Does there exist an integer $N$ such that every set of $\geq N$ points in $\mathbb R^4$
contains six distinct points which are vertices of two intersecting triangles?


More generally, given dimensions $d\_1,\dots,d\_k$ such that generic affine subspaces
of $\mathbb R^d$ of dimensions $d\_1,\dots,d\_k$ intersect in a point, does every large 
enough (but finite) set of $\mathbb R^d$ contain $k+\sum\_i d\_i$ distinct points defining 
$k$ simplices of dimension $d\_1,\dots,d\_k$ intersecting in a point?


There are many variations on this problem. For example: Given an integer $k$,
does every large enough subset of $\mathbb R^4$ contain the vertices of $k$
triangles (with all $3k$ vertices distinct) such that all triangles intersect 
pairwise? The corresponding planar problem has a positive answer: 
By the Erdős Szekeres theorem, every large enough subset of $\mathbb R^2$ (in generic
positition) contains $2k$ points in convex position. This defines $k$ "diagonals"
which are all pairwise intersecting.


(The Erdős-Szekeres Theorem shows of course that it is enough to consider points in convex position for all questions mentioned above.)



An Erdős-Szekeres-type question
The answer to the first question is yes (with $N = 7$). Consider a $2$-dimensional simplicial complex $K$ with 7 vertices and with all possible triangles. Assume there are 7 points in $\mathbb R^4$. You can construct a linear mapping $f \colon |K| \rightarrow \mathbb R^4$ ($|K|$ denotes the geometric realization of the complex). It is well known that $K$ does not embed into $\mathbb R^4$. More precisely, so called Van Kampen obstruction of this complex is nonzero, and this implies that for any continuous map $f'\colon |K| \rightarrow \mathbb R^4$ there are two vertex disjoint simplices $\sigma, \tau$ of $K$ whose images $f(|\sigma|), f(|\tau|)$ intersect. If you aply this result on $f$ you get the desired conclusion. In an unlikely case that $\sigma$ or $\tau$ have smaller dimension then 2, you simply extend them to triangles (this may occur only if the points are not in generic position).


This reasoning extends to the case $k = 2$, $d\_1, d\_2 = m$ for some parameter $m$, $d = 2m$ and $N = 2m + 3$. (At the moment, I am not able to answer the general question, however.)


