An affine set $C$ contains every affine combinations of its points Show that an affine set $C$ contains every affine combinations of its points. 
Proof by induction:
From the definition of an affine set, we know that 
$\forall x_1,x_2\in C \text{ and } \theta_i\in R \text{ such that } \theta_1+\theta_2=1,\text{ we have }\theta_1 x_1+\theta_2 x_2\in C$. Thus the base case of the induction is verified. 
Suppose $\forall x_1,x_2,\dots,x_n\in C$, and $\theta_i\in R,s.t. \sum\theta_i=1$,we have$\sum_{i=1}^n\theta_i' x_i\in C$. 
Now we need to show that $\forall x_1,x_2,\dots,x_{n+1}\in C$, and $\theta_i''\in R,s.t. \sum\theta_i''=1$,we have$y = \sum_{i=1}^{n+1}\theta_i'' x_i\in C$.
We know that $z=\theta_n'x_n+(1-\theta_n')x_{n+1}\in C$. Substitute this in the induction hypothesis to get:
$$y = \theta_1'x_1+\dots+z+(\theta_n'-1)x_{n+1}$$ 
This is a combination of arbitrary n+1 points in $C$ and the parameter $\theta_i'$ sums to 1. Hence, we have showed that an affine set $C$ contains every affine combinations of its points. 
Is my proof correct?
 A: The base case involving two points follows from the definition of an affine set, as you've shown.
Your induction step doesn't look right. Here's my solution:
For induction, assume that any affine combination $ \sum_i^{k-1} \theta_i x_i \in C$.
Now, $\forall x_1,...,x_k \in C$, and $\forall \theta_1,...,\theta_k \in \mathbb{R}, s.t. \sum_i^k \theta_i = 1$, using the fact that $\sum_i^{k-1}\theta_i = 1-\theta_k$,  we can write
$$  \sum_i^k \theta_i x_i = (1-\theta_k) (\sum_i^{k-1} \frac{\theta_i}{1-\theta_{k}} x_i) + \theta_{k} x_{k} \quad \in C$$
where we have $\sum_i^{k-1} \frac{\theta_i}{1-\theta_{k}} x_i\in C$ by inductive hypothesis.
The proof for the convex case is essentially the same.
A: i think it is not correct or not written down very well:
you wrote that you want/need to show that $\forall x_1,x_2,\dots,x_{n+1}\in C$, and $\theta_i''\in R,s.t. \sum\theta_i''=1$,we have$y = \sum_{i=1}^{n+1}\theta_i'' x_i\in C$.
But in the end you "only" show that $$y = \theta_1'x_1+\dots+z+(\theta_n'-1)x_{n+1} \in C$$ 
which is something different ( and it is unclear, if it covers really all possibilities for the $n+1$ points and coefficients). it would be nice if you end up with the stuff you wanted to show.
