# Proof that it's possible to colour all elements in set, that all subsets will be bicolored

(For my easy understanding, let me rewrite the question. The author should feel free to remove my edit or... accept it; I am leaving the original formulation at the end intact).

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REFORMULATION

An ordered pair $(A\ B)$ of subsets of the ring of integers $\mathbf Z$ is called globally ordered if $\ \Leftarrow:\Rightarrow\ \ a \le b\$ for every $a\in A$ and $b\in B$.

Let $k>1$ be an arbitrary natural number. Let $\ S\$ be a finite family of k-element subsets of $\mathbf Z,\,$ each two of which have a non-empty intersection. Is it true that if no two members of $S$ form a globally ordered pair then it's possible to color $\mathbf Z$ in such a way that every member of $S$ is bicolored ?

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PS. I'd like to ask the author to be a bit more detailed and explicit about the definitions related to the coloring.

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THE ORIGINAL QUESTION:

Assume we have a set of numbers $\{ 1, 2, ..., n \}$ and a set of subsets $\{ M_1, ..., M_s \}$, such that $|M_i| = k$ and $\forall i,j\ (i\ne j\implies|M_i \cap M_j| = 1)$.

Let's define 2-chain -- it is two subsets, $M_i, M_j$ with one common element - $x$ and for every element $y \in M_i$ true that $y < x$, and and for every element $z \in M_j$ true that $z > x$. It's increasing 2-chain and in the same way we could define decreasing 2-chain.

We make a random permutation of numbers.

Proof that if there is exist such permutation that there is no 2-chains, that it's possible to colour all numbers in a way that all $M_i$ will be bicolored.

I think that we should use Lovász local lemma, but I can't build proper probability space.

• @bof , of course, for $i \neq j$, sorry for mistake in formulation. – DislocatedShoulder Dec 21 '15 at 5:37
• Accidentally, I omitted the condition about the 1-point intersections. Now I have inserted the non-empty intersections and (I hope) it seems to be fine. – Włodzimierz Holsztyński Dec 21 '15 at 5:44
• I feel that the globally ordered together with the later appearance of the non-empty intersection are after all ok. The 1-point follows from these conditions. – Włodzimierz Holsztyński Dec 21 '15 at 5:48
• @Włodzimierz Holsztyński yes, thank you for reformulation. I think I'd like to remain both formulation of problems, cause your smoother and more general, and mine, cause it's original problem in form that I have heard it. – DislocatedShoulder Dec 21 '15 at 5:53
• Why can't you just, for each element of $S$ (or each $M_i$), colour the top element red and the bottom element blue? What am I missing? – bof Dec 21 '15 at 6:51

Yes. Let $S$ be a family of finite subsets of some linearly ordered set $L.$ Suppose that each member of $S$ has at least two elements, and that no two members of $S$ form a "globally ordered pair". Then we can color every element of $L$ red or blue so that, for each $X\in S,$ the top element of $X$ is red and the bottom element of $X$ is blue.