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?
Best Answer
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.