I am having trouble understanding certain parts of the proof of Kirchberger's Theorem, as presented here. Specifically, I am having problems understanding the proof of the following combinatorical lemma mentioned in the link:
Let $A,B$ be sets in $\mathbb{R}^n$ whose convex hulls intersect. Then, there exist subsets $A'\subset A,B'\subset B$ such that $|A'|+|B'|\leq n+2$ whose convex hulls intersect.
I will give details about what parts of the proof I don't understand, but in general, any proof of this lemma will suffice for me.
The problematic parts are:
- There is a sequence of at least $n+2$ points, $a_2,...,a_m,b_1 ,....,b_p$, and the book concludes that because of the size of the sequence, there exists scalars $\alpha _2,...,\alpha _m,\beta_1,...,\beta_p$, at least one of them negative, such that: $$\alpha _2a_2+...+\alpha _ma_m=\beta_1b_1+...+\beta_pb_p$$$$\alpha _2+...+\alpha _m=\beta_1+...+\beta_p$$ I don't understand why this is necessarily possible.
- The way $A_0$ and $B_0$ are constructed in the proof lacks explanations as to why $x$ is in the intersectionof their convex hulls, and in general the entire construction of the smaller sets is very confusing.
Any help will be greatly appreciated.
There is a typo: $\beta_m$ should be $\beta_p$. The essence is that $a_1$ is omitted. If all weights were nonnegative, you can divide by the sum of $\alpha_i$ and $\beta_i$ to get: $$\frac{\alpha_2}{\sum_{i=2}^m \alpha_i} a_2 + \ldots + \frac{\alpha_m}{\sum_{i=2}^m \alpha_i} a_m = \frac{\beta_1}{\sum_{i=1}^p \beta_i} b_1 + \ldots + \frac{\beta_p}{\sum_{i=1}^p \beta_i} b_p$$ which implies that you can start over with $m+p$ reduced by 1 (since you have two convex combinations that are equal, but without $a_1$).
$x \in \text{conv}A_0$ because you can divide both equations above by $\psi$. That gives a set of nonnegative weights that sum to 1. For the same reason, $x \in \text{conv}B_0$.