Justifying a claim about sets of points whose affine subspaces have empty intersection The claim is given in page 5 of this: http://wiki-math.univ-mlv.fr/gemecod/lib/exe/fetch.php/barany_lecture_1.pdf. It states that if a finite subset of points in general position $S$ of $R^d$ is partitioned into $r$ pairwise disjoint subsets $S_1, \ldots S_r$, then if the intersection of all affine subspaces $aff S_i$ is empty, we have that $|S| \leq (r-1)(d+1)$.
The proof (or at least a "counting argument") is given in the document, but I don't really understand it; how can $r(d+1) - |X| \geq r(d+1) - (r-1)(d+1)$? What is $|X|$?
 A: I agree that this argument could be written better. Here's how I understand it:
First of all, observe that every (non-empty) affine subset $A$ of $\Bbb{R}^d$ can be written as the intersection of at most $d - \operatorname{dim}(A)$ hyperplanes in general position. We can do this by choosing one hyperplane at a time so that $\operatorname{dim}\left(H_{k+1} \cap \bigcap_{j=1}^k H_j\right) < \operatorname{dim}\left(\bigcap_{j=1}^k H_j\right)$ (see this).
Therefore each $\operatorname{aff}S_i$ can be written as the intersection of exactly $n_i$ hyperplanes in general position and, up to a slight rotation of some of them, we can consider a collection of hyperplanes in general position $\{H_1,\dotsc,H_{n_1+\dotsc+n_r}\}$ such that
$$
\operatorname{aff}S_i = \bigcap_{j=1}^{n_i} H_{j+n_{i-1}}
$$
where we set $n_0 = 0$. It follows that
$$
\bigcap_{i=1}^r \operatorname{aff}S_i
= \bigcap_{i=1}^r \bigcap_{j=1}^{n_i} H_{j+n_{i-1}}
= \bigcap_{j=1}^{n_1+\dotsc+n_r} H_j.
$$
Since this intersection is empty by hypothesis, the choice of hyperplanes in general position implies that $n_1+\dotsc+n_r \geq d+1$.
On the other hand, as mentioned in that argument, if the points of $S_i$ (which are at most $d+1$ by hypothesis) are in general position we know that $\operatorname{dim}\left(\operatorname{aff}(S_i)\right) = \lvert S_i \rvert - 1$, i.e. $n_i = d+1 - \lvert S_i \rvert$. Hence we just proved that
$$
d+1 \leq \sum_{i=1}^r n_i = \sum_{i=1}^r \left(d+1 - \lvert S_i \rvert\right)
= r(d+1) - \lvert S \rvert.
$$
Finally, rearranging gives $\lvert S \rvert \leq (r-1)(d+1)$.
