Some resource says that a set $A \subset V$ is affine if $\forall x,y \in A, t \in F$, $tx+(1-t)y \in A$ while others say that $A$ is affine if $\forall x_1,\dotsb,x_n \in A, a_1, \dotsb a_n, \in F, a_1+\dotsb+a_n = 1$, $a_1x_1+\dotsb+a_nx_n \in A$.
I am trying to show that this two is equivalent, suppose that I have the first one holds. Then I want to show $\forall x_1,\dotsb,x_n \in A, a_1, \dotsb a_n, \in F, a_1+\dotsb+a_n = 1$, $a_1x_1+\dotsb+a_nx_n \in A$. I first consider a simple case where there are only three elements, ${x_1,x_2,x_3}$,but I do not think it is viable. So maybe the first one is wrong? Because the second definition includes the interior of a triangle while the first does not.
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$\begingroup$Note that the second definition is a generalisation of the first. A set is affine iff it contains all lines through any two points in the set (hence, as a trivial case, a set containing a single point is affine).
(Thanks to @McFry who caught a little sloppiness in my original answer.)
Use induction: Suppose it is true for any collection of $k \le n-1$ points (it is trivially true for $n=1$) and consider the point $\sum_k a_k x_k$.
If $a_k = 1$ for all $k$, we must have $n=1$ (since $\sum _k a_k = 1$), so we are finished. Hence we can assume that $a_k \neq 1$ for some $k$. By renumbering, we can assume that $a_n \neq 1$.
So, suppose $a_n \neq 1$, then you can write $\sum_k a_k x_k = \sum_{k<n}a_k x_k + a_n x_n = (1-a_n) \sum_{k<n}{a_k \over 1-a_n}x_k + a_n x_n$, and note that $\sum_{k<n}{a_k \over 1-a_n} = 1$.
The other direction is immediate.
$\endgroup$ 2 $\begingroup$Given points $x_1, \ldots, x_n$, you have by assumption all the lines from $x_i$ to $x_j$ contained in your space. Now you could argue that the inside of the shape is just lines between points on the lines that form the edges.
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