Quotient space example: following this definition  of wikipedia, I was trying to prove that this example is correct:
Let $V = \left\{ (x_1,x_2,x_3) \right\}_{x_i \in \mathbb{R}}$ and $W = \left\{ (x_1,0,0) \right\}_{x_1 \in \mathbb{R}}$. For given $x \in V$ the coset should be
$$
[x] = \left\{ x + u \;:\; u \in W\right\} = \left\{(x_1,x_2,x_3) +(u_1,0,0) \;:\; u_1 \in \mathbb{R} \right\} = \left\{ (v,x_1,x_2) \;:\; v \in \mathbb{R} \right\},
$$
Namely
$$
[x] = \left\{ (v,x_1,x_2) \;:\; v \in \mathbb{R} \right\}
$$
Question 1. Is it correct to write: 
$$
V/W = \left\{ [x] \right\}_{x \in V}
$$
Question 2. To prove that $V/W$ is isomorphic to $\mathbb{R}^2$ it means to find a linear isometry, right? i.e. A linear transformation, invertible, that preserve the norm right? If yes, the transformation I would use is this one:
$$T([x]) = (x_1,x_2)= y $$,
I guess it is easy to show that it is linear, also it is invertible. I'm puzzled about to proove that it is an isometry, because I'm not sure how to formalize it.
 A: For Question 1, there's nothing wrong from a set theoretic perspective in writing $V/W=\{[x]\}_{x \in V}$, as long as you keep in mind that this is not a 1--1 indexing of the elements of $V/W$. In other words, you can have two elements $x \ne x' \in V$ such that $[x]=[x']$, for example $x=(1,3,4)$ and $x'=(2,3,4)$. 
For Question 2, isomorphism of vector spaces does not mean linear isometry. The definition of vector spaces talks only about the operations of vector sum and scalar multiplication, it does not talk about metric structure. Of course, sometimes we also apply extra metric structures on vector spaces, such as a norm or a positive definite inner product. But if all you have to do is to prove that the $V/W$ and $\mathbb{R}^2$ are isomorphic as vector spaces, you simply need to produce a linear isomorphism, meaning a linear bijection that preserves vector sum and scalar multiplication. Isometry is irrelevant.
A: 1.) By definition, $V/W = \{[x]\mid x \in V \}$, so that you are correct.
2.) Vector spaces $A$ and $B$ are isomorphic iff there exists a bijective linear map $T: A \rightarrow B$
