I am trying to show the following;
Show that the smallest affine subspace $ \mathcal{B} $ of an affine space $\mathcal{A}$ that contains the points $P_{1} , \dots , P_{r} $ of $ \mathcal{A} $ consists of the points $ a_{1} P_{1} + \dots + a_{r} P_{r} $ where $ a_{1} + \dots + a_{r} = 1 $.
The definitions and notations I am working with are as follows. An affine space is a set $ \mathcal{A} $ acted on by a vector space $V$ over a division ring $K$. The vector $ \overrightarrow{OQ} \in V $ is the unique vector such that for points $O, Q \in \mathcal{A} $ we have $ O + \overrightarrow{OQ} = Q $.
The point $ a_{1} P_{1} + \dots + a_{r} P_{r} $ represents the point $ O + a_{1} \overrightarrow{ OP_{1} } + \dots + a_{r} \overrightarrow{ OP_{r} } $, which I already have shown is independent of the choice of $O$.
An affine subspace is a set of points $ \mathcal{C} $ of the form $O + w $ where $O$ is a point and $w$ runs through the vectors in a vector subspace $W$ of $V$.
Now with all that out explained, I have been trying to approach the problem like this: I can see that any affine subspace must contain all points $ a_{1} P_{1} + \dots + a_{r} P_{r} $. So I just need to show that the set $ \mathcal{D} $ containing all such points is an affine subspace.
If $ \mathcal{D} $ is an affine subspace, then each point can be written as $O + w $ for some $O \in \mathcal{D} $ and $w $ in some vector subspace $ W$.
So for $w_{1} , w_{2} \in W $ and $k \in K$ we must have $w_{1} + w_{2} \in W $ and $k w_{1} \in W $. This means that $ O + ( w_{1} + w_{2} ) $ and $ O + k w_{1} $ must be points of $ \mathcal{D} $ because $W$ must be closed under addition and scalar multiplication. By the definition of $\mathcal{D} $, $w_{1} $ and $w_{2} $ must have the form
$w_{1} = a_{1} \overrightarrow{ OP_{1} } + \dots + a_{r} \overrightarrow{ OP_{r} } $ where $ a_{1} + \dots + a_{r} = 1 $
and
$w_{2} = b_{1} \overrightarrow{ OP_{1} } + \dots + b_{r} \overrightarrow{ OP_{r} } $ where $ b_{1} + \dots + b_{r} = 1 $.
What I want to ask is how to show that $ w_{1} + w_{2} $ and $k w_{1} $ are of the form $ c_{1} \overrightarrow{ OP_{1} } + \dots + c_{r} \overrightarrow{ OP_{r} } $ where $ c_{1} + \dots + c_{r} = 1 $?
Fix $b_1,b_2,...,b_r$ such that $b_1+b_2+...+b_r=1$. Let $x=\sum\limits_{k=1}^{r} b_kP_k$. Define $W$ as $\{\sum\limits_{k=1}^{r} c_kP_k: \sum\limits_{k=1}^{r} c_k=0\}$. It is easy to check that $W$ is a subspace. Let us check that $\{\sum\limits_{k=1}^{r} a_kP_k: \sum\limits_{k=1}^{r} a_k=1\}=x+W$. If $\sum\limits_{k=1}^{r} a_k=1$ then $\sum\limits_{k=1}^{r} a_kP_k=\sum\limits_{k=1}^{r} (a_k-b_k)P_k+\sum\limits_{k=1}^{r} b_kP_k=\sum\limits_{k=1}^{r} (a_k-b_k)P_k+x$ and the first term belongs to $W$. This gives LHS $\subset$ RHS. On the other hand any point in RHS is of the type $\sum\limits_{k=1}^{r} c_kP_k+x$ with $\sum\limits_{k=1}^{r} c_k=0$. From the definotion of $x$ this becomes $\sum\limits_{k=1}^{r} (c_k+b_k)P_k$ and this belongs to LHS because $\sum\limits_{k=1}^{r} (c_k+b_k)=0+1=1$.