To obtain the promised conclusion of the lemma 7, as we can just apply a suitable bijection X->Y to map the pair (X,P’) to a pair (Y,P). (And rename Y to X in the current notation of the Lemma.)

]]>Regarding the number of points: the construction in Lemma 7 creates a convexity space with |N| points where N is the nerve. The nerves in question are HUGE though. More precisely, look at the proof of theorem 2 in Section 2. Let K=3(k-1)+1. There are K of A-families, 3*binom(K,4) B-families and binom(K,2) C-families. We then apply the closure operator (denoted by hat) to each of these families, to obtain A-hat, B-hat, C-hat families. Then N consists of these families and **their subfamilies**. For example, each C-hat-family has size 2^K-(binom(K,2)-1)-K-1=2^K-K(K+1)/2, and so there are 2^{2^K-K(K+1)/2} subfamilies of each C-hat-family. Each of them gives a point in the resulting convexity space. For k=3, we thus get 2^100 points from each C-set alone. That is a lot of points!

It is very likely that one can significantly reduce the size of these counterexamples. Back when I wrote the paper, I believed I could reduce the number of points somewhat by making the construction somewhat messier, but I saw no way to reduce the number of points to polynomial in k, say.

I hope this answers your questions. Let me know if you have more.

]]>First of all, in your definition of nerves, there are some obvious holes: you never define the universe of your sets, however this is not deep and I can guess what you meen in each case. Then, in property (n1) you say that nerves are “downsets”… it is obvious that a Tverberg 3-partition implies three Radon 2-partitions… but, ¿what does it means a 1-partition? ¿is the conv({})≠{}?

On lemma 7, where “the magic is done”, you define X=N and claim that P \subset X, but N \subset 2^P so in the best case, P \element X… So, at least, the proposition as written is not what you claim to prove…

Next, in the counterexample, I was trying to follow the details for k=3 (n=7), the first non-trivial example, but I got nowhere😦

In my humble opinion there should be an argument of the following style:

“let X={x1, … ,x7} be 7 arbitrary points and A,B,C an arbitrary 3-partition of X. Then, since bla bla bla, conv(A)\cap conv(B) \cap conv(C) = {}”

… but I cannot see such argument… it may be helpful if you can clarify some questions for this specific example:

¿how many points does the example has? ¿how many subsets of these are convex?

Finally, I agree with Gil: the other results would do a nice separate paper…

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