What does $V + x_0$ mean if $V$ is a vector space and $x_0$ is a point? From Rotman's Introduction to Algebraic Topology:

If $A$ is the affine set in $R^n$ spanned by an affine independent set
  $\{ p_0, ..., p_m\}$, then $A$ is a translate of an $m$-dimensional sub-vector-space V of $R^n$, namely
  $A = V + x_0$ for some $x_0 \in R^n$.
PROOF: Let $V$ be the sub-vector-space with basis $\{p_1-p_0, ..., p_m-p_0\}$' and
  set $x_0=p0$

What does the notation $V + x_0$ mean if $V$ is a vector space and $x_0$ is a point?  If this just means a translate, then what is a translate?  What's an analytical description for what $V+x_0$ represents?
 A: Intuitively, this denotes the space shifted by the vector $x$. Imagine moving the space, preserving its orientation, such that the origin is moved to $x_0$. Formally, if $V\subseteq \mathbb{R}^n$ and $x_0\in \mathbb{R}^n$, then
$$V+x_0 = \{v+x_0 \, \vert \, v\in V\}$$
A: In general,if you have a vector space $V$ and a subspace $W\subseteq V$, you can define the translates of $W$ by any point $x\in V$ setting
$$W+x:=\{w+x\mid w\in W\}$$
Of course $W+x=W$ if and only if $x=0$, namely no translation is operated.
Note also that the translate is not (in general) a vector space, since closure up to finite sum is violated. Geometrically, you are shifting the entire subspace by a fixed vector, without changing the orientation or modifying the topology.
An example to understand. Let
$$W:=\{(t,t+1,s)\in k^3 \mid t\in k\}$$
You may see that this is a plane in the space, but it is not a vector space. Indeed
$$ W = \{(t,t,s)\in k^3 \mid t\in k\}+ (0,1,0)$$
and the first set is exactly a $2$-dimensional vector space. So $W$ looks like a whole $k^2$ moved up in the second coordinate.

Further details
Following your last comment, I'm adding some details about the result you cite. Recall that giving an affine independent set $\{p_0,p_1,\ldots ,p_n\}\subseteq A$ is equivalent to say that the vectors $p_1-p_0,p_2-p_0,\ldots,p_n-p_0$ are linearly independent. Let us call $$V:=\langle p_j-p_0\mid j=1,\ldots ,n\rangle$$
the $n$-dimensional vector space spanned by those vectors and let us prove that
$A=V+p_0$. 
If $x\in V+p_0$ then $x=v+p_0$ where 
$$v=\sum_{j=1}^{n} \alpha_j (p_j-p_0)$$ 
for some scalars $\alpha_j$. But then 
$$x=\left (1-\sum_{j=1}^n \alpha_j\right ) p_0+\sum_{j=1}^{n} \alpha_j p_j$$
is an element of $A$ because $p_j$ belong to the affine independent set and
$$\left (1- \sum_{j=1}^n\alpha_j\right )+\sum_{j=1}^n\alpha_j=1$$
Conversely, if $x\in A$ then there are scalars $\alpha_j$ such that $\sum \alpha_j=1$ and 
$$x=\sum_{j=0}^n \alpha_j p_j$$
But then 
$$x=x-p_0+p_0=x-\left (\sum_{j=1}^n \alpha_j\right ) p_0 + p_0 = \sum_{j=1}^n \alpha_j (p_j-p_0) + p_0\in V+p_0$$
exactly as we wished.
