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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?

msteve
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user3382078
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2 Answers2

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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.

Yibo Yang
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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.

supinf
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