imagine the same thing but with countably infinitely many origins.
Now try to show that any open set around one of the origins intersects any open set around the other origins and so it can't be Hausdorff. As for pseduometric, try to prove that the different origins are topologically distinguishable, i.e., the space is $T_0$. Then if it were pseudometric space, it would actually be a metric space which contradicts the fact it's not Hausdorff.