Let $V$ be a vector space and $S$ a nonempty subset of $V$. I want to show that $S$ is an affine subset (a translated subspace of the form $\{v\} +$ $U$, for $U$ a subspace of a vector space $V$ and fixed $v \in V$) of $V$ if and only if (let's call this condition [*]) for every list of points $(x_1, ... , x_m)$ in $S$ and every list of scalars $(\lambda_1, ... , \lambda_m)$ in the field of $\mathbb{R}$ with $\sum_{i=1}^{m} \lambda_i = 1$, (then) the sum $\sum_{i=1}^{m} \lambda_i x_i \in S$.
This seems very easy to do in the first direction (starting with $S$ as an affine subset of $V$):
Suppose that $S$ is an affine subset of $V$. This means that there exists some subspace $S'$ of $V$ such that $\{\{v\} + S' \} = S$ for some $v \in V$. Checking if the sum $\sum_{i=1}^{m} \lambda_i x_i$ (remember that since $S$ is an affine subset of $V$, $x_i = v + s'_i$) lies in $S$ with condition [*]:
$$\sum_{i=1}^{m} \lambda_i (v + s'_i) = (\lambda_1 + ...+\lambda_m)v + (\lambda_1 x_1 + ...+\lambda_m x_m)$$
and we see that $(\lambda_1 + ...+\lambda_m)v = v$, and the RHS of the above equality is equivalent to $v + (\lambda_1 x_1 + ...+\lambda_m x_m) \in S$ since $S'$ is a subspace.
However, I do not see how to go in the other direction; I've been thinking about how to show that a nonempty subset $S \subseteq$ $V$, whose elements can be represented as linear combinations of elements in itself restricted by condition [*], is an affine subset of $V$ - but things are escaping me! Thanks in advance for your thoughts!
EDIT: Specifically, I am trying to internally construct some subspace $S'$ of $V$ out of elements of $S$, because in order to show that $S$ is an affine subset of $V$, we have to show that everything in $S$ can assume the form $v + s', s' \in S'$. So I tried defining every $s \in S$ in terms of $v$ and an arbitrary $s'_i$. But where is the space-structure for $S'$ going to even come from?
Some hints:
1) If $v \in S$ and $S$ has [*], it seems that it shouldn't be hard to show that $S - v$ has [*], and so we can assume WLOG that $0 \in S$. This means that we don't have to worry about how to shift our affine space, which is convenient.
2) If $0 \in S$, then given any $v \in S$, we can do something nice by applying [*] to the finite subset $\{0,v\}$. Once you notice what it is, it's easier to get the vector subspace structure on $S$.