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?