Problems with the definition of vectors as directed line segments in $\mathbb{R}^3$ First I'll say where I'm working: The vectorial spaces $\mathbb{R}^2$ and $\mathbb{R}^3$.
Then I'll define a vector of this spaces as the following:
$\textbf{Definition. }$ A vector $\vec{v}$ is the set of all equal directed line segments. 
Now suppose that $$\underbrace{\overrightarrow{AB}}_{\mbox{directed line segment}} \in \vec{v},$$ which is a correct notation, by definition. So why do we write: $$\overrightarrow{AB} = \vec{v}$$   
So I'm a little bit confused. If we understand this two objects as sets, from my point of view:


*

*$\overrightarrow{AB}$ as a directed line segment, it's a set of points in space(plane).

*$\vec{v}$ as a vector, it's a set of directed line segments, with infinite elements.  


$\textbf{Question. }$How can these two objects coincide as sets? Which implies $\overrightarrow{AB} = \vec{v}$.
 A: This is a reasonable question. 
The point is that "equality of directed line segments" is an equivalence relation, so every directed line segment is art of some "vector" (where a vector is an equivalence class). There's even a function --- let's call it $p$ --- that sends a directed line segment to its equivalence class (but there's no function in the other direction! Too many equivalent directed-segments for each vector!). So properly speaking, we should write:
"Let $\vec{v} = p(AB)$"
to indicate the association of a directed segment to a vector. But no one ever does this. I suppose the reason is that the association is pretty much a natural one, and carrying around an extra name for it would be a pain in the neck. 
We do the same thing in other contexts, too. For instance, one can define rational numbers as equivalence classes of pairs of integers under a certain equivalence relation. An integer $n$ is then represented, as a rational number, as the equivalence class of the pair $(n, 1)$. But we simply write things like "$n \in \mathbb Q$" rather than distinguishing $n$ from the equivalence class of $(n, 1)$. 
A: $\overrightarrow v$ is a class of equivalence, so you take one element there to represent all the others, that way there is no problem in stating $\overrightarrow v = \overrightarrow {AB}.$
A: If you take as the first point $A$ the origin $O$ of $\mathbb R^n$ you have a canonical representant of the vector $\overset{\to}{v}=p(AB)$, according to the notation of John Hughes:
$$
\overset{\to}{v}=p(AB)\cong OB'
$$
for a suitable $B'$ (which is  denoted, by the way, as $B'=B-A$)
A: Technically, $\vec v$ is the equivalence class $\{(P,Q) \in  \Bbb R^n: Q-P = B-A\}$ and, even though it is technically incorrect, we usually say
$$\text{Let $\vec v = \overrightarrow {AB}$}$$
as a shorthand way of making the assignment.
It's kind of like saying the circle $(x-1)^2 + (y-3)^2 = 25$ when you mean
$$\text{The circle $\{(x,y) \in \Bbb R^2 : (x-1)^2 + (y-3)^2 = 25\}$}$$
