Intuition or guide on this question regarding subspaces/vector spaces So I had an exam two days ago and admittedly, I couldn't do this last question very well. I was wondering if someone could give me a guide on how to do this or how to think about it? I'll post what I did in the exam.  
Question
Let $V$ be a vector space over a field $\mathbb{F}$ and $U$ a subspace of $V$. For any $\textbf{v}$ in $V$ we define the set
$$\textbf{v} + U = \{\textbf{v} + \textbf{u}|\textbf{u}\in U\}.$$
We write $V/U$ for the set of all of these sets,  $$V/U = \{\textbf{v} + U|\textbf{v} \in V\},$$
and we define addition and scalar multiplication on $V/U$ by
$$(\textbf{v}_1 + U) + \textbf{v}_2 + U) = (\textbf{v}_1 + \textbf{v}_2) + U \quad \text{and} \quad \alpha(\textbf{v}+U) = (\alpha \textbf{v})+U.$$
You may assume these operations are well defined.
i) Find the identity element for addition in $V/U$. Prove that your answer is correct.
ii) Prove that the vector distributive law holds in $V/U$. (That is, the one involving a scalar and two vectors, not two scalars and a vector.)
iii) Let $\textbf{v}_1,\textbf{v}_2$ be in $V$. Show that $\textbf{v}_1 + U = \textbf{v}_2 + U$ if and only if $\textbf{v}_1 - \textbf{v}_2$ is in $U$.
iv) $\textbf{You may now assume}$ that $V/U$ is a vector space over $\mathbb{F}$, with operations defined as above. Prove that if $V$ is finite-dimensional, then $V/U$ is also finite-dimensional and
$$\dim V/U = \dim V - \dim U.$$  
My attempts
Firstly, I'm not sure what the "standard" name for these type of subspaces are, as I've never encountered them before, so I was lost on what it even meant to "add" an element to another subspace (referring to $V/U$).
i) I claimed that the identity element is $U$ since for any $\textbf{w} = \textbf{v}_1 + U$ in $V/U$, the additive inverse is $-\textbf{w} = -(\textbf{v}_1 + U) = -\textbf{v}_1 + U$ (by definitions of the addition/scalar multiplication).
$\textbf{w} + (-\textbf{w}) = U +(-1)U = (1-1)U = 0U = U$.
ii) I used the same reasoning (and the definitions to show that $\lambda (\textbf{w}_1 + \textbf{w}_2) = \lambda \textbf{w}_1 + \lambda \textbf{w}_2$.
iii) & iv) I had no clue how to start or think about.  
Thinking about it again, I'm confused how $U$ can be an identity element of a set... when $U$ is itself a set (with elements in it)..
 A: The sets $\mathbf{v} +U$ are not subspaces technically (they are not closed under addition), but the standard name for them is "cosets" (of $U$). Your intuition for $(i)$ and $(ii)$ are correct.
For $(iii)$, suppose first that $\mathbf{v_0} = \mathbf{v}_1 - \mathbf{v}_2 \in U$. Then for any $\mathbf{u} \in U$, we have $\mathbf{v}_2+\mathbf{u} = \mathbf{v}_1 + (\mathbf{u}-\mathbf{v_0})$, which shows that $\mathbf{v}_2 + U \subseteq \mathbf{v}_1 + U$, and also $\mathbf{v}_1 + \mathbf{u} = \mathbf{v}_2 + (\mathbf{u}+\mathbf{v}_0)$, which shows that $\mathbf{v}_1+U \subseteq \mathbf{v}_2+U$. Conversely, suppose that $\mathbf{v}_1+U=\mathbf{v}_2+U$. Take any $\mathbf{u} \in U$. Then $\mathbf{v}_1+\mathbf{u} \in \mathbf{v}_1+U$, and so also by assumption, it is in $\mathbf{v}_2+U$. So there is some other $\mathbf{u}' \in U$ so that $\mathbf{v}_1+\mathbf{u} = \mathbf{v}_2+\mathbf{u}'$. Then $\mathbf{v}_1-\mathbf{v}_2=\mathbf{u}'-\mathbf{u} \in U$.
Part $(iv)$ is a well-known result that is known by the name "Rank-Nullity Theorem". A more abstract version is also known by the name "First Isomorphism Theorem". You should be able to find a proof by just searching on Google.
A: These questions are on fundamental properties for quotient spaces, as defined above by $V/U = \{ v + U \mid v \in V \} $.
i) The additive identity of $V/U$ is $0 + U$ or $U$.
ii) I am assuming that you are referring to the distributive property of $a(u + v) = au + av$ for $\forall a,b \in F$ and $\forall u,v \in V$.
This can be simply proven by using the definition of addition and scalar multiplication on $V/U$.
For $v_1, v_2 \in V$ and $\lambda \in F$, $\lambda((v_1 + U) + (v_2 + U)) = \lambda ((v_1 + v_2) + U)$ by definition of addition.
By definition of scalar multiplication, $\lambda ((v_1 + v_2) + U) = \lambda (v_1) + \lambda(v_2) + U = \lambda (v_1) + U + \lambda(v_2) + U$. Thus, the vector distributive law holds in $V/U$.
iii) If $v_1 + U = v_2 + U$, let $u_1, u_2 \in U$ s.t. $v_1 + u_1 = v_2 + u_2$. Then, $v_1 - v_2 = u_2 - u_1 \in U$. Thus, $v_1 - v_2 \in U$.
If $v_1 - v_2 \in U$, for $u \in U$,
$ v_1 + u = v_2 + ((v_1 - v_2) + u) \in v_2 + U$
$v_2 + u = v_1 + ((v_2 - v_1) + u) \in v_1 + U$
Thus, $v_1 + U \subset v_2 + U$ and $v_2 + U \subset v_1 + U$ and $v_1 + U = v_2 + U$.
iv) Indeed, $V/U$ is a vector space over $F$. As $V/U \subset V$, if $V$ is finite-dimensional, $V/U$ is also finite-dimensional (basis for $V$ spans $V/U$). We can prove this by using the quotient map $\pi: V \rightarrow V/U$ which is defined such that $\pi(v) = v + U $ for $v \in V$.
$\textrm{null }\pi = U$ since if $v \in \textrm{null } \pi$, $\pi(v) = v + U = 0 + U$, so $v - 0 \in U$, and $ v \in U$. Thus, $U = \textrm {null } \pi$. $\textrm{range } \pi = V/U$ by definition of $V/U$.
By using the Rank-Nullity theorem, we can note that $\textrm{dim } V = \textrm{dim } U + \textrm{dim } V/U$. Thus, $\textrm{dim } V/U = \textrm{dim } V - \textrm{dim } U$.
