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consider the set $\{1,...,n\}$, we want to decompose it into sets $S_1,....,S_t$ such that $\vert S_i\vert \geq k$ for all i and $\vert S_i \cap S_j \vert \leq 1$ for all $i\neq j$.

Is there an upper bound on $t$? Obviously depending on n and k.

Clearly for $k=1$ we get no bound. And for $k=n$ we have $t=1$.

This problem comes from Lemma 2.1 in "On the lattice property of the plane and some problems of Dirac, Motzkin and Erdős in combinatorial geometry" by Jozsef Beck, 1983. There for $\sqrt{2n}<k \leq n$ it is claimed that $t < \frac{2n}{k}$, but I dont see it. The proof seems to use that $t \leq \frac{2n}{k}$ holds, if I assume that this is correct I can follow that claim.

Thanks in advance!

Edit: My idee is to show that $2n \geq \sum_{i=1}^t \vert S_i\vert \geq t \cdot k$. The right inequality is clear, but I miss an argument for the left one.

Mike Earnest
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Peter K
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  • It is well known that $t < \frac{n(n-1)}{k(k-1)}$, but this is not enough to prove $t<2n/k$ in the range $\sqrt{2n}<k<n$. – Mike Earnest Feb 14 '22 at 19:36
  • That is part i) of the lemma, the part I'm stuck in is ii). – Peter K Feb 14 '22 at 20:46
  • I edited back in the link; turns out it works just fine, it was just a problem on my end. The paper gives a proof of $t<2n/k$ in their proof of lemma 2.1, is there are a particular step of their proof that confused you? I'd rather not write out the whole proof if you are just stuck on one step. – Mike Earnest Feb 14 '22 at 20:58
  • I'm stuck, where they assume that $t=\lceil 2n/k \rceil$. I mean sure if you assume that you get a contradiction, but why is it enough to show that this does not hold? What is with the case that $t > 2n/k$? – Peter K Feb 14 '22 at 21:02

1 Answers1

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If you prove $t=\lceil 2n/k\rceil$ is impossible, that automatically implies $t$ is impossible whenever $t > \lceil 2n/k\rceil$. This is because if there existed $S_1,\dots,S_t$ satisfying the constaints, then you could just ignore some of the sets to get $S_1,\dots,S_{\lceil 2n/k\rceil}$ satisfying the constraints, so the latter being impossible implies the former is as well.

Mike Earnest
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