Understanding Quotient spaces - Shrinking down I am looking at Page 57 of Kreyszig's Functional Analysis, and I have been given an exercise:
Let $X=\Bbb R^3$ and $Y=\{(\xi_1,0,0)| \xi_1 \in \Bbb R\}$
1) Find $X/Y$:
So $X/Y=\{[x]:x+Y,\forall x\in X\}$. Now if $Y=\{(1,0,0)\}$ I would think of $X/Y$ as compressing $\Bbb R^3$ into the space $\{(x,y,z):x,y,z\in\Bbb R, 0\leq x\leq 1\}$. That may be wrong already, is it? But when $\xi_1\in \Bbb R$, it looks like we are setting $x=0$ in the quotient space always, is that correct?
2) Find $X/X$: I imagine this gives us only $\{(0,0,0\}$
3) Find $X/\{0\}$: This gives us back $\Bbb R^3$?
 A: *

*If $Y=\{(1,0,0)\}$, then $Y$ is not a vector space, this $X/Y$ is not defined. 

*You are almost correct. In fact, $X/X$ is a set of equivalence classes. Therefore, any element of $X/X$ will be $[x]$ for some $x\in X$. In the case of $X/X$, the space is equal to $X/X = \{[(0,0,0)]\}$ (note the square brackets you don't have, so the only element of $X/X$ is not $(0,0,0)$, which is an element of $X$, but $[(0,0,0)]$). You are correct, however, that $X/X$ is isomorphic to $\{(0,0,0)\}$

*Again, as in $2$, you are almost correct. $X/\{0\}$ is not $\mathbb R^3$, but it is isomorphic to $\mathbb R^3.$

A: You have $\mathbb R^3/\{(x_1,0,0); x_1\in\mathbb R\} \cong \mathbb R^2$. (As vector spaces and also as normed linear spaces - it is a bit unclear what you are asking about, but the answer is the same; I will deal with vector spaces.)
Here are two possible ways how to look at this:


*

*You can notice that $[(0,x_2,x_3)]$ for $x_2,x_3\in\mathbb R$ are all equivalence classes (and none of them is listed twice here). So $[(0,x_2,x_3)]\mapsto(x_2,x_3)$ is the isomorphism between these two spaces.

*For vector spaces you can apply the first isomorphism theorem to the surjective linear map $\mathbb R^3\to\mathbb R^2$ given by $(x_1,x_2,x_3)\mapsto(0,x_2,x_3)$.

